document resume mathematics, vol. 2. institutim · document resume. se 014 116. mathematics, vol....

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TITLEINSTITUTIM

REPORT NOPUB DATENOTE

EDRS PaIcEDESCRIPTORS

IDENTIFIERS

DOCUMENT RESUME

SE 014 116

Mathematics, Vol. 2.Department of the Navy, Washington, D.C. Bureau ofNaval Personnel.NavPers-10071-B68276p.

MF-$0.65 HC-$9.87*Calculus; *College Mathematics; Geometry;*Instructional Materials; *Mathematics; Physics;TrigonometryLogarithms; Vectors

ABSTRACTThe second of three volumes of a mathematics training

course for Navy personnel, this document contains material primarilyfound at the college level. Beginning with logarithms andtrigonometry, the text moves into vectors and static equilibrium(physics). Coordinate geometry, conic sections, and the tangents,normals, and slopes of curves follow. Four chapters are devoted tolimits, differentiation, and integration; the text concludes with twochapters on combinations, permutations, and probability. Relateddocuments are SE 014 115 and SE 014 117. (JM)

4

U S DEPARTMENT OF HEALTH.EDUCATION Et WELFAREOFFICE OF EDUCATION

THIS DOCUMENT HAS BEEN REPRODUCED EXACTLY AS RECEIVED FROMTHE PERSON OR ORGANIZA:IONINATING IT POINTS OF VIEW OR OPINIONS STATED DO NOT NECESSARILYREPRESENT OFFICIAL OFFICE OF EDUCATION POSITION OR POLICY

:.:.

0.:0. .

:::

t .

MA.HERIA.I,S VOL..

f

BUREAU OF NAVAL PERSONNELNAVY TRAINING COURSE NAVPERS 10071-B

PREFACE

2

The purpose of this Navy Training Course is to aid those personnelwho need an extension of the knowledge of mathematics gained fromMathematics Vol. 1, NavPers 10069-C. To serve the wide varietyof needs, the text is general in nature and is not directed, therefore,toward any one specific specialty.

The definitions and notations of logarithms followed by computationswith logarithms occur early in the text. Trigonometric ratios andanalysis and applications along with aids to computations occur next.Vectors and static equilibrium are followed by trigonometric identitiesand equations.

Straight lines, conic sections, tangents, normals, and slopes precedethe introduction to differential and integral calculus.

The introduction to calculus is intended as a survey course priorto a more rigorous study of the subject. The limit concept is followedby a discussion of derivatives and integration. Basic integration .irnrilasfollow this discussion.

The last chapter of the course covers combinations and permu-tations, and gives an introduction to probability.

Numerous examples and practice problems are given throughout thetext to aid the understanding of the subject matter.

This training course was prepared by the Navy Training PublicationsCenter, Memphis, Tennessee, for the Bureau of Naval Personnel.

1968 Edition

Stock Ordering Number0500-025-0100

For sale by the Superintendent of Documents, U.S. Government Printing Office, Washington, D.C., 2:0102 - Prise $1.76

0:0i'../

THE UNITED STATES NAVY

GUARDIAN OF OUR COUNTRYThe United States Navy is responsible for maintaining control of the seaand is a ready force on watch at home and overseas, capable of strongaction to preserve the peace or of instant offensive action to win in war.

It L upon the maintenance of this control that our country's gloriousfuture depends; the United States Navy exists to make it so.

WE SERVE WITH HONOR

Tradition, valor, and victory are the Navy's heritage from the past. Tothese may be added dedication, discipline, and vigilance as the watchwordsof the present and the future.

At home or on distant stations we serve with pride, confident in the respectof our country, our shipmates, and our families.

Our responsibilities sober us; our adversities strengthen us.

Service to God and Country is vir special privilege. We serve with honor.

THE FUTURE OF THE NAVY

The Navy will always employ new weapons, new techniques, andgreater power to protect and defend the United States on the sea, unc'erthe sea, and in the air.

Now and in the future, control of the sea gives the United States hergreatest advantage for the maintenance of peace and for victory in war.

Mobility, surprise, dispersal, and offensive power are the keynotes ofthe new Navy. The roots of the Navy lie in a strong belief in thefuture, in continued dedication to our tasks, and in reflection on ourheritage from the past.

Never have our opportunities and our responsibilities been greater.

CONTENTS

Chapter

1. Logarithms

2. Computation with logarithms

3. Trigonometric measurements

4. Trigonometric analysis

5. Oblique triangles

6. Vectors

7. Static equilibrium

8. Trigonometric identities and equations

9. Straight lines

10. Conic sections

11. Tangents, normals, and slopes of curves

12. Limits and differentiation

13. Derivatives

14. Integration .

15. Integration formulas

16. Combinations and permutations

17. Probability

Appendix

I. Logarithms of trigonometric functions

U. Common logarithms of numbers

III. Natural sines and cosines

IV. Natural tangents and cotangents

Index

1

Page

1

9

17

31

55

77

92

103

122

135

165

179

193

206

220

234

244

257

258

259

264

270

iii

ACTIVE DUTY ADVANCEMENT REQUIREMENTS

REQUIREMENTS*

-

El to E2 E2 to E3#t E3

to E4#E4to ES

t E5

to E6 t E6 to E7 t E7 to E8 t (8 to E9

SERVICE

4 mos.service--

Orample,tion ofrecruit

training.

6 mos.as E-2.

6 mos.as E-3.

12 mos.as E-4.

24 mos.as (-5.

36 mos.as E-6.8

we!'

enlisted

"Mk.'

36 mos.as E-7.8 of 11yearstotal

servicemust beenlisted.

I

24 mos.as E-8.10 of 13

yearstotal

servicemust beenlisted..

SCHOOLRecruit

Training. ,DT3,

Class Afor PR3,

P13.AME 3,HM 3

Class 8

for AGCMUC,

MNC.

PRACTICAL

FACTORS ,

Locally

PreParidchock-

Records of Practical Factors, NavPers 1414/1, must be

completed for E-3 and all PO advancements.

PERFORMANCE

TEST

,

, -

Specified ratings must completeapplicable performance tests be-fore taking examinations. ,

ENLISTED

PERFORMANCE

EVALUATION

As used by COwhen approving

advancement.

Counts toward performance factor credit in ad-vancement multiple.

EXAMINATIONS**

Locally__A

P "Phtests." See

below,

Navy-wide excaninations required

for all PO advancements,

Navy-wide, ,

selection board.

7

NAVY TRAININGCOURSE (INCLUD-

ING MILITARY

REQUIREMENTS)

.Required for E-3 and all PO advancements

e, '- unless waived because school comple-<., <,:,

. , Non, but need not be repeated if identicalcourse has already beer. completed. See:

< -, y ,,:, NavPers 10052 (current edition).,

Correspondencecourses andrecommendedfeeding. see

NavPers 10052(current edition).

AUTHORIZATION

,

CommandingOfficer

US. Naval ExaminingCenter

Bureau of Naval Personnel

* All advancements require commanding officer's recommendation .t 1 year obligated service required kr E-5 and E-6; 2 years for E-6, (-7, E-8 and E-9.

* Military leadership exam required for E-4 and E-5.** For E-2 to E-3, NAVEXAMCEN exams or locally prepared tests may be used.

iv

4

INACTIVE DUTY ADVANCEMENT REQUIREMENTS

REQUIREMENTS* El to E2 E2 to E3 E3 to E4 E4 to ES ES to E6 E6 to E7 ES E9

TOTAL

TIME

IN

GRAD:

4 mos. 6 mos. 15 mos. 1 i mos. 24 mos. 36 mos. 36 mos. 24 mos.

TOTAL

TRAINING

DUTY IN

GRADE t

14 days 14 days 14 days 14 days 28 days 42 days 42 days 26 days

PERFORMANCE

TESTS

Spetilled ratings must complete apOmealeperformance tests before taking exami-nation.

DRILL

PARTICIPATION Satisfactory par4dpation as a member of a drill unit.

PRACTICAL FACTORS

(INCLUDING MILITARY

REQUIREMENTS)

Record of Practical Factors, NavPers 1414/1, must be completedfor all advancements.

NAVY TRAINING

COURSE (INCLUDING

MILITARY REQUIRE-

MENTS)

Completion of applicalge course or courses must be enteredin service record.

EXAMINATION Standard

Exam

Standard

Exam

or

Rating

Training.

Standard Exam

required for all PO

Advancements.

Standard Exam,

Selection Board.

Aka Pass MILeadership

Exam for (-4and E-S. 4P4

AUTHORIZATION CommandingOfficer

U.S. Naval Examining

CenterBureau of Naval

Personnel

Recommendation by commanding officer required for all advancements.t Active duty periods may be substituted for training duty.

5

CHAPTER 1

LOGARiTHMS

The basic definitions and terminology as-sociated with the study of logarithms were dis-cussed in Mathematics, Vol. 1, NavPers10069-C. Some of these basic topics are re-viewed in the fcllowing paragraphs, followed bydiscussion of the use of log tables and naturallogarithms.

REVIEW OF DEFINITIONS

The most important definition to remember,concerning logarithms, is that everylogarithm isan exponent. For example, since 3' is equal to9, the logarithm of 9 to the base 3 is 2. In orderto state a logarithmic relationship, a base mustbe stated or implied; the various exponents whichdesignate powers of the base are logarithms tothat base.

The usual method of expressing the basicdefinition oi logarithms in symbols is as follows:

If bx = a, then x = logb a

The two forms shown in the foregoing expres-sion are defined as follows:

EXPONENTIAL FORM: bx = a

LOGARITHMIC FORM: x = logs aEXAMPLE: Change the expression 23 = 8 to

logarithmic form.SOLUTION: If bX = a, then logs a = x.

Let b = 2, x = 3, a = 8Substituting,

log2 8 = 3

EXAMPLE: Change the expression logn 100= 2 to exponential form.

SOLUTION: If logs a = x, then bx = aSubstituting,

102 = 100

1

PRACTICE PROBLEMS:1. Change 103 = 1,000 to logarithmic form.2. Change ex = N to logarithmic form.3. Change log2 4 = 2 to exponential form.4. Change log10 3.16 = 1/2 to exponential

form.ANSWERS:1. log10 1,000 = 3

2. loge N = x3. 22 = 44. 101/2 = 3.16

RULES FOR CALCULATION

Numerical calculation by means of loga-rithms is performed by using 10 as the base.Therefore, in the discussion which follows, nobase designation is used. The expression log Ais understood to mean the base 10 logarithm ofA.

Two important abilities are necessary forlogarithmic calculation, as follows:

1. Recognition of logarithms as exponents.2. Knowledge of the rules for exponents in

multiplication and division of algebraic quan-tities.

The first of these abilities was discussedin the foregoing section. The second is the sub-ject of the following paragraphs.Multiplication

Suppose that we wish to multiply A and B, andwe know the following:

A = 10m

B = lOn

The product AB then isAB = 10m x 10n

= 10(m +

MATHEMATICS, VOLUME 2

In logarithmic form,

log A = m

log B = n

log AB m + n

Assuming that we can find AB if we know itslugarithm, the procedure then may be stated asfollows:

To multiply two numbers by means of log-arithms, add their logarithms and find the num-ber whose logarithm is the sum.

EXAMPLE: Multiply 100 times 1,000 bylogarithms.

SOLUTION:

100 = 102 :. log 100 = 2

1,000 = 103:. log 1,000 = 3

log (100 x 1,000) = 2 + 3

= 5

:.100 x 1,000 = 105

= 100,000

Divioion

If A is to be divided by B, and A and B arethe same numbers as in the foregoing discus-sicn4 then we have

A 10mB n 10(1n

10

In logarithmic form,

log A = m

log B nAlogs=m-n

This may be stated in words as follows:To divide B into A by logarithms, subtract

log B from log A and find the number whoselogarithm is the difference.

EXAMPLE: Divide 1,000 by 100 by log-arithms.

2

SOLUTION:

log 1,000 = 3

log 100 = 2

log 112° = 1

14000 1

100 10

= 10

Powers

In order to calculatc a power such as A3 bylogarithms, we observe that

A3 =AxAxA

:. log A3 = log A + log A + log A

= 3 log A

Stated in words, the power rule is as follows:

To find An by logarithms, first express logAn as n log A. The number whose logarithm isn log A is the desired power, An.

EXAMPLE: Find the value of 1002 by usinglogarithms.

SOLUTION:

log 1002 = 2 log 100

= 2(2)

= 4

Therefore,

1002 = 104

= 10,000

Boots

The calculation of roots by logarithms iseasily accomplished by first changing the formof the expression to a fractional power. Thenthe power rule is used.

=AMPLE: Find 11100 by logarith.ns.

Chapter 1LOGARITHMS

SOLUTION:

loo1/21log 1001/2 = 1- log 100

4 (2)

USING LOG TABLES

Tables of logarithms could be constructedusing any number as a base. For purposes ofcalculation, the most logical numuer for a baseis 10, the base of the decimal number system.Logarithms to the base 10 are called COMMONLOGARITHMS.

COMMON LOGARITHMS

Most of the numbers encountered in variouscalculations are not integral (whole number)powers of 10. For example, the number 316can be expressed as a power of 10 only if weresort to fractional powers. We find that

316 = 105/2

= 102 1/2 (approximately)

= 102'5

(The exact value of105 is very close to 316.23.)

Therefore, in logarithmic form,

log 316 = 2.5 (approximately)

Every logarithm consists of an integral partand a fractional part. Thus the logarithm of 316is

2.5 = 2 + 0.5

The integral part is the CHARACTERISTIC; inthis example, the characteristic is 2. The frac-tional part is the MANTISSA; in this example,the mantissa is 0.5.

3

Integers

The characteristic for the logarithm of aninteger may be determined by inspection. Forexample, if the integer is between 1 and 10, itis equal to a power of 10 behveen 0 and 1. Thisconcept is explained fully in Mathematics, Vol-ume 1, NavPers 10069-C.

The numbers in the following list serve toillustrate how the characteristic is determinedby the size of the number:

log 3,6 = 0.5563

log 36 = 1.5563

log 360= 2.5563

log 3,600= 3.5563

Since log 1 is 0 and log 10 is 1, we expect thelogaritlun of 3.6 to be a numberbetween 0 and 1.Therefore, its characteristic is 0. On the otherhand, 3,600 is greater than 1,000 and less than10,000. Therefore its logarithm is between log1,000 and log 10,000, and its characteristic is 3.In the foregoing tabulation, we find that thecom-plete logarithm of 3,600 is 3;5563.

Scientific notation provides a convenientmethod for determining the characteristic. Forexample, 3,600 is written as 3.6 x 103 in scien-tific notation. Thus we have

log 3,600 = log (3.6 x 103)

= log 3.6 + log 103

The characteristic pf log 3.6 is 0, told the char-acteristic of log 10 is 3. Therefore, the char-acteristic of log 3,600 is 3, the sum of thecharacteristics of the two separate logarithms.Any expression written in scientific notationconsists of a number between 1 and 10, multi-plied by a power of 10. Since the characteristicof a number between 1 and 10 is 0, the power of10 determines the characteristic of the log-arithm.

The exponent that we obtain as the powerof 10 in scientific notation is indicated by thenumber of digits between the actual position ofthe decimal point in the original number andthe standard position of the decimal point. Thestandard position is immediately after the firstnonzero digit in the number. For example, inthe number 3,600, the decimal point is understood

8


to be after the second 0 in the original number.This is 3 digits to the right of standard position,so that the exponent of 10 for scientific notationis 3. This exponent is also the characteristic forlog 3,600. If the decimal point in the originalnumber had been to the left of standard position,the exponent of 10 (and therefore the character-istic) would have been negative.

Fractions

When the logarithm of a fraction is obtained,a negative characteristic occurs. For example,

log 04036 = log (3.6 x 104)

= log 3.6 + log 10-2

= log 3.6 + (-2)

The mantissa for log 3.6 is 0.5563. There-fore,

log 0.036 = 0.5563 - 2

Since logarithm tables do not list negativemantissas, we do not subtract 2 from 0.5563to obtain the final form of log 0.036.

Some tables handle the problem of negativecharacteristics by placing a negative sign abovethe characteristic as follows:

log 0.036 =1-.5563

Observe that an entry such as

-2.5563

would be misleading. It would be interpreted asif the mantissa, as well as the characteristic,were negative.

Perhaps the most universal form for nega-tive characteristics is as follows:

log 0.036 = 8.5563 - 10

This form is numerically equal to

0.5563 - 2

but it has the advantage of presenting the char-acteristic, as well as the mantissa, as a posi-tive number.

Negative numbeis and 0 do not have loga-rithms. In the case of 0, none is needed for

purposes of calculation. However, it is oftennecessary to multiply and divide negative quan-tities using logarithms. In order to use log-arithms for this purpose, we first determinethe sign of the final answer. Then all numbersare treated as positive, and the predeterminedsign is affixed.

PRACTICE PROBLEMS:Determine the characteristic of the loga-

rithm for each of the following numbers:1. 322. 4763. 0.254. 0.0074ANSWERS:1. 1

2. 23. -1 or 9 - 104. -3 or 7 - 10

MantissaTables of logarithms normally contain only

mantissas. Since these are understood tobe thedecimal parts of the logarithms which theyrepresent, the decimal points are often omittedin the printed table. Appendix ll of this courseis constructed in this way, and the user is ex-pected to supply the characteristic and the deci-mal point with each mantissa. Table 1-1 is anexcerpt from appendix II.

Observe that the table of logarithms hasheadings consisting of the abbreviationNo. (rep-resenting Number) and the digits 0 thrm ,h 9.The first two digits of any number whose loga-ritlun we seek are found in the No. column. Thethird digit is found as one of the column head-ings, 0 through 9. The mantissa for the loga-rithm of any three-digit number is found op-posite the first two digits, which are located inthe No. column, and below the third digit.

EXAMPLE: Find the mantissa for the loga-rithm of 124.

SOLUTION:1. In the No. column, table 1-1, find the

number 12.2. Move to the right, staying in the row of

mantissas opposite 12, until you reach the columnwith the digit 4 as a heading.

3. Read the mantissa, .0934. NO% thatwe supply the decimal point.

EXAMPLE: Find the complete logarithm Gf13.7.

4-.

9

Chapter 1-LOGARIltIMS

Table 1-1. -Example of common logarithm table.

No. 0.

1,

2 3 4 5 6 7 8.

9

10 0000 0043

_

0086 0128 0170 0212 0253 0294 0334 0374

11 0414 0453 0492 0531 0569 0607 0645 0682 0719 0755

12 0792 0828 0864 0899 0934 0969 1004 1038 1072 1106

13 1139 1173 1206 1239 1271 1303 1335 1367 1399 1430

14 1461 1492 1523 1553 1584 1614 1644 1673 1703 1732

SOLUTION: Always determine the char-acteristic before entering the table.

1. The characteristic of 13.7 is 1. Wemay now disregard the decimal point and enterthe table with just the digits 137.

2. In the No. column, table 1-1, find thenumber 13.

3. Move to the right, staying in the 13 row,until you reach the 7 column.

4. The mantissa is .1367.5. The logarithm of 13.7 is 1.1367.

PRACTICE PROBLEMS:

Find the logarithms of the following num-bers:

1. 1182. 342 (See appendix II.)3. 14.64. 5.48ANSWERS:

10

NUMBERS MANTISSAS

r-- 1120 .04925

1125

1130 .0531

.0039

We analyze the foregoing tabulation in termsof the difference between the numbers and thedifference between the mantissas. The largebracket on the numbers indicates a differenceof 10, and the small bracket shows that our

5number is 10of the way between the two num-

bers in the table. Therefore, the mantissa cor-a

responding to our number should be 10of the

way between the mantissas in the table.

1.2.3.4.

2.07192.53401.16440.7388

5 (.0039) = .00195

=I .0020 (to 4 places)10--

INTERPOLATION

Interpolation is the process of calculatingthe mantissa for the logarithm of a numberhaving one more digit than the table entries, Forexample, to find the logarithm' of 1125 it wouldbe necessary to interpolate in table 1-1.

The logarithm of 1125 is halfway betweenthe logarithms of 1120 and 1130. Therefore,we find the mantissas for the logarithms ofthese two numbers and then determine themantissa that is halfway between them. Thework is arranged as follows:

Adding .0020 to .0492, we obtain the mantissacorresponding to 1125; it is .0512. Therefore,

log 1125 = 3.0512

EXAMPLE: Find log 25.67.

SOLUTION:

1. The characteristic is 1.

2. The number lies between 2580 and 2570.Therefore the mantissa lies between .4082 and.4099.

jo


3. Tabilate:

10

[2560

2567

.4082

.0017

2570 .4099

74. Our number is fr) of the way between2560 and 2570. Therefore, the mantissa is-1-1 of the way between .4082 and .4099.10

5. j (.0017) = .00119

= .0012 (to 4 places)6. .4082 + .0012 = .4094

:.log 25.67 = 1.4094PRACTICE PROBLEMS:Find the logarithms of the following numbers:1. 0.23452. 5.4323. 473.64. 9817ANSWERS:1. 9.3702 - 102. 0.73523. 2.67544. 3.9919

TRIGONOMETRIC FUNC TIONS

Logarithms of trigonometric functions maybe obtained by first looking up the decimal valueof the natural function and then finding the loga-rithm of this decimal number. However, theprocess is laborious, and it is rendered unneces-sary by the existence of tables of logarithms ofthe trigonometric functions.

Appendix I of this training course is a samplepage from a typical tale of trigonometric log-aritiuns. Its construction is similar to that ofappendix II, except that 10 is added to eachmantissa. Thus the entry for log cot 38° is10.10719, with the quantity "4 0" understood.The complete logarithm is then

log cot 38° = 10.10719 - 10

log sin 380 = 9.78934 - 10

ANTILOGARITHMS

The procedure of finding a number when weknow its logarithm is called "finding the anti-logarithm.° The word "antilogarithm" is ab-breviated "antilog," and a symbol sometimesused to indicate the antilog is log-1. The -1 ina symbol of this kind tends to be confusing, sinceit is not an exponent. It is an indicator whichemphasizes the INVERSE relationship betweenlogs and antilogs.

The antilogarithm is easily found when thecorresponding mantissa is an exact table entry.

EXAMPLE: Find the antilogarithm of1.1271.

SOLUTION:1. Find the mantissa .1271 in appendix II.

(The decimal point is understood.)2. The column in which .1271 is found de-

termines the third digit of the antilog. The thirddigit is 4.

3. The row in which .1271 is located de-termines the first two digits of the antilog. Inthat row, and in the No. column, we find thedigits 13. Thus the digits of the antilog are 134.

4. Since the characteristic of the originallogarithm is 1, the antilogarithm must be anumber between 10 and 100. Thus the decimalpoint must be placed between the 3 and the 4.

5. We conclude that

antilog 1.1271 = 13.4

Interpolation must be used when the mantissawhose antilogaritim we seek is not an exactentry in the table.

EXAMPLE: Find the antilogarithm of8.5124 - 10.

SOLUTION:1. Find the mantissas nearest to .5124 in

appendix I. These are .5119 and .5132. Since.5124 is between .5119 and .5132, its antilog isbetween the antilogs corresponding to these twomantissas.

2. Tabulating the results of step 1, andletting N represent the antilog corresponding to

22 0.10719

The addition of 10 to each table entry makesthe correct form automatic in those cases wherea "9 - 1 " type of format is involved. For ex-ample, the character stic for log sin 38° is -1,with the following result:

.5124, we have

NUMBER

.0325

.0326

MANTISSA

.5119

.5124 J

.5132

.0005 .0013

6

Chapter 1-LOGARITHMS

(The numbers .0325 and .0326 are decimalsbe-cause of the -2 characteristic.)

3. Our tabulation shows that we use .0005parts out of .0013 ingoingfrom.0325to N. Thisis the same as 5 out of 13, so that we use5 of the total difference inthe mantissa. Thus,

5N should be 15 of the way from .0325 to .0326.

4. Multiplying to findAof the difference

between the numbers, we have

1 x . 0001 = .384 x .000113

= .00004 (approximately)

5. The difference between .0325 and N isadded to .0325 to obtain N.

N = .0325 + .00004

= .03254

O. We conclude that

antilog 8.5124 - 10 = .03254

PRACTICE PROBLEMS:Find the antilogarithms of the following log-

arithms:1. 2.70302. 9.3830 - 103. 1.84514. 0.3842ANSWERS:1. 504.72. 0.23103. 70.004. 2.422

NATURAL LOGARITHMS

Natural logarithms are so named because thenumber.e, the base of the natural logarithm sys-tem, is involved in the law of nature governinggrowth and decay. This law is stated in symbolsas follows:

A = A ert

In the foregoing equation?, A represents thetotal amount after a perioc! of growth, and Aorepresents the amount at the beginning of the

7

growth period. The letter r represents thecontinuous rate of growth, and t represents thetime period during which growth occurs. Thesame remarks apply for a period of decay.

By means of higher mathematics, the num-ber e is found to have the value

e = 2.71828 (approximately)

This number is the base of the natural loga-rithm system.

CHANGING BASES

The relationship between the common loga-rithm of a number and its natural logarithm isas follows:

In N 2.3028 log N

Observe that the special abbreviation, In N, isused to represent loge N.

The derivation of the foregoing equation isdescribed in the following paragraphs.

Let ex = N, where N is any annber. Takingthe natural logarithms of both sides, we have

xlne=1,01

Since ln e means loge e, In e is the same as 1.Therefore,

x = In N

This result is also obtainable from the basicdefinition of logarithms.

Taking common logarithms on both sides inthe expression

ex N

we have the following:

log ex = log N

x log e =logN

=08 e

Equating the two expressions which we haveobtained for x, we have

In N isaLtiog e


From the table of common logarithnis, we findthat log 2.71828 is 0.4343. Finally,

70 log N111 = 6.4343

1. 152. 803. 294. 35

Since the raciprocal of 0.4343 is 2.3026,ln N = 2.3026 log N

ANSWERS.

1. 2.7080PRACTICE PROBLEMS: 2. 4.3820Find the natural logarithms of the following 3. 3.3673

numbers: 4. 3.5553

.1=

8

CHAPTER 2

COMPUTATION WITH LOGARITHMS

The use of log tables and an introduction to Exponential solution:natural logarithms were discussed in chapter 1.Included in that chapter were the rules for nu-merical calculations involving logarithms anda review of the laws of exponents.

Additional mention of the laws of exponentswill be given in the following paragraphs followedby discussions of the useof logarithms in numer-ical computations.

Six rules of exponents are shown in table 2-1for reference and review.

The purpose of the study of logarithms is toenable us to shorten computations with numbers.In many computations involving multiplication,division, powers, roots, or combinations ofthese, the solutions may be reached more easilyby replacing ordinary arithmetical processeswith logarithmical processes.

Appendix II gives the common (base 10) loga-rithms of numbers to four places. All calcula- INTERPOLATION:tions by means of logarithms in this chapter use10 as the base. In accordance withthe convention NUMBERestablished in chapter 1 the expression log A isunderstood to mean the base 10 logarithm of A. 980

386 x 254 = 102.5866 x 102.4048

= 102.5866 + 2. 4048

= 104. 9914

Logarithmic solution:

log (386 x 254) = log 386 + log 254

= 2.5866 + 2.4048

= 4.9931

Antilog 4.9914 = 980 + (interpolation correc-tion)

MULTIPLICATION

The logarithm of the product of two or more 981

members is the sum of the logarithms of theseparate numbers.

EXAMPLE: Use logarithms to find the prod-uct:

386 x 254

SOLUTION:

Logarithmicequation

Exponentialequation

log 386 = 2. 5866 386 = 102. 5866

log 254 = 2.4048 254 = 102. 4048

MANTISSA

. 9912. 0002

1 . 9914

. 9917

. 0002. 4

. 0005x 1 =

Antilog of .9914 = 980 + .4 = 980.4

. 0005

When combined with the characteristic of 4 toplace the decimal

Antilog 4.9914 * 984040:.386 x 254 =

The exponential solution shown in the exampleis not a part of normal calculations involvinglogarithms. It was shown in this first exampleproblem solely for the purpose of reemphasizing

9

14


Table 2-1. Laws of exponents.

OPERATION ALGEBRAIC NUMERIC

MULTIPLICATION

DIVISION

POWER OF A

POWER

POWER OF A

PRODUCT

POWER OF A

QUOTIENT

TRANSPOSING

NEGATIVE

EXPONENTS

aman= am+ n

-nam= am

a

(am) n = amn

(ab)m = ambm

(On1 = m

a-m a

-n 1a =an

42.43

o-7-r . 35

(123) 2

= 42+3

7-3= 5

= 123°

=45

4= 5

2 = 126

2)4 54 34 24

2 2

=412"

1 =525-2

A-3 1- 34

the relationship between exponents and loga- antilog 3.7236= - 5,291rithms.EXAMPLE: Use logarithms to find the prod- Therefore,uct of (126) x (-42).SOLUTION: Recall from chapter 1 that nega-

tive numbers do not have logarithms. In usinglogarithms to solve problems that involve nega-tive numbers, we first determine the sign of uct of 1.73 x 0.0024 x 0.08.the final answer. After this sign is determined,the indicated operations are performed treat-ing all numbers as positive quantities, and thepredetermined sign is affixed to the answer.

In our example, dealing first withonly signs,we determine the answer to be negative. Thatis, (+) x (-) results in (-) answer. At this pointthe problem can be restateduse logarithms tofind the product of

(126) x (-42) = - 5,291

EXAMPLE: Use logarithms tofind the prod-

- (126 x 42)

- log (126.42) -(log 126 + log 42)

= - (2. 1004 + 1. 6232)

3.7236

10

15

SOLUTION:

log (1.73 x 0.0024 x 0.08)

log 1.73 + log 0.0024 + log 0.08

log 1. 73 = 0. 2380

log 0.0024 7.3802 - 10

log 0.08 = 8.9031 - 10

sum = 16.5213 - 20this can be adjusted to

sum = 6.5213 - 10

Chapter 2-COMPUTATION WITH LOGARITHMS

or

log (1.73 x 0.0024 x 0.08) = 6.5213 - 10

antilog 6. 5213 - 10 = .0003322

PRACTICE PROBLEMS:

Use logarithms to find the product of thefollowing:

1.73 x 0.0024 x 0.08 = .00033221. 53 x 76 x 0.021 x 153

2. 1.02 x 109 x 4.76 x 10-3

It is in problems such as this example thatthe relationships of logarithms, exponents, andthe scientific notation can be used. Go back tothe step where

3.

4.

0.00432 x 0.00106 x 15

0.102 x 103.5 x 76.2

log (1.73.0.0024.0.08) = 6.5213 - 10

Perform the calculations required to go from thisstep to the next one in which

antilog 6.5213 - 10 = 0.0003322Proper decimal placement is often more diffi-cult in a problem of this type than in a problemwhich does not involve a negative characteristic.The difficulty may be reduced if the relation-ships of logarithms, exponents, and scientificnotation are utilized. After the step

log (1.73.0.0024.0.08) = 6.5213 - 10reference to the tables shows that the antilogof the mantissa is between 332 and 333. Inter-polation and rounding off produces the digits3322 as the antilog of the mantissa.

In chapter 1 it was shown that a character-istic could be determined by expressing a num-ber as a number between 1 and 10, multipliedby a power of 10. An antilog can be deter-mined by the converse of this procedure if thedigits which represent the antilog of the man-tissa are written as a number between 1 and 10and multiplied by a power of 10. The particularpower of 10 to be used is equal to the character-istic of the logarithm in question.

Apply this to the example problem where thedigits are 3322 and the logarithm is 6.5213 -10.The power of 10 is -4 since the characteristicis 6 - 10 or -4. Expressing the digits as anumber between 1 and 10 multiplied by thispower of 10 yields 3.322 x 10-4, and the anti-log of 6.5213 -10 equals 3.322 x 10-4. In-spection of the example problem indicates theantilog is 0.0003322 and since

0.0003322 = 3.322 x 10-4

the results of both methods are the same.

ANSWERS:

1. 12,942

2. 4,855,555

3. 6.87 x 10-5

4. 804.4

DIVISION

The logarithm of the quotient of two numbersis the logarithm of the dividead minus the loga-rithm of the divisor. As with multiplication,this rule is simply an application of the law ofexponents. For example,

105 4- 103 = 105-3 = 102

EXAMPLE: Find the quotient of 344 by useof logarithms.

11

SOLUTION:

log (37. 4 1.7) = log 37.4 - log 1.7

log 37.4 = 1. 5729

- log 1.7 = 0.2304

1. 3425

log (37.4 + 1.7) = 1.3425

antilog 1.3425 22 22

37. 41.7 = 22

16.3EXAMPLE: Find the quotient of


SOLUTION:

log (16.3 4- 0.008) = log 16.3 - log 0.008

log 16.3 = 1. 2122

log 0.008 = 7.9031 - 10

In order to prevent the complication of sub-tracting the characteristic, 7, from the smallercharacteristic, 1, we add 10 to and subtract 10from, the logarithm of the dividend. Note thatthis does not change the value of the logarithm.Thus,

log 16.3 = 11.2122 - 10

- log 0.008 = 7.9031 - 10

3.3091

antilog 3.3091 = 2037

16.32037=

0. 008

PRACTICE PROBLEMS: Use logarithms tosolve the following problems:

1. 635.6 4. 25.4

2. 0.26 0.061

3. 0.126 4- 0.00542

4. 874 .÷- 26.3

ANSWERS:

1. 25.03

2. 4.263

3. 23.25

4. 33.23

COLOGARITHMS

Dividing one number by another may beaccomplished logarithmically by addition ratherthan subtraction, if the cologarithm is employed.The cologarithm of a number is the logarithmof the reciprocal of the number. The reasonthat the cologarithm may be added is explainedin the following example:

Evaluate 1512

log 1-1 = log 15 (IQ12

Thus,

1= log 15 + log (T)

15log - = log 15 + colog 1212

The cologarithm may be easily derived bysubtracting the logarithm of the numberfrom thelogarithm of 1. Thus,

colog 12 = log (A)

= log 1 - log 12

but log 1 = 0

Therefore,

colog 12 = 0 - log 12

= 0 - 1. 0792

Writing 0 as 10 - 10, we have

colog 12 = (10.0000 - 10) - log 12

colog 12 = (10.0000 - 10) - (1.0792)

= 8.9208 - 10

Since writing the colgarithm is almost as simpleas writing the logarithm, it is sometimes advan-tageous to use cologarithms in complicated prob-lems and thus make the problem one of addition.

Returning to the original problem

15log i-2 = log 15 + colog 12

15 1. 1761log D = 8. 9208 - 1010. 0969 - 10

15log yi = O. 0969

antilog 0. 0969 = 1. 25

Chapter 2-COMPUTATION WITH LOGARITHMS

15and 1. 25 = D

EXAMPLE: Evaluate the following by use oflogarithms and colgarithms.

343. 8592 x O. 76

or

SOLUTION:

343. 8log 592 x O. 76

= log 343. 8 - log 592 - log O. 76

= log 343. 8 + colog 592 + colog 0. 76

log 343. 8 = 2. 5363

colog 592 = 7. 227 - 10

colog 0. 76 = O. 1192

antilog

.

9. 8832 - 10

343. 8

9. 8832 - 10

= O. 7642.

= 0' 7642592 x O. 76

It should be understood that this problemcould be solved without using cologarithms byfinding the logarithms of the two numbers in thedenominator, adding them and then subtractingtheir sum from the logarithm of the number inthe numerator. Since both methods are equallyaccurate, the selection of either method becomesone of convenience.

PRACTICE PROBLEMS:

Evaluate the following numbers by use oflogarithms and cologarithms:

210 x 4. 11. 33 x O. 8754 x 1. 7

14 x 0.27 x 36.162' 11 x 8 x 17 x 6. 76

ANSWERS:

1. 17.54

2, O. 0135

-..

RAISING TO A POWER

To find the log of the power of a number,multiply the logarithm of the number by theexponent of the power. This rule is based onthe law of exponents for finding the power of apower. For instance,

3 2 3 3 3+3 6(4 ) = 4 x 4 = 4 = 4

(43)2 = 43x2= 46

(102. 8)2 = (102.01205)2

= 104.02410

13

18

or

Also

EXAMPLE: Find the value of (18. 53)5.

SOLUTION:

log (18. 53)5 = 5 log 18. 53

5 log 18. 53 = 1.2679 x 5

= 6. 3395

antilog 6. 3395 = 2,185,263

(1ft. 53)5 = 2,185,263

TAKING A ROOT

To find the log of the root of a number,divide the logarithm of the number by the indexof the root.

We recall the law of exponents for taking aroot. For instance,

Also,

3[8-i= (82)1/3 82/3

5 Ir.bili(5-0- =51405.0126

5 0126 1/5= (10-* )

= 101.0025


EXAMPLE: Find the value of 5(521.1

SOLUTION:

log 5N1i27.I = log 327. 6

log 327.6 = 2.5153

1 log 327. 6 = 2. 5153 55

= 0.5031

antilog O. 5031 = 3. 185

. 51/521-776- = 3.185

1.

2.

3.

4.

ANSWERS:

1.

2.

3.

4.

(3.276)3

(0.00468)2

64:T0867

5;11177

35.15

0. 0000219

0.4532

2.987

ROOTS OF QUANTITIES WITHNEGATIVE CHARACTERISTICS

When a logarithm with a negative charac-teristic is to be divided, it is advisable to addand subtract a number that will, after dividing,leave a minus 10 at the right. This is done tokeep the logarithm in standard form. Forexample, if the problem were 51r0-ar-018, we wouldhave

I 1log 5V O. 0018 = 5 log 0. 0018

1= -5- (7.2553 - 10)

Here, to keep a minus 10 in the final logarithm,we must add and subtract 40 before dividing.Thus,

1log

5 ((DMZ = 5 (47.2553 - 50)

47. 2553 - 05log54(T0T.8

= -I--

log5Nrea018 = 9. 4511 - 10

PRACTICE PROBLEMS: Evaluate the fol-lowing by the use of logarithms,

14

ALGEBRAIC OPERATIONS

This chapter has demonstrated the use oflogarithms in numerical calculations. Practicalapplications in many fields involve calculationsincluding algebraic expressions in which loga-rithms are useful. In these problems both thelaws of algebra and the laws of exponents orlogarithms hold true and the use of logarithms inalgebraic operations is valid. For example:log [ (x+ 2)(x + 5)] = log (x + 2) + log (x + 5)

EXAMPLE: Simplify the following.

x2 - 5x - 6x + 1

SOLUTION: (Using only laws of logarithms.)2

"5 x- 5x

1- 6 log (x2 - 5x - 6) - log (x + 1)

+

SOLUTION: (Using logarithms and laws ofalgebra.)

x2log - 5x - 6X + = log

= log

- 6) (x +Cx + 1)

6EtVl.= log (x- 6)

EXAMPLE: Solve for x.

23x = 250

19

Chapter 2COMPUTATION WITH LOGARITHMS

SOLUTION: Take the logarithmof both sidesof the equation.

log 23x log 250

Then, x log 23 = log 250

log 250x log 23

2. 39791. 3617

= 1. 761

In complicated problems it may not be pos-sible to solve for the unknown as directly as wedid in the example. In that case we can continueto use our knowledge of logarithms. Return to

2.3979the step where x = 1.3617. Againtake the loga-

rithm of both sides of the equation.

log x = log 21: 33697197

= log 2. 3979 - log 1. 3617

= O. 3798 - 0. 1341

= 0. 2457

Take the antilog of both sides of the equation.

x = antilog 0. 2457

x = 1. 761

EXAMPLE: Solve for x.

x2 = 729

3

log x = log 729

3 log x = log 729

3 log x = 2 log 729

log x 22 2 lo 729

15

20

2(2. 8627)

5. 72543

log x = 1. 9084

x = antilog 1. 9084

x = 81

PRACTICE PROBLEMS: Use logarithms tosolve for x in the following problems.

1. 17x = 31

ANSWERS:

1. O. 000121

8

2. x = 6. 3496

2. 2. 0

APPLICATIONS

The use of logarithms can simplify the solu-tion of many problems encountered in mathe-matics, science, andengineering. By applicationof the operations described in this chapter, manycomplicated equations can be reduced to additionand subtraction problems.

EXAMPLE: Find the volume of a circularcone having a height of 3.71 units and a baseradius of 2.71 units.

SOLUTION: The formulatorvolume of a cir-17r2 hcular cone is v = , where v = volume, r =

radius, and h = height.Take the logarithm of both sides of the equa-

tion as the first step in the solution.

log v = log (21-2--U)3

32 log IT + log r2 + log h - log 3

= log ir + 2 log r + log h - log 3

= O. 4972 + 2(0. 4330) + O. 5694 - O. 4771

= 1. 4555

v = antilog 1. 4555

v = 28. 5


For simplicity, physical units of measure-ment were not included in the example. Whenproblems involving physical units are solved,it is often simpler to solve the problem sepa-rately for the units and attach the proper unitsto the answer. For example, if the base andheight in the example problem were given ininches, solution for the proper units in theanswers could be as follows:

nr2hv =

3

Since pi and 3 are not involved with physicalunits they are ignored in this solution and theproblem is expressed as

v = inches squared x inches

= inches cubed

Therefore, in the solution; v will be expressedin cubic inches.

Many electronic problems can be simpli-fied by using logarithms, and other elec-tronic problems include common logarithmsin the basic formulas. An example of a formulathat includes a logarithmic expression is theformula for finding gain in decibels where

P1decibels = 10 log

Engineering and electronic problems fre-quently deal with numbers in the millions anddecimal fractions in the millionths. Thesevalues are easily expressed as exponentiP:!: tothe base ten and common logarithms are tivma natural and convenient means of simplifyingthese problems.

EXAMPLE: Find the numerical value of 'CCin a circuit where f = 22,000,000 cycles per

second, C = .0000000015 farads, and 'CC =1

2/71C'

16

SOLUTION: The formula is X =C 22rfC

1

Taking logs on both sides,

log Xc = log 1 + colog (277fC)

= 0 + colog 6. 28 + colog (2. 2 x107) (+ colog 1.5 x 10-9)

= (9. 2020 - 10) + (2. 6576 - 10) +(8.8239)

= 20. 6835 - 20 = O. 6835

Xc = antilog O. 6835

Xc = 4. 825

PRACTICE PROBLEMS: Use logarithms tosolve for the numerical value of the unlaiown inthe following problems.

1. Find the volume (v) of a sphere having aradius of 7.59. The formula for the volume ofa sphere is

47rr3v = 3

2. Find the value of I in the formula

P = I2R

when P = 217 and R = 550,000,

ANSWERS:

1. 1830 2. 0.0198

CHAPTER 3

TRIGONOMETRIC MEASUREMENTS

This is the first of several chapters in thiscourse which deals with the subject of trigo-nometry. Reference to the table of contentsshows that chapters 4, 5, and 8 also dealdirectly with triangles and trigonometry. Ad-ditionally, chapters 6 and 7 deal with vectorsand their application to statics. The study ofvectors is so closely related to trigonometrythat it is normally included in a trigonometrycourse, and in this course it is included in thesame area.

Mathematics, Vol. 1, NavPers 10069-C, in-troduces numerical trigonometry and some ap-plications in problem solving. However, trig-onometry is not restricted to solving problemsinvolving triangles; it also forms a foundationfor some advanced mathematical concepts andsubject areas. Trigonometry is both.algebraicand geometric in nature, and in this courseboth of these qualities will be utilized.

MEASURING ANGLES

In Mathematics, Vol. 1, it was pointed outthat angles are formed when two straight linesintersect. Before proceeding with measure-ment of angles, an extension of the concept ofangles is required. In this courze, an angle isconsidered to be generated when a line havinga set direction is rotated about a point. Figure3-1 depicts the generation of an angle.

Lay out the line AO, as shown in figure 3-1,as a reference line having a set direction. Useone end of the line as a pivot point and rotatethe line from its initial position OA to anotherposition OB, as in opening a door. As the lineturns on a pivot point, it is generating the angleAOB. Some of the terminology used in this andsubsequent chapters is given in the following:

1. Radius vectorthe line which is rotatedto generate the angle.

17

2. Initial positionthe original position ofthe radius vector; corresponds to line OA infigure 3-1. (Also called the initial side of theangle.)

3. Terminal sidethe final position of theradius vector; corresponds to line OB in figure3-1.

4. Positive anglean angle generated byrotating the radius vector counterclockwisefrom the initial position.

5. Negative anglean angle generated byrotating the radius vector clockwise from theinitial position.

The convention of identifying angles by useof Greek letters is followed in this text. Whenonly one angle is involved it will be calledtheta (0). Other Greek letters will be usedwhen more than one angle is involved. The ad-ditional symbols normally used will be phi (),alpha (a), and beta (13).

One unit of angular measure familiarto mostpeople is the revolution. However, this unit istoo large for many uses, and three other unitsare discussed in following paragraphs.

DEGREE SYSTEM

This is the most common system of angularmeasurement. In this system a completerevolution is divided into 360 equal parts calleddegrees (360°). For accurate work each degreeis divided into 60' (minutes), and each minuteinto 60" (seconds). In many cases degreemeasurements are expressed in degrees andtenths of a degree.

For convenience in working with angles,the 360° is divided into four parts of 90° each,similar to the rectangular coordinate system.The 900 sectors (calledquadrants) are numberedaccording to the convention shown in figure 3-2.

When the radius vector (the line gener-ating the angle) has traveled less than 90°from its starting point in a counterclockwise


0 A

Figure 3-1.Generation of an angle.

2 mdQUADRANT

letQUADRANT

3rd

QUADRANT4th

QUADRANT

Figure 3-2. Quadrant positions.

direction (or, as we conventionally call it, in apositive direction), the angle is in the firstquadrant. When the radius vector lies between90° and 1800, the angle is inthe second quadrant.Angle between 180° and 2700 are said to lie inthe third quadrant, while anglesgreater than 2700and less than 360° are in the fourth quadrant.

When the line generating the angle passesthrough more than 360°, the quadrant in whichthe angle lies is found by subtracting from theangle the largest multiple of 360 that the anglecontains, and determining the quadrant in whichthe remainder falls. The original angle liesin the same quadrant in which the remainderangle falls.

EXAMPLE: In which quadrant is the angle850°?

SOLUTION: The largest multiple of 360°contained in 850° is 720°. Then 850° - 720°= 130°. Since 130° is in quadrant 2, 850° alsolies in the second quadrant. This relationshipis shown in figure 3-3.

RADIANS

There is another and even more fundamentalmethod of measuring angles. The unit for thistype of measurement is the radian. It has cer-tain advantages over the degree method, for itrelates the length of arc generated to the size

18

+11111I

of the angle. Radian measurement also greatlysimplifies work with trigonometric functions incalculus. Assume that an angle is generated, asshown in figure 3-4. It we impose the conditionthat the length of the arc (s) described by theextremity of the line segment generating theangle must equal the length of the line (r), thenwe would describe an angle exactly one radianin size; that is, for 1 radian, s = r.

Recall from plane geometry that the circum-ference of a circle is related to the radius bythe formula

C = 2:rr

This says that the length of the circumferenceis 2ir times the length of the radius. From therelationship of arc length, radius, and radiansin the preceding paragraph, this can be extendedto say that a circle contains 2ir radians.

Since the arc length of the circumference is21 radians and the circumference encompasses3600 of rotation, it follows that

2ir radians = 360°

ir radians = 180°

By dividing both sides of this equation by ir wefind that

180°1 radian = = 57.2959°IT

In this course we shall use the followingconversion factors:

1. 1 radian = 57°17'45" 57.3° (approxi-mately)

2. 10 = 0.017453 radians (approximately)If absolute accuracy is desired in a con-

version, use the following:1. To

v., 180°

2. To

bY 1110°It is customary to indicate degrees by the

symbol (° ) and to indicate radians as a purenumber with no name or symbol attached. Forexample, sin 3 should be understood to re-present sine of 3 radians, whereas the sine of3 degrees would be written sin 3°.

Certain angles occur so frequently in trig-onometric problems that it is worthwhile to

convert radians to degrees multiply

convert degrees to radians multiply

Chapter 3TRIGONOMETRIC MEASUREMENTS

( A ) (8)

Figure 3-3.Angle generation. (A) 1300; (B) 850°.

learn the degree and radian equivalences. These MILSare shown in the following list:

Radians Degrees

v/6 30

v/4 45

v/3 60

r/2 90

180

3v/2 270

2 7r 360

Figure 3-4. Radian measure.

19

24

The mil is a unit of angular measurementwhich is not widely used but has some militaryapplications in ranging and sighting. A mil isdefined as 1/6400 of the circumference of acircle. This is equavalent to 3'22.5" or 0.00098radians.

The importance of the mil in practical ap-proximation is due to the fact that it is ap-proximately 1/1000 of a radian. A circulararc whose length is 1/1000 of the radius willsubtend an angle of 1 mil. For very smallangles the arc (a) and the chord (c) are nearlyequal as shown in figure 3-5.

Since the arc and chord are very nearlyequal (ratio of chord to arc nearly 1) for verysmall angles, we shall consider them as equalfor our purposes and develop a method of ap-proximating range (R) to a target of known size(c). Recall from the previous section that aradian is the ratio of arc length to length ofradius. In figure 3-5 (with c essentially equalto a), we will consider c as arc length and r aslength of radius. Then the size of angle m inradians is

m = rThen the length (r) is expressed as

r = --m


Figure 3-5.Relationship of arc, chord, and radius.

If we express the angle m in mils (1/1000 of aradian), the formula for range is

cm

r 1000

r = 1000c111

If the range is known and it is desirable to findthe length (c) of a target, this formula can betransposed to

rmc 1000

These formulas yield good approximationsfor angles up to several hundred mils and makerapid estimates of range to an object of knownsize possible.

EXAMPLE: A building known to be 80 feetlong, perpendicular to the line of sight, sub-tends an arc of 100 mils. What is the approxi-mate range to this building?

SOLUTION:

1000 x 80 ftr = 100

r = 800 ft

PRACTICE PROBLEMS:Determine the quadrant in which each of the

following angles lies.1. 260°2. 290°3. 800°4. 1,930°Express the following angles in degrees; the

angles are expressed in radians.5. 20vjpg 5iru

6

-

7. A tower 500 feet away subtends a verticalangle of 250 mils. What is the height of thetower?

ANSWERS:1. 3rd2. 4th3. 1st4. 2nd5. 3,600°6. 150°7. 125 ft

MEASURES WITH RADIANS

The radian measure of an angle was intro-duced in previous paragraphs. It was pointedout that radian measure of angles was of par-ticular use when working with trigonometricfunctions in calculus.

Because of the relationship of the radian toarc length, it has some special applications inmeasurements of angular velocity and sectorarea. It was pointed out earlier that in the anglesho in figure 3-4, 0 = 1 radian when s = r.This relationship can be generalized by theformula

s = r0

when 0 is expressed in radians. This relation-ship is convenient for solving many types ofproblems. A few sample problems are includedin this section.

EXAMPLE: The human eye cannot clearlydistinguish objects if they subtend less than0.0002 radians at the eye. What is the maximumdistance at which a submarine periscope 6inches in diameter can be picked out from itssurroundings? (Of course, this problem dis-regards the wake made by the periscope).


SOLUTION:From this information we know that 0 = 0.0002

radian and s = 6 in. = 0.5 ftSubstituting in

s = r0

0.5 = 0.0002r

5000 2 500 feet

ANGULAR VELOCITY

gives

or

Another type of problem which radian mea-surement simplifies is that which connects therotating motion of the wheels of a vehicle to itsforward motion. Here we will not be dealingwith angles alone but also with angular velocity.Let us analyze this type of motion.

Let us consider the circle at the left infigure 3-6 to indicate the original position of awheel. xs,s the wheel turns it rolls so that thecenter moves along the line CC' where CI is thecenter of the wheel at its final position. Thecontact point at the bottom of the wheel movesan equal distance PP'; but as the wheel turnsthrough angle 0, the arc s is made to coincidewith line PP', so that

s = PP'

(,$) /j/Z///// / ff,e/71

or the length of arc is equal to the forwarddistance the wheel travels. But since

= rO

the forward distance d that the wheel travels is

d = r0

Dividing both sides of the equation (2) by t,we have

d=

0

The forward velocity v of the vehicle is

equal to and the angular velocity w is equalt

to the angle divided by the time required todescribe the angle. Thus,

v = rw

if w is measured in radians per unit time.

EXAMPLE: Determine the distance a truckwill travel in 1 minute if the wheels are 3 feetin diameter and are turning at the rate of 5revolutions per second.

SOLUTION: In this problem, first convertangular velocity from revolutions per secondto radians per second by multiplying the

- NOM, M=1=0 111M.M.1=Mli

Figure 3-6.Angular rotation.

21

26


revolutions by Mr. (There are 2w radians in onerevolution (3600).)

2r = 6.2832

5 rev x 6.2832 = 31.416 rad/sec

3, 520 ft/minw 4 ft

w = 410560

3

Then, radians per second are converted to w = 2,640 radians per minuteradians per minute as follows:

rad/min = rad/sec x 60

rad/min = 31.416 x 60

rad/min = 1,884.96

Thus, after 1 minute 0 = 1,884.96 radians and

d = r 03d = y ft x 1,884.96

d = 2,827.44 ft

EXAMPLE: A car is traveling 40 milesper hour. If the wheel radius is 16 inches,what is the angular velocity of the wheels (a)in radians per minute and (b) in revolutionsper minute?

SOLUTION:(a) First, convert miles per hour to feet

per minute.

Thus,

ft/min mph x 528060

4040 mph = .1-6 x 5,280 ft/min

40 mph = 3,520 ft/min

Then change the radius to feet and1r = 13

ft

Then,

r = 4 ft3

v = rw

22

radians per minute(b) rpm2w

2840 rad/minw -2w

w = 420. 1 rpm

AREA OF A SECTOR

From plane geometry we find that the areaof the sector of a circle is proportional to theangle enclosed in the sector.

Consider sector AOB of the circle shown infigure 3-7, If 8 is increased to a full 360° (or2r radians), it encoMpasses the entire circleand the area (A) of the sector equals the areaof the circle which is given by the formula

A = ir r2

Multiplying both sides of the equation by 0 gives

AO = eirr2

We are dealing with a complete circle where0 = 2w. Thus, we can substitute 2w for 9 in theleft member, obtaining

2ir A = 8ir r2

Simplifying this by dividing both sides by 2wgives

n 2 1A =

2 2

Therefore, the area of a sector of a circle canbe found by the formula

1A = -i- r20

with 8 expressed in radians.


Figure 3-7.Sector of a circle.

if the radius and arc length are known, thearea can be found directly by the formula

1A = r s

This formula is found by the following process:

1A = r29

1A = y rr9

Since s = rO, we substitute in the formulaabove and have

1A = rs

Of course, if r and s are given, the area couldbe found by computing 0 and using the formulafor area in terms of r and O.

EXAMPLE: Find the area of a sector of acircle with radius of 6 inches having a centralangle of 600.

SOLUTION: First convert 60° to radians:

radians = degress x 1800

= 60° x 180°ff

9 =3

1 irA = x (6 inches)

2 x

36v inches2A 6

A = 6r sq in.

.

This answer can be converted by multiplying6 by 3.14159. However, in trigonometry and

higher mathematics the conversion is, in manycases, not carried out. In many instances it isless cumbersome to carry n through computa-tions rather than convert it to a decimal number.

PRACTICE PROBLEMS:1. How far does a car travel in 1 minute if

the radius of the wheels is 18 inches and theangular velocity of the wheels is 1,000 radiansper minute?

2. A car travels 2,000 feet in 1 minute. Theradius of the wheels is 18 inches. What is theangular velocity of the wheels in radians perminute?

3. What is the diameter of a circle if asector of this circle has an arc length of 9inches and an area of 18 square inches?

ANSWERS:1. 1,500 feet per minute2. 1,333 radians per minute3. 8 inches

PROPERTIES OF TRIANGLES

Mathematics Vol. 1 contains information onthe trigonometric ratios and other properties oftriangles. This section reviews the trigono-metric functions for acute angles and restatessome of the properties of triangles for reviewand reference.

PYTHAGORLAN THEOREM

This theorem states that in any right tri-angle, the sum of the squares of the aides ad-jacent to the right angle is equal to the squareof the hypotenuse (side opposite the right angle).In the triangle shown in figure 3-8 this rela-tionship is expressed as

X2 + y2 = r2

This relationship is useful in solving many

A = 1 r2 problems and in developing other trigonometric

2 concepts. The following are examples of the

23

28


application of this relationship as applied to thetriangles in figures 3-9 and 3-10.

EXAMPLE: The legs of a right triangle,shown in figure 3-9, which are adjacent to theright angle are 3 and 4 inches in length. Howlong is the hypotenuse?

SOLUTION:

r2 = X2 + y2

r2 = 42 (inches2) + 32 (inches2)

r2 = 16 (inches2) + 9 (inches2)

r2 = 25 (inches2)

r = V25 (inches2)

r = 5 inches

This, as pointed out in Mathematics, Vol. 1,is the 3-4-5 triangle, whose side relationshipshold also for multiples of 3,4, and 5.

EXAMPLE: Figure 3-10 shows a plot ofground in the shape of a right triangle. Thelongest side of the plot is 40 feet. One of the

Figure 3-8.Pythagorean relationship.

Is

Figure 3-9.--Right triangle withhypotenuse unknown.

other sides is 10 feet long. Howremaining side (x)?

SOLUTION:

2 2 2x + y = r2 2 2x = r - y

x2 =.40 2 (ft2) - 102 (ft2)

x2 = 1,600 (ft2) - 100 (ft2)

x2 = 1,500 (ft2)

x =1/ 1,500 (ft2)

x = 38.7 ft

SIMILAR TRIANGLES

long is the

Another relationship of triangles that isuseful in trigonometry concerns similar tri-angles. Whenever the angles of one triangleare equal to the corresponding angles in an-other triangle, the two triangles are said tobe similar.

For example, triangle (A) in figure 3-11 issimilar to triangle (B). Since the two triangles

et yal0feet

Figure 3-10.Right triangle with oneside unknown.

24 .

(A)

Figure 3-11.--Similar triangles.


are similar by definition, the following propor-tion involving the lengths of the sides is true:

a b ca' b' c'

This relationship can be used to find thelength of unknown sides in similar triangles.The following is an example of this methodusifig the triangles shown in figure 3-12.

EXAMPLE: Triangles (A) and (B) iii figure3-12 are similar with lengths as shown. Findthe length of side b' and side c'.

SOLUTION:

a b ca' b'

10 11.18 57 b' c'

Solve the first two of the equal ratios for W.

10 11.18'7 b'

b' = 11' 18 x10

78.26b' 10

b' = 7.826

Solve for side c' using the first and third of theequal ratos.

10 5c'

5 x 7c' - 10

c' = 3.5

NOTE: Side c' could have been determinedby using the second and third of the equal ratios,and the reader should verify this. The selec-tion of the first and third ratios in the examplewas only because it was obvious thatthe numer-ical calculations would be simpler withoutdecimals.

25

30

( A)

10

(8)

Figure 3-12.Similar triangles,solution example.

Similar Right Triangles

Recall from plane goemetry that the sum ofthe interior angles of any triangle is equal to180°. Using this fact it follows that two trianglesare similar if two angles of one are equal totwo angles of the other. The remaining anglein any triangle must be equal to 1800 minus thesum of the other two angles. Keep theseprinciples in mind during the following dis-cussion of similar right triangles.

Two RIGHT TRIANGLES are similar if anacute angle of one triangle is equal to an acuteangle of the other triangle. If these angles areequal, the triangles are similar because the rightangles in the two triangles are also equal toeach other. Whenever one acute angle of aright triangle is given, the other acute anglemay easily be found. Assuming that the acuteangle given is 25 degrees, the other acute anglewill be

180° - 90° - 25° = 65°But 65° and 25° are complements of each other.Thus, if one acute angle of a right triangle isequal to 0, the other acute angle will be (90° - 0).

Many of the practical uses of trigonometryare based on the fact thattwo right triangles aresimilar if one acute angle of one triangle isknown to be equal to one angle of the othertriangle.

Thus, in figure 3-13, we have two similarright triangles, so we may write

x y


(A) ( B )Figure 3-13.Similar right triangles.

or interchanging the mean terms in the propor-tion, we have

and in like manner

Z. .4 x'r r' a'a'4 r r'

This is one of the main principles of numericaltrigonometry.

PRACTICE PROBLEMS: Refer to figure3-14 in solving the following problems.

1. Use the Pythagorean theorem to cal-culate the missing sides in triangles (A) and (B).

2. Find sides a and b of triangle (D), as-suming that triangles (C) and (D) are similartriangles.

ANSWERS:

1. (a)4(b)

2. (a) 3 3/4(b) 1 1/2

TRIGONOMETRIC FUNCTIONSAND TABLES

The properties of triangles given in theprevious section provide a means for solvingmany practical problems. Certain practicalproblems, however, require knowledge of righttriangle relationships other than the Pythagoreantheorem or the relationships of similar triangles

26

before solutions can be found. Rumples of twoof the problems which require this additionalknowledge are as follows:

1. Find the value of the missing sides andangles in a right triangle when the value of oneside and one acute angle are given.

2. Find the value of the missing side and thevalue of the angles in a right triangle when twosides are known.

The additional relationships, between thesides and angle's-of-a right triangle, are calledtrigonometric functions or trigonometric ratios.These ratios were introduced in Mathematics,Vol. 1, and are reviewed in the following para-graphs. The basic foundations of trigonometryrest upon these functions.

RATIOS FOR ACUTE ANGLES

There are six ratios of the sides of a righttriangle; these ratios form the six trigono-metric functions. We use the acute angle 0 infigure 3-15 and define the trigonometric ratios.

In figure 3-15 the three sides x, y, and r areused two at a time to form six ratios. Theseratios and the trigonometric function name as-sociated with each are as follows:

1. -Y- is the sine of 0, written sin 0

2. 7( is the cosine of 0, written cos 0

3. Y is the tangent of 0, written tan 0

4. is the cosecant of 0, written csc 9

5. - is the secant of 0, written sec 0

6. is the cotangent of 0, written cot 0

The functions of a right triangle in any posi-tion are made easier to remember by the con-vention of naming the sides, as in figure 3-16,and defining the functions by means of thesenames.

In any right triangle the side y is the sideopposite the angle whose function we are seeking.The side x is always adjacent to the angle. Thus,we can summarize the information in this section

Chapter 3TRIGONOMETR1C MEASTIREMENTS

(A) (B)4

(C)

Figure 3-14.Triangles for practice problems.

Firmre 3-15.Right triangle fordetermining ratios.

in the following manner by reference to figure3-16:

opposite sidesin hypotenuse

x adjacent sidecos 0 r hypotenuse

y opposite sidetan 0 x adjacent side

. hypotenusecsc 0 y opposite side

=r = hypotenusesec 0 x adjacent side

cot = x - adjac ent side0 y opposite side

W-aen the lengths of the sides of a right tri-angle are known, as in figure 3-17 (A), the sixtrigonometric ratios can be computed directly.

The values of all the trigonometric functionsof the angle 0 in the triangle in figure 3-17 (A)are as follows:

27

: .

3(D)

Is

4441

'41

411 la

SIDE ADJACENT TOCABsX

Figure 3-16.Names of sides of aright triangle.

3sin 0 = rx 4cos 9 = 5

3

= 0.600

= 0.800

tan 0 = = = 0.750

= 1.2505sec 0 = =x 4


4

(A)

3(8)

Figure 3-17.-Practice triangles.

csc

cot 0

5

43

= 1.667

= 1.333

EXAMPLE: Give the values of all thetrigonometric functions of the angle 0 in thetriangle of figure 3-17 (B).

SOLUTION: Here only two sides are given.To find the third side, use the Pythagoreantheorem.

X2 + y2 = r2

2 2 2y = r x

y2 = 62 - 32 = 36 - 9 = 27

y = = Nr9-573 = 3 VS. = 5.196

Now, using these values of x, y, and r

sin

cos e

tan

csc 0

sec

cot = -Y

5.196 = 0.88606

3= - = 0.5006

5.196 = 1.732

5.1966

1.1547

6 = 2.000

35. 196

28

TABLES

Trigonometric tables are lists of the numer-ical values of the ratios of sides of right tri-angles. If we desire to know the ratio of twosides of a right triangle containing a knownacute angle 0, we look for the angle ti in atable and thus find the desired ratio. The ta. lein appendix III provides the sine and cosine ofangles from 0° to 90°. Appendix IV gives naturaltangents and cotangents of angles from 0° to90°.

The tables in the appendixes give the tri-gonometric functions in degrees and minutes.The format used in these tables is fairly standardfor trigonometric tables. Refer to appendix IIIand the first page of values. Tofind the sine orcosine of angles less than 5°, enter the table atthe appropriate degree value listed at the top ofthe table. The minute columnwhichisused withthese values is the first column at the left of thetable and is read from top to bottom.

Values of the functions between 85° and 90°are also listed on this page. To determine thesine or cosine of an angle in this range, enterthe table at the degree value at the bottom of thepage. The minute values which correspond tothese angles are found in the last column of thetable, and are read from bottom to top. Thecolumn headings (sin and cos) are seen tochange from the top to the bottom of the column.The correct name of the function is the onewhich appears at the degree value in use. Thetables of the tangent and cotangent ratios inappendix IV are laid out in the same formatas the sine and cosine tables.

Most tables list the sine, cosine, tangent,and cotangent of angles from 0° to 90°. Veryfew give the secant and cosecant since theseare seldom used. When needed, they may befound from the values of the sine and cosine.The reciprocal of the sine gives the value ofthe cosecant.

r 1 1csc v = - = -y y sin 0

The reciprocal of the cosine gives the value ofthe secant.

33

sec 0 = = I =x x cos 0


The other functions, the tangent and co-tangent, can also be expressed in terms of sineand cosine as follows:

r= sillx x cos 9

cot 9 r cos 9Y Y sin 9

In addition, the cotangent can be determineddirectly as the reciprocal of the tangent.

cot 9 = = - 1Y y- tan

TE

These relationships are the fundamentaltrigonometric identities and will be used ex-tensively in solving the more complex iden-tities which are the topic of chapter 8.

PRACTICAL USE OF THETRIGONOMETRIC RATIOS

Proper use of the trigonometric ratios andthe other principles of triangles furnish power-ful tools for use in problem solving. The knowl-edge of when (and how) to use which of the ratiosis an important part of the problem. This knowl-edge comes with experience and practice; how-ever, each of the functions is more applicable tocertain type problems than to others. Thissection points out some of the situations in whicha specific function is used most advantageously.

20(A)

Tangent of an Angle

1 opposite. sideThe ratio . of be usedx adjacent side `"1"whenever one of these sides and an acute angleare given and it is desired to find the otherside.

EXAMPLE: Find the length of side y infigure 3-18 (A).

SOLUTION: From the figure

tan 35° =

From the tables

tan 35° = 0.70021

are two expressions for tan 350, soHere therethat

= 0.70021

= 14.0042

Sine of an Angle

To find y or r, when either of these and anacute angle are given, use the sine to find theunknown.

EXAMPLE: Fine the value of r in figure3-18 (3).

SOLUTION: From the figure

(E)

Figure 3-18.Practical use of ratios.

29'

5sin 65° =

(C)


From the tables

sin 65° = 0.90631

Thus,

5 = 0.90631

r = 5.517

Cosine of an Angle

To find x or r, when either of these and anacute angle are given, use the cosine to findthe desired part of the triangle.

EXAMPLE: Find the value of x in thetriangle of figure 3-18 (C).

SOLUTION:

(C)

= cos 66°5

x = 5 cos 66°

= 5(0.40674)

= 2.0335

526

(S)

(

15

Figure 3-19.Triangles for practiceproblems.

PRACTICE PROBLEMS: Refer to figure3-19 in working problems 1 through 4.

1. a. Find the values of the sine, cosine,and tangent of 0 in triangle (A).

b. What is the value of 0 to the nearestdegree?

2. Using the sine function, find the value ofy in triangle (B).

3. Using the cosine, find the value of x intriangle (C).

4. Using the tangent, find the value of y intriangle (D).

5. A navigator on a ship notes that twopoints on either side of a strait are 5 milesapart and subtend an angle of 40° as shown infigure 3-20. How far from the strait is theship- if it is equidistant from both points?NOTE: The 40° angle can be divided into two20° angles, and this will form two righttriangles.

ANSWERS:

1. a. sin 0 = 3/5 = 0.60000cos 0 = 4/5 = 0.80000tan 0 = 3/4 = 0.75000

b. 0 = 36°2. 294::. 12.124. 8.48665. 6.87 miles

5 MILES

MI

Figure 3-20.Ship approaching strait.

CHAPTER 4

TRIGNOMETRIC ANALYSIS

This chapter is a continuation of the broadtopic of trigonometry introduced in chapter 3.The subject is expanded in this chapter to allowanalysis of angles greater than 900. The chap-ter is intended as a foundation for analysis ofthe generalized angle, an angle of any numberof degrees. Additionally, the chapter introducesthe concept of both positive and negative angles.

RECTANGULAR COORDINATES

The rectangular or Cartesian coordinatesystem introduced in Mathematics, Vol. 1,NavPers 10069-C, is used here. In Vol..1 thecoordinate system was used in solving equations;in this chapter it is used for analyzing the gen-eralized angle. The following is a brief reviewof important facts about the coordinate system:

1. The vertical axis (Y axis in fig. 4-1) isconsidered positive above the origin and nega-tive below the origin.

2. The horizontal axis (X axis) is positiveto the right of the origin and negative to the leftof the origin.

3. A point anywhere in the plane may belocated by two numbers, one showing the dis-tance of the point from the Y axist and the othershowing the distance of the point from the Xaxis. These points are called coordinates.

4. In notation used to locate points, it isconventional to place the coordinates in paren-theses and separate them with a comma. TheX coordinate is always written first. Thus,point P in figure 4-1 would have the notationP(4,-5). The general form of this notation isP(x,y).

5. The X coordinate is positive in the firstand fourth quadrants negative in the second andthird. The Y coordinate is positive in the firstand second quadrants negative in the third andfourth, The signs of tlie coordinates are shownin parentheses in figure 4-1. The algebraic

31-

36

0 7

6

5

4

3

2

X-AXIS X

-8 -7 -6 -5 -4 -3 -2 -1 0 I 4 5

-2

-3

-4

-5-6-7

P(41-5)

p-,-) 0072

Figure 4-1.Rectangular coordinate system.

signs of the coordinates of a point are used inthis chapter for determining the algebraic signsof trigonometric functions.

6. The quadrants are numbered in the man-ner shown in chapter 3 of this course. Thequadrant numbers are shown with the signs ofkhe coordinates in the figure.

ANGLES IN STANDARD POSITION

An angle is said to be in standard positionwhen certain conditions are met. To constructan angle in standard position, first lay out arectangular coordinate system. The angle (8)is then drawn with the vertex at the origin ofthe coordinate system, and the original sidelying along the positive X axis as shown in fig-ure 4-2. In standard position, the terminal side

4

i


Figure 4-2.Standard position.

(radius vector) of the angle may lie in any ofthe quadrants, or on one of the axes which sep-arate the quadrants. When the terminal sidefalls on an axis a special case exists, in whichthe angles are called quadrantal angles and arediscussed separately later in this chapter.

The quadrant in which an angle lies is de-termined by the terminal side. When an angleis placed in standard position the angle is saidto lie in the quadrant which contains the tenni-ral side. For example, the negative angle 0,shown in standard position in figure 4-3, issaid to lie in the second quadrant.

Coterminal Angles

When two or more angles in standard posi-tion have their terminal sides located at thesame position they are said to be coterminatIf 0 is any general angle then 0 plus or minusan integral multiple of 3500 yields a coterminalangle. For example, the angles 9, 0, and a infigure 4-4 are said to be coterminal angles. If0 is 450 then

= e - 360°

= 45 - 360°

= - 315°

32

Figure 4-3.Negative angle in quadrant U.

Figure 4-4.Coterminal angles.

Chapter 4TRIGONOMETRIC ANALYSIS

and,

a = 8 + 3600

a = 450 + 3600

a = 405°

The relationship of coterminal angles canbe stated in a general form. For any generalangle 09 measured in degrees, the angles 0 co-terminal with 0 can be found by the following

= 0 + n(360° )

where n is any integer, positive, negative orzero; that is,

n = 0, *1, *2, *3,...

The principle of coterminal angles is usedin developing other trigonometric relationshipsand in other phases of trigonometric analysis.An expansion of this principle states that thetrigonometric functions (ratios) of coterminalangles have the same value. This fact is thebasis for part of a discussion in a later sectionof this chapter.

PRACTICE PROBLEMS: Which of the fol-lowing sets of angles are coterminal?

1. 60°, -300°, 420°.2. 735°, -345°, -705°.3 450, -450, 34504. 0°, 360°, 180°.

ANSWERS:

I. Coterminal2. Coterminal3. Not coterminal4. Not coterminal

Definition Of Functions

The trigonometric functions (ratios) are de-fined in chapter 3 of this course in two formsas follows:

I. By means of the sides labeled x, y, and r.2. By means of the names of the sides: op-

posite side, adjacent side, and hypotenuse.In this chapter a third form is introduced,

using the nomenclature of the coordinate sys-tem. The three systems are not different sys-tems, they merely define the same functions byusing different terminology. Defining the same

33

38

ratios in three different sets of terms no doubthas the appearance of complicating a relativelysimple operation. However, as progress ismade through this course and in advance mathe-matics courses, particular problems or situa-tions arise where it is natural to think of theratios in the specific terms used in one of thedefinitions. It is the intention of this course tointroduce the three terminology groups andshow that the three definitions are synonymousand interchangeable.

"o arrive at the definitions, construct anangle in standard position in respect to a co-ordinate system as shown in figure 4-5 (A).Choose a point P with coordinates (x,y) as apoint on the radius vector. The distance OP isdenoted by the positive number r (radius vector).

NOTE: In this chapter the conventionaldesignation of the radius vector as alwayspositive is followed.

The trigonometric functions of the generalangle 0 in figure 4-5 (A) are defined as follows:

sin e ordinater radius vector

x abscissacos 0 - -r radius vector

tan 0 = ordinatex abscissa

The remaining functions are reciprocals of theones given and can be written in this terminol-ogy by inverting the given functions.

CSC 0 =1 radius vector

sin- ordinate

1 radius vectorsec e = cos e abscissa

1 abscissacot e = tan e ordinate

The values of the functions are dependent onthe angle 8 alone and are not dependent uponthe selection of a particular point P. If a dif-ferent point is chosen, the length of r, as wellas the values of the X and Y coordinates, willchange proportionally and the ratio will beunchanged.

Comparing (A) and (B) in figure 4-5 it isseen that the X and Y values of the point P in(A) correspond to the lengths of sides xand y in


(A)Figure 4-5.Functions of general angles.

(B). Therefore, the ratios defined here are thesame ratios defined in chapter 3 of this course;the only change is in terminology. The proce-dure given here allows for finding the functionsof an angle when only a point on the radius vec-tor is given.

EXAMPLE: Find the sine and cosine of theangle shown in figure 4-5 (A) when the point P(x,y) has the value P(3,4).

SOLUTION: To determine the sine and co-sine it is necessary to find the value of r.Since the values of the X and Y coordinatescorrespond to the lengths of the sides x and yin figure 4-5 (B), we can determine the lengthof r by use of the Pythagorean theorem or byrecalling from Mathematics, Vol. 1 the 3-4-5triangle. In either case, the length of r is 5units. Then,

sin 0

sin 0

cos o

cos 0 =

ordinateradius vector

45

abscissaradius vector

34

(B)

NOTE: In the remainder of this chapter allangles are understood to be in standard position,unless specifically stated to the contrary.

PRACTICE PROBLEMS: Without using ta-bles, find the sine, cosine, and tangent of theangles whose radius vectors pass through thepoints given below.

1. P(5,12)2, P(1,1)3. P(1, Ira)4. P(3,2)NOTE: In problems such as these it is often

helpful to construct a graphic illustration of theproblem similar to figure 4-5.

ANSWERS:

121.

5cos =

12tan = -5-

2. sin =1 If

2


itC 0 S = -2-

tan = 1

153. sin = -2-

tan = I-3-

4. sin = 2 24513

cos = 3-Ts-

, 2Lan

QUADRANT SYSTEM

The quadrants formed in the rectangularcoordinate system are used to determine thealgebraic signs of the trigonometric functions.The quadrants in figure 4-6 show the algebraic

ISINE (+)COS ()TAN ()

ISINE (4)CAS (4)TAN (+)

ILE

SINE ()COS (i.)TAN (+)

33C

SINE ()COS (4)TAN ()

Figure 4-8.Signs of functions.

4035

signs of the sine, cosine, and tangent in thevarious quadrants.

A ready recall of the algebraic signs of thefunctions in particular quadrants will be of as-sistance in many problems. It is not manda-tory, however, that the information in figure4-6 be memorized. The information will belearned in some cases by extensive usage and,in addition, the following section on referencetriangles supplies a means for rapidly deter-mining the algebraic signs of all of the func-tions for any angle in standard position.

The algebraic signs of the remaining func-tions, while not shown on the figure, can be de-termined from the signs of the given functions.In all quadrants the cosecant has the same signas the sine, the secant has the same sign as thecosine, and the cotangent has the same sign asthe tangent.

The last group of practice problems in-volved angles in the first quadrant only, whereall of the functions are positive. When thesigns that the functions take on in each quad-rant are known, more complicated problems ofthis type can be solved.

EXAMPLE: Find all of .the trigonometricfunctions of 0 if tan 0 = i , sin 0 < 0, andr = 13,

SOLUTION:

Reference to figure 4-6 shows that an anglewith a positive tangent and a negative sine canoccur only in the third quadrant.

Then, since

and

ordinate y,tan 0 =1-Ficria x

5tan e = 12

it is possible to determine a point P(x,y) whichlies on the radius vector of this angle. Thevalue of the Y coordinate is the value of theordinate and the value of the X coordinate is thevalue of the abscissa. Thus, it appears that thepoint is P(12,5). However, L 44e Cartesiancoordinate system, both of the coordinates arenegative in the third quadrant so the point isP(-12,-5). This does not conflict with the valueof the tangent that was given, however, since


-5 = 5-12 12

and the tangent is positive in a quadrant whereboth coordinates are negative.

Figure 4-7 shows the angle 8 constructedusing the following information which was de-rived from an analysis of the previous para-graph. The point

P(x,y) = P(-12,-5)

lies on a radius vector in the third quadrantwith r = 13.

Now the functions can be read from thefigure.

sin 8

COS 8

tan

cot e

ordinate -5 5radivs vector T5

abscissaradius vector

-12 1213

ordinate -5 5abscissa

--12 12

abscissaordinate

-13 12-5 5

0taA

14-12,-5)

Figure 4-7.Finding the trigonometricfunctions for a third-quadrant angle.

36

radius vector 13 13sec 8 abscissa 7-11 12

radius vector 13 13csc 0 = ordinate -5

Reference to figure 4-6 shows that the func-tions have the correct algebraic signs for thethird quadrant. The solution also meets thespecifications of the problem for tan 8 = it andand sine 0 < 0 as the sine of 0 is .

PRACTICE PROBLEMS: Without using ta-bles, find the sine, cosine, and tangent of 0 un-der the following conditions:

31. tan 0 = , r = 5, and 0 is not in the firstquadrant.

-212. tan 0 r = 29, and cos 0 > 0.20

ANSWERS:

1.

2.

sin =

cos =

tan 0 =

sin =

-3

-4

3

-2121

20cos 8 =

-21tan 0 =

Reference Angle

The reference angle for any angle 0 instandard position is the smallest positive anglebetween the radius vector of 0 and the X axis,In general the reference angle for 0 is nir ± 0,where n is an integer. Expressed in anotherform 0' = n(180°) * 0, where O'is the referenceangle for 0 and again n is some integer.

In trigonometric analysis the reference an-gle is used to form a reference triangle. Thistriangle is used to find the functions of an anglewhen less information is given than was avail-able in the problems of the previous section.


For a geometrical explanation of the ref-erence triangle construct an angle in standardposition, such as the angle 9 in figure 4-8. IfP(x,y) is any pout on the radius vector a per-pendicular from P to the point A on the X axisforms a right triangle with sides OA, OP, andAP. Then the X coordinate equals * the dis-tance OA, the Y coordinate equals ± the dis-tance AP, and r equals the distance OP. Thesigns of x and y depend on the quadrant in whichthe radius vector falls. The distances OA andAP are not only the X and Y coordinates, theyare also the lengths of the sides of the refer-ence triangle AOP.

In the previous section it was possible tofind the functions if the tangent, value of r, andsome means of determining the quadrant wereknown. Using the reference triangle, the rangeof solvable problems of this type is extended.

EXAMPLE: Find the sin, cos, and tan of 017when csc 0 = - 15 , cos 0 < 0.

SOLUTION: The cosecant is negative wherethe sine is negative; i.e., quadrants 3 and 4.The cosecant and cosine are both negative onlyin quadrant 3. Construct an angle in standardposition in the third quadrant as in figure 4-9.Drop a perpendicular to the negative X axis andlabel the angle formed by the X axis and theradius vector 0'.

P(x,y)

A

Figure 4-8.Reference triangle.

X= 1,517Ri

-1517

Figure 4-9.Reference triangle in quadrant 3.

It can be seen in figure 4-9 that angles 0 and9' have the same terminal side. Therefore, thefunctions of 9 and0' are identical, and csc 0'=

r hypotenuseSince cscy opposite 'the hypotenuse is

labeled 17 and the opposite side is labeled -15.17NOTE: The fraction - indicates that15

either the numerator or denominator is nega-tive, but not both. In this case, we know thatthe denominator is the negative member fromthe following:

1. By convention, r is always positive.2. The Y coordinate is negative in quadrant

3.X From the PythagorPan theorem the value of

x is found to be 8 and the negative x value is -8.From this information (y = -15, x = -8, r = 17)the functions can be written as follows:

42

37

sin 9 =sin17

8cos 9 = cos 0' =

-tan o = tan 0 = 15=

15-8 8


EXAMPLE: Find the value of the six trigo- 7

nometric functions of 0 if sec 9° = - and25..;0 < 0 < Ir. 24

rants 2 and 3; since 0 must lie between 0 I nd 7

cos 0 . - ,SOLUTION: The secant is negative in mad- c5

radians, or 00 and 1800, the angle must lie in - 7tan 0 =quadrant 2. Construct and label the reference 24triangle as shown in figure 4-10 (A). Use thePythagorean theorem to determine the y value, 24cot 0

y = Jr2 - x2

y = 1/252 - (-24)2

y = 114-25

y =

7

25sec 0 z4

25csc 0 =7

EXAMPLE: Find the sine, cosine, and tan-gent of 0 when cot 0 = 1.

SOLUTION: This example requires analysisof two points that were not encountered in pre-

= 7 vious examples. First, with only the cotangentgiven, how is it possible to determine which of

With the reference triangle labeled as in two quadrants (where the cotangent is positive)figure 4-10 (B), the functions can be read will contain the angle? The cotangent is posi-directly from the figure. tive in quadrants one and three; from the given

25

-24 X

(A) (B)

Figure 4-10.Reference triangles in quadrant 2.

38

X


information it is not possible to determine inwhich quadrant the angle falls, In this examplewe are dealing with two angles. The relation-ships of the functions of these two angles shouldbe observed when they are determined.

The second new point to consider is thevalueof the function. There are an infinite number ofside lengths that give a value of 1 for the co-tangent. Any right triangle which has equal ad-jacent and opposite sides will result in thissituation. This does not really present a prob-lem, however, because the angle with a cotan-gent of 1 has the same value whether the sidesare 1 and 1, 14 and 14, or any other equal sidelengths. For simplicity, we consider the func-

1tion to be the ratio of y in this problem andconstruct reference triangles as in figure 4-11.

From the figure the functions of 01, and 02can be written directly.

1sin 01

=-7"

cos 9 =-2-

1/2

Figure 4-11.--45° reference angles.

44

and

1tan 91

= = 1

1 Ifsi.n.

2

-1cos 02 =

v2

-a

A ." 1tan u == 12 -1

It is shown here that the functions for theseangles are identical, or differ only by algebraicsign. This is as expected, for the angle 01the reference angle 03 are equal. The solutiot,to this problem would normally consider thegeneral angle 0 and be given in the followingform:

sin 0 =

cos0=± 2

tan 0 = 1

Use of Tables

The tables of trigonometric functions nor-mally contain only values for the functions ofangles between 00 and 900. Use of referenceangles provide a means of using the tables tofind the values of the functions for anglesgreater than 90°. The principle involved here.

X is the following: Consider 0 as any angle and9' as its reference angle. Any trigonometricfunction of 0 is equal to the same function of 0,with the sign of the quadrant in which 9 termi-nates attached.

This is shown geometrically in the followingmanner. Construct the angle 0 and the refer-ence triangle with inclosed angle 9' as shownin figure 4-12. The triangle AOB is the refer-ence triangle and it can be seen that the point Bis also a point on the radius vector of 0. Wederive the functions of the two angles to showthat they are equal except perhaps for thealgebraic sign.

39


8=145°

Figure 4-12.-Angles 0 and '.

X

First, consider the triangle AOB. The lengthsof the sides of this triangle are OA and AB, andthe length of the hypotenuse is OB. If we dis-regard any signs and treat the angle 61' as anypositive acute angle the sine function is deter-mined as follows.

si opposite ABn 9hypotenuse OB

Considering angle 0, we determine the co-.o-dinates of point B to be x = OA and y = OB (ifwe disregard signs). Considering the signs, itshould be clear that the coordinates of B aregiven by x = *OA and y = i0B, where the signsare determined by the quadrant in which 8 ter-minates. This allows the sine of 0, for any sizeof 0, tc., be determined as

sth Z= tAB ABr OB OB

By definitiort; :he radius vector OB is alwayspositive.

Hence, we have determined the sine of 9 toOA ABbe * and the sine of 0' to be 0--B- . TheseOB

functions values differ only by algebraic sign asstated earlier.

4540

EXAMPLE: Find the six trigonometricfunctions of 145°.

SOLUTION: The angle 145° in standard po-sition is shown in figure 4-13. The referenceangle is the smallest positive angle between theterminal side (or ray) and the X axis or, in thiscase, an angle of 35°. Since 145° lies in thesecond quadrant the sine and cosecant are posi-tive and all other functions are negative.

Referring to the tables in appendixes III andIV, and utilizing the principle explained in thischapter we have

sin 145° = sin 35° = 0.57358cos 145° = - cos 350 = -0,81915tan 145° = - tan 350 = -0.70021cot 145° = - cot 350 = -1.42815

Recall that earlier in this chapter it waspointed out that certain functions were recipro-cals of other functions. Using this we can de-termine

1csc 145° = csc 350 = sin135°

0.57358 - 1'7434

1sec 145° = -sec 350= -cos 35" -0.81915

-1.2208

Figure 4-13.-14V angle and reference angle.


EXAMPLE: Find the sine and cosine of an 4. In which quadrant must 0 fall when:angle of 690°.

SOLUTION: An angle of 6900 is coterminal a. sin 0>0, col 0<0with an angle of 3300

b. cos 0<0, tan 0<06900 - 360° = 330°

c. sec 0>0, csc 0<0and both have a reference angle of 30° as shownin figure 4-14.

The angle terminates in the third quadrantso the sine is negative and the cosine positiveand

sin 6900 = - sin 30° = -0.5000

cos 690° = cos 30° = 0.86603

PRACTICE PROBLEMS: Without using ta-bles, find the six trigonometric functions of 0under the conditions given in problems 1, 2,and 3.

1. cos 0 = - 3, 0 not in quadrant 2.

52. sin

83. tan 0 = ir<0<217.15 '

Figure 4-14.-690° angle.

4641

5. Use tables to find the sine, cosine, andtangent of 281°.

ANSWERS:

41. sin =

3cos a = -

tan 8 =

cot 0

3

4

A 5sec 17 =

5csc 9 = -

52. sin

12cos 0 = 13

5tan 0 =

12cot 0 = -3-

13sec 0 = *T2

CSC 0 = - 135

83. sin

15cos 0 =


8tan 9 =

cot 0 =.

17sec 0 = - rd-

17csc a = - 8-

4. a. 2b. 2c. 4

5. sin 281° = - 0.98163cos 281° = 0.19081tan 281° = 5.14455

SPECIAL ANGLES

There are two groups of angles consideredin this section. The first group considered con-tains angles which occur so frequently in prob-lems that their functions are normally consid-ered separately.

The second group considered contains thoseangles whose radius vectors fall on one of thecoordinate axes. These angles cannot be con-sidered as falling in one of the quadrants, andthe group is treated as a special case.

FREQUENTLY USED ANGLES

As stated previously, the approximate valuesof the trigonometric functions for any angle canbe read directly from tables or can be determinedfrom the tables by the use of principles statedin this course. However, there are certainfrequently used simple angles for which theexact function values are often used becausethese exact values can easily be determinedgeometrically. In the following paragraphs thegeometrical determination of these functions isshown.

30° - 60° Angles

To determine the functions of these anglesgeometrically, first construct an equilateraltriangle with the side lengths of 2 units. Thefunctions to be determined are not dependent onthe lengths of these sides being two units; this

42

size was selected for convenience. The ratiosfor given values of angles will be constant forall side lengths. Triangle OYA in figure 4-15is a triangle constructed as described above.

If a perpendicular is dropped from angle Yto the base at point X, two right triangles areformed. Consider the right triangle YOX,formed by the perpendicular, which alsobisectsangle Y forming a 30° angle. This triangle con-tains a 60° and a 300 angle. It is seen in figure4-15 that the side adjacent to the 60° angle isone-half the length of the hypotenuse, or in thiscase 1. Using the Pythagorean theorem withright triangle YOX, the length of the side oppo-site the 60° angle (the length of the perpendicu-lar) is found to be 4.

Figure 4-16(A) shows the right triangle YOXof figure 4-15 tranderred to a rectangular co-ordinate system, with the 60° angle positionedat the origin of the coordinate system.

The triangle now is set as a reference tri-angle, and inspection of figure 4-16 shows thatthe adjacent side is one-half the length of thehypotenuse. This relationship holds for any 60°triangle, regardless of the lengths of the sides.From the figure it is seen that the cosine of 60°is 1 The remaining functions can be read from2 .the figure.

Consider now the right triangle XYA of fig-ure 4-15. If this triangle is placed on a coordi-nate system with the 30° angle positioned at theorigin, the situation is as shown in figure 4-16(B). In this figure it is seen that the side opposite

47 -

Figure 4-15.Equilateral triangle.

Y

1

X


the 30° angle is one-half the length of the hy-1potenuse. Therefore, the sine of 30° is T .

(A)

X

(8)

Figure 4-16.Triangle on coordinate system.

In dealing with 30°-60°-90° triangles it is notnecessary to memorize the functions of the 30"and 600 angles. These angles will occur inpairs in right triangles and if one remembersthat the side opposite the 30° angle is one-halfthe length of the hypotenuse it is a simple mat-ter to construct a reference triangle with sidelengths of 1, 2, and ITS to derive the functions.

45° Angles

For a geometrical derivation of the functionsof 45° angles construct a square, one unit lengthon a side, as shown in figure 4-17. If a diago-nal is drawn from point 0 to point Y, two righttriangles are formed. Inspection of the figureshows that triangle XOY contains two 45° angleswith side lengths of 1 unit. These conditionsare also met in the reference triangle in figure4-18. Here it is seen that the two legs of a 450triangle must be equal. This relationship istrue of all 45° triangles and is not altered by thelengths of the legs.

Reference to figure 4-18 allows the functionsfor 45° to be written directly. These functions,like those for 30° and 60°, need not be memo-rized. It is only necessary to remember thatthe legs of a 45° triangle are equal; from this,it is a simple matter to construct a referencetriangle with sides of unit length and all of thefunctions can be read from the reference figure.

A

I

0

1

///////

////

////

45 //45"

r1

Figure 4-17.Determining 45° angles.

43

48,


2

45.

,I)

2

Figure 4-18.-45° reference triangle.

The trigonometric functions for 30°, 60°, and450 are summarized in table 4-1 for ready ref-erence. The function values shown are alsoapplicable for any angle which has one of thesefor a reference angle, upon proper considera-tion of the appropriate algebraic signs of x andy in the various quadrants.

EXAMPLE: Find the six trigonometricfunctions of 300°.

SOLUTION: With the angle drawn in stand-ard position as shown in figure 4-19, choose thepoint (1, - arS) on the radius vector with r = 2.The functions of 3000 are then

Figure 4-19.-300° angle in standard position.

vsin 3000 = =

cos 300° = i= 1

tan 300° = = -

cot 300°Y 3

Table 4-1.--Trigonometric functions of special angles.

0 sin 0 cos 0 tan 9 cot 0 sec 0 csc o

300 12

Vii sri-3- la

2 4.5 23

60. v-§-i

12 it 45

.5 22,5

3

450 ri2

ViI 1 1

44

49


sec 3000 = = 2

csc 3000 = L=Y 3

QUADRANTAL ANGLES

In previous sections of this chapter thoseangles which are exact multiples of 90° havebeen ignored, in order to simplify the explana-tions. The functions of angles which are exactmultiples of 90° (0°, 90°, 180°, 270°, -90°, etc.)are considered a special case and the anglesare called quadrantal angles. By definition, anangle which has its radius vector falling alongone of the coordinate axes, when the angle is instandard position, is a quadrantal angle.

The trigonometric functions of the quadrantalangles are defined in the same manner as forother angles, except for the restriction that afunction is undefined when the denominator ofthe ratio is zero.

To derive the functions of quadrantal angleswe choose points on the terminal sides wherer = 1, as shown in figure 4-20. In each angle inthis figure it is seen that r = 1. Then either xor y is zero and the other one is *1.

Consider the case where B = 90° in figure4-20 (A). To derive the functions use the pointP (0, 1) and r = 1.

rs1

sin r

x 0cos 90° = 1= 0

0

rx1(-IA)

(A) (B)

tan 90°

cot 90°

sec 90°

csc 90°

_ Y /- is undefined

r 1= = is undefinedx u

r 1= =1

=Y

The functions for the other quadrantal anglescan be determined from the other parts of fig-ure 4-20. The functions are summarized intable 4-2 for ready reference. The values ofthe functions of the special angles, quadrantaland 30°-60°-45°, are used frequently and for thatreason are important. However, additionalimportance is attached to the quadrantal anglesfor they serve as key values in the graphs oftrigonometric functions.

PRACTICE PROBLEMS: Without using ta-bles, determine the trigonometric functions of0 in problems 1 through 4.

1. 0 = 210°

2. 0 = 360°

3. 0 = 585°

4. 0 = - 180G

5. Without reference to tables or drawingsdetermine the value of 0 described in the fol-lowing, where x and y are points on the radiusvector r.

rz I

270°

X

(0O)

(C)

Figure 4-20. Functions of quadrantal angles.

45

50

0

es 06(I 0)

r:1

CD)


Table 4-2. Functions of quadrantal angles.

0sin 0 cos 0 tan 0 cot 0 sec 0 csc 0Deg. Rad.

0° 0 0 1 0 =defined 1 undefined

900 i 1 0 undefined 0 undefined 1

180° 1r 0 - 1 0 undefined -1_

undefined

270° 31T2-1 0 undefined 0 undefined -1

a. x y = 3

b.

c. y = -r, x = 0

2. sin 360° = 0

cos 360° = 1

tan 360° = 0

Without using tables, determine the nurneri- cot 3600 = undefinedcal value of the functions of indicated angles andverify the statements in problems 6 through 8. sec 360° = 1

6. sin2 150Q + cos2 15042

-= 1 csc 360° = undefined

7. sin 340° = 2 sin 170° cos 170°3. sin 585° = -

8. cos 270° = cos2 135° - sin2 135°

NOTE: Terms such as sin2 0 are used to cos 585° -indicate (sin 0)2.

11. sin 210° = -

cos 210° = -

tan 210° =3

cot 210° =

sec 210° = -

csc 210° = -2

2

46

51.

tan 585° = 1

cot 585° = 1

sec 585° = -

csc 585° = -

4. sin (-1800) = 0cos (-180°) = -1

tan (-180°) = 0cot (-180°) is undefinedsec (-180°) = -1csc (-180°) is undefined


5. a. 0 = 45°

b. 9=600

C. 8=2700

FUNCTIONS OF ANGLESGREATER THAN 90°

In previous sections of this chapter a meansof dealing with angles greater than 900, by useof reference triangles, was developed. Whenusing reference triangles, it was found that anyfunction of an angle is, except possibly for thealgebraic sign, numerically equal to the samefunction of the reference angle.

In addition to the reference triangle, thereare formulas for determining the signs of func-tions of any angle. These are normally calledreduction formulas. This section shows thegeometrical development of some of the mostcommonly used reduction formulas. In general,reduction formulas provide a means of renucingthe functions of any angle to an equivalent ex-pression for the function in terms of a positiveacute angle. In the discussion in the followingparagraphs, this acute angle is designated 0.The reduction formulas can be used in the solu-tion of some trigonometric identites and inother applications which require analysis oftrigonometric functions.

SINE OF AN ANGLE

The numerical value of the sine of an angleis equal to the projection of a unit radius on theY axis. Figure 4-21 shows a circle with a unitradius. According to the definition of the sineof an angle

sin 6 =

But r in figure 4-21 is equal to 1 so that

sin 0 = y

Note that y is equal to y', which is the projec-tion of the unit radius on the Yaxis. Therefore,

sin 6 = y'

does not contradict the definiaon of the sine ofan angle less than 90°.

47

Figure 4-21. Circle diagram, sine of an angle.

Now, consider an angle in the second quad-rant. In figure 4-21 we have constructed theangle (180° - 0). The sine of this angle is evi-dently the same as the sine of the angle in thefirst quadrant or

sin (1800 - 0) = sin 0

In the third quadrant the projection of unitradius on the Y axis becomes negative. Thesine of the angle (180° + 0) is the same in mag-nitude but of opposite sign to that of an angle inthe first quadrant, so that

sin (180° + 0) = - sin 0

These formulas give the value of the sine ofany angle between 900 and 270° in terms of thesine of an angle less than 90°. The sine of anangle in the fourth quadrant is found from theformula

sin (380° - 0) = sin (-0) = - sin 0

COSINE OF AN ANGLE

The numerical value of the cosineof an angleis equal to the projection of a unit radius on theX axis.

52 Cr'.


Using the same method of analysis that wasused for the sine, we develop the formulaswhich are used to evaluate the cosine of anyangle.

In figure 4-22, notice that the projections ofthe unit radius on the X axis, or in other words,the cosine of the angles 0 and 1800 - 9 are thesame length but of different sign.

cos (1800 - 0) = - cos 0

Mso, the cosine of an angle in the thirdquadrant is found from the relation

cos (180° + 0) = - cos 0

And lastly, the cosine of an angle in the fourthquadrant is found from the relation

cos (360° - 0) = cos (-0) = cos 0

TANGENT OF AN ANGLE

The value of the tangent of an angle is equalto the lcngth of that part of the tangent to theunit circle at 00 between y .=-0-amd the intersec-tion of the continuation of the unit radius withthe tangent line.

Figure 4-22.Circle diagram, cosine of anangle.

48

53

Let us clarify this definition by studyingfigure 4-23. The line MN is the tangent to theunit circle at 0°. The continuation of the unitradius CD cuts the tangent at M.

According to this definition, MA is the tan-gent of the angle 0. This new definition doesnot contradict the previous rule where we foundtan 8 from triangle CMA as follows:

tan 0 = MACA

Therefore, because CA is equal to 1

tan 0 = MA

For angles in the second quadrant we write

tan (180° - 0) = - tan 0

For angles in the third quadrant we have

tan (180° + 0) = tan 0

Figure 4-23.Circle diagram, tangent of anangle.


For angles in the fourth quadrant the relationis

tan (360° - 0) = tan (-0) = - tan 0

PRACTICE PROBLEMS: Express each ofthe following as trigonometric functions of O.

1. sin (180° - 0)

2. cos (360° - 0)

3. cos (720° - 0)

4. tan (1800 + 0)

5. tan (-0)

ANSWERS:

1. sin 0

2. cos 0

3. cos 0

4. tan 0

5. - tan 0This section has developed reduction formu-

las for the sine, cosine, and tangent of angles.The reduction formillas apply as well to theother three functions. The formulas developerlin this section and the corresponding formulasfor the remaining functions are summarized inthe following paragraphs.

Functions of 1800 - 0:

sin (1800 - 0) = sin 0

cos (1800 - 0). -cos 0

tan (180° - 0). -tan 0

cot (1800 - 0) = -cot 0

sec (180° - 0) = -sec 0

csc (180° - 0) = csc 0

Functions of 180° + 0:

sin (1800 + 0) = -sin 0

cos (1800 + 0) = -cos 0

49

54

tan (180° + 0). tan 0

cot (180° + 0). cot 0

sec (180° + 0) = -sec 0

csc (180° + 0). -csc 0

Functions of 360° - 0:

sin (360° - 0) = -sin 0

cos (360° - 0) = cos 0

tan (360° - 0) = -tan 0

cot (360° - 0) = -cot 0

sec (360° - 0)= sec 0

csc (360° - 0)= -csc 0

Functions of -0:

sin (-0)= -sin 0

cos (-0)= cos 0

tan (-0) = -tan 0

cot (-0) = -cot 0

sec (-0) = sec 0

csc (-0)= -csc 0

EXAMPLE: Use reduction formulas andtables to find the sine of 220°.

SOLUTION: 220° is in the third quadrantand can be considered a function of 180° + 9.

4

sin (180° + 0) = -sin 0

sin 220° = sin (180° + 40°)

sin 220° = -sin 400

sin 220° = -0.64279

EXAMPLE: Find the tangent of -350°.SOLUTION:First,

tan (-0) = -tan 9tan -350° = -tan 350°


then,

tan (360° - 0) = -tan 0

tan 3500 = tan (3600 - 3500)

tan 3500 = -tan 100

combining the two,

tan -350° = -tan 3500

tan 350° = -tan 100

therefore,

tan -350° = -tan 3500 = -(-tan 10°)

tan -350° = tan 10°

tan -350° = 0.17633

COMPOSITE CIRCLE DIAGRAM

All the trigonometric functions can be shownas lengths of lines in a circle diagram as il-lustrated in figure 4-24. The circle has a unitradius. A number of things may be learnedfrom this diagram. For example, the threesides of the right triangle OAB are sin 0, cos

, and 1, so that from the Pythagorean theorem

sin20 + cos 20= 1

In the same way, in triangle ORM

1 + tan20 = sec20

and in triangle ONP

1 : cot20 = csc20.

These three equations are true for any angle.These equations are trigonometric identities

and will be used in chapter 8 of this course.PRACTICE PROBLEMS: Use reduction

formulas and tables to find the values of thesine, cosine, and tangent of 0 in the followingproblems.

1. 0 = 137°

2. 0 = 2i4°

50

Figure 4-24.Composite circle diagram.

3. 0 = 325°

4. 0 = -290°

ANSWERS:

1. sin 137°

cos 137°

tan 137°

2. sin 214°

cos 214°

tan 214°

3. sin 325°

cos 325°

tan 325°

= sin 43° = 0. 68200

= -cos 43° = -0.73135

= -tan 43° = -0.93252

= -sin 34° = -O. 55919

= -cos 34° = -0.82904

= tan 34° = 0. 67451

= -sin 35° = -O. 57358

= cos 35° = 0. 81915

= -tan 35° = -0.70021


4. sin -290° = -sin 2900 = sin 700 = 0.93969

cos -290° = cos 2900 = cos 70° = 0.34202

tan -290° = -tan 290° = tan 70° = 2.74748

PERIODICITY OF FUNCTIONS

The trigonometric functions exhibit a prop-erty which is also possessed by some othermathematical functions. These functions ex-hibit a regular repetition of the values whicheach function has in a certain range.

The importance of the functions which areperiodic is that once the values are known forone period of the variable, the values of thefunctions are then known for all values thevariable takes on in its range.

GRAPH OF THE SINE

Figure 4-25 shows the graph of the sinefunction. The angle is plotted on the horizontalaxis, increasing to the right, and the corre-sponding value of the sine function is plotted onthe vertical axis. Two complete revolutionsare plotted. It can be seen onthe graph that thevalue of the sine varies between +1 and -1 andnever goes beyond these limits as the angle

1

varies. The graph also shows that the she in-creases from 0 at 00 to a maximum positive at90° (I) and then decreases back to 0 at 1800 (r).The value of the sine continues decreasing to a

31Tmaximum negative at 2700 (-2--) and then in-

creases to a value of 0 at 360°(2v). If the sec-ond revolution is analyzed it is found to repeatthe variations of the first revolutIon for thecorre: mnding points. Therefore, the period ofthe sine function is 360° or 271. radians.

GRAPH OF THE COSINE

The cosine also has a period of 271, as seenin figure 4-26. The range of values which thecosine can take on also lies between +1 and -1.However, as seen on the graph, the cosine va-ries from a value of 1 at 0° to a value of 090° and uontinues decreasing to reach a maxi-mum negative value at 180°(v). The cosine in-creases from the 180° point and reaches amaximum positive value (equal to 1) at the 360°(2 ir) point.

GRAPH OF THE TANGENT

The graph of the tangent, figure 4-27, showsa special kind of discontinuity called an infinite

Figure 4-25.Graph of the sine.

51


Figure 4-26.Graph of the cosine.

Figure 4-27.Graph of the tangent.

discontinuity, %len the angle is slightly lessthan 90°, (i), the value of the tangent will bevery large and positive. When the angle is ex-actly 4 the tangent curve and the Y axis areparallel. Thus, the tangent of is plus orminus infinity. When the angle becomes slightlygreater than the tangent assumes very large

52

negative values. Thus, the tangent goes fromlarge positive values through infinity to largenegative values at i. The same occurs at ,

This is called an infinite discontinuity, Theperiod for this function is 180° (7r radians).

PRACTICE PROBLEMS: In the problemplisted below use the graphs of figures 4-2514.:N.-26, and 4-27 to answer the questions; then usethe tables in appendixes III and W to verify theanswers.

For what values of 91. is cos 9 ircreasing if 0 0

2. do sin 9 and cos 9 decrease0-0 as- 2 rr?

3. do cos 9 and tan 9 increase

1__3; ?

4. do sin 0, cos 9, and tan 9 increase to-,gether if 0 5-9:5-2 Tr?

?together if

together if

ANSWERS:

1. None

2. 12150 te_g

3. r < 0 <2

3r4. 0 27T


r .'vUNCTIONS AND COMPLEMENTARYANGLES

Inspect the two triangles in figure 4-28.Triangle (B) is exactly the same as triangle (A)except that it has been turned on the side.

(A)

cos 0 = sin (90° - 9)

tan 0 = cot (90° - e)

sec 0 = csc 90° - 0)

csc 0 = sec (90° - 0)

cot 8 = tan (90° - 0)

The trigonometric function of an angle isequal to the cofunction of its complement. Thesix trigonometric functions consist of threepairs of cofunctions, The functions are arrangedin pairs so that the name of one can be obtainedfrom the other by adding or deleting the prefix"co." For example: sine, cosine; tangent, co-tangent; secant, cosecant.

90°-0 The cofunction principle accounts for theformat of tables of trigonometric functionssimilar to the one in appendix III. Refer to thepage of this table that contains the function of21'. Enter the table at the top of the sin 21°

Figure 4-28.Complementary angles, column and go down the column to 21°30' asdetermined by the minute column en the left.The value found for sin 21°30' (or 21.5°) is

In triangle (A) In triangle (B) 0.36650. Looking now to the bottom of the pageand reading up the minute column on the ex-treme right, find the cosine of 68°30' (or 68.5°).

sin e cos (90° - 0).1. It is seen that cos 68°30' is also 0.36650. Since

(B)

cos 021°30' + 68°30' = 90°

T sin (90° -21°30' and 68°30' are complementary angles and

tan 0 .the cofunctions are numerically equal. If the

=Z.cot (90° - 0) Y- relationship of complementary angles and co-functions did not hold true, a table of values offunctions from 0° to 90° would require 90 col-

sec 0 = xcsc (90° - 0)=f; umns each for the sine and cosine instead of

the 45 columns that are required.r PRACTICE PROBLEMS: Express the fol-csc = sec (90° - ) = lowing as a function of the complementary angle.

1. sin 27°

cot 0 .x tan (90° - 0) =2. tan 38°17'3. csc 41°4. cos 16°30'22"Express the following as an acute angle less

The ratio in triangle (A) is equal to ratio than 45°.5. sec 79°37'16"

1, in triangle (B); therefore 6. cos 56°7. sin 438°

sin 9 = cos (90° - 0) 8. tan 48°ANSWERS:1. cos 63°

In the same way, 2. cot 51°43'

53


3. sec 49° 6. sin 34°4. sin 73029'38" 7. cos 12°5. csc 10°22'44" 8. cot 42°

CHAPTER 5

OBLIQUE TRIANGLES

The tv.o previous chapters considered right-1.-Igles and angles which could be calculated byusing triangles which included right angles.This chapter considers oblique triangles whichare, by definition, triangles containing no rightangles.

In chapter 19 of Mathematics, Vol. 1, NavPers10069-C, a method for solving problems involv-ing oblique triangles was introduced. Thismethod employed the procedure of dividing theoriginal triangle into two oimore right triangles,and using the right triangles to solve the probleminvolved. It was also pointed out at that timethat there were direct methods of dealing withoblique triangles.

This chapter develops two methods of dealingdirectly with oblique triangles. These twomethods, or laws, are developed in the firstsection of the chapter as aids to calculations.The chapter also contains example and practiceproblems for solving oblique triangles con-sidered in four standard cases. In this chapter"solving a triangle" is defined as finding thethree sides a, b, and c and the three angles A,B, and C of an oblique triangle, when some ofthese six parts are given.

Also included in this chapter are some prob-lems using lo.rithms in solving oblique trian-gles (where another law is idroduced) and prob-lems concerning the area of a triangle whichcombine the area formula of plane geometrywith the laws developed in this chapter.

AIDS TO CALCULATION

The aids to calculations, or aids in solvingoblique triangles, developed in this chapter aretwo theorems, known as the law of sines and thelaw of cosines. This section is concerned withthe development and proof of these laws; sub-

55

sequent sections will be concerned with usingthem in calculations.

LAW OF SINES

The law of sines states that the lengths ofthe sides of any triangle are proportional to thesines of their opposite angles. If a triangle isconstructed and labeled as shown in figure 5-1the law of sines can be written

a b c= sin B =

PROOF: For the proof of the law of sines weredraw the triangle of figure 5-1 and drop aperpendicular from A to the oppc,:ite side asshown in figure 5-2 (A).

Reference to the triangle shown in figure 5-2(A) shows that the perpendicular from A to theopposite side has divided the triangle into tworight triangles and the trigonometric functionspreviously developed are used here. Consider-ing these two right triangles we obtain

and,

sin B = or h = c sin B

hsin C - b or h = b sin C

Here we have two expressions for h which areequal to each other, so

c sin B =b sin Cor in another form

c bsin C sin B

In figure 5-2 (B) the triangle is redrawn witha perpendicular from C to an extension of the


Figure 5-1.Triangle ABC.

(Al

h' = b sin A

to form

a sin B = b sin A

or

a

(S)

a bsin A sin P

Figure 5-2.Proving the law of sines.

opposite side. Considering the right triangleBCD thus formed, it is seen that

h'sin B = or IV = a sin Ba

and in triangle ACD

sin (180° - A) =If or h' = b sin (180° - A)

From chapter 4 of this training course, recallthat

sin (1800 - 8) = sin 0

SO

sin (180° - A) = sin A

then

h'sin A = or h' = b sin A

Now equate

h' = a sin B

and

56

(e)

Two separate triangles were used inprovingthe law of sines simply for clarity of explana-tion. If the two triangles are combined in onefigure, as shown in figure 5-2 (C), it is seenthat the two laws could be derived from this oneillustration. Here it is obvious that the angle Bwhich appears in both of the ratio pairs is thesame angle; thus the ratios

c bsin C sin B

and

asin A sin B

can be combinqd to form the law of sines

a b csin A sin B sin C

agi previously stated.

LAW OF COSINES

The second of the laws to be developed inthis section is the law of cosines which states:In any triangle the square of one side is equalto the sum of the squares of the other two sidesminus twice the product of these two sides

Chapter 5.0BLIQUE TRIANGLES

multiplit. the cosine of the angle between which expands toFor .he triangle in figure 5-3 (A), the

law of cosines can be stated as h2 = a2 - b2 + 2bx x2

= b2 + c2 - 2bc cos A

(6)

Substituting again for x the value given in (2)

PROOF: Consider the triangle in figure 5-3 results in(B) with a perpendicular dropped from B toside b to form two right triangles. To prove

h2 = a2 - b2 + 2bc cos A - c2 cos2 A (7)

a2 = b2 + c2 - 2bc cos A Equatnig the two values of h2 in (4) and (7) gives

(8)consider first the triangle ABD and note that

c2 - c2 cos2 A = a2 - b2 + 2bc cos A - c cos2 A

or

and also

cos A = (I) Canceling like terms and rearranging,

x = c cos A (2)

h2 = c 2 - x2 (3)

2 2 2 2 2c - e1----= a - b + 2bc cos A -

-a2 = -b2 c2 + 2bc cos A

a2 = b2 + c2 - 2bc cos A (9)

The proof is thus complete.Substituting in (3) the value of x given in (2) To prove another form of the law of cosines

results in

h2 = c2 - c2 cos2 A

In triangle BDC

h2 = a2 - (b - 42

(4)

(5)

c2 = a2 + b2 - 2ac cos C

refer to figure 5-3 (C) and note that in the righttriangle CAD

b

Figure 5-3.Proving law of cosines.

5762

cos C =1-

(C)


or

and also

y = b cos C

h2 = b2 - y2

Substituting for the value of y gives

h2 = b2 - b2 cos2 C

In triangle ABD

h2 = c2 - (a - 3)2

which expands to

2 2 2h = c - a + 2ay - y2

With additional substitution

2 2h = c - a2 + 2ab cos C - b2 cos2C

Equating the two h2 values which are representa-tive of the two right triangles results in

b2 - b2 cos2C

= c2 - a2 + 2ab cos C - b2 cos2C

Canceling and rearranging yields

c2 = a2 + b2 - 2ab cos C

completirk, the proof.The same procedures can be applied to prove

the remaining form of the law of cosines. Insummary, the three forms of the law of cosinesare

a2 = b2 + c2 - 2bc cos A

b2 = a2+ c2 - 2ac cos B

c2 = a 2 + b2 - 2ab cos C

The trigonometric functions or other methodspreviously noted are normally more effectivein dealing with right triangles; however, applica-tion of these laws can be used in an analysis ofsome trigonometric principles and identities.

EXAMPLE: Show that

c2 = a2+ b2 - 2ab cos C

holds true in the right triangle shown in figure5 -4.

.SOLUTION: In the figure it is shown thatC = 900. Recall from the graph of the cosine inchapter 4 of this course (or from appendix III)that cos 900 = 0. Therefore, the formula

c 2 = a 2 + b2 - 2ab cos C

can be reduced to

c2 = a2 + b2 - 2ab cos 90°

c2 = a2+ b2 - (2ab) (0)

c2 = a2 + b2

Reference to the figure shows that c is the hypo-tenuse al a right triangle and from the Pytha-gorean theorem

c2 = a2 + b2

The law of sines and the law of cosines areused mainly to solve oblique triangles, as willbe shown in the following sections. In addition,these laws also hold true for right triangles. Figure 5-4.Example problem, law of cosines.

58

Chapter 5 OBLIQUE TRIANGLES

Thus, the law of cosines would provide a propersolution to the right triangle.

In this problem the law of cosines reduces tothe Pythagorean theorem. However, in worldngwith oblique triangles remember that while thelaw of cosines applies to all triangles, the Pytha-gorean theorem can only be used when dealingwith right triangles.

PRACTICE PROBLEMS:1. Refer to figure 5-5 and prove that

b2 = a2 + c2 - 2ac cos B.2. Assume that the triangle in figure 5-5 is

such that a.b.c=2. Transpose the formula inproblem 1 and solve for cos B, then refer to thetable of functions in the appendix and verify thatB = 60°.

FOUR STANDARD CASES

It was stated previously that the solution of atriangle consists of finding the six parts (sidesa, b, and c; and the angles A, B, and C) whensome of these values are known. If three ofthese parts are known, at least one of which is thelength of a side, the remaining parts can normallybe calculated by one of the methods discussed inthe following paragraphs. For convenience, themethods for solving oblique triangles are devel-oped by considering the triangles in four catego-ries as follows:

1. Two of the angles and one of the sides areknown.

2. The three sides are known.3. Two of the sides and the angle between

them are known.

Figure 5-5.Practice problem, law of cosines.

59

4. Two of the sides and an angle that is notbetween them are known.

The last situation described in the precedinglist is known as the AMBIGUOUS CASE for, undercertain conditions, two triangles which are notcongruent can contain the same three knownparts.

Recall f rom plane geometry that two trianglesare congruent (having the same shape and size)if one of the following conditions is met:

1. Three sides of one triangle are equal tothe corresponding sides of a second triangl,

2. Two sides and the included angle of onttriangle are equal to-the corresponding parts o,a second triangle.

3. Two angles and a side of one triangle areequal to the corresponding parts of a second tri-angle.

It is seen here that the ambiguous case isthe only one that does not parallel a plane geom-etry theorem for congruent triangles. The firstand fourth (ambiguous) cases (or categories) oftriangles will employ the law of sines in the solu-tions and cases 2 and 3 will be solved by using thelaw of cosines.

TWO ANGLES AND ONE SIDE

When two angles and a side are known, the re-maining angle can be determined so easily thatthis case could be assumed to be one in which oneside and all angles are known. (The third angleis equal to the difference between 180° and thesum of the known angles.) The law of sines isthen used twice to find the length of the remainingsides. To find either of the unknown sides, selectthe ratio pair which includes the ratio involvingthe unknown side and the one which considers theknown side.

EXAMPLE: Using the law of sines, find thelength of the lettered sides in the triangle infigure 5-6 (A).

SOLUTION: From the law of sines

sin C = sin B

Since

5sm 97.5 = sin 30

sin 0 = sin (180° - 0),

sin 9750 = sin 82.5° = 0.99144

MATHEMATICS, VOLUME 2 .

Also

sin 30° = 0.5000

SO,

5 _ b 2.50000 s0.99144 0.5000 %°4 0.99144

or

b = 2.5216

Angle A is equal to

180° - B - C = 52.50

Again from the law of sines,c a

sin C sin A

5 a

sin 52.5° = 0.793355 a

0.99144 0.79335

5(0.79335)a0.99144

a = 4.001EXAMPLE: Figure 5-6 (B) shows a flagpole

standing vertically on a hill which is inclined15degrees with the horizontal. Aman climbing thehill notes that at one point his line of sight to thetop of the pole makes an angle of 40° with thehorizontal. At another point, 200 feetfurtherupthe hill, this angle has increased to 55°. Howhigh is the flagpole? Solve using only the law ofsines.

SOLUTION: First, define all the angles in thetriangles OAB and OBD. In triangle OAB

BAO = 40° 15° = 25°

1-OBA = 180 - (550 - 15°)

= 180° - 40° = 140°

L AOB = 180° - 140° -25° = 15°

60

In triangle OBD

LDBO = 55° - 15° az. 40°

LBDO = 90° + 13° = 105°

BOD = 90° - 55° = 35°

These two triangles have OB as a commonside. We can use the law of sines to find BOin triangle OAB and then apply the law again intriangle OBD to find the length of side OD whichis the height of the flagpole. Thus,

AB OBsin AOB sin BAO

200 OBsin 15 - sin 25

OB 200 sin 25°= 326.57 ftsin 15-

And in triangle OBD

OB ODsin BDO sin DBO

326.57 _ODiiiios° si71-417

OD326.57 sin 40°sin 105

sin 0 = sin (180° - 0)

sin 105° = sin 75°

326 57 sin 40°ODsin 75

OD = 217.3 ft

PRACTICE PROBLEMS: Refer to figure 5-7in solving the following problems where thefigures (A), (B), and (C) are to be used respec-tively with problems 1, 2, and 3. Use the law ofsines in solving these problems.

1. Find a and b using the values given in thetable of functions of special angles (chapter 4of

Chapter 5-0BLIQUE TRIANGLES

_this -course). Leave answers in radical form ANSWERS:

where applicable.2. Find aides d and f to two decimal places.3. Find the length of ato two decimal places.

(A)

(8)

Figure 5-6.Case 1, example problems.

(A)

1. a=3b = 3V-3-

2. d = e.o7f 3.96

3. 8.39

THREE SIDES

When the three sides of a triangle are given,the triangle can be solved by three successiveapplications of the law of cosines. Each applica-tion yields the value of one angle. The order ofdetermining the angles is not important; any ofthe three angles may be determined first. Aparticular angle is found by using the form ofthe law of cosines in whichthe cosine of the anglein questions appears. When the three angles havebeen found, the solution is checked by verifyingthat A + Fs + C 180°.

EXAMPLE: Solve the triangle ABC, givena = 7, b = 13, and c 14. Determine the size ofthe angles to the nearest degree.

SOLUTION: To simplify the procedure solvethe law of cosines algebraically for cos A.

a2 = b2 + c2 - 2bc cos A

2 bc cos A = b2 + c2 - a2

cos A = b2 + c2 - a2

10

(B)

Figure 5-7.Case 1 practice problems.

61

66.

2bc

10

(C)


The remaining forms of the law can besolved in the same manner and the results are

and

cos B a2 + c2 - b22ac

cos C a2 + b2 - c22ab

22cos C =182

cos = 0.12088

C = 83°

Checking the calculations gives

Now in the given problem A + B+ C = 300+ 670 4 830 = 1800

Then

and

cos A b2 + c2 - a22bc

cos A 132 + 142 - 722 x 13 x 14

cos A 169 + 196 - 49364

cas A - 316364

cos A = 0.86813

A = 30°

ccs B a2 + c2 - b22ac

=cos Jo 49 + 196 - 1692 x 7 x 14

It may appear that tile best method to use insolving a triangle when three sides are givenwould be to calculate two angles and find the thirdangle by subtracting the sum of the twofrom 180°.While this method shortens the computation, italso destroys the check on the calculations, andis not recommended.

EXAMPLE: Solve the triangle ABC whena = 8, b = 13, and c = 17. Express the angles tothe nearest degree.

SOLUTION:

Then

b2 + c2 - a2cos A

2bc

cos 4k 169 + 289 - 64442

cos A = 394442

cos A = 0. 8914e,

A = 27°

76 cos B 64 + 289 - 169cos B = 5-6 272

cos B = 0.67647cos = 0.38776

B = 67°

and finally

cos C = a2 + b2 - c22ab

B = 47"

cos C 64 + 169 - 289208

,., 56cos C = 49 + 169 - 196 cos k. -182 208

62

67

Chapter 5-0BLIQUE TRIANGLE8

cc,s- C = -0.26927

cos C = -cos 740

-cos C = cos 740

a2 = 49

2a = 74

a274

+ 25 - 70 cos 19°

- 70(0. 94552)

- 66.1864reference to reduction formulas gives

-cos C = cos (180° - C)

and

C = 106°

then, checking,

A + B+ C = 27° + 47° + 106° = 1800

PRACTICE PROBLEMS: The side lengths oftriangle ABC are given in the following problems.Use the law of cosines to determine the sizesof the angles to the nearest degree.

1. a = 3, b = 4, c = 52. a= 2, b=3, c =43. a 7, b = 14, c = 11ANSWERS:1. A = 37°

B = 53°C = 90°

2. A = 29°B = 47°C = 104°

3. A = 29°B = 100°C = 51°

TWO SIDES AND THEINCLUDED ANGLE

Where two sides and the angle between themare given, the triangle is solved most easily byrepeated use of the law of cosines. Using thegiven parts, solve first for the unknown side.Then, with three sides known, solve for the re-maining angles in the same manner "s in casetwo.

EXAMPLE: Using the law of cosines, solvethe triangle ABC shown in figure 5-8, angle ac-curacy to the nearest degree.

SOLUTION: First find the unknown side.

a2 = b2 + c2 - 2bc cos A

/ 2 2a"- = 7 + 5 - 2(5) (7) cos 19°

63

68

a2 = 7. 8136

a = 7.8136

a = 2. 795

To compute the angles, round the values givenabove to

and

cos B = a2 + c2 - b22ac

25 - 492 x 2.8 x 5

cos B = - 16' 228

cos B = -0.57857

B = 125°

2 2 9

COsqC = a + b2ab

7.8 + 49 - 25 31.839.2

= 179°

cos C = 2 x 2.8 x 7

cos C = O. 81633

C = 35°

then, checking,

A + B+ C =19° + 125° + 35°

A+ B+C a 180°This is acceptable with the accuracy requiredhere.

EXAMPLE: Two ships leave port at thesame time; one (ship A) sailed on a course of050° at a speed of 10 knots, the second (ship B)sails on a course of 110° at 12 knots. How farapart are the two ships at the end of 3 hours?


e a 5

Figure 5-8.Case 3, example problem.

SOLUTION: A good rule in any problem ofthis type is to first draw a picture to show theproblem. In figure 5-9 a coordinate systemoriented to compass headings is constructed, andthe given information is plotted. From the figureit is seen that the desired answer is the distanceAB opposite the angle labeled C, where C = 1100- 500 = 600 .

Using the law of cosines

c 2 = a 2 + b2 - 2ab cos C

c 2= 362 + 302 - 2(30 x 36) cos 600

1c2 = 1296 + 900 - (1080) x

c 2 = 2196 - 180 = 1116

c =

c = 33.5 approximately

The ships are approximately 33.5 miles part atthe end of 3 hours.

PRACTICE PROBLEMS: Use the law ofcosines to solve the triangles described below.Express angles to nearest degree mid sides totwo decimal places.

1. a = 10, b = 7, C = 25°2. b = 11, c = 17, A = 20°3. a = 12, c = 26, B = 140°ANSWERS:1. c = 4.69

A = 116°B = 39°

2. a = 7.65B = 29°C = 131°

6964

Figure 5-9.Plot of ship's courses.

3. b = 36A = 12°C = 28°

TWO SIDES AND ANOPPOSITE ANGLE

When two sides and a nonincluded angle aregiven, the triangle falls in the ambiguous cate-gory and one of the following cases will exist:

1. There is no solution.2. There are two solutions.3. There is one solution.The category is called ambiguous for the

given parts cannot alvays establish the shape andsize of one triangle. There may be two triangleswhich are not congruent, but still contain thegiven parts. The ambiguity of this category canbe seen if we assume that three parts (B, b, andc) are given, and we attempt to construct thetriangle from this data. We consider the possi-bilities as follows:

1. If angle B is obtuse as in figure 5-10 (A),the side b must be larger than side c for atriangle to exist. In this case, b>c, there isonly one triangle which exists and only one solu-tion.

2. If B is a right angle, as infigure 5-10 (B),side b must be larger than side c for a triangleto exist and there is only one triangle and onesolution.


3. Figure 5-11 shows the situations whichcan exist if B is an acute angle. In (A) a figureis constructed with A <900 and b < c; that is, aline drawn from vertex A is too short to reachthe line BC. In this case, no triangle exists andthere is no solution.

4. In figure 5-11 (B), the line from A (sideb) is exactly the, distance from A to the line BC.In this situation, only one triangle exists and-itis a right triangle with one solution.

5. Figure 5-41 (C) shows a triangle wherethe line from A touches the line BC in two places.In this case the sides b and b' are longer thanthe side b in figure 5-11 (B), but still shorterthan line c. In this category there are twotriangles, BAC' and BAC, which contain thegiven parts. With the law of sines two solutionscan be found for this possibility.

6. The last possibility considered is oneshown in figure 5-11 (D). In this figure the lineb is longer than c and again would touch the lineBC in two places to form two triangles. How-ever, only the triangle ABC is considered Sincethe triangle ACC' does not include angle B asan interior angle. This situation is consideredto nave only one solution. If b = c, then c and c'coincide and there is only one triangle.

Ce rtain relationships of angle, side, and func-tion values can be found to determine in advancewhich of the possibilities previously listed existsfor a given triangle. However, this knowledgeis not required before the solutionris attempted.It can be determined in the process of attempt-ing a solution, as will be shown in the exampleproblems, or a drawing can be made from thedata given.

The triangles presented in this section havehad value sizes only in relation to eachother or

- in relation to 90°. Figure 5-12 points out theambiguity which can exist when a triangle isdescribed by giving two sides and an angle op-posite one of these sides. Thisfigure shows twotriangles constructed with given data of A=30°,a = 4-9*-and c = 6. Solution of these two triangleswill show that B, C, and b are not the same forthe two, and this is a case where two solutionsarise.

A good approach to solving triangles whentwo sides and an opposite angle (ambiguouscategory) are given is to first use the law ofsines to find the unknown angle opposite a givenside. Then the third angle can. easily be deter-minEd and the law of sines can be used again,to compute the unknown sicid. In the followingexamples and practice' problems desired

( A)

)

Figure 5-10.Ambiguous case, B > 900.

accuracy is in degrees to the nearest minuteand sides to two decimal places.

EXAMPLE: Solve the triangle (or triangles)ABC when B = 45°, b = 3, c = 7.

SOLUTION; First use the law of sines .tofind angle C

c bsin C sin B

65

70

sin C = c sin B

sin C = 7 sin 4503

sin C 7 x 0.707113

sin C = 4.949173

sin C = 1.64992


A

( A )

(C)

The calculations show that

sin C >1

A

Figure 5-11.Ambiguous case, B < 900

However, reference to tables or a graph of thesine function shows that the sine isnever greaterthah 1, so this is the case where no triangle orsolution ex1st:3.

EXAMPLE: Solve the triangle (or triangles)ABC when A = 22°, a = 5.4, c = 14.

SOLUTION: Apply the law of sines to deter-mine angle C.

c asin C sin A

a

66

( D

sin C 14 sin 22°5. 4

sin C 14 x 0.374615114

sin C = 0.97121

C = 76° 13'

Since the side opposite the known angle is smallerthan the other given side, there are two anglesto consider. Since

sin (1800 - C)= sin C

the other angle (C') is 103° 47'.Contimie the solution considering two trian-

gles, ABC and A'B'C'. In ABC;

Chapter 5--OBLIQUE TRIANGLES

( A )

(B)

Figure 5-12.--,Two different triangles derivedfrom identical given ciata.

b = 14.27

This completes the solution of triangle ABC.In the second triangle, angle B' isfound first

by using values previously calculated

8'

13'

8'

8'

=

=

=

=

180° - + C')

1800 - (22° + 103°

180° - 125° 47'

540 13'

47')

Then from the law of sines

a' sin 8'sin A'

b' 5.4 x sin 540 13'sin 22°

5 4 x 0.811230.37461

b' = 11.70

A = 22°, a = 5.4, c = 14, and C = 76° 13' This completes the solution for lx)thpossibletriangles. Figure 5-13 shows a scale drawing ofthe triangle of this example. The angle A was

In A'B'C'; A' = A, a' = a, = v, and constructed with one side of length 14 unitster-minating at B, andthe other line toform an angleof 22°. A compass was set to 5.4 units and an arc

C' = 103° 47 '.was struck using B as the center. As shown inthe figure, the arc intersected the line fromA at two points. Therefore, there are two

Solving ABC first, angle B is found by triangles which satisfy the data.

B = 180° - (A + C)

8 = 1800 - (22° + 76° 13')

B = 180° - 98° 13'

B = 81° 47'

Then by the law of sines

a sin Bsin A

b = 0.374615.4 X 0. 98973

EXAMPLE: Solve the triangle (or triangles)when A = 350, a = 10, b = 8.

SOLUTION: From the law of sines

sin B = b sin A

Then

41; 4

a

sin B = 8 sin 35010

sin B = 0.45886

B = 27° 19'

C = 1800 - (A + B)

it


A= 22°

OC14II4

Figure 5-13.Scale drawing of example triangles.

C = 1800 - (350 + 270 19')

C = 1170 41'

Apply the law of sines again to find c

c - a sin Csin A

10 sin 117° 41'C = sin 35°

C =10 x 0.88553

O. 57358

c = 15.44

but

SO

35° + 270 19' + 620 19' =124° 38'

A + B + C' # 1800

and ABC' is not a triangle described by thegiven data.

PRACTICE PROBLEMS: In the followingproblems use the given data to solve the tri-angle or triangles involved.

1. C = 100°, c = 46, b = 302. A = 40°, a = 25, b = 303. B = 42°, b = 2, c = 44. B = 30°, b = 10, a = 10ANSWERS:

This solves one triangle and since the side 1. B = 39° 58', A = 40° 02', a = 30.04opposite the given angle is larger than the other 2. B = 500 29', C = 89° 31', c = 38.89 andgiven side, there should be only one solution. B = 1290 31, C = 100 29', c = 7.08However, if this point is overlooked and the 3. No solutionsolution is continued in an attempt to find a 4. A = 30°, C 120°, c = 17.32second solution as in the previous example

sin C' = sin (180° - C)

sin C' = sin (180° - 117° 41')

sin C' = sin 620 19'

C' = 62° 19'

Now if this angle is contained in a secondtriangle described by the given data, it isknown that

A + B + C' = 180°

-

68

LOGARITHMIC (;OLUTIONS

Computations involving triangle solutions areoften concerned with the multiplication or divi-sion of trigonometric functions, which containvalues given in four or five decimal places, andother values which may also contain numerousdigits. There are many opportunities for arith-metic errors in these computations and, inmanycases, the errors are the result of the multipli-cation and division by large decimals. In


logarithmic solution the computations are re-duced to addition problems, and the number oferrors is frequently reduced.

By the combined use of tables of trigonome-tric functions and tables of logarithms one couldsolve the triangles by first finding the value forthe function and converting this to a logarithmor by converting the logarithm of a function valueto a decimal and then convertingthis to an angle.However, the logarithm equivalents of the princi-pal natural functions have long since been workedout, and are available in tables, so that the neces-sary multiplication or division may be performedby the use of logarithms.

A table of "common logarithms of trigonome-tric functions° usually lists the logs for the sine,cosine, tangent, and cotangent of angles from0° through 180°. Appendix I shows a sample pagefrom such a table; reference to the appendixshows that both the characteristic and the man-tissa are listed. In addition, for each valuelisted, a characteristic of -10 at the end of thelog is understood.

Take the log listed for the sine of 38° 00'00", for example. This is listed as 9.78934.What this actually means is 9.78934-10, which inturn means that the log of this function is actu-ally -1+.78934. On the other hand, the log listedfor the tangent of 51° 10' 00" is 10.09422. Whatthis means is 10.0S422-10; in other words, thelog of this function is 0.09422. The logs areprinted in this manner simply to avoid the neces-sity for printing minus characteristics. Notethat, even when a characteristic is minus, themantissa is considered as plus.

A complete table of the logarithms of trigono-metric functions is not included in this course.For purposes of the examples and practice prob-lems in this course, a short table of values forthe sine, cosine, and tangent is given in table 5-1.The values in this table are given for each 5°from 0° to 90°. The complete logarithm (bothcharacteristic and mantissa) is given in thistable. The problems in this section will beworked with an accuracy of side lengths to threedigits and angles to the nearest 5°.

The solution of oblique triangles was con-sidered in four cases. In logarithmic solutionsthe solutions are also considered in four cases.In cases 1 and 4 the law of sines was used forsolutions. Since the law of sinea fits well withlogarithmic solutions, cases 1 and 4 alsouse thelaw of sines for logarithmic solutions. Cases 2and 3 were solved using the law of cosines; how-ever, the law of cosines involves addition and

749

subtraction and does not lend itself to logarithmicsolutions. For cases 2 and 3, we will usemethods other than the law of cosines for log-arithmic solutions.

CASES 1 AND 4

As previously stated, the solution of these twocases by the law of sines adapts readily to log-arithmic solution since the law of sines involvesmultiplication and division. Example solutionsof the cases are given in the following para-graphs.

EXAMPLES: Use logarithms to solve the tri-pee ABC when A = 1.10°, B = 25°, and c = 125.

SOLUTION: Find the unknown angle as thefirst step

C = 180° - A - B

C = 180° - 1100 - 25°

C = 45°

Using the law of sines with ratios involvinga and c find the value of a.as follows:

a csin A sin C

a = sin C x sin A

Taking the logarithm of both sides of theequation gives

log a = (log c - log sin C) + log sin A

The logarithms of trigonometric fulititionsgiven in table 5-1 include only angles from 0° to90° so the equation above becomes

log a = (log c - log sin C) + log sin (180° - A)

log a = (log 125 - log sin 45°) + log sin 70°

Refer to table 5-1 and appendix II and convertthe values to logarithms. One method of simpli-fying the computation is to convert each log-arithm to one with an end characteristic of -10and use the following procedure

log 125 = 12. 0969 - 10log sin 45° = 9. 8495 - 10 subtract


Table 5-1. -Logarithms of trigonometric functions.

Degrees Log sin Log cos Log tan

0 0000

5 8.9403 - 10 9.9983 - 10 8.9420 - 10

10 9.2397 - 10 9.9934 - 10 9.2463 - 10

15 9.4130 - 10 9.9849 - 10 9.4281 - 10

20 9.5341 - 10 9.9730 - 10 9.5611 - 10

25 9.6260 - 10 9.9573 - 10 9.6687 - 10

30 9.6990 - 10 9.9375 - 10 9.7614 - 10

35 9.7586 - 10 9.9134 - 10 9.8452 - 10

40 9.8081 - 10 9.8843 - 10 9.9238 - 10

45 9.8495 - 10 9.8495 - 10 0.0000

50 9.8843 - 10 9.8081 - 10 0.0762

55 9.9134 - 10 9.7586 - 10 0.1548

60 9.9375 - 10 9.6990 - 10 0.2386

65 9.9573 - 10 9.6260 - 10 0.3313

70 9.9730 - 10 9.5341 - 10 0.4389

75 9.9849 - 10 9.4130 - 10 0.5720

80 9.9934 - 10 9.2397 - 10 0.7537

85 3.9983 - 10 8.9403 - 10 1.0580

90 . 0000

log (c/sin C) = 2.2474

log (125/sin 450 ) = 12.2474 - 10 add

log sin 70° = 9.9730 - 10

log a = 22.2204 - 20

log a = 2.2204

70

Taking the antilog,

a = 166

To complete the solution, find side busing thesame procedure

b asin B sin A

b = x sin B


log b = (log a - log sin A) + log sin B Evaluation of the logarithms gives

log b = (log 166 - log sin 700) + log sin 25° log b = 10. 6021 - 10

log sin A . 9.8081 - 10 add

log 166 = 12.2204 - 10 log (b x sin A) = 20.4102 - 20

log sin 70° = 9.9730 - 10 subtract

log (a/sin A) = 2.2474

log (a/sin A) = 12.2474 - 10

log sin 25° = 9.6260 - 10

log b = 21.8734 - 20

log b = 1.8734

b = 74.7

log (b x sin A) = 20.4102 - 20

log a = 10.4771 - 10 subtract

log sin B = 9.9331 - 10

We note that log sin B is less than 0, sothere are two argles to consider, say B and IV,

add where sin (180° - B) = sin B'.

Now

log sin B = 9.9331 - 10

B = 60°

B' = 120'

This completes the logarithmic solution for Then the corresponding angles, C and area triancle in case 1.

EXAMPLE:. Solve the triangle (or tri- C = 180° - (A + B)angles) when A = 40°, a = 3, b = 4. C = 180° - (40° + 60°)

This is an ambiguous case and with the .sideopposite the given angle smaller than the other C = 80°given side there are two possibilities: eitherthere are two solutions or side a is too short andto reach the baseline and there are no solutions.Recall from earlier examples that there is no C' = 1800 - (A + B')solution when the sine of the angle upposite thesecond given side (angle B in this case) is C' = 1800 - (40° + 120°)greater than one. In logarithmic solutions thecondition for no solution is when the log sin of C' = 20°the angle is greater than zero; this correspondsto a sine greater than one. Reference to table5-1 shows that all of the values listed for log To complete the solution find sides c and c'.sin are less than zero (negative characteristic). Consider first side c and use the following form

SOLUTION: Solve first for the unknown of the law of sinesangle opposite a given side using the law of sines

b sin Asin B = a

this in logarithmic form becomes

log sin B = log b + log sin A - log a

7671

a sin Cc = sin A

or, in logarithmic form,

log c = log a + log sin C - log sin A


Evaluate the logarithms

log a = 10. 4711 - 10

log sin C = 9.9934 - 10

log (a x -in C) = 20. 4705 - 20

log sin A = 9.8031

log c = 10. 6624

Taking the antilog,

c = 4. 6

Finally, to find use

log e' =

log a

log sin C'

, a sin C'c sin A

add

- 10 subtract

- 10

log a + log sin C' - log sin A

. 10. 4771 - 10

= 9. 5341 - 10

log (a x sin C') = 20. 0112 - 20

log (a x sin C') = 10. 0112 - 10

log sin A = 9. 8081 - 10

log c' = O. 2031

Therefore,

= 1. 6

add

subtract

PRACTICE PROBLEMS: Use logarithms tosolve the triangle (or triangles) described by thefollowing data.

LAW OF TANGENTS

The law of cosines does not lend itself to loga-rithmetic solutions. The two cases in which weused the law of cosines are solved by loga-rithms using two different methods. The firstmethod to solve triangles in case 3, where twosides and the included angle are given.

The law of tangents is expressed inwords asfollows:

any triangle the difference between twosides is to their sum as the tangent of half thedifference of the opposite angles is to the tangentof half their sum.

For any pair of sidessuch as side a and sidebthe law may be expressed as follows:

a - b tan 1/2 (A -a + b tan 1/2 (A + B)

The law may be expressed in a form that includesother combinations of sides and angles by sys-tematically changing the letters in the formula.

In solving case 3 by the law of tangents. selectthe formula which includes the given sides, saya and b; then angle C is also given. The sum ofthe unknown angles, A + B, is found as

A + B = 180° - C

and the law of tangents is used to find A - B.After the sum and difference of A and B aredetermined, the angles themselves can be found.With the angles known, the law of sines is usedto find the unknown side.

EXAMPLE: Solve the triangle ABC whenA = 25°, b = 10, c = 7.

SOLUTION: With b and c given and b > c usethe law of tangents in the form

b - c tan 1/2 (B - C)b + c tan 1/2 (B + C)

First, determine the sum of the unknown angles

1. A = 70°, B 100°, c = 50 B + C = 180° - A2. A = 80, a = 11, b = 183. A = 40°, a = 25, b = 30 B + C = 180° - 25°ANSWERS:

B + C = 155°1. C = 10°, a = 271, b = 2842. No solution Then 1/2(B + C) = 1/2(155°)3. B = 50°, C = 90°, = 39.84. B' = 130°, C' = 10°, c = 6.75 1/2(B + C) = 77 5°

72- -

77


and

b - c = 10 - 7 = 3

b + c = 10 + 7 = 17

At this point, the only unknown in the formulafor the law of tangents is tan 1/2(13 C). Thenext step is to transpose the formula and solvefor this unknown.

b - c tan 1/2(B - C)b + c tan 1/2(B + C)

(b - c) (tan 1/2(B + C)) =(tan 1/2(B - C)) (b + c)

(b - c) (tan 1/2(B + C)) tan 1/2(B - C)b + c

tan 1/2(B - C) = 3 tan 77 5017

Rounding the angle to 80° for use with the giventable, the following logarithmic equation can bewritten:

log tan 1/2(B - C) = log 3 + log tan 80° - log 17

then

log 3 = 10. 4771 - 10

log ts.n 80° = 10. 7537 - 10

log (3 x tan 80°) = 21. 2308 - 20

log 17 = 11.2304 - 10

log tan 1/2(B - C) = 10. 0004 - 10

log tan 1/2(B - C) = O. 0004

1/2(B - C) = 45°

B - C = 90°

(+)

(-)

There are now two equations for B and C,(B + C) and (B - C), these are solved simulta-neously to find B and C.

First the two are added to find B

B + C = 155°

B - C = 90°

28 = 240°

B = 120°

Next, subtract the two to find C

B + C = 155°

-B + C = -90°

2C = 65°

C = 32. 5°

To fit the given table round C to 350, then

A + B + C = 25° + 120° + 35° = 180°

In the final part of the solution, use the law ofsines to find side a.

73

78

asin A sin E

b sin Aa = sia B

log a = log b + log sin A - log sin B

log a = log 10 + log sin 25° - log sin 120°

log 10 = 11. 0000 - 10

log sin 25° = 9. 6260 - 10

log (b sin A) = 20. 6260 - 20

log sin B = 9. 9375 - 10

log a = 10. 6885 - 10

log a = O. 6885

a = 4. 88

(+ )


There are six forms of the law of tangents,two of which have been shown. The remainingfour are formed, as previously stated, by re-placing the corresponding letters inthe formula.

PRACTICE PROBLEMS: Use logarithms andthe law of tangents to solve the triangles de-scribed by the given data.

1. a = 4, b = 3, C = 60°2. a = 0.0316, b = 0.0132, C = 500ANSWERS:1. A = 70°, B = 50°, c = 3.692. A = 105°, B = 25°, c = 0.0239

CASE 2

The logarithmic solution of oblique triangleswhen three sides are given involves formulasderived from the law of cosines. These formulasare called half-angle formulas and, in these solu-tions, are expressed in terms of the semiperi-meter of the triangle and the radius of a circleinscribed in the triangle.

The half-angle formulas expressed in $(semiperimeter) and r (radius of inscribed cir-cles) are, not presented in this course. Thelogarithmic solutions of oblique triangles inthiscourse are limited to cases 1, 3, and 4.

AREA FORMULA

In this section two formulas for finding thearea of oblique triangles are given. Theseformulas are used to find the area of trianglesin cases 1 and 3 from the given parts.

Recall from plane geometry that the area ofa triangle is found by the formula

A 4bhwhere b is any side of the triangle and h is thealtitude drawn to that side. To avoid confusionbetween A for area and A as an angle in the tri-angle the word "area* will be used inthis chap-ter. Then the formula is stated as

1area = bh

Reference to figure 5-3 (B) shows that thelength of the altitude h can be found by

h = c sin A

79'74

Substituting this value of h in the area formularesults in

1area = bc sin A

bc sin Aarea - 2

This area formula is stated in words asfollows:the area of a triangle is equal to one-half theproduct of two sides and the sine of the anglebetween them. This fornrila is used for solutionsof triangles when two sides and the included angleare given.

A second formula for area can be derivedfrom the law of sines and the previous formula.From the law of sines

sin B sin C

cb sin B- sin C

Substituting this value of b in the area formula

results in

bc sin Aarea =

area = (c sin 13) (c sin A)sin C, 2

c2 sin a sinarea = 2 sin C

T his formula can be used to solve triangles whentwo angles and one side are given since, whentwo angles are given, the third angle can be founddirectly. It can be seen that the area formulascan be easily adapted to logarithmic solutions,as well as to normal solutions.

PRACTICE PROBLEMS: Derive the areaformulas most applicable when the following aregiven.

1. a, b, CP. a, c, B3. A, C, b4. B C. a

1. area =ab sin C-2-

Chapter 5:.:41.-iQUE TRIANGLES

2. area - ac sin B2

b2 sin A sin C3. area 2 sin B

4. area - a2 sin B sin C2 sin A

EXAMPLE: Find the area of triangle ABCwhen A = 40°, b = 13, and c = 9.

SOLUTION: Use the first area formulagiven in this section:

area

area

bc sin A2

13 x 9 x sin 400

13 x 9 x 0.64279area= 2

area = 37.6 (square units)

This formula (as well as the other areaformula) adapts easily to logarithmic solutions.In a logarithmic solution the formula

area - bc sin A2

is written

log area = log b + log c + log sin A - log 2

EXAMPLE: Find the area of triangle ABCwhen A = 25°, B = 105°, c = 12.

SOLUTION: First determine angle C.

C = 180° - (A + B)

C = 180° - 130°

C = 50°

then apply the area formula

2 A sin Bsin

area = 111212 sin 25° sin 105°1 2 sin 50°

area - 144 x 0.42262 x 0.965932 x 0.76604

(1)

With the specific values involved in this prob-lem, a logarithmic solution should simplify thearithmetical process. To write the logarithmicequation go back to equation (1). Recall that thelogarithm of 124 is 2 log 12 and that sin (180° -

105°) = sin 75° and write the equation as

log area = 2 log 12 + log sin 25°

+ log sin 75° - (log 2 + log 50°)

2 log 12 = 2.1584

log sin 25° = 9.6260 - 10

log sin 75° = 9.9849 - 10 add

log (c2 sin A sin B) = 21.7693 - 20

log (c2 sin A sin B) = 11.7693 - 10

andlog 2 = 0.3010

log sin 50° = 9.8843 - 10 add

log (2 sin C) = 10.1853 - 10

subtracting these sums

log (c2 sin A sin B) = 11.7693 - 10

log (2 sin C) = 10.1853 - 10 subtract

log area = 1.5840

area = 38.37

PRACTICE PROBLEMS: Find the area ofthe triangles described by the given data.

1. Find the area of the triangle given in the

second example problem (A = 25°, B = 105°,

c = 12), without using logarithms.

So'15

MATEMATICS, VOLUME 2

2. A = 25°, B = 450, c = 24 ANSWERS1. 38. 369

3. A = 42°, b = 4. 4, c = 3 2. 91. 63. 4. 4

4. A = 120°, b = 8, c = 12 4. 41.5

CHAPTER 6

VECTORS

Vector quantities, which we will now concernourselves with, are different from scalar quan-tities. Navigation involves the use of vectorquantities, surveyors use vectors in their work,as do structural engineers, and electrical andelectronic technicians. Many of the applica-tions of electricity and electronics involve theuse of vector quantities.

DEFINITIONS AND TERMS

In this section we will make a distinctionbetween a scalar quantity and a vectbr. We

will also define the coordinate systems used inworking with vectors and show some of thesymbols used.

SCALARS

Heretofore, we have been concerned withscalar quantities, which are measurements orquantities having only magnitude, in the appro-priate units. Examples of scalar quantitiesare: 10 pounds, 4 miles, 17 feet, and 28.2pounds per square inch.

VECTORS

A vector, in contrast to a scalar, has di-rection as well as magnitude. Examples ofvector quantities are: 6 miles due north, 9

blocks toward the west, and 250 knots at 30°.Notice that the vector quantities have both amagnitude and a direction.

1

SYMBOLS

The letters A, B, C, and D have been pre-viously used to represent scalar quantities inalgebra. In vector algebra, a notation is usedto denote scalar symbols in relation to vectorymbolo. A dash over the letters, for exaMple

A and 15, denotes vectors.

A vector can conveniently be representedby a straight line. The length of this straightline represents the magnitude, and its positionin space represeats the direction of the vectorquantity. In figure 6-1 the vefctors A , B, and C.

are equal because they have the samemagnitudeand direction and vector 15 is not equal to

either vector X, or C because although ithas the same magnitude it does not have thesame direction.

In navigational problems, a coordinate sys-tem is used in which the compass points serveas indicators of direction, and magnitude isgiven by lengths of the lines. For example,using the origin of the coordinate system asthe point of departure from the harbor, figtire6-2 represents two ships heading out to sea.Vector represents a ship bearing 450 fromdue north at a speed of 20 knots, and vectorrepresents a ship bearing 60° from due northat a speed of 25 knot% Notice that directionsin this coordinate system are measured clock-wise from due north.

When we use the trigonometric system indesignating angles or giving a direction to amagnitude we use the Cartesian coordinatesystem which includes the abscissa (measure-ment on the X axis) and the ordinate (measure-ment on the Y axis). Directions in this coordi-nate system are measured counterclockwisefrom the X axis. For example, using the originof the coordinate system as the point of de-parture from the harbor, figure 6-3 representsa ship heading out to sea bearing 30° at a speedof 25 knots. This is represented by vector laand is the same representation as vector B infigure 6-2 but is shown on a different coordi-nate system.

Angle measurements will be referencedfrom the vertical in the compass coordinatesystem and will be referenced from the hori-zontal in the Cartesian coordinate system.

77

82


Figure 6-1.Comparison of vectors.

Figure 6=2.Compass coordinate system.

COMBINING VECTORS

If a vector A represents the displacementof a particle or the force acting on the particle,it is convenient to let -A represent the dis-placement of a particle in the opposite direc-tion or to rffresent a force in the directionopposite to A. Thus, vectors A and -A areequal in magnitude but are opposite in direc-tion.

0(ORIGIN)

(ORDINATE)

25 KNOTS

X ( ABSCISSA)

Figure 6-3.Cartesian coordinate system.

ADDITION

The resultant of two vectors acting in thesame direction or acting in opposite directionsis the algebraic sum of their magnitudes. Anexample of this is walking due east four stepsand then walking due west one step. The re-sultant is three steps due east. If one travelsfrom his home to his place of employment hemay have several choices for his route. Wewill assume two of these routes as indicated infigure 6-4. He ?nay move east to point A thennorth to point B or he may move north topoint C then east to point B. In either case hearrives at his place of employment. If weassume his travel in both directions as forcesacting on him, we can call the direct distancefrom home to place of employment, in figure6-4, the RESULTANT of these two forces andrefer to it as vector W. Notice that eitherpath taken results in vector R.

We may now state that ol -I- AB = 4-10

CB = W. The symbol (4) is used to indicatethat vector ra is added vectorially to vectorX. From this it is apparent that vectors maybe added in either order with the same results.

If several vectors A, IN, and W are to beresolved into components, lc, Tat A, and Yr,Vat Yb are used to denote these components,as figure 6-5 portrays.

In the addition of vectors, the initial pointof vector Ti must be placed directly on the

78

Chapter 6VECTORS

ers.0 A E

PLACEOFEMPLOYMENT

terminal point of vector A, and so on for anynumber of vectors. Then vector whichjoins the initial point of vector A with theterminal point of the last vector /It is theresult of adding vectors A and and C throughN vectorially. Hence A 4-e 1§. r 4-*

N= W.. Here it may be shown that thecommutative and associative principles apply,which means that it makes no difference whichvector is used first and which order is followedwhen adding vectors. (The commutative andassociative laws may be reviewed in Mathe-matics, Vol. 1, NavPers 10069-C.)

SUBTRACTION

Subtracting a vector is defined as adding anegative vector:

It follows that if

X-4-13=0Then

Figure e-4.Right angle vectors. =

IMMO OM 1. MEM IMOD IMIN

Vr ifb

r ILD.M.NoMIMO =No imM i MOM

Figure 6-5.Components of vectors.

84 79


A careful study of figure e-4 will show thatif

Then

VECTOR SOLUTIONS

In the previous sections we agreed that:

X-I-- . #N=RApproaching this relation from a graphicalstandpoint, one can understand exactly whatthis means.

GRAPHICAL

As an example of the graphical method, sixvectors may be used to represent the pathtaken by a man looking for a lost golf ball. Hestands at position Po, hits the ball and doesnot notice where the ball went. Vectors Athrough r in figure 6-6 represent the path hetakes in an attempt to find the ball. Pf is theposition where the ball is found and we willcall it the termination point. Figure 6-7 showsanother of the many different arrangements ofthe six vectors. The dotted lines from Po to Ptindicate the resultant vector, and this resultanthas the same magnitude and direction regard-less of the arrangement of the vectors we use.This method of graphically solving vector

Figure 6-6.Polygon example number 1.

Figure 6-7.--Polygon example number 2.

problems is called the polygon method, and isused in civil engineering problems involvingstructures such as bridges; it is also, used inlogic problems of everyday living.

If two vectors are to be resolved into asingle resultant, this may be done graphicallyby the parallelogram method. Given any twovectors A and g lying in a plane (fig. 6-8)form a parallelogram by projecting t onto W,initial point to terminal point, and A onto B,initial point to terminal point, thus forming'a parallelogram which has as a diagonal theresultant vector

This process can be reversed in order tofind the components of a vector as shown infigure 6-9. Vector N is given and the problemis to find the rectangular components of this

80

Figure 6-8.Parallelogram example 1.

Chapter 6VECTORS

Fikure 6-9.Parallelogram example 2.

A

x 901b

vector. In this case the parallelogram is arectangle and the projections of Ti on the Xaxis and the Y axis show the components.Generally, the graphical method of resolvingvectors will be used to check the validity of ananalytical method of solution.

ANALYTICAL

The trigonometric functions are used tosolve vector problems analytically.

EXAMPLE: Find the resultant of twovectors at right angles to each other. Vector

represents 90 pounds of force and vector 13represents 60 pounds of force.

SOLUTION: Vector A" is directed verticallyand B lies on the reference line, as shown infigure 6-10. In this case the angle 6' is un-known and the resultant is required. ThePythagorean theorem is sufficient to solve forthe magnitude of the resultant. This is thecase only if we establish a right triangle fromour vector and its components.

Since the magnitude of If is a scalar quan-tity we will designate it by r. As you recallfrom Mathematics, Vol. 1, NavPers 10069-C,the Pythagorean tiworem of right trianglesstates: x2 + = r4. We apply this to ourfigure and find that:

r2= (90)2 + (60)2

r = V1 (90)2 + (60)2

= Iii,7oo

= 108.2 pounds

Figure 6-10.--Vector sum.

If we desire the angle 8 of the resultantvector IT we may use the trigonometric functionfor the tangent of an angle; that is,

Then,

tan 0 = ix = = 1.50000

= 56° 19'

EXAMPLE: Resolve the vector IT into itscomponents. In figure 6-11, If represents 50mph at 30°. We are to find the Yr and Yrcomponents.

SOLUTION: Recalling that the trigonometricfunctions for sin 8 and cos 8 are:

and

sin 8 = Z

XCOB 8 = r


mIn . ow, mml as, ens wil

30°

Xr

Figure 6-11.Vector resolution.

I

I

we use these and find that

then

and

then

sin 300 = -Y-50

50 sin 300 = yy = 25 mph

xcos 30° = 50

50 cos 30° = xx = 43.3 mph

Vector components acting in the same direc-tion or in opposite directions may be added orsubtracted algebraically. Vector componentsin the form li = Yr + Yr fulfill this statement.

EXAMPLE: Add the following vectors givenby their rectangular components.

SOLUTION:

If

Then

13- = ri - T

X + li = IT + I

Observe that the first component of eachpair is the X component, and the second is theY component. If vector X is to be added to W,the X component of the resultant is the sum of

the IC components of A and S. The samereasoning applies to the Y components. Noteespecially that an X component is never addedto a Y component or vice versa.

PRACTICE PROBLEMS: Add the followingre :tangular form vectors.

1. n - 3 and I + 72. 5:5-6 + ER and 1:121 + Ur3. ir.T + CS and U-.72. - 3-.14. 11 2- + SW and 78- - 111

ANSWERS:

1. IF - 22. KIT + 'KO3. TB. + ET4.EXAMPLE: Subtract the following vectors

given by their rectangular components.

If

X = 47 + IT

Then subtract X from rt.

Thus

PRACTICE PROBLEMS: Subtraez the fol-lowing rectangular form vectors.

1. Subtract KY + 17. from Kr +2. Subtract FT + El from WS - 233. Subtract 22:2 - 3:2 from 13:2 + 113:24. Subtract -ta + U from -IT - F.-6

ANSWERS:

1. Kg + rf2. 46- - g'S3. -TT + 111:54. TIT ''. 33The notation up to this point has involved

the regular rectangular coordinates. The formW = X. + Yr implies that a number of hori-zontal units and a number of vertical unitscombine to determine the end point of a vector.A second method commonly used describesa vector in terms of polar coordinates.

Chapter 6VECTORS

POLAR COORDINATES

If the length of the vector is known, allthat is required to locate the vector is theangle through which it has been rotated. Mea-sured from the reference line, the notationused is

= r

where r is the magnitude and 9 defines thedirection. Thus, r is a scalar quantity. Forinstance, a vector 10 units long at 30° wouldbe written 10/300.

If Yr and Yr are known, the scalar quantityr can be found by using the Pythagoreantheorem:

r = (xr)2 (yr)2

The angle can then be found by using the tan-gent, thus:

Yrtan 0 = --

x

Now we have a method whereby we can changefrom the rectangular form to the polar coordi-nate form when working with vectors.

EXAMPLE: Change the vector 1. + T intopolar form. (NOTE: xr is always placed first.)

SOLUTION:

If

Then

r = = 5

4tan 9 = -5 . 1.33333

9 = 53° 8'

R = "a +

= 5/ 530 8'

PRACTICE PROBLEMS: Change the rec-tangular form into polar form,

1. T +2.

83

S. ,PY + T4. ro 2:11ANSWERS:1. 2.24/63° 26'2. 10/36° 52'3. 2/30°4. 18.5/8° 42'This method may be reversed and it is

possible to change a vector from polar torectangular coordinates.

EXAMPLE: Change the vector 30/65° intorectangular form.

SOLUTION:

If

R = 30/65°

= r sin 0

Then

r = 30 sin 650

= 27.18930

And

= 30 cos 65°

Xr = r cos 0

= 12.67860

Thus

R = 12. 68 + WTI

PRACTICE PROBLEMS: Change the polarform to rectangular form.

1. 5/25°2. 83/72°3. 20/63°4. 8.2/31° 23'ANSWERS:1. T:51" + ITT2. 2376- + 1133. 175- + 17.824. 7.00 + 4:21-If we are to combine two vectors, proceed

as follows:EXAMPLE: Find the resultant of two vec-

tors X and IN if X = 12/102° and 13 = 5/12° infigure 6-12.


l's Y'r 12ta.n 0 = =5

= 2 40000I x't r1

Therefore/i , 0 = 67° 23'

%

%

%

r12 #

%

%' Now, the direction of the resultant is 67° 23'I

,% from the X' axis but the X' axis is 12° from

% the X axis so the resultant 13 67° 23' + 12° or% 790 23' from the X axis.

%

% a .................Ior % - X

---,5

12°

Figure 6-12.A new reference.

Another approach to this problem is byX resolving each of our vectors into their X axis

and Y axis components and then adding thesecomponents algebraically. We have 5/126whichresolves into the following (fig. 6-13):

SOLUTION: In this problem we do not haveeither vector to coincide with the X axis or theY axis. We may choose a new frame of refer-ence, that is, X' and Y' to determine themagnitude of the resultant. Because K differsin direction by 90° from II, we may still usethe properties of right triangles.

Therefore

And

sin 0 = Z

= Z5

= 5 sin 12°

= 5(0. 20791)

= 1. 03955

Therefore r2 = (12)2 + (5)2 COS 0 = r

And r = (12)2 + (5)2 =

= Therefore

= 13 i = 5 cos 12°

We now have the magnitude of our resultant = 5(0. 97815)and need only to find its direction. As we are

=concerned with only two vectors we may ap- 4. 89075proach this probleth in either of two ways. Wemay find our direction from our new referencethen add the angle our new reference makeswith the standard X axis and Y axis reference. We have now determined the x axis andIn the new frame of reference we find the Y axis components of one of the vectors. Weresultant to be: will proceed to find the components of the

Chapter 6VECTORS

12°

Figure 6-13.Components for 5/12°.

other vector, 12/102°, as shown in figure6-14. The Y axis component is as follows:

If

Then

And, since

Then

Therefore

sin 8 =

sin 0 =12

= 12 sin 102°

sin (180° - 0) = sin

sin 102° = sin (180° - 780 )

= sin 78°

= 12 sin 78°

Figure 6-14.Components for 12/102°.

We now find the X axis components as follows:

If

Then

And, since

Then

= 12 cos 102°

cos (180° - 0) = - cos 8

cos 1020 = cos (180° - 78°)

Therefore

= - cos 78°

= 12(0.97815) = 12(-0.20791)

= 11.73780 = - 2.49492

85


We must now add the X axis componentsand the Y axis components of the two vectorsalgebraically as follows:

And

= 1.03955 + 11.73780

= 12.77735

= 4.89075 + (-2.49492)

= 2.39583

We now have r and "Ir in rectangular form andmay use the Pythagorean theorem to find theresultant in scalar measurement as shown infigure 6-15. This is as follows:

If

Then

r2 = (2. 39583)2 + (12. 77735)2

r = V 5.74030 + 163.25173

= V 168.99203

= 13

This is in agreement with the result found byusing the method of finding the scalar resultantof two vectors.

We must now find the direction of if, asfollows:

Since

Therefore

tan =

12. 777382. 39583

= 5. 33480

0 = 790 23'

This direction agrees with the direction foundwhen we used the first method of finding thedirection of the resultant of two vectors.

86

91

12.77735

2.39583

Figure 6-15.-Resultant of 7 and V.

PRACTICE PROBLEMS: Find the resultantof two vectors at right angles to each other.

1. X 5/0°13- = 10/90°

2. X = 7.5/90°13- = 6.3/180°

3. X = 131/185°= 00/275°

4. X = 65/45°F = 120/135°

ANSWERS:1. 11.18/63° 26'2. 9.8/130° 2'3. 144.1/209° 36'4. 136.5/106° 36'Let us examine a problem of adding several

vectors. We will use the method last de-scribed. The method may be used to find theaddition of any number of vectors. We willconsider a problem of the addition of severalvectors, as follows:

EXAMPLE: Find the resultant of the vec-tors in figure 6-16, analytically.

Chapter 6VECTORS

Figure 6-16.Resultant of several vectors.

SOLUTION: The vectors are given as fol-lows:

X is 50a°g is 100/30°

is 75/90°TY is 50/143° 8'

E is 70.7/2.15°is 55/315°

We will use the method of resolving eachvector into its component X axis and Y axiscoordinates. We set up the coordinate systemand place each vector so that it radiates fromthe origin. Then we find

ib x 100 cos 30°= 100(0. 86603)

= 86. 6

Tib = 100 sin 30°= 100(0. 50000)

= 50

= 75 cos 90°= 75(0)

= 0= 75 sin 90°= 75(1)

= 75

id = 50 cos 143° 8'= 50(-cos 36° 52')= 50(-0. 80003)

= - 40. 002= 50 sin 143° 8'= 50(sin 36° 52')= 50(0. 59995)

= 30

= 70. 7 cos 225°= 70. 7 (-cos 45°)= 70. 7 (-0. 70711)

= - 50

if

= 70. 7 sin 225°= 70. 7(-sin 45°)= 70. 7(-0. 77011)

= - 50

= 55 cos 315°= 55 (cos 45°)

ia = 50 cos 0° = 55 (O. 77011)

= 50(1) = 38. 9

= 50 = 55 sin 315°

-37a.= 50 sin 00 = 55(-sin 450)= 50(0) = 55(-0. 77011)

= 0 = - 38. 9

9287


We now collect the X axis components and the rectangular form vectors as complex numbers,Y axis components, as follows: as follows:

Vector it I-

X 50 0T3- 86. 6 50

0 75

-ff - 40 30E - 50 - 50it 38. 9 - 38. 9

Adding the X axis components and the Y axiscomponents, we find the magnitudes of r andY as follows:

If

Then

And

= 3 + 41

rt2 = 8 + 5i

3 + 418 + 51

= 85. 5 24 + 321

+ 15i + 2012

The magnitude of the resultant rt is

r = V (85.5)2 + (66.1)2

11680. 67

= 108

The direction is given by using the tangentfunction, as follows:

66. 1tan o =

= 0. 77309

Therefore

e = 370 42'

MULTIPLICATION

Before we discuss the mechanics of multi-plication and division of vectors, in polar form,we will multiply and divide vectors in rec-tangular form. This will serve as an intuitiveexplanation of why the mechanics of polar formmultiplication and division may be used.

As explained in Mathematics, Vol. 1, Nav-Pers 10069-C, we may express the following

Thus

24 + 471 + 2012= 24 + 471 + 20(4)= 4 + 47i

Mi);)We now find,_ as shown previously, the polarform of + ri which is as follows:

(4)2 (47)2

=

= 47.2And

47tan 0 = = 11. 75000

0 = 85° 8'Then

(Tti)(rt2) = 47. 2a52LE

In multiplying vectors i and if2, in polarform, we first change to polar form as follows:

If

88

93

Chapter CIVECTORS

ThenFt

1= 5/53° 8'

irt2 = 9. 43Zar

The multiplication of R1 and it2 results in aproduct which we will label 11192. The follow-ing rule will be used:

To multiply two vectors find the product ofthe scalar quantities and the sum of the anglesthrough which they have been rotated.

In our example

rt1

= 5/53° 8'

and

The product

1-12 = 9. 43/32°

of the scalar quantities is

(5)(9. 43) = 47. 2

the angles isand the sum of

(53° LE) + (32°) = 85° 8'

We now have the product of VI and if2 which isTh,2 and is equal to 47.2/85°- 8' This resultis the same as the result of multiplying thevectors in rectangular form and we intuitivelyunderstand why the mechanics of polar multi-plication may be used.

DIVISION

We will now divide vector by rI2 inrectangular form as follows:

If

Then

= 40 + 3

II1

= 40 + 301

= 8 + 51

89

94'

Thus

And

If

Then

(40 + 30i) (8 - 51)8 + 5i -877E

320 + 401 - 15012

64 2512

470 + 40189

470 40 4

42 89 ge.

= 5. 28 + O. 449i

= 5.28 + URI

r = 11-(-5:78)2 + (O. 440)2

= 07.7)= 5. 3

tan 8 = 0' 4495. 28

= O. 08504

0 = 4° 52'

In dividing vector by No in polar form,we first change to polar iorm as follows:

If

It1

= 475 + 'NU

Then

rt1

= 50/36° 52'

= 9. 43g2°_

In division of vectors in polar form we willuse the following rule:

To divide two vectors, in polar form, findthe quotient of their scalar quantities and thedifference between the angles through whicththey have been rotated.


Thus

rti 50/36° 521r2

=9. 431320

= 5. 3/4° 52'

This result is the same as the result obtainedby dividing vector R1 by R, in rectangularform, and we intuitively uncTerstand why themechanics of polar division may be used.

PRACTICE PROBLEMS: Multiply the fol-lowing vectors:

1. (5/100) (10a)2. (8.3Pr) (1.1/73°)3. (6.2/52°) (8/200°)4. (100/45°) (30/20°)

ANSWERS:

1. W/15°

2. 9.13/19°3. 49.052°

4. 3000/650

PRACTICE PROBLEMS: Perform the indi-cated division:

64/24°1.

8/24°

300/24°2.

20/154°3.

5/142°

64/18°4.

16/27°

ANSWERS:

1. 8/_01

2. 15alt3. 124/12°4. 4/-9°

95

90

It follows that a vector can be raised to anyintegral or fractional power. To square avector, square the scalar quantity and multiplythe angle by 2,

EXAMPLE: Square the vector 8/32°

SOLUTION: (8/12° )2

= (8)2/320 (2)

= 64/64°

To cube a vector, cube the scalar quantityand multiply the angle by 3.

EXAMPLE: Cube the vector 3Zir

SOLUTION: (3/21° )3

(3)3/40 (3)

= 271120

To find the square root of a vector, extractthe square root of the scalar quantity anddivide the angle by 2.

EXAMPLE: Find the square root of thevector 16/70°.

SOLUTION: V 16110° or (16/70° )1/2

= V-176/70° 4 2

= 4Zg3

To find the cube root of a vector, extractthe cube root of the scalar quantity and dividethe angle by 3.

EXAMPLE: Filx1 the cube root of thevector 27133°.

SOLUTION: 34 27/33 6 or (27/330)1/3

3i-Fo3 4. 3

= 3/11°

Chapter 6VECTORS

PRACTICE PROBLEMS: Perform the indi- ANSWERS:cated operations:

1. (10290)2 1. 100ZI0°

2. (4/10°)3 2. 134/300

3. (64Z9_02)1/2 3. 8/45°

4. (64/9(01/3 4. 4L3g0

CHAPTER 7

STATIC EQUILIBRIUM

Statics is a branch of physics that deals withbodies at rest. In this chapterwe will make useof the previous investigation ofvectors to estab-lish the mathematical basis necessary for anunderstanding of static equilibrium. Sinceforces acting upon bodies have magnitude anddirection, they may be represented by vectors.

DEFINITIONS AND TERMS

The following paragraphs include definitionsand terms which will be used in this chapter.The definitions used will clarify the meaningsof the discussions on static equilibrium.

EQUILIBRIUM

If a body undergoes no change in its motion,it is said to be in a state of equilibrium. We willdiscuss a body at rest as indicated by the termstatic equilibrium. Balanced forces may actupon a body in static equilibrium,but no motion,neither translatory nor rotary, will occur. Inorder for bodies to be in static equilibrium, twoconditions are required. These two conditionsare (1) the body must not have translator),motion, and (2) the body must not have rotarymotion,

TRANSLATION

Translation, as defined, is motion independ-ent of rotation. Attention is called to the factthat translation involves magnitude and directionof motion and hence can be describedby vectors.

ROTATION

Rotation, as defined, is the turning motion ofa body, such as a wheel turning. .Rotation isindependent of translation.

92

TRANSLATIONAL EQUILIBRIUM

If a body at rest is acted upon by an un-balanced force, it will be set into motion. Thismotion is called translation. All particles ofthe moving body will have at any instant the samevelocity and direction of motion. Two forcesacting in the same line upon a body must beequal in magnitude, but opposite in direction,if the body is to remain in equilibrium. Wewill consider our translations to be confinedto the XY plane.

FIRST CONDITION

The first condition of equilibrium may bestated as follows: For a particle to be in equi-librium the sum of the vectors (forces) actingin any direction upon that particle must equalzero.

In figure 7-1, an iron block is resting on atable. The weigt.t of the block isdirected down-ward; thus the table must exert a force equal andopposite to the weight of this iron block. Theblock has weight, W, and the table exerts apush, P, upward against the block. Since thebodies are in equilibrium, there can be no un-balanced force. Thus,

WrPandW=PThe weight of the block can be represented

by a vector because we know the force anddirection exerted by the weight. The magnitudeand direction of the push (P) by the table isalso known, and it can be represented by a vec-tor. The vectors W and 7 are shown in figure7-1. The weight is in equilibrium because thesum of the vectors acting upon it is equal tozero. We call these two forces parallel con-current forces. We will also call P the equi-librant of W. ft is relatively easy to find theequilibrant of two or more vectors which are

Chapter 7STATIC EQUILIBRIUM

acting upon the same point. We first find theresultant of the vectors. The equilibrant ofthe resultant will have the same magnitude lqutwill be opposite in direction. In figure 7-2 theresultant of X and fl is W. The magnitude of 7is 18 and the direction is 56°18'. The magnitudeof U, the equilibrant of -1:11.is 18 and the directionis 236°18'. The sum of P and is zero; there-fore, a point 0 is in equilibrium.

Figure 7-1.Table and weight.

Figure 7-2.The equilibrant.

93

In figure 7-2, the vectors X and arecalled nonparallel concurrent vectors. Allvectors can be resolved into horizontal and vert-ical components. Since the sum of all forcesacting on a particle must be equal to zero, tosatisfy the condition of equilibrium, we can saythat the sum of all vertical components mustequal zero, and the sum of all the horizontalcomponents must equal zero. The symbols forthis condition of equilibrium are:

EX = 0and

EV = 0The symbol z is the Greek letter, sigma,

and means °the sum of.. Thus, ZX equals0 means that the sum of the vectors along theX axis equals zero.

We may show graphically that a particle Pis in equilibrium while being acted upon byA, 11, C, 15, and r, as in figure 7-3, by drawinga polygon of forces. If the polygon of forces isclosed, there is no resultant force acting uponparticle 7, and that particle is in equilibrium.

We will now examine the condition of equilib-rium of a point which is acted uponby nonparal-lel concurrent forces.

EXAMPLE: We are to find the force of 7i, infigure 7-4, in order that point 0 will remain inequilibrium while being acted upon by IN and C.

SOLUTION: We are looking_for the equilib-rant of the resultant of 13 and U. The resultantof it and C, called V, is found as follows:

yb = 5 sin 30°

a 5(0.50000)

ic 2.5

rcb = 5 cos 30°

5(0.88603)

= 4.3

= 8 sin 270°

= 8(-1.00000)

= .8

= 8 cos 270°

= 8(0.00000)

= 0

S8


Figure 7-3.--Polygon of forces.

Figure 7-4.Resultant and equilibrant.

We now add the X axis components and theaxis components and find that

Y = 2.5 + (-8)

= -5.5

= 4.3 + 0

= 4.3

and

thus

and

-5.5tan = 4.3

= -1.27907

0 = 308° 1'

r = (4.3)2 4- (-5.5)2

=

= 6.9

therefore our resultant is

6.9/3080 I'and the equilibrant is

6.9Z128° V

In some cases we are given the vectors bythe problem and can easily find our solution.

EXAMPLE: A boy in a swing, as shown infigure 7-5, weighs 70 pounds and ispulledback-ward with a force of 30 pounds. Find the forcethe ropes exert on the swing; also find the anglethe ropes make with the horizontal axis.

SOLUTION: We must find the resultant of thetwo vectors given and then find the equilibrantof the resultant. This is done as follows:

Yia = -30

Tcb = 0

Sfb = -70and

thus

and

94

99

tan() = = 2.33333-30

0 z.. 246° 48'

002 (..70)2

=

=I 76.2


therefore the resultant is76.2/246° 48'

and the equilibrant is76.2/660 48'

Thus the ropes exert a 76.2-pound66°48' on the swing.

PRACTICE PROBLEMS: Find the equilibrantof the following vectors.

1. 35/0.° and 60/90°

2, 7/35° and 9/125°3. 12/15° and 7/25°4. 9/55° and 10/100°

ANSWERS:

1. 69.4/239° 45'2. 11.4/267° 7'3. 18.7/198° 43'4. 17.5/257° 16'

FREE BODY DIAGRAMS

One of the distinct advantages of vectors isthat a vector may be substituted for the cableor member of a mechanism it is going to repre-sent. As seen in figure 7-6 vectors may besubstituted for the cables holding the weight W.

force at Starting at point 0, a vector representing thetension in cable MO can be drawn, and vectorsmay also be drawn for the tonsion in cables NOand OW. This will give us the forces actingon particle 0. Figure 7-6 (B) is called a freebody diagram, and vector A represents thetension in cable MO. Vectors B and W repre-sent the tensions in NO and WO, respectively.

Figure 7-5,Boy in a swing.

Figure 7-6. Single weight.

(81

Free body diagrams are very important inmechanics and the student should learn to drawthese diagrams with ease. In a free bodydiagram, a member of a mechanism is replacedby a vector representing the force in that mem-ber and acting in the same direction as themember. The student should pay particularattention to the magnitude of the vector whichrepresents a member of the mechanism. Infigure 7-6, notice that the vector representa-tion for MO is longer and the vector representa-tion for NO is shorter in the free body diagram(B) than they appear in the pictorial view (A).

We may use the free body diagram to graph-ically verify our mathematical solution to aproblem. (Refer again to fig. 7-5.) We find theboy in the swing to be in equilibrium and we willuse a free body diagram to verify this. Wedraw our diagram as shown in figure 7-7 (A)where vector r is the equilibrantof the resultant.We draw the vectors W, IT, and C., initial pointto terminal point, as shown in figure 7-7 (B).If the vectors form a closed loop, we have thesum of the vectors equal to zero and havepresent a state of equilibrium.

95

log


1111.10 .7§ X

(8 )

Figure 7-7.Closed loop.We have discussed parallel concurrent forces

and nonparallel concurrent forces. In the follow-ing paragraphs we will discuss noncurrent paral-lel forces, remembering that we are still underthe requirements for the first condition ofequilibrium.

96

In figure 7-8 (A), we find a board balancedon and supported by a fulcrum. Draw the freebody diagram as shown infigure 7-8 (B). Assumethe board weighs so little that it is insignifi-cant. Consider forces in a downward directionto be negative (-) and those upward to bepositive (+). For equlibrium,'we must haveZX equal to 0 and zY equal to 0. We have no Xaxis components, therefore zX equals 0. TheZY equals 0 because we have a state of equi-librium. Therefore

-X-15+ U=oA= 42 lbs

= 18 1 b s

= 60 lbs

42 POUNDS

and

therefore

18 POUNDS

Fl

( A)

0

B)

Figure 7-8. Parallel nonconcurrent forces.


Figure 7-9.Free body practice problems.

97

PRACTICE PROBLEMS: Draw the iree bodydiagrams for the member indicated inthe follow-ing figures (emphasize direction and not magni-tude):

1. Figure '7-9 (A) (Bar)2. Figure 7-9 (B) (Boom)3. Figure 7-9 (C) (Bar)4. Figure 7-9 (D) (Point 0)ANSWERS:1. Figure 7-10 (A) 3. Figure 7-10 (C)2. Figure 7-10 (B) 4. Figure 7-10 (D)

CABLES

gz/1:6)%o

1 WEIGHT],/

Wi W2

fe'

Figure 7-10.Free body answers.

(A)

(B)

(C)

(0)


ROTATIONAL EQUILIBRIUM

The first condition of equilibrium guaranteedthat there would be only translatory motion. Itwas stated that there was a distinction betweenthe motion of translation and rotation. Thesecond condition of equilibrium concerns theforces tending to rotate a body.

SECOND. CONDITION

Figure 7-11 shows a body acted upon by twoequal and opposite forces, F1 and F2. The sumof the forces in the horizontal direction equalszero, and there is no translatory motion. It isclear that there will be rotation of the body.These two equal and opposite forces not actingalong the same line constitute a couple and causea moment to be produced. The term couple isdefined as two equal forces acting on a body butin opposite directions and not along the sameline. For a body acted upon by a couple toremain in equilibrium, it must be acted upon byanother couple equal in magnitude but oppositein direction.

The magnitude of a couple is the perpendic-ular distance between the forces multiplied byone of the forces. This product is called themoment of the couple. We willuse M to indicatea moment and we can say, to fulfill the condi-tions of equilibrium, that EM equals 0. That is,the sum of all the moments acting upon a bodymust equal zero to maintain equilibrium. Clock-wise moments, such as in figure 7-11, arepositive and counterclockwise moments arenegative. Our statement that the sum of all themoments acting upon a body must be zero, thatis, IM equals 0, is called the second conditionfor equilibrium.

Assume that the body shown in figure 7-11will rotate about a point halfway between thetwo forces. A moment, defined as a force actingon a lever arm L, is present for both of theforces. The moment acting onthebody in figure7-11 will be

Since

then

F1 (4) + F2 (4)

F1

= F2

M = F1 (2) + F1 (t)= F

1L

If it were assumed that the body were to rotateabout the point upon which F2 acted, then thelever arm would be L for F1 and zero for F2.And again M equals F1L. Hence, themoment ofa couple is one of the forces multiplied by thedistance between them. The definition formoment of a couple holds. The dimensions ofa moment will include a distance as well as aforce. The effect of a force upon the rotationis the perpendicular distance from the rotationpoint to the line of action of the force. In theEnglish system the most used term is foot-pounds.

EXAMPLE: Calculate the momentof a coupleconsisting of two forces, F1 (equal to 20 pounds)and F2 (equal to 20 pounds), acting directlyopposite to each other at a distance of 3 feet.The moment of this couple is M equals FL or

M = (20)(3) = 60 ft-lb

Notice that in this example there is nobalance of moments; that is, ZM does not equal0, and the conditions for equilibrium are not met.

We now put to use the first and second con-ditions for equilibrium. That is,

EY = 0

EX = 0

z M =

98

103


In figure 7-12 (A) we have aboardbalanced on afulcrum and we are to find the weightW2 and theforce F on the fulcrum if W1 equals 12 poundsand the distances are as shown. There are nohorizontal forces; therefore EX equals 0. Wedraw the free body diagram as shown in figure7-12 (B) and find the solution as follows:If

then

Therefore

Thus

and

W1

= 12 lb

L1 = 2 ft

W2 = unknown

L2 = 8 ft

F = unknown

LIW1 = L2W2

(12 lb) (2 ft)

= 24 ft-lb

L2W2 = 24 ft-lb

(8 ft) (W2) = 24 ft-lb

W2 = 3 lb

W1 + W2 - F = 0

Thus

A verythe second

F = 12 lb + 3 lb

= 15 lb

EY = 0useful theorem that originates fromcondition of .,1:luilibrium states that: If

12 lbs

a'

At

12 lbs

IF e'

(AI

Figure 7-12.Forces and moments.

three nonparallel forces acting upon abody pro-duce equilibrium, their lines of actionmust passthrough a common point. In other words, thethree conditions EX equals 0, EY equals 0, andEM equals 0 must be satisfied for equilibrium;and, in order for three nonparallel forces toproduce zero moment, the lines of action of theforces must pass through a common point, thushaving zero lever arm.

CENTER OF GRAVITY

The earth's gravitational field attracts eachparticle in a body and the weight of that body isregarded as a system of parallel forces actingupon each particle of the body. All of theseparallel forces can be replaced by a singleforce equal to their sum. The point of applica-tion of this single force is called the center ofgravity (or C. G.) of the body. For bodies ofsimple shape and uniform density, the C. G. isat the geometric center and can be found byinspection.

The C. G. of an irregularly shaped body canbe found by suspending the body from three dif-ferent points on the body. In each case the bodywill come to rest (equilibrium) with its C. G.directly beneath the point of suspension. Theintersection of any two of these lines will de-termine the C. G. The third line should alsopass through this intersection and thus may beused to check the result. Figi.re 7-13 (A)shows this simple system for finding the C. G.Figure 7-13 (B) shows that in some cases theC. G. may fall outside of the body.

99


(Al

Figure 7-13. Center of gravity.

APPLICATIONS

The greatest difficulty encountered in solvingproblems dealing with equilibrium is finding allof the forces acting upon a body. The use of afree body diagram will aid in eliminating thisdifficulty.

The procedure recommended for solvingstatic equilibrium problems is as follows:

1. Sketch the system, taking into account allknown facts, and assign symbols to all of theknowns and unknowns.

2. Select a member that involves one ormore of the unknowns and construct a free bodydiagram.

3. Write the equations obtained from EXequals 0, EY equals 0, and EM equals 0.

4. Solve these equations for the unknowns.5. Continue the process from one side of a

structure to the other side.EXAMPLE: Consider the ladder standing

against a building in figure 7-14 (A) and makingan angle of 600 with the ground. The ladder is16 feet long and weighs 50 pounds.

SOLUTION: We sketch the free body diagramas shown in figure 7-14 (B) and assign symbolsto the known and unknown forces. The arrowsindicate the directions of the forces and h and vrepresent horizontal and vertical components ofa force. The frictional force f holds the ladderfrom slipping, h is the horizontal force of thewall pushing against the ladder, and v is thevertical force which the ground exerts on theladder. We assume all the weight of the ladderto be located at the center of gravity and assignthe letter W to indicate this weight. The ladderis in a state of equilibrium and we have thefollowing:

100

LX = 0

LY = 0

LM = 0

therefore

f - h = 0W - v =0

We use trigonometry to find AB and BC, asfollows:

and

AB = r cos 0

= 16 cos 60°

= 16 (0.50000)

= 8

BC = r sin

= 16 sin 600

= 16 (0.86603)

= 13.85

Using similar triangles we find that W is locateddirectly above the midpoint of AB.

Figure 7-14.Ladder problem.

Next, take the moments about the bottomof the ladder and in that way the two forces(f and v) have zero lever arm and are eliminated.The moments (clockwise) are as follows:

Chapter 7--STATIC EQUILIBRIUM

LM =

= 4W - h(16 sin 60°)Notice that we used the perpendicular distancesfrom point A to where the forces were applied.

We now have equations as follows:

f - h = 0

W - v = 0

4W - h (16 sin 600) = 0

and substituting known values, then solving, wefind

and

and

W - v = 0

50 lb - v = 0v = 50 lb

4W - h (16 sin 60°) = 0

4 (50 lb) - h (13.85) = 0

h = 14.4 lb

f - h = 0

f - 14.4 = 0

f = 14.4 lbEXAMPLE: Determine the forces acting upon

the members of the A-type frame as shown infigure 7-15 (A). The horizontal surface is con-

(A)

sidered smooth and no horizontal force can beexerted on the legs of the frame. A weight of1,000 pounds hangs from the crossbar. Theframe is considered as having no weight.

SOLUTION: Draw the free body diagrams andassign symbols as shown in figures 7-15(B) and(C). Since the system is symmetrical, thereactions at A and C are equal, and each isequal to 500 pounds (each carries half the load).Thus, Av equals 500 pounds and Cy equals 500pounds. The forces Dv and Ev can be foundfrom the diagram of the crossbar in figure 7-15(B) by taking EM about D.Thus

EM = 3W Ev(6) = 0

= 3(1000) - Ev(6)

Ev = 500 lb

and from symmetry

Dv = 500 lb

These forces Ey and Dv are upward to opposethe weight on the bar; thus this member mustexert the same forces downward on the inclinedmember.

It is apparent that BC pushes upward againstAB. This force is unknown, but it does havea vertical and horizontal component.

Using the two conditions of equilibrium wefind the following:

z5e = Dh - Bh = 0

ZM = 5Av - 3Dv - (6 sin 60°)Dh = 0

--IP Eh61 .1

301Ov Ev I

1000Ibs 6111I 1

A v 5$

(0) (c)

Figure 7-15.A-type frame.

101

IC6

49Bh


Thus

z17 = 500 + B - 500 = 0v

Bv = 0

LMb = 5(500 lb) - 3(500 lb) - 6(0.86603)

Dh = 192 pounds

fil = 192 lb - Bh = 0

Bh = 192 lb

From symmetry we find the following:

= 0

Eh = Dh = 192 pounds

E = D = 500 poundsv v

The magnitude and direction of the X axisand Y axis forces may be used tofind the forcesand their directions.

One important thing to remember whentaking ZM equal to 0 is to take the momentsabout some point that will eliminate one ormore of the unknowns. In the last example,the moment equation was taken about point Band eliminated the forces Bb and Bv. Themoment equations can be taken about any point

and more than one moment equation can betaken, if necessary.

PRACTICE PROBLEMS: Find the requiredinformation in the following:

1. Two boys pull a wagon, each with a forceof 35 pounds. The angle between the ropes onwhich the boys are pulling is 30°. What is theresultant pull on the wagon?

2. A large portrait weighs 100 pounds, andis supported by a wire 10 feet long which ishooked to the picture at two points 5 feet apart.Find the tension in the wire.

3. A 180-pound man is standing half-way upa 20-foot, 20-pound ladder. The bottom of theladder is 4 feet from the base of the verticalwall it is leaning against. Find the forces ex-erted on the ladder. (Use same symbols asshown in fig. 7-14 (B).)

4. A bar of uniform weight, 12 feet long andweighing 7 pounds, is supported by a fulcrumwhich is 4 feet from the left end. If a 10-poundweight is hung from the left end, find the weightneeded at the right end to hold the bar in equi-librium and find the force with which the ful-crum pushes against the bar.

ANSWERS:1. 67.6 pounds2. 57.8 pounds3. f = 20.4 pounds, h = 20.4 pounds, v = 200

pounds4. 3.25 pounds and 20.25 pounds

CHAPTER 8

TRIGONOMETRIC IDENTITIES AND EQUATIONS

This is the final chapter in the section deal-ing directly with trigonometry and trigonometricrelationships. This chapter includes the basicidentities, formulas for additional identitiesinvolving multiples of an angle, and formulasfor identities which involve more than oneangle. Methods and examples of the use of theidentities in simplifying expressions are given,and practice problems in simplification areincluded.

Also included in the chapter are methodsfor solving equations involving trigonometricfunctions. In many cases, the verification orsimplification of an identity is an integral partof the solution of an equation. An additionaltopic considered in the chapter is the inversetrigonometric functions. Examples and prob-lems involving equations and the inverse func-tions are also given.

FUNDAMENTAL IDENTITIES

In earlier chapters it was shown that all ofthe trigonometric functions of an angle couldbe determined if one function or certain relatedinformation was given. This seems to indicatethat there are certain special relationshipsamong the functions. These relationships arecalled identities and are independent of anyparticular angle. Many of the identities whichwill be considered in this section were estab-lished in earlier chapters and will be used hereto change the form of an expression. In manyproblems, especially in calculus and otherbranches of mathematics, one particular methodof expressing a function is more useful thanany of the others. In these instances, theidentities are used to put the expression in thedesired form.

An equality which is true for all values ofan unknown is called an identity. Identities are

familiar in algebra, although they are notalways specifically identified as such. A fac-toring process such as

(x2 - 1) = (x 1) (x + 1)

involves expression of an identity since it istrue for all values of the variable. In trigo-nometric identities, the same situation musthold; that is, the equality must be true for allvalues of the variable.

Problems in identities are often given asequalities, and the identity is established bychanging either one or both sides of the equalityuntil both sides are the same. The fundamentalrule in proving identities is as follows: NEVERWORK ACROSS THE EQUALITY SIGN. Thealgebraic rules for cross multiplication arenever used. In this course all problems will besolved by working on only one side of theequality; that is, one side of the equality willbe reduced or expanded until it is identicalto the other side.

There are no hard ana fast steps or methodsto use in solving identities. However, thereare some basic procedures or hints which willnormally prove helpful in verification of' theidentities. Some of these hints are as follows:

1. Reduce the complex expressions to simpleexpressions, rather than Wilding up from asimpler to a more complex one.

2. When possible, change the expressionto one containing only sines and cosines.

3. If the expression contains fractions, itmay help to change the form of the fractions.

4. Factoring an expression may suggest asubsequent step.

5. Keep the other member of the equalityin mind. Since we are striving to change oneexpression to another, the form of the desiredexpression may suggest the steps to be taken.


RECIPROCALS AND RATIOS

One group of identities was presented in partin chapter 4 of this course. Recall that

Since

and

CSC 0 = 1sin 0

csc 0 =y

the cosecant function can be written as

and simplified to

csc 9 = sin 01

There are six reciprocal identities, one foreach function. The numbers assigned these andsubsequent identities in this chapter do not con-stitute any rules as to order or precedence.They are used only to simplify the explanationof steps in the example problems. The reciprocalformulas follow directly from the definitions ofthe trigonometric functions

1sin 9 CIO-a

1co s 0 sec 01tan 0 = cot 01csc 9 = sin 0

sec 0 =co1s 0

1cot 0 = tan

There are also identities involving the sine,cosine, tangent, and cotangent of an angle whichare sometimes called ratio identities. Theseidentities also result directly from the functions,and one of these expresses the tangent in termsof the sine and cosine, as follows:

Since

and

sin 0tan 0 = cos 0

sin 0 =

n xcos cr = r

substituting these values in (7) gives

tan 0

which can be simplified to

tan 0 =

(7)

This is the definition of the tangent functiongiven in an earlier chapter. The followingidentity for the cotangent,

cos 0cot 0 - sin 0 (8)

can be shown to reduce identically to the defini-tion of the cotangent function in terms of x, y,and r. The reciprocal and ratio identities are

(1) used to simplify trigonometric expressions asshown in the following example problems.

(2) EXAMPLE: Simplify the expression

(3)sin 0 cos 0 tan 0

to an expression containing only the sine function.(4) SOLUTION: One method of accomplishing

this is to apply identity (7) to the given expres-(5) sion; then it becomes

(6) sin 0 cos 0 (sin 19)0cos

104

1C9

Chapter 8TRIGONOMETRIC IDENTITIES AND EQUATIONS

or

sin 0 cos 0 sin 0cos 0

Simplifying the cos 0 terms in both the numera-tor and denominator of the fraction results in

or

sin 0 sin 91

sin20

This is the desired form and is identical, for allvalues of 8, to the original expression.

EXAMPLE: Use fundamental identities toverify the identity

ncsc + cot fa = 1 4" cossin 0

SOLUTION: Since the right-hand member ofthe identity contains sines and cosines, use (4)and (8) to change the left member to sines andcosines.Then

1 cos 0 1 + cossin e + sin e sin 0

Change the sum of fractions in the left memberto a single fraction as in the following

1 + cos 0 1 + cos 9sin 0 sin 0

and the identity is verifiedObserve that the right-hand member of the

identity was not altered throughout the entireprocess. This is in accordance with our statedintention of working on just one side of theequality sign.

If we desire to verify this identity by retain-ing the left member and operating on the rightmember, the following steps may be used.

1 + cos 9csc 0 + cot e =

Change the fraction in the right member to thesum of two fractions

csc 0 + cot 0 = 1 cossin e + sin e

105

Next, apply (4) and (8) to the right member

csc 0 + cot 0 = csc 9 + cot 0

and the identity is verified.

SQUARED RELATIONSHIPS

Another group cd fundamental identities in-volves the squares of the functions. These, insome texts are called Pythagorean identitiessince the Pythagorean théorern is used in theirdevelopment. Consider the Pythagorean theorem

X2 + y2 = r2

and divide both sides by r2,

x2 v2

2 4. r1

Write this in the form42 /1\ 2

cr7 kr)and consider that

and

If cos 0then

cos 0 = r

sin 0 =

1

x yand sin 0 are substituted for I.-

and IT

(cos 0)2 + (sin 9)2 = 1

This is rewritten as

cos2 + sin2 9 = 1

which is a fundamental squared or Pythagoreanidentity.

NOTE: The practice of writing .an expres-sion such as (sin 0)2 in the form sin4 0 is com-mon, and is the preferred method.

In the same manner, dividing both sides ofthe equation

110

2x + y2=


by x2 (where x is not equal to 0) gives EXAMPLE: Verify the identify2 2

1 + =2 2 FIET 1x x

or

sin 0 cos 0

2 21 + = (9

Then, since

and

substitution gives

or

tan 0 = Y

sec 0 = r

1 + (tan 0)2 = (sec 0)2

1 + tan2 0 = sec2 0

which is another fundamental identity.The identity

1 + cot2 0 = csc20

(10)

SOLUTION: Reduce the leftmember to equalthe right member. First, change each functionto sin( or cosines as follows:Apply (4) to the denominator of the firstfractionto obtain

sin 0 cos 9 ,1 + =

sin 0

Simplification of the first fraction gives

sin2 0 cos 0 = 11 sec 0

Applying (5) to the remaining fraction gives

sin2 0 + cos 0 - 11

C OS

Simplification gives

sin29 + cos2 0 = 1

(11) Then applying (9) to the left member results in

is derived in a similar manner. 1 = 1The three squared identities can be trans-

posed algebraically to other forms with the and the identity is verified.following results:

cos20 = 1 - sin20

sin2 0 = 1 - cos20

tan2 0 = sec20 - 1

sec20 - tan29 = 1

cot2 0 = csc 20 - 1

csc29 - cot2 0 = 1

(12)

(13)

(14)

(15)

(16)

(17)

EXAMPLE: Verify,2,

1 + COL QC

SOLUTION: As a firstterm cot22x.Then

cos22x

the following identity:1

sin 2x

step, apply (8) to the

1

sin22x sin22x

In addition to the fundamental identities, Combine the left term into a single fractionthere are many complicated identities involving with a denominator of sin22xthe trigonometric functions. In the majority ofcases, these identities can be proved by use . 2sin 2x + cos22xof the laws of algebra and the fundamental

=identities. sin22x sin 2x

106

Chapter 8TRIGONOVETRIC IDENTITIES AND EQUATIONS

Applying (9) to the numerator of the left membergives

1 1

2sin 2x sin 2x

and the identity is verified.PRACTICE PROBLEMS: Verify the follow-

ing identities:

= cos2x1

tan x + 11.

2. csc x - sin x = cos x cot xsina8

3. 1 - cos 01 + cos 0

4. cos 0 (sec 0 - cos 0) = sin20

5. tan2 x (1 - sin2 x) =

6. sin3x - 1 - cos2x

1

CSC X1

7. 2 + cot2x 2csc2x -

REDUCTION FORMULAS

cot2x

In chapter 4 of this course, reduction for-mulas were developed for dealing with anglesgreater than 900. These reduction formulas canbe combined into a general category of iden-tities which also includes the formulas developedin chapter 4 for dealing with cofunctions andcomplementary angles. The formulas, of thetYPe

or

sin(90° - 0) = cos 0

sec(180° + 0) = -sec 0

are listed in chapter 4 and the listings will notbe repeated in this chapter.

The formulas from chapter 4 will be used tosimplify expressions, in the same manner as theother identities, in the following examples.

EXAMPLE: Simplify the expression

sin(180° - 0)tan(90° - 0)cot(180° - 0)

into an expression containing functions of 0alone.

SOLUTION: From chapter 4, the followingformulas are chosen

sin(180° - 0) = sin 0

tan(90° - 0) = cot 0

cot(180° - 0) = -cot 0

Substitution of these values in the expression

sin(180° - 0)tan(90° - )cot(180° - 9)

results in the expression

sin 0 cot 0 (-cot 9)

Rewrite this in the form

-sin 0 cot 0 cot 0

and apply identity (8) to one of the cot 9 factors.Then,

107

cos-sin u111

COLS 0

results and this can be simplified to

-cos 0 cot 0

to complete the problem.EXAMPLE: Express the following as an ex-

pression containing the least possible number offunctions of 9.

sin(360° - )tan(90° - )csc

SOLUTION: The following formulas are givenin chapter 4:

and

sin(360° - 0 ) = sin(-0) = -sin 0

tan(90° - 0) = cot 0

Substitution of these values in the original ex-pression results in

-sin 0 cot 0 csc 0

Rewrite this as

-cot 0 sin 0 csc 0

1112


and apply identity (1) to the factor sin 0 to arrive orat

-cot 0 cst. 1csc 0

or-cot u

which is an expression in terms of one functionof 0.

PRACTICE PROBLEMS: Express the fol-lowing as expressions containing the least num-ber of functions of 0.

1. sin (1800

2. sin (180°

3. cos (360°

ANSWERS:

1. sin2O

2. tan 0

3. tan

- 0) sin 0

+ 0) sec (180° - 0)

- 0) cot (90° - 0) csc (90°

INVERSE TRIGONOMETRICFUNCTIONS

= tan lA

In the last notation we do not mean -1 to rep-resent an algebraic exponent and tan-1A does

1 1denote if we meant tan-1 equalstan A tan A°1we would have written (tan A)-1 equals tan A'

In this course, the preferred notation isarctan A.

PRINCIPAL VALUES

For any angle there is one and only one func-tion which corresponds to it; but to any value ofa trigonometric function, there are numerous

0) angles which will satisfy the value. For instance,

In this section we will discuss the definitionswhich apply to the inverse trigonometric func-tions along with the principal values of thesefunctions. Relations among these functions willbe examined by the use of examples andpracticeproblems.

DEFINITIONS

It is often convenient and useful to turn atrigonometric function around so that instead ofwriting

tan 9 = A

we write

0 = the angle whose tangent is A

Rather than write out the last statement, math-ematicians use either of the followingnotations:

0 = arctan A

0 = arctan 1can be written

tan 0 = 1but 1 is the tangent of many angles such as 450,225°, 405°, 585°, and others. Any angle 0 whichsatisfies (45° + n . 180°), where n is an integer,satisfies the expression

tan 0 = 1For any inverse trigonometric function there

are two angles less than 360° which satisfy it.Thus,

= arccos(0.500) refers to 600 and 300°0 = arccos(-0.500) refers to 120° and 240°

= arcsin(0.707) refers to 45° and 135°= arcsec(2.000) refers to 60° and 300°Since a given inverse trigonometric function

has many values, one of these values is selectedas its principal value.

To denote principal values, we will capitalizethe first letter in the name and we will use theranges for principal values as follows:

-900 5 Arcsin x -45 900

0° s Arccos x 180°

-90° < Arctan x < 90°-901' 5- Arccsc x 5 90°

0° s Arcsec x 5- 180°0° < Arccot x < 180°

108

113

Chapter 8-TRIGONOMETRIC IDENTITIES AND EQUATIONS

All principal values lie between -90° and1800 proceeding counterclockwise form -90°.The principal values for positive numbers arebetween 00 and 900. For negative numbers,principal values of the inverse sine, tangent,and cosecant lie between -90° and 0°, whileprincipal values of the inverse cosine, cotangent,and secant lie between 900 and 180°. We willuse, for understanding, the examples whichfollow.

EXAMPLE: Find the principal value of theangle in the function

0 = Arccos (0.4472)

SOLUTION: Ush:F, the trigonometric tables,we find the angle whose cosine is 0.4472 is 63°26' or 296° 34'. We choose 63°26' as this is thefirst quadrant angle and is the principal value.We reject 296° 34' because it is afourth quadrantangle and the principal values for the Arccosfunction are limited to the first and secondquadrants.


0 Arccos (-0.5000)

SOLUTION: Using the trigonometric tables,we find the angle whose cosine is (-0.5000) is120° or 240°. We choose 120° because 240° is inthe third quadrant and does not satisfy the valuewe agreed on as the principal value for the cosinefunct ion.


Arctan 1

SOLUTION: The angle whose tangent is 1 is45° or 225°. We reject 225°, a third quadrantangle, and select 45° because it is a first quad-rant angle.


0 = Arcsec 2.236

SOLUTION: If the trigonometric tables donot list secant values, the function may bechanged by the following:Since

sec 0 cos1

then

1Arcsec (2.236) = Arccos

= Arccos (O. 4472)

and we find that the angle whose cosine is 0.4472is 63° 26' and the prindpal value of the anglewhose secant is 2.236 is 63° 261.

PROBLEMS: Find the principal values of theangles in the following functions:

1. 0 = Arccos (0.9135)2. 0 = Arcsin (0.8829)

3, e = Arctan (11.430)4. 0 = Arccot (-0.1169)5. 0 = Arcsec (1.0075)

6. 0 = Arctan (-0.1228)ANSWERS:

1. 24°

2. 62°

3. 85°

4. 96° 40'

5. 7°

6. -7°

In dealing with the inverse trigonometricfunctions, we may be presented the problem offinding the rincipal value of the function0 = Arcsin --9. In this case we could solve it

2as follows:Draw a right triangle as shown infigure 8-1.

This expression 4 can be rewritten as a- bythe following steps:

12611\ 2 1

-271

Recall that sin 0 equalsi. The triangle isa 45°-90° triangle as shown in Mathematics,Vol. 1, NavPers 10069-C. Now, the function

0 = Arcsin -§-

109

.114


Figure 8-1.-45° - 900 triangle.

becomes

10 = Arcsin

and we find that

= 4 5°

Use this same approach to answer the followingquestions.

4. 30°

5. 80°

6. 120°

If we are to find, using the principal values,the value of the expression Arctan - Arctan

in radians, we proceed as follows:

Arctan 45- = 60°

and

1Arctan = 300

thus

and

then

Arctan i3 - Arctan = 60° - 30°

10 radians180

30°019-) radians

= radians6

PROBLEMS:anglesthe twoematics,

1.

2.

3.

4.

5.

6.

Find the principal values of thein the following functions. (Hint; draw

special triangles as shown in Math-Vol. 1, NayPers 10069-C):

0(112

Arccos

PROBLEMS:give theradians:

1.

2.

3.

4.

ANSWERS:

1.

Using the principal values,values of the following expressions in

1 1Arcsin - Arccos

Arccos

_ =

9 = Arctan (a)

9 = Arccot (-4)

0 Arccos

Arcsin

Arctan 1 - Arctan

Arctan - Arcsin

r

=

0 Arcsin=

0 = Arccot

ANSWERS:

1. 45°

2. 60°

3. 150°

110.f .115

Chapter 8-TRIGONOMETRIC IDENTITIES AND EQUATIONS

RELATIONS AMONGINVERSE FUNCTIONS

In order to understand the relations amongthe inverse functions, we will start by drawinga triangle. If 9 equals Arcsin x, we can writesin 9 equals x. We now draw a triangle whichcontains the angle whose sine is x and assumethe hypotenuse equal to one. The remainingside of the triangle will be, from the Pyth-agorean theorem, Nil - x2 . This is shown infigure 8-2.

Now, we can write all of the functions andinverse functions of the angle 9 in terms of thesides of the triangle as follows:

sin 9 = x, or 9 = Arcsin x

cos 9 = 1,17..--x2 , or 9 = Arccos d.----72

x___X___.577tan 9 -2

or-

9 = Arctan17/1. x N2

1 1csc 9 = ,- or 9 = Arccsc-xx1 1

sec 9 = , or 0 = ArcsecNIT=T12 if.77-c2

cot 0 = N/1 - x29 or 0 = /1 - x2

All of the inverse functions are equal to 9; there-fore, they are equal to each other. We will usethis type of analysis to solve a few problems.

EXAMPLE: Using principal values, find1r3the tangent of the angle whose sine is -2-; that is,

tan Arcsin =

SOLUTIOI:rk Draw the trianglethe Arcsin 2 as shown in figureremaining side will be given by

X = 22 (15)2

Using the tangent ratio we havetan 9 = 15

and

= 60°

containing

8-3. The

Figure 8-2.-Triangle containing Arcsin x.

EXAMPLE: Using principal values find

4sin (Arccos 3 - Arcsin -5)

SOLUTION: Draw the triangle as shown infigure 8-4. The missing side of the triangle

3containing Arccos -5- is given by

.152 32 = 16 = 44Notice that this triangle also contains Arcsin-5-,

so that

3 4Arccos 5 = Arcsin -5

and the original expression becomes

sin 0°

which is zero.

PRACTICE PROBLEMS:lowing expressions:

1. sin (Arccos

2. tan (Arcsin 1+0

3. cot (Arcsin

4. sec (Arctan -2151-)x 1

5. cos (Arc sin 1X" 1

111

.1116

Evaluate the fol-


Figure 8-3. Triangle containing Arcsin --VI2-.

ANSWERS:

41. 512.

1/Zi77---i3. x

422E4 - 2x2 + 14.x2 - 11

xfx2 - 2x2 - 1

FORMULAS

In this section we will discuss the trigono-metric formulas for addition and subtractionof angles, half angles, double angles, and trans-cendental functions. We will use examples forbetter understanding and in some instances wewill derive formulas.

112

Figure 8-4.Triangle containing Arccos 4.

ADDITION AND SUB-TRACTION FORMULAS

Four formulas express the sine or cosine ofthe sum and difference of two angles as afunction of the sines and cosines of the singleangles. They are very important because theyare the basis for much of trigonometric analysis.From the following four formulas we may deriveall of the formulas in the following sections:

sin (A + B) = sin A cos B + cos A sin B (1)

cos (A + B) = cos A cos B - sin A sin B (2)

sin (A - B) = sin A cos B - cos A sin B (3)

cos (A - B) = cos A cos B + sin A sin B (4)

We will prove these four formulasfor angleswhose sum is less than 900. They are actuallytrue for all angles.

211


In figure 8-5 we have indicated the sum oftwo angles, A and B. The hypotenuse of thetriangle containing angle B has been set equalto 1 so that the legs of this triangle have thevalues sin B and cos B.Thus,

1

and

PL-T = sin B

OL = cos B1

In the triangle containing angle A, we can seethat

Therefore,

and

Therefore

ON ONcos A = OL cos B

ON = cos A cos B

COSA MKS

SIN.A SIN.13

L

kl NSlN.A SINAI

Figure 8-6.--Sum of two angles, part two.

terms of the sine and cosine of angles A and B.First, let us prove that angle A is equal to angleA'.

Triangle OMF is similar to triangle PFL, andin A = NL

=NL angles A and A' are corresponding angles.s OL cos B Therefore, angle A equals angle A'. In triangle

PRL

NL = sin A cos Band

cos A' = PRsin B

Now, let us add a construction to the figure, PR = cos A' sin Bas in figure 8-6, and calculate more lines in

SIN. II

Also,

and

= cos A sin B (7)

sin A' = RLsin* B

RL = sin A' sin B

= sin A sin B (8)

SIN. A COS. 8 Furthermore,

RL = MN = sin A sin B (9)

Now, in triangle OMP we can writeCOS. A COS. El

Figure 8-5.--Sum of two angles, part one.

up s

PMsin (A + B) = -r- = PR + RM (10)


but

RM = LN (11)

SO

sin(A+ B) = PR + LN = LN + PR (12)

Also

cos(A + B) = = ON - MN (13)

Substituting from equations (6) and (7) intoequation (12)

sin(A + B) = sin A cos B + cos A sin B (14)

Substituting from equations (5) and (9) intoequation (13)

cos(A + B) = cos A cos B - sin A sin B (15)

EXAMPLE: Use the addition formulas to findcos 75°.

SOLUTION: Use cos(A + B) equal to cos Acos B - sin A sin B. From this we write cos75° equals cos (450 + 30°) and substitute asfollows:

cos(45° + 300) = cos 45° cos 300 - sin 45° sin 30°

=(171)(1)- (4)(1)

(1)-4

The subtraction formulas can be derived fromthe addition formulas by substituting (-B) for(+B). Now, we have

sin(A - B) = sin A cos(-B) + cos A sin(-B) (16)

and

cos(A - B) = cos A cos(-B) - sin A sin (-8) (17)

But, the cosine of a negative angle is equalto the cosine of the angle. The sine of a negativeangle, however, is equal to minus the sine of theangle; that is,

cos(-B) = cos B

sin(-B) = -sin B

Substitution of these values in equations (16) and(17) gives

sin(A - B) = sin A cos B - cos A sin L (20)

and

cos(A - B) = cos A cos B + sin A sin B (21)

EXAMPLE: Show that

cos 9 - sin 9sin (45° - 0) =

SOLUTION: Applying equation (20)

sin(45° - 0) = sin 45° cos 0 - cos 45° sin 0but

sin 45° = cos 450= 4

Substituting these values we havecos9 sin 0sin (45° - 0) = v2

cos 9 - sin 0Nri

We can use these four formulas, (14), (15),(20), and (21), to drive a number of other im-portant ones. One of these is the tangent ad-dition formula.

tan A + tan Btan (A + B) = (22)1 - tan A tan BIn order to prove this formula, proceed asfollows:

Taking the ratio of equalities (14) and (15),we have

sin(A + sin A cos B + cos A sin B(23)cos(A + B) cos A cos B - sin A sin B

Dividing both numerator and denominator ofthe right side of this equation by cos A cos Bwe have

sin Acos Bsin B

tan (A + B) = cos A1 sin A sin B

cos A cos B

which reduces to

tan A + tan Btan (A + 1 - tan A tan B

(24)

(25)


Another important formula is the subtractionformula for the tangent. To find tan(A -replace B by (-13). Therefore,

tan A + tan(-B)tan(A - (26)1 - tan A tan(-B)tan A - tan B1 + tan A tan B (27)

EXAMPLE: Use subtraction formulas tofind tan 15°

SOLUTION: Use the special triangles aspreviously discussed.

and

tan 15° = tan(45° - 300)

tan 450 - tan 30°tan (45° - 300, 1 + tan 45° tan 30°

Irs

-1+(-1753-

Therefore,

= 4 + 2/5

= -(2 + 4-5)

= -3.732

tan .105° = -3.732

PROBLEMS: Use the addition and subtnc-tion formulas to find the values of the followingwithout tables:

1. sin 75°2. cos 15°3. tan 75°4. cot 165° Hint: recall cot(180 - 0) =

1-cot 0 and cot 0 =

ANSWERS:3 - +5775 1. 4

3+71 1773.3 - - 2.

+4

9 - 615 4-75 3. 2 +

= 2 -

EXAMPLE: Given tan 45° equals 1 and tan60° equals 15. Find the tangent of 105°

SOLUTION: Applying this knowledge to equa-tion (22), we have

1tan(45° + 60°) = tan 105° =

It is easy to evaluate a fraction in this form byrationalizing the denominator.

Multiply numerator and denominator by the In equations (14) and (15), if B equals A, we can

same numbers as are in the denominator but writeconnected by the opposite sign.

(1 + /5)2 1 + 215 + 3(1 - 1[1) (1 + V3) 1 - 3 cos 2A = cos A cos A - sin A sin A

14. ur 473 - 2

DOUBLE ANGLE FORMULAS

The addition formulas may be used to derivethe double angle formulas.

sin 2A = 2 sin A cos A

cos 2A = cos2A - sin2A

2 tan Atan 2A =tan" A

(28)

sin 2A = sin A cos A + cos A sin A (29)

115

120


from which we obtain

and, using

then

sin 2A = 2 sin A cos A

sin2 A + cos2 A = 1

cos 2A = cos2 A - sin2 A

The solutions of this equation may be obtainedby setting the factors equal to zero and making

(30) use of inverse trigonometric functions. We maywrite

cos 2x = 0

2x = Arccos 0

= 90°

x = 45°and

= 2 cos2 A - 1 (31) 2 sin 2x - 1 = 0= 1 - 2 sin2 A 1sin 2x =

Substituting A for B in equation (25), we obtain 12x = Arcsin2 tan Atan 2A (32) 2x = 30°, 150°

1 - tan2 A

EXAMPLE: Evaluate sin 15° cos 15°.SOLUTION: Since

and

then

2 sin A cos A = sin 2A

1sin A cos A = 2 sin 2ADividing all the angles in equation (31) by 2

we obtain

x = 1 5°, 750

The equation has three solutions, x equals 1 5°,45°, and 750 Notice that in writing down thevalues of the inverse functions, it was necessaryto include 1 50° since, when divided by 2, thisgives an angle in the first quadrant.HALF ANGLE FORMULAS

2 A 2 A-2-1 cos A = cos -2- - sin (33)sin 15° cos 15° = sin 2(15°)

Using equation (33), we can derive two useful1= sin 30° 2 Aformulas. Adding and subtracting sin on the

=GMright side of equation (33) we have

cos A = (cos2 4+ sin2 - 2 sin2 t (34)1

71- Observe that the methods necessary forsimplifying trigonometric identities and equa-EXAMPLE: Find the three fin::: quadrant tions often include operations which may at firstangles which satisfy the trigonometric equationappear to be pointless. In the preceding sentence

sin 4x = cos 2x we referred to gadding and subtracting sin2 t onthe right side of equation (33).° The advantage ofSOLUTION: From the double angle formulas, 2 Awe can write adding sin becomes obvious when we group it

A2 sin 2x cos 2x = cos 2x with cos2 T. The expression (-sin2-2A.) is added

or to the right-hand side along with (sin2 4)in

cos 2x(2 sin 2x - 1) = 0 order to avoid changing the overall value.

116

121

Chapter 8-TRIGONOMETRIC DENTITIZS AND EQUATIONS

The quantity in the parentheses in equation(34) is equal to 1, so

cos A = 1 - 2 sin2 A (35)2

Rearranging equation (35) and taking the squareroot of both sides, we have

A /1 - cos Asin = 2 (36)

AAdding and subtracting cos2 -f on the right sideof equation (33), we have

cos A = 2 cos2 2(cos2 + sin2 -41) (37)2 2

but, the quantity within the parenthesis is equalto 1 so tha

Rearranging

2 Acos A = 2 cos (38)

equation (38) and taking the squareroot of both sides gives us

COSA / 1 + cos AI v 2

Taking the ratiohave

, ALan -2-

(39)

of equations (36) and (39), we

ri - cos A= v (40)1 + cos ANotice that the use of (+) or (-) is dependentupon the quadrant of argle termination.

EXAMPLE: Find cos 15°, if cos 30° equalsNr30.866 or

SOLUTION: From

30°cos -2- = cos 15° =

Thus

equation (39) we have

/1 + 0.8662

cos 15° =

The solution using -I f

0. 9659

30°+ 11-52

cos = cos 15° =2 2

+2

PRACTICE PROBLEMS: Use the half angleformulas to find the exact value of the following:

1. sin 15°2. cos 135°3. tan 22.5°4. tan 195°ANSWERS:

2 Nrg1. ± 2

2. ±2

3. ±

4. ± -7-4T1

TRANSCENDENTAL FUNCTIONS

To define transcendental functions we statethat any functions other than algebraic func-tions will be classified, for the purposes of thiscourse, as transcendental functions. This groupof transcendental functions includes such func-tions as trigonometric functions, inverse trigo-nometric functions, exponential functions, andlogarithmic functions.

Later in calculus we will prove that the sineand cosine of an angle can be calculated from thefollowing series, if the angle is expressed inradian measure:

and

x3 x5 x7sin x = x -1

+51 71

- + . . .3

COS X = 12 4 x6x x

41 61

NOTE: 31 is read 3 factorial and is equal to1 x 2 x 3 or 6. 51 is read 5 factorial and isequaltolx 2 x 3 x4x 5 or 120.

At the same time the expansion of the func-tion ex where e is the number 2.71828. . . , thebase of the system of natural logarithms, is asfollows:

117

and

x2 x3 x4ex = 1 + x + + + + .

21 31 TI

-x x2 x3 x4e 31 41

122


Moreover, by using the notation i = 1 twosimilar expressions are obtained:

eiaL +x2 1x3 x4 ix5 x6 1x7- 2T ST

and

= 1- ix - x2 + 12(3 x4 1x5 x6 ix7-ET 31+

41-

51 61 +71 +.01ixAdding eix to e- term by term and dividing by

two we obtain an expression equal to cos x.

eix + e -ix2 - cos x

Subtracting e-ix from eix term by term anddividing by 2i we have an expression equal tosin x.

ixe - = sin x

In calculus, trigunometric and logarithmicfunctions are grouped together and called trans-cendental functions, partly because they cannotbe expressed by a simple algebraic formula,and partly because they both are related to thenumber e. This relation becomes important inderivations in advanced and applied calculus.

EQUATIONS

A trigonometric equation is an equality whichis true for some values but may not be true forall values of the variable. The principles andprocesses used to solve algebraic equations maybe used to solve trigonometric equations. Theidentities and reduction formulas previouslystudied may also be used in solving trigono-metric equations. There are so many differentapproaches to solving these equations that wewill use several examples for better under-standing.

MULTIPLE SOLUTIONS

We will use the following examples andprac-tice problems to show the multiple solutions ofa trigonometric equation.

EXAMPLE: Find the value of 0 if

tan 0 = 1

SOLUTION: We must find the angle or angleshaving a tangent equal to 1. Using the inversetrigonometric functions, we may write

118

tan 9 = 1

arctan 1

= 45°

and recalling that

tan(180° + 0) = tan 0then

Thus,

and

tan(180° + 45°) = tan 450

tan 2250 = tan 45°

arctan 1 = 45°

arctan 1 = 225°

Therefore, the solutions for Cava equation are 450and 225°. The 450 angle is a first quadrantangle and the 225° angle is a third quadrantangle and both have a positive sign and thesame trigonometric value; thus we have amultiple solution.

Notice that the two solutions of the equationdiffer from the investigation of the inversetrigonometric functions in which we were re-quired to find the PRINCIPAL value. ThePRINCIPAL value solution was found to be in aparticular quadrant. In multiple solutions wewill use the term PRIMARY to indicate that weare searching for solutions which are restrictedto the range

00 s 9 < 360°Notice also that if we remove the restric-

tion of the term PRIMARY we may write

tan 0 = tan(O + n . 3600)

and we find, if n is an integer, that there aremany solutions to the equation.

EXAMPLE: Find the primary solutions tothe equation


SOLUTION: We first use the inverse trigo- 1. tan 0 = -1nometric function to write

then

but

2. sec 0 = 2, IScos II = a

2 3. sin 0 = T14. cos 0 =

,./ ANSWERS:0 = arccos 2= 300 1. 135° and 315°

2. 600 and 3000cos 0 = cos (3600 - 0)

3. 240° and 300°cos 30° = cos 330°

4. 600 and 300°Therefore, the solutions are 30° and 330°.

EXAMPLE: Find the primary solutions tothe equation

csc 0 = 2

SOLUTION: As in the previous example, wewrite

The following examples show how to find thesolution of more difficult equations.

EXAMPLE: Find the primary values of theequation

sin 0 + sin 0 cot 0 = 0

SOLUTION: We factor the equation andcsc 0 = 2 find that

and sin 0 + sin 0 cot 0 = 0

1 andcsc 0 = sin 0 sin 9 (1 + cot 0) = 0

Setting each factor equal to zero, we have thetwo equations

sin 0 = 0

then and

1 1 + cot 9 = 00 = arcsin -2-

= 30°Solve each as follows:

but

sin 0 = sin (180° - 0)

sin 30° = sin 1500

then

andTherefore, the solutions are 30° and 150°

PROBLEMS: Find the primary values of 0in the following equations:

119

sin 9 = 0

9 = arcsin 0

= 0° and 1800

1 + cot 9 = 0

cot 0 = -1

MATHEMATICS, VOLUME 2-

Therefore, Therefore, the solutions are 30°, 90°, and 1500.PRACTICE PROBLEMS: Find the primary

value of the following equations:= 135° and 315° 1. 2 cos20 = 3 cos 0 - 1

Therefore, the solutions to the equation are 0°, 2. tan20 = 3135°, 180°, and 315°.EXAMPLE: Find the primary values of the 3. 2 sin20 - 3 sin 0 = -1equation

0 = arccot -1

2 sin20 + sin 0 - 3 = 0SOLUTION: Factor the equation and then set

each factor equal to zero, as follows:

2 sin20 + sin 0 - 3 =and

(2 sin 9 - 1) (sin 9 - 1) = 0

then

2 sin 9 - 1 = 0

and

sin 0 - 1 = 0

Solve each as follows:

2 sin 0 - 1 = 0

then

and

Also,

then

2 sin 0 = 11sin 0 = 2

. 10 = arcsm

0 = 30° and 1500

sin 0 - 1 = 0

sin 0 = 1

= arcsin 1

= 90°

ANSWERS:

1. 00, 600, 300°

2. 60°, 120°, 240°, 3000

3. 30°, 900, 150°

LIMITED SOLUTIONS

We will consider two types of possiblesolutions to fall within the category of limitedsolutions. The following examples show both ofthese types.

The first type of limited solution occurs whenan equation is solved and one of the solutions isnot true upon inspection.

EXAMPLE: Find the primary values of 0 inthe equation

sin2O sec 0 = sec 0

SOLUTION: We first rearrange, and thenfactor the equation as follows:

sin20 sec 0 = sec 0

sec 0 - sin20 sec = 0

sec 0 (1 - sin20) = 0

Set each factor equal to zero

1 - sin20 = 0

and

sec 0 = 0

Solving the first equation

1 sin20 = 0

sin29 = 1

sin 0 = ±1

120

125


then

= arcsin *1

= 900 and 1800

Solving the second equation

sec 0 = 0

then

These seem to be the values of the equation,but we squared both sides of the equation andwe must now substitute these values into theoriginal equation to verify the values. Uponsubstituting we find

tan 9 - sec 0 + 1 = 0

This implies that

tan 0° - sec 00 + 1 = 0

= arcsec 0 and since

However, there is no angle for which sec 0 0 - 1 + 1 = 0equals zero and we reject this false solution.Therefore, the primary solutions for the original the value 00 holds true.equation are 90° and 180°.

For 180° we findThe second type of limited solutions occurwhen introducing a radical into an equation by tan 0 - sec 0 + 1 = 0substitution or by squaring both members of anequation in solving the equation. These solutions tan 180° - sec 180° + 1 = 0are called extraneous roots. Solutions of thistype must be substituted into the original equa- buttion for verification.

EXAMPLE: Find the primary values of 9 for 1 -(-1) + 1 0the equation

tan 0 - sec 0 + 1 = 0 and we say 180° is an extraneous root.SOLUTION: We first rearrange the equa- PRACTICE PROBLEMS: Find the primarytion to read

values for 0 in the following equations:tan 0 + 1 = sec 0

1. tan 0 cos28 = sin29Square both sides:

2. cos29 sin 0 = sin 0 + 1tan29 + 2 tan 0 + 1 = sec29

Rearrange again: 3. 2 sec 9 + 1 - cos 0 = 0

2 tan 0 = sec29 - (tan20 + 1) 4. cot 0 - csc 8 - i3 = 0

This givesANSWERS:

2 tan 0 = 0and

tan 9 = 0Therefore,

o = arctan 0= 0° and 180°

1.

2.

3.

4.

0°, 45°,

270°

180°

240°

180° , 225°

CHAPTER 9

STRAIGHT LINES

The study of straight lines provides anexcellent introduction to analytic geometry.As its name implies, this branch of mathematicsis concerned with geometrical relationships.However, in contrast to plane and solid geometry,the study of these relationships in analyticgeometry is accomplished by algebraic analysis.

The invention of the rectangular coordinatesystem made algebraic analysis of geometricalrelationships possible. Rene Descartes, aFrench mathematician, is credited with thisinvention, and the coordinate system is oftendesignated as the Cartesian coordinate systemin his honor.

Recalling our study of the rectangular coor-dinate system in Mathematics, Vol. 1, NavPers10069-C, we review the following definitionsand terms:

1. Distances measured along, or parallel to,the X axis are ABSCISSAS. They are positiveif measured to the right of the origin; theyare negative if measured to the left of theorigin. (See fig. 9-1.)

2. Distances measured along, or parallel to,the Y axis are ORDINATES. They are positiveif measured above the origin; they are negativeif measured below the origin.

3. Any point on the coo;dinate system is des-ignated by naming its abscissa and ordinate.For example, the abscissa of point P (fig. 9-1)is 3 and the ordinate is -2. Therefore, thesymbolic notation for P is

P(3, -2)

In using this symbol to designate a point, theabscissa is always written first, followed bya comma. The ordinate is written last. Thusthe general form of the symbol is

P(x, y)

122

4. The abscissa and ordinate of a point areits COORDINATES.

DISTANCE BETWEEN TWO POINTS

The distance between two points, Pi andP2, can be expressed in terms of their coor-dinates by using the Pythagorean Theorem.From our study of Mathematics, Vol. 1, Navrers10069-C, we recall that this theorem is statedas follows:

In a right triangle, the square of the lengthof the hypotenuse (longest side) is equal to thesum of the squareS of the lengths of the twoshorter sides.

Let the coordinates of Pi be (x111) and letthose of P2 be (x2,y2), as in figure 9-2. Bythe Pythagorean Theorem,

d =i(P1N) + (P

2N)

2

where d represents the distance from P1 to P2.We can express the length of PIN in terms ofxi and x2 as follows:

Likewise,

PIN =x2 - x1

P2N=y

2-y1

By substitution in the formula developedpreviously for d, we reach the following con-clusion:

d = Ni(x2 - x1)2 + (y2 - y1)2

Although we have demonstrated the formulafor the first quadrant only, it can be proved

Chapter 9STRAIGHT LINES

Figure 9-1. Rectangular coordinate system.

for all quadrants and all pairs of points.EXAMPLE: In figure 9-2, xi = 2, x2 = 6,

3,1 = 2, and y2 = 5. Find the length of d.

SOLUTION: d = 1(6 - 2)2 + (5 - 2)2

= /42 + 32

= /16 + 9

=

= 5

This result could have been foreseen by ob-serving that triangle P1NP2 is a 3-4-5 tri-angle.

EXAMPLE: Find the distance between P1(4, 6) and P2(10, 4).

SOLUTION' d = 1(10 - 4)2 + (4 - 6)2

d = i136 -7-1- 4

d = 2 rfo

DIVISION OF A LINE SEGMENT

Many times it becomes necessary to findthe coordinates of a point which is someknown fraction of the distance between Pi andP2.

123

In figure 9-3, P is a point lying on the linejoining Pi and P2 so that

P P1 kPIP2

If P should lie one-quarter of the way betweenPi and P2, then k would equal 1/4.

Triangles PiMP and P1NP2 are similar.Therefore,

P1M p1 p

P1N P1P2

PIPSince 12 is the ratio that defines k,

PP1M

P = kN

1

Therefore,

P1M k (P1N)Referring again to figure 9-3, observe that

PiN is equal to x2 - Likewise, P1Mis equalto x - xi. Therefore, replacing P1M and PiNwith their equivalents in terms of x, the foregoingequation becomes

x - x1

= k(x2

- x 1)

x = x1 + k(x2- x 1)

is y milimmmirmimmwomminiWE 11.111111111101111111111-1110111-M11111IIIUIMIIIINEIMIThroimrMmusommassamplaB imiumlI busimisminsilimarummr. wemi Emmil Elmweliwoommipm MMMMM pm. lim01111mThr uonas uormwommumo-mmr!.No milNummuiAullum SOsm. Nummummum .-.2onom74M1111 EUS4110151111111..11110MOIN Allal-011 MNIM111 diII upsMom mrIriiiiiird.miwi muumuu.IiiII011 11111111W.MuiiiNEF8-ui.. MIME 1111111"

EN MINNINOr.11.111....111.01111.1111P011Elli 111111M115111.111R111111111ASII! MINA111AllUllMIUrn 111 1111111,1111111III 11111ENIIIII 11101111 IOWatm soilliggiumaimainimum11.NEN

Nomumwol 161111111 1111111.111M

11.11 MIN1111N11.1111MM .1111

II1 ritar11111!suommmm110111111.111111111.MW11111111rnillIN111111Figure 9-2. Dietance between two points.


X1

Xx2

Figure 9-3. Division of a line segment.

By similar reasoning,

y1 142 yl)The x and y found. as a result of the foregoing

discussion are the coordinates of the desiredpoint; whose distances from P1 and from P2 aredetermined by the value of lc

EXAMPLE: Find the coordinates of a point1/4 of the way from P1 (2,3) to P2 (4,1).

SOLUTION:1k = , x2 - = 2, y2 - y1 = -2

1 1 5x = 2 + .71 (2) = 2 + =

y = 3 + d-1 (-2)

The point P is (5i 9-1).

FINDING THE MIDPOINT

1 5=0

When the mi.dpoint of a line segment is to befound, the value of k is 1/2. Therefore,

124

1x = x1 + (x2 - x1)

1 1

xl + 2 x2 2 xl

1 1= xi + x2

1

(x1 + x2)

By similar reasoning,1

Y 2 (Y1 + Y2)

EXAMPLE: Find the midpoint of the linebetween P1 (2,4) and P2 (4,6).

SOLUTION: k =2

3

1y = (4 + 6)

= 5

The midpoint is (3,5).

INCLINATION AND SLOPE

A line drawn on the rectangular coordinatesystem and crossing the X axis forms a positiveacute angle with the X axis. This angle, shownin figure 9-.4 as angle a, is called the angle ofinclination of the line.

The slope of any line, such as AB infigure 9-4, is equal to the tangent of its angleof inclination. Slope is denoted by the letterm. Therefore, for line AB,

m = tan a

If the axes are in their conventional positions,a line sloping upward to the right has a positiveslope. A line sloping downward to the righttali; a negative slope.

Chapter 9STPAIGHT LINES

A

Figmrp 9-4,Angle of inclination.

Since the tangent of a is the ratio of P2Mto P1M, we can relate the slope of line ABto the points P1 and P2 as follows:

P2Mm = tan a =

Designating the coordinates of P1 as (x1ty1),and those of P2 as (x2,y2), we recall that

P2M Y2 yl

P1M = x2 -1

121 =

The quantities (x2 - xi) and (Y2 - yl)represent changes thai oCcur in the values ofthe x and y coordinates as a result of changingfrom P2 to Pi on line AB. The symbol usedby mathematicians to represent an incrementof change is the Greek letter delta (A). There-fore, x means "the change in x" and Ay means"the change in y.1 The amount of change in

the x coordinate, as we change from P2 toPi, is x2 - x. Therefore,

zix = x2 - x1

y2

We use this notation to express the slope ofline AB, as follows:

m = -4Y-Ax

EXAMPLE: Find the slope of the lineconnecting P? (7,6) and Pi (-1,-4).

SOLUTION:

m = AxAy = y2 yl = 6 - (-4) = 10

ax = x2 - x1 = 7 - (-1) 8

10 5

It is important to realize that the choiceof labels for P1 and P2 is strictly arbitrary.If we had chosen the point (7,6) to be Pi inthe foregoing example, and the point (-1,-4)to be P2, the following calculation would haveresulted:

m el

Ay = Y2 Y1 =

x2 xl =

- 6

- 7

=

=

-10

-8

-10 5m = -8 =

This is the same result as in the foregoingexample.

The slope of 5/4 means that a point movingalong this line would move vertically +5 unitsfor every horizontal movement of +4 units.This result is consistent with the previously

125

'130


stated meaning of positive slope; i.e., slopingupward to the right.

If line AB in figure 9-4 were parallel to theX axis, yi and y2 would be equal and the dif-ference (y2 - yi) would be 0. Therefore,

0m - 0-x2 xl

Thus we conclude that the slope of a horizontalline is 0. This conclusion can also be reachedby noting that angle a (fig. 9-4) is 0 when the lineis parallel to the X axis. Since the tangent ofOa is 09

m = tan a = 0

The slope of a line that is parallel to theY axis becomes meaningless. The tangent ofthe angle a increases indefinitely as a ap-proaches 90°. It is ^nmetimes said thatm 00

(m approaches infinity) when a approaches 900.

PARALLEL AND PERPENDICULARLINES

If we are given two lines that are parallel,their slopes must be equal. Each line willcut the X axis at the same angle a , so that

m1

= tan a, In2 = tan a

Therefore, ml m2We conclude that two lines which are parallelhave the same slope.

Suppose that two lines are perpendicular toeach other, as lines Li and L2 in figure 9-5.The slope and inclination of Li are ml and ai,respectively. The slope and inclination of L2are m2 and a 2, respectively. Then the followingis true:

m = tan a1 1

m2 = tan a2

It can be shown geometrically that a 2 (fig.9-5) is equal to al, plus 90°. Therefore,

Figure 9-5. Slopes of perpendicular lines.

tan a2 tan (al 90°)

= - cot al

1_tan al

Replacing tan a i and tan a2 by their equivalentsin terms of slope, we have

126

r112 =1

ml

We conclude that, if two lines are per-pendicular, the slope of one is the negativereciprocal of the slope of the other.

Conversely, if the slopes of two lines arenegative reciprocals of each other, the linesare perpendicular.

EXAMPLE: In figure 9-6, show that lineLi is perpendicular to line L2. Line Li passesthrough points Pi (0,5) and P2 (-1,3). LineL2 passes through points P2 (4,3) and P3(3,1).


Figure 9-6. Proving lines perpendicular.

SOLUTION: Let in and m2 represent theslopes of lines Li and L2, respectively. Then

we have

ANGLE BETWEEN 'IWO LINES

When two lines intersect, the anglebetweenthem is defined as the smallest angle throughwhich one of the lines must be rotated to makeit coincide with the other line. For example,the angle (I) in figure 9-7 iL the angle betweenlines Li and L2.

Referring to figure 9-7,

45 = a2 -

It is possible to determine the value of di-rectly from the slopes of lines Li and L2, asfollows:

M1 =

m =2

5 - 32

-2 1.

tani an (a2 cv1)

tan a2 al0 - (-1)

1 - 31+ tan a

1tan a

2

3 - (-1) 4

Since their slopes are negative reciprocalsof each other, the lines are perpendicular.

PRACTICE PROBLEMS:1. Find the distance between P1 (5,3) and

P2 (6p 7).2. Find the distance between P1 (1/2,1)

and P2 (3/2,5/3).3. Find the midpoint of the line connecting

Pi (5,2) and P2 (-1,-3).4. Find the slope of the line joining P1

(-2,-5) and P2 (2,5).ANSWERS:1.

2. VII3

3. (2, .4)

4. 5

1 27

132

This result is obtained by use of the trigono-metric identity for the tangent of the dif-ference between two angles. Trigonometric

Figure 9-7.--Angle between two lines.


identities are discussed in chapter 8 of thistraining course.

Recalling that the taLgent of the angle ofinclination is the slope of the line, we have

tan al = m1 (the slope of L )

1

tan a2 = m2 (the slope of L2)

Substituting these expressions in the tangentformula derived in the foregoing discussion,we nave

m2 mltan = 1 + m m1 2

If one of the lines were parallel to the Yaxis, its slope would be infinite. This wouldrender the slope formula for tan 0 useless,because an inainite value in both the numerator

m2 - mland denominator of the fraction ' pro-'1 2duces an indeterminate form. However, if oneof the lines is known to be parallel to the Yaxis the tangent of 0 may be expressed byanother method.

Suppose that L2 (fig. 9-7) were parallelto the Y axis. Then we would have

a2 = 90°

0 = 90° - altan 0 = cot al

PRACTICE PROBLEMS:1. Find the angle between the two lines

which have m1 = 3 and m2 = 7 for slopes.2. Find the angle between two lines whose

slopes are mi = 0, in2 = 1. = Osig..nifies that line Li is horizontal and the formulastill holds).

3. Find the angle between the Y axis and aline with a slope of m = -8.

4. Find the obtuse angle between the X axisand line with a slope of m = -8.

128

ANSWERS:1. 10°18'2. 45°3, 7°7'4. 97°7'

EQUATION OF A STRAIGHT LINE

In Mathematics, Volume 1, NavPers 10069-C,equations such as

2x + y = 6

are designated as linear equations, and theirgraphs are shown to be straight lines. Thepurpose of the present discussion is to studythe relationship of slope to the equation ofa straight line.

POINT-SLOPE FORM

Suppose that we desire to find the equationof a straight line which passes through a knownpoint and has a known slope. Let (x,y) re-present the coordinates of any point on theline, and let (xi,y1) represent the coordinatesof the known point. The slope is representedby m.

Recalling the formula defining slope interms of the coordinates of two points, wehave

111 =

. . y - y1 = ni(x x1)

EXAMPLE: Find the equation of a linepassing through the point (2,3) and having aslope of 3.

SOLUTION:

x1 = 2 and y1 = 3

y - y1 = m(x - x1)

. y - 3 = 3(x - 2)

y - 3 = 3. -6

y - 3x = -3


The point-slope form may be used to findthe equation of a line through two known points.The values of xi, 2r22 yl) and y2 are firstused to find the slope of the line, and then eitherknown point is used with the slope in the point-slope form.

EXAMPLE: Find the equation of the linethrough the points (-3,4) and (4,-2).

SOLUTION:

Y2 ylm =x2 - x1

-2 - 4 -64 + 3 7

Letting (x,y) represent any point on the line,and using (-3,4) as a known point, we have

-6y - 4 = -,--f [x (-3)]

7(y - 4) = -6(x + 3)

7y - 28 = -6x - 18

7y + 6x= 10

SLOPE-INTERCEPT FORM

Any line which is not parallel to the Y axisintersects the Y axis in some point. The xcoordinate of the point of intersection is 0,because the Y axis is vertical and passesthrough the origin. Let the y coordinate ofthe point of intersection be represented by b.Then the point of intersection is (0,b), asshown in figure 9-8. The y coordinate, b,

Figure 9-8.Slope-intercept form.

This is the standarda straight line.

EXAMPLE: Findthat intersects the Yand has a slope of 5/3.

SOLUTION:

slope-intercept form of

the equation of a lineaxis at the point (0,3)

v mx + b5

y x + 3

3y = 5x + 9

PRACTICE PROBLEMS:

Write equations for linesslopes as follows:

is called the y intercept. 1. P(3,5), m = -2

The slope of the line in figure 9-8 is 4,4',1cY.

The value of 4 y in this expression is y -b,where y represents the y coordinate of any

2. P(-2,-1), m =

point on the line. The value of x is equalto the x coordinate of 13(x,y), so that

3. Pi(2,2) and P2(-4,-1)

= y - bAX x

mx = y - b

y = mx + b

4. Y intercept = 2, in

ANSWERS:

1. y = -2x + 11

2. 3y = x - 1

129 .

having points and


3. 2y = x + 2

4. y = 3x + 2

NORMAL FORM

Methods *or determining the equation ofa line usually depend upon some knowledgeof a point or points on the line. We nowconsider a method which does not requireadvance knowledge concerning any of the line'spoints. All that is known about the line isits perpendicular distance from the originand the angle between the perpendicular andthe x axis.

In figure 9-9, line AB is a distance p awayfrom the origin, and line OM forms an angle 0with the X axis. We select any point P(x,y)on line AB and develop the equation of lineAB in terms of the x and y of P. Since Prepresents ANY point on the line, the x andy of the equation will represent EVERY pointon the line and therefore will represent theline itself,

PR is constructed perpendicular to OB atpoint R. NR is drawn parallel to AB, andPN is parallel to OB. PS is perpendicularto NR and to AB. Since right triangles OMBand RSP have their sides mutually perpen-dicular, they are similar; therefore, anglePRS is equal to 0 Finally, the x distanceof point P is equal to OR, and the y distanceof P is equal to PR.

In order to relate the distance p to x andy, we reason as follows:

ON = (OR) (cos )

= x cos

PS = (PR) (sin())

= y sin

OM = ON + PS

p = ON + PS

p = x cos 0 + y sin 0

This final equation is the NORM.AL FORM.The word "normal" in this usage refers to theperpendicular relatimisnip between OM andAB. "Normal" frequently means "perpen-dicular" in mathematical and scientific usage.

The distance p is considered to be alwayspositive, and 0 is any angle between 0° and360°.

EXAMPLE: Find the equation of a linethat is 5 units away from the origin, if theperpendicular from the line to the origin formsan angle of 300 with the positive side of theX axis.

SOLUTION:

p = 5; 0 = 300

p = x cos 0 + y sin

5 = x cos 30° + y sin 30°

1(12)

y

PARALLEL ANDPERPENDICULAR LFNES

The general equation of a straight line isoften written with capital 1 etters for co-efficients, as follows:

Ax + By + C = 0

These literal coefficients, as they aro called,represent the numnical coefficients encoun-tered in a typical linear equation.

Suppose that we are given two equationswhich are duplicates except for the constantterm, as follows:

Ax + By + C = 0

Ax+ By+ D=0

By placing these two equations in slope-interceptform, we can show that their slopes areequal, as follows:

ii) C.)B/

13/

Thus the slope of each line is - A/13.Since lines having equal slopes are parallel,

we reach the following conclusion: In any two

130

135


Figure 9-9.Normal form.

X

linear equations, if the coefficients of the x and y A situation similar to that prevailing withterms are identical in value and sign, then the parallel lines involves perpendicular lines.lines represented by these equations are par- For example, consider the equationsallel.

EXAMPLE: Write the equation of a lineparallel to 3x - y - 2 = 0 and passing throughthe point (5,2).

Ax + By + C = 0

Bx - Ay + D = 0SOLUTION: The coefficients of x and y in

the desired equation are the same as those in Transposing into the slope-intercept form, wethe given equation. Therefore, the equation is have

3x - y + D = 0

Since the line passes through (5,2), the valuesx = 5 and y = 2 must satisfy the equation. Sub-stituting these, we have

3(5) - (2) + D = 0

D = -13

Thus the required equation is

3x - y - 13 = 0

Y = (2) x 4. (-A-

D)

Since the slopes of these two lines are nega-tive reciprocals, the lines are perpendicular.

The conclusion derived from the foregoingdiscussion is as follows: If a line is to beperpendicular to a given line, the coefficientsof x and y in the required equation are foundby interchanging the coefficients of x and y

131

136

MATHEMATICS , VOLUME 2

in the given equation and changing the signof one of them.

EXAMPLE: Write the equation of a line per-pendicular to the line x + 3y + 3 = 0 and havinga y intercept of 5.

SOLUTION: The required equation is

3x - y + D = 0

Notice the interchange of coefficients and thechange of sign. At the point where the linecrosses :ne Y axis, the value of x is 0 andthe value of y is 5. Therefore, the equationis

3(0) - (5) + D = 0

D = 5

The required equation is

3x - y + 5 = 0

PRACTICE PROBLEMS:

Find the equations of the following lines:

1. Through (1,1) and parallel to 5x -3y = 9.

2. Through (-3,2) and perpendicular to+ y = 5.

3. Through (2,3) and perpendicular to 3x -2y = 7.

4. Through (2,3) and parallel to 3x2y = 7.

ANSWERS:

1. 5x - 3y = 2

2. x y = -5

3. 2x + 3y = 13

4. 3x - 2y = 0

DISTANCE OF A POINTFROM A LINE

It is frequently necessary to express thedistance of a point from a line in terms ofthe coefficients in the equation of the line.In order to do this, we compare the two rormsof the equation of a straight line, as follows:

General equation: Ax + By + C 0Normal form: x cos 0 + y sin 0 - p = 0

The general equation and the normal formrepresent the same straight line. Therefore,A (the coefficient of x in general form) isproportional to cos 0 (the coefficient of x inthe normal form). By similar reasoning, B isproportional to sin 9 , and C is proportionalto -p. Recalling that quantities proportionalto each other form ratios involving a constantof proportionality, let k be this constant.Thus we have

cos = kA

sin 0= k

cos 0 = kA

sin 0 = kB

^

Squaring both sides of these two expressionsand then adding, we have

IN;:1

cos2 0 + sin2 0 = k2(A2 + 82)

1 = k2(A2 + B2

k2 1

A2 + B2

k = 1

jA2 B2


The coefficients in the normal form, ex-pressed in terms of A, B, and C, are as follows:

cos 0 =

sin 0 =

A

/A2 B2

11 A2 + B2

*jA2 + B2

The sign of 1A 2 + B2 is chosen so as to make p(a distance) always positive.

The conversion formulas developed in theforegoing discussion are used in finding thedistance from a point to a line. Let p representthe distance of line L from the origiri. (Seefig. 9-10.) In order to find d, the distanceof point Pi from line L, we construct a linethrough Pi and parallel to L. The distanceof this line from the origin is OS, and thedifference between OS and p is d.

Figure 9-10.Distance from a point to a line.

We obtain an expression for d, based on thecoordinates of P1, as follows:

OS = x1 cos 0 + y1 sin e

d = OS - p

= xi cos e + y1 sin 0 - p

Returning to the expressions for s:n cos 0and -p in terms of A, B, and C (the coefficientsin the general equation), we have

d =A

)B

jA2+ B2 yl jA2 132

±IA2 + B2

The denominator in each of the expressionscomprising the formula for d is the same.Therefore we may combine as follows:

d -xlA + yl B + C

A2 B2

We use the absolute value, since d is a dis-tance, and thus avoid any confusion arisingfrom the * radical.

EXAMPLE:. Find the distance from the point(2,1) to the line 4x + 2y + 7 = 0.

SOLIJ ION:

d(4) (2) + (2) (1) + 7

'42

133

slaS

8 + 2 + 7

-7q


PRACTICE PROBLEMS:

In each of the following problems, find thedistance from the point to the line:

ANSWERS:

1. 19 JR10

1. (5,2), 3x - y + 6 = 0 2. 1/-6-

2. (3,-5), 2x + y + 4 = 0 3. 2

3. (3,-4), 4x + 3y = 10 4. 1

4. (-2,5), 3x + 4y - 9 = 0

CHAPTER 10

CONIC SECTIONS

This chapter is a continuation of the studyof analytic geometry. The figures presented inthis chapter are plane figures which are in-cluded in the general class of conic sections orsimply "conics."

Conic sections are so named because theyare all plane sections of a right circular cone.A e!-rlf. can be formed by cutting a cone perpen-dicular to its axis. An ellipse is produced whenthe cone is cut obliquely to the axis and the sur-tace. A hyperbola results when the cone isintersected by a -.)lane parallel to the axis, anda parabola Is the result when the intersectingplane is parallel to an element of the surface.These are illustrated in figure 10-1.

When the curve produced by cutting the coneis placed on a coordinate system it may be de-fined as follows:

A conic section is the locus of a point thatmoves so that its distance from a fixed point isin a constant ratio to its distance from a fixedline. The fixed point is the focus, and the fixedline is the directrix.

The ratio r3ferred to in the definition iscalled the eccertricity. If the eccentricity (e)is less than one, the curve is an ellipse. Ife is greater than one, the curve is a hyperbola.If e is equal to 1, the curve is a parabola. Acircle is a special case having an eccentricityequal to zero, and may be defined by the dis-tance from a point. It is actually a limitingcase of an ellipse in which the eccentricity ap-proaches zero. Thus, if

e = 0, it is a circle

e < 1, it is an ellipse

e = 1, it is a parabola

e 1, it is a hyperbola

The eccentricity, focus, and directrix areused in the algebraic analysis of conic sectionsand the corresponding equations. The conceptof the locus of an equation also enters into ana-lytic geometry; this concept is discussed beforethe individual conic sections are studied.

THE LOCUS OF AN EQUATION

In chapter 9 of this course, methods foranalysis of linear equations are presented. Ifa group of x and y values (or ordered pairs, P(x,y)) which satisfy a given linear equation areplotted on a coordinate system, the resultinggraph is a straight line.

When higher ordered equations such asx2 + y2 = 1 or y =

are encountered, the resulting graph is not astraight line. However, the points whose coordi-nates satisfy most of the equations inx and y arenormally not scattered in a random field. If thevalues are plotted they will seem to follow a lineor curve (or a combination of lines and curves).In many texts the plot of an equation is calleda curve, even when it is a straight line. "7hiscurve is called the locus ot the equation. Thelocus of an equation is a curve containing thosepoints, and only those points, whose coordinatessatisfy the equation.

At times the curve may be defined by a setof conditions rather than by an equation, thoughan equation may be derived from the given con-ditions. Then the curve in questlonwould be thelocus of all points which fit the conditions. Forinstance a circle may be said to be the locus ofall points in a plane which lie a fixed distancefrom a fixed point. A straight line may be de-fined as the locus of a point that moves in a planeso that it is at all times equidistant from twofixed points. The method of expressing a set

135


HYPERBOLA CIRCLE

PARABOLA ELUPSE

Figure 10-1.Conic sections.

of conditions in analytical form gives an equation.Let us draw up a set of conditions and translatethem into an equation.

EXAMPLE: What is the equation of the curvewhich is the locus of all points which arlequidistant from the two points (5, 3) and (2,1)?

SOLUTION: First, as in figure 10-2, choosesome point having coordinates (x, y). Recall fromchapter 9 of this course that the distance betweenthis point and (2, 1) is given by:

/(y - 1)2 + (x - 2)2

The distance between pAnt (x, y) and (5, 3) willbe given by

3)2 5)2

Equating these distances, since the point is tobeequidistant from the two given points, we have

.1(y 1)2 2)2 3)2 5)2

136

Squaring both sides

(y - 1)2 + (x - 2)2 . (y - 3)2 + (n- 3)2

Expanding2

y - 2y + 1 + x2 - 4x + 4

= y2 - 6y + 9 + x2 - 10x + 25Canceling and collecting terms:

4y + 5 = -6x + 34

4y = -6x + 293y = - x + 7. 25

This is the equation of a straight line with a slopeof minus 3/2, and a Y intercept of +7.25.

EXAMPLE: Find the equation of the curvewhich is the locus of all points which areequidistant from t a line x = -3 and the point(3, 0).

SOLUTION: The distance from the point (x,y) on the curve to the line will be (x (-3)) or

141

xly)

513)

X

Figure 10-2.Locus of points equidistantfrom two given points.

Chapter 10CONIC SECTIONS

(x + 3). Refer to figure 10-3. The distance frcmthe point (x, y) to the point (3, 0) is

- 0)2 + (x - 3)2

Equating the two distances,

x + 3 = 1y2 + (x - 3)2

Squaring both sides,

x2 + 6x + 9 = y2 + x2 - fix + 9

Canceling and collecting terms,

y2 = 12x

which is the equation of a parabola.EXAMPLE: What is the equation of the curve Since

the locus of which is apointwhich moves so thatat all times the ratio of its distance from thepoint (3, 0) to its distance from the linex = 25/3 is equal to 3/5? Refer to figure 10-4.

SOLUTION: The distance from a point (x,y) to the point (3, 0) is given by

d1

= - 3)2 + (y - 0)2

The distance from the same point (x, y) to theline is

25d = x2 -7

then

Figure 10-4. Ellipse

d1 3 3

`A2

= or d1 = d2

3)2 _2 3 (25 -"Squaring both sides and expanding,

x2 y2 59 (x2 530 6295)

x2 - 6x + 9 + y2 x 6x + 259 2

Collecting terms and transposing

16 2 225 x + y = 16

Dividing through by 16

2 2x25 16 a

Figure 10-3.Parabola. This is the equation of an ellipse.

137

142

MATHEMATICS, VOLIME 2

PRACTICE PROBLEMS: Fi ad the equationwhich is the locus of the point which moves sothat it is at all times:

1. Equidistant from the points (0, 0) and(5, 4),

2, Equidistant from the points (3, -2) and(-3, 2).

3. Equidistant irom the line x = -4 and thepoint (3, 4).

4. Equidistant from the point (4, 5) and theline y = 5x -4. HINT: Use the standard distanceformula to find the distance from the point P(x, y) and the point P (4, 5). Then use the fnrmulafor finding distance from a point to aline, givenin chapter 9 of this course, to find the distancefrom P (x, y) to the given line. Put the equationof the line in the form Ax + By + C = 0.

ANSWERS:

1. y = 1. 25x + 418

2. 2y = 3x

3. y2 - 8y = 14x - 9

4. x2 + 10xy + 25y2 - 168x - 268y + 1050 = 0

THE CIRCLE

A circle is the locus of a point which is al-ways a fixed distance from a fixed point calledthe center.

The fixed distance spoken of here is the radiusof the circle.

The equation of a circle with its center at theorigin (figure. 10-5) is, from the definition:

frx- 0)2 + (y - 0)2 = r,

where (x, y) is a point on the circle and r is theradius and replaces d in the standard distanceformula. Then

or2 2x + y2 = r

(1)

If the center of a circle, figure 10-6, is atsome point x = h, y =k, the distance of the mov-

138

143

ing point from Lhe center will be constant andequal to

or

./(x - h)2 + - k)2 = r

- h)2 + - k)2 = r2 (2)

Equations (1) and (2) are the standard forms forthe equation of a circle. Equation ;1) is merelya special case of equation (2) in which h and kare equal to zero.

The equation of a circle may also be ex-pressed in the form:

x2 + y2 + Bx + Cy + D =

where B, C, and D are constants.

(3)

THEOREM: An equation of the second de-gree in which the coefficients of the x2 and y2terms are equal, and there is no (xy) term, re-presents a circle.

Whenever we find an equation in the form ofequation (3), it is best to convert it to the formof equation (2), so that we have the coordinatesof the center of the circle and the radius as partof the equation. This may be done as shown inthe following example problems.

Figure 10-5.Circle with centerat the origin.


Figure 10-6.Circle with centerat (h, k).

EXAMPLE: Find the coordinates of thecenter and the radius of the circle which isdescribed by the following equation:

x2 + y2 - 4x - 6y + 9 = 0

SOLUTION: First rearr2nge the terms

x2 - 4x + y2 - 6y + 9 = 0

and complete the square in both x and y. Com-pleting the square is discussed in the chapter onquadratic solutions in Mathematics, Vol. 1,NavPers 10069-C, The procedure consists i4asi-cally of adding certain quantities to both sidesof a second degree equation to form a perfectsquare trinomial. When both the first and seconddegree members are known, the square of one-half the coefficient of the first degree term isadded to both sides of the equation. This willallow the quadratic equation to be factored intoa perfect square trinomial. To complete thesquare in x in the given equation

x2 - 4x + y2 - 6y + 9 = 0

add the square of one-half the coefficient of xto both sides of the equation

x2 - 4x + (2)2 + y 2 - 6y + 9 = 0 + (2)2

then

(x2 - 4x + 4) + y2 - 6y + 9 = 4

(x-2)2 + y2 - 6y + 9 = 4

completes the square in x.For y then

2)2 y2 (3)2

(x - 2)2 + (y2 - 6y + 9) + 9 = 4

(3)2

+ 9

- 2)2 + - 3)2 + 9 = 4 + 9

completes the square in y.Transpose all constant terms to the right-

hand side and simplify

(x - 2)2 + (y - 3)2 = 4 + 9 - 9

(x - 2)2 + (y - 3)2 = 4

and the equation is in the standard form ofequation (2). This represents a circle with thecenter at (2, 3) and with a radius equal toif or 2.

EXAMPLE: Find the coordinates of thecenter and the radius of the circle given by theequation

2 2 1 27x + y +.2. X - .3Y

SOLUTION: Rearrange and complete thesquares in x and y

2 1 2 27x +-2- x+y -3y it)

2 1 1 2 9 27 1 9

Transposing all constant terms to the right-handside and adding,

2 2 9tic 2 + 18 4. 3y + i) 4

139

1144

...=111,


Reducing to standara Thrm

1 2 3 2 2(x + + (y - = (2)

Thus, the equation represents a circle with itscenter at (-1/4, 3/2) and a radius equal to 2.

PRACTICE PROBLEMS: Find the coordi-nates of the center and the radius for the circlesdescribed by the following equations.

1. x2 - 29y2 - 4y + = 025

2. x2 + 6x + y2 - 14y = 23

3. x2 - 14x + y2 + 22y = -26

2 2 2 2 24. x + y

5. x2 + y2 - 1 = 0

ANSWERS:

21. Center (-6., 2), radius

2. Center (-3,7), radius 9

3. Center (7,-11), radius 12

1 1 2 VT.3-4. Center (- -0- , radius

5. Center (0,0), radius 1

THE CIRCLE DEFINEDBY THREE POINTS

In certain situations it is convenient to con-sider the following standard form of a circle

x2 + y2 + Bx + Cy + D = 0

as the equation of a circle in which the specificvalues of the constants B, C, and D are to bedetermined. In this problem the unknowns to befound are not x and y, bit the values.of the con-stants B, C, and 1). The conclitionswhich define

140

I 45

the circle .are used to form algebraic relation-ships between these constants. For example, ifone of the conditions imposed on the circle isthat it pass through the point (3,4) then the stan-dard form is written with x and y replaced by 3and 4 respectively; thus

x2 + y2 + Bx + Cy + D = 0

is rewritten as

(3)2 + (4)2 + B(3) + C(4) + D = 0

38 + 4C + D = -25

There are three independent constants intheequation of a circle; therefore, there must bethree conditions given to define a circle. Eachof these conditions will yield an equationwithB,C, and D as theunknowns. These three equationsare then solved cimultaneously to determine thevalues of the constants which satisfy all of theequations. In an analy- is, the number of inde-pendent constants in .11 J general equation of acurve indicate how many cnnditions must be setbefore a curve can be completely defined. Also,the number of unknowns in an equation indicatesthe number of equations which must be solvedsimultaneously to find the values of the unknowns.For example, if B, C, and D are unknowns in anequation, three separate equations involvingthese variables are required for a solution.

A circle may be defined by three noncollinearpoints, that is, by three points which do not lieon a straight line. There is only one possible cir-cle through any three noncollinear points. To findthe &nation of the circle determined by the threepoints substitute the x, y values of each of thegiven points into a general equation to form threeequations with B, C, and D as the unknowns.These equations are then solved simultaneouslyto find the values of B, C, and D in the equationwhich satisfies the three given conditions.

The solution of simultaneous equations intwovariables is discussed in Mathematics Vol. 1.Systems involving three variables use an exten-sion of the same principles, but with three equa-tions instead of two. Step-by-step explanationsof the solution will be given in the example prob-lems.

EXAMPLE: Write the equation of the circlewhich passes through the points (2,8), (5,7), and(6, 6).

SOLUTION: The method used in this solutioncorresponds to the addition-subtraction method


used for solution of equations in two variables.However, the method or combination of methodsused will depend on a particular problem. Noone method is best suited to all problems.

First, write a general equation of the form

x2 + y2 + Bx + Cy + D = 0

for each of the given points, substituting the givenvalues for x and y and rearranging

For (2, 8)

For (5, 7)

For (6,6)

4 + 64 + 28 + 8C + D = 0

2B + 8C + D = -68

25 + 49 + 5B + 7C + D = 0

5B + 7C + D = -74

36 + 36 + 6B + 6C + D = 0

6B+ 6C+ D=-72

To aid in the explanationwe number the threeresulting equations

2B + 8C + D = -68

5B + 7C + D = -74

6B + 6C + D = -72

The first step is to eliminate one of the unknownsand have two equaticns and two unknowns remain.The coefficient of D is the same in all three equa-tions and is the one most easily eliminated byaddition and subtraction. This is done in thefollowing manner. Subtract (2) from (1)

2B + 8C + D = - 68

5B + 7C + D = - 74

(1)

(-) (2)

-3B + C = 6 (4)

Subtract (3) from (2)

5B + 7C + D = - 74 (2)

68 + 6C + D = - 72 (3)

-B + C (5)

141

(1-46

This gives two equations, (4) and (5), in two un-knowns which can be solved simultaneously,Since the coefficient of C is the same in bothequations it is the most easily eliminated vari-able.

To eliminate C, subtract (4) from (5)

-B+ C =-2

-38 + C = 6

28 = -8

B = -4

To find the value of C substitute the valuefound for B in (6) in equation (5)

-B + C = -2

-(-4) + C = -2

C = -6

(5)

(7)

Now the values of B and C can be substitutedin any one of the original equations to determinethe value of D.

If the values are substituted in (1)

28 + 8C + D = -68

2(-4) + 8(-6) + D = -68

-8 - 48 + D = -68

D = -68 + 56

(1)

D = -12 (8)

The solution of the system of equations gavevalues for three independent constants in thegeneral equation

x2+ y2+ Bx+ Cy+ D=0

When the constant values are substituted theequation takes the form of

x2 + y2 - 4x - 6y - 12 = 0Rearranging and completing the square in x and y,

(x2 - 4x + 4) + (y2 - 6y + 9) - 12 = 4 + 9

(x - 2)2 + - 3)2 = 25

z


which corresponds to a circle with the centerat (2, 3) with a radius of 5. This is the circledescribed by the three given conditions and isshown in figure 10-7 (A).

The previous example problem showed onemethod for determining the equation of a circlewhen three points are given. The next exampleshows another method for solving the same prob-lem. One of the most important things to keepin mind when studying analytic geometry is thatmany problems may be solved by more than onemethod. Each problem should be analyzed care-fully to determine what relationships exist be-tween the given data and the desired results ofthe problem. Relationships such as distance fromone point to another, distance from a point to aline, slope of a line, the Pythagorean theorem,etc., will be used to solve various problems.

EXAMPLE: Find the equation of the circledescribed by the three points (2, 8), (5, 7), and(6, 6). Use a method other than that used in theprevious example problem.

SOLUTION: A different method of solvingthis problem results from the reasoning in thefollowing paragraphs.

The center of tne desired circle will be theintersection of the perpendicular bisectors ofthe chords connecting points (2, 81 with (5, 7)and (5, 7) with (6, 6), as shown in Zigure 10-7(B).

The perpendicular bisector of the line con-necting two points is the locus of a point which

( A )

moves so that it is always equidistant from thetwo points. Using this analysis we can get theequations of the perpendicular bisectors of thetwo lines.

Equating the distance formulas which de-scribe the distances from a point (x,y), which isequidistant from the points (2,8) and (5,7),gives

or

j(x _2)2 8)2 i(c 5)2 7)2

Squaring both sides glves

(x - 2)2 + & - 8)2 = (x - 5)2 + - 7)2

2 2- 4x + 4 + yx - 16y + 64 =

x2 - 10x + 25 + y2 - 14y2 + 49

Canceling and combining terms results in

or6x - 2y = 6

3x - y = 3

( )

Figure 10-7.Circle described by three points.

142

Chapter 10 CONIC SECTIONS

Follow the same procedure for the points(5, 7) and (6, 6).

vfix 5)2 27 1/(X - 6)2 + (y - 6)2

Squaring each side gives

(x - 5)2 + (y - 7)2 = (x - 6)2 + (y - 6)2

x2 - 10x + 25 + y2 + 14y + 49 =

x2 - 12x + 36 + y2 - 12y + 36

Canceling and combining terms gives a secondequation in x and y.

Or

2x - 2y = -2

x - y = -1

Solving the equations simultaneously will givethe coordinates of the intersection of the twoperpendicular bisectors; this is the center ofthe circle.

3x - y = 3

x - y = -1 (Subtract)

2x = 4

x = 2

Substitute the value x = 2 in one of the equa-tions to find the value of y.

= 3

Thus, the center of the circle is the point (2,3).The radius will be the distance between the

center (2,3) and one of the three given points.Using point (2, 8) we obtain.

r = 1(2 - 2)2 + (8 - = Ar2-3 = 5

143

La,

The equation of this circle is

(x - 2)2 + - 3)2 = 25

as was found in the previous example.If a circle is to be defined by three points

the points must be noncollinear. In some casesit is obvious that the three points are non-collinear. Such is the case with points (1,1),(-2, 2), and (-1, -1), since the points are in quad-rants 1, 2, and 3 respectively and cannot be con-nected by a straight line. However, there aremany cases in which it is difficult to determineby inspection whether or not the points are colli-near, and a methel for determining this ana-lytically is needed. In the followng example anattempt is made to find the circle described bythree points, when the three points are colli-near

EXAMPLE: Find the equation of the circlewhich passes through the points (1,1), (2, 2),(3,3).

ZIOLUTION: Substitute the given values of xand y in the standard form of the equation of acircle to get three equations in three unknowns.

x2 + y2 + Bx + Cy + D = 0

For (1,1) 1 + 1 + B + C + D = 0

B + C + D = -2 (9)

For (2, 2) 4 + 4 + 2B + 2C + D = 0

2B + 2C + D = -8 (10)

For (3,3) 9 + 9 + 3B + 3C + D = 0

38 + 3C + D = -18 (11)

To eliminate D, first subtract (9) from (10).

2B+ 2C + Drs -8

B + C + D = -2 (Subtract)

B + C = -6

Next subtract (10) from (11).

3B + 3C + D = -18

2B + 2C + D = -8

(12)

B + C = -10 (13)

4


Then subtract (13) from (12) to eliminate one ofthe unknowns.

B + C = -6

B + C = -10

0 + 0 = 4

ANSWERS:

1. x2 + y2 - 10x + 4y = 56

2. x2 + y2 - 6x - 14y = 7

3. No solution; the given points describethe straight line y = x.

0 = 4 4. x2 + y2 - 2x + 14y = 75This solution is not valid and there is no cir-cle through the three given points. The readershould attempt to solve (12) and (13) bythe sub-stitution method. When the three given pointsare collinear an inconsistent solution of sometype will result.

If we attempt to solve the problem by eli-minating both B and C at the same time (tofind D) another type of inconsistent solution re-sults. With the given coefficients it is notdifficult to eliminate both A and B at the sametime. First, multiply (10) by 3 and (11) by-2 and add the resultant equations.

6B + 6C + 3D = -24

-6B - 6C - 2D = 36 (+)

D = 12

Then multiply (9) by -2 and add the resultantto (10)

-2B - 2C - 2D = 4

2B + 2C + D = -8

-D = -4

D = 4

This gives two values for D and is in-consistent since each of the constants must havea unique value consistent with the given con-ditions. The three points are on the straightline y = x.

PRACTICE PROBLEMS: In each of the prob-lems below find the equation of the circle whichpasses through the three given points.

1. (14,0), (12,4), and (3,7)2. (10,3), (11,8), and (7,14)3. (1,1), (0,0), and (-1,-1)4. (12,-5), (-9,-12), and (-4,3)

144

THE PARABOLA

The parabola is the locus of all points whichare equidistant from a fixed point, called thefocus, and a fixed line called the directrix.In the parabola shown in figure 10-8, the pointV, which lies halfway between the focus of thedirectrix is called the VERTEX of the parabola.In this figure and in many of the parabolasdiscussed in the first portion of this section,the vertex of the parabola will fall atthe origin;however, the vertex of the parabola, like thecenter of the circle, can fall at any point in theplane.

In figure 10-8, the distance from the point(x, y) on the curve to the focus (a, 0) is

I(xa)-2+ y2

The distance from the point (x, y) to the di-rectrix is

x + a

Since by definition these two distances are equalwe may set them equal

- a)2 + y2 = x + aSquaring both sides

a)2 y2 a)2

Expanding

2 - 2ax + a2 2+ y = x2x + 2ax + a2Canceling and combining terms bve an equa-tion for the parabola

y2 = 4ax

Chapter 10--CC.NIC SECTIONS

For every positive value of x in the equa-tion of the parabola there are two values of y.But when x becomes negative the values of yare imaginary. Thus, the curve must be en-tirely to the right of the Y axis when the equa-tion is in this form and a is positive. If theequation is

y2 = - 4 ax

(a negative) the curve lies entirely to the leftof the Y axis.

If the form of the equation is

x2 = 4ay

the curve will open upward and the focus willbe a point on the Y axis. For every positivevalue of y there will be two values of x and thecurve will be entirely above the X axis. Whenthe equation is in the form

x2 = -4ay

Figure 10-8.The parabola.

the curve will open downward, be entirely be-low the X axis, and have as its focus a point onthe negative Y axis. Parabolas which are rep-resentative of the four cases given here areshown in figure 10-9.

When x is equal to a in the equation

it follows that

and

y2 = 4ax

2= 4a2

y

y = 2a

This value of y is the height of the curve at ttefocus or the distance from the focus to pointD in figtire 10-8. The width of the curve at thefocus is the distance from point D to point D'in the figure and is equal to 4a. This width iscalled the LATUS RECTUM in many texts; how-ever, a more descriptive term is FOCAL CHORDand both terms will be used in this course. Thelatus rectum is one of the properties of a par-abola which is used in the analysis of a par-abola or in the sketching of a parabola.

EXAMPLE: Give the value of a, the lengthof the focal chord, and the equation of the par-abola which is the locus of all points equi-distant from the point (3,0) and the line x -3.

SOLUTION: First plot the given informationon a coordinate system as shown infigure 10-10(A). Reference to figure 10-8 shows that thepoint (3, 0) corresponds to the position of thefocus and that the line x = -3 is the directrix

X of the parabola. Figure 10-8 also shows thatthe value of a is equal to one half of the distancefrom the focus to the directrix or, in thisproblem, one half the distance from x = -3 tox = 3. Thus, the value of a is 3.

The second value required by the problemis the length of the focal chord. As statedpreviously, the focal chord length is equalto 4a. The value of a was found to be 3 sothe length of the focal chord is 12. Refer-ence to figure 10-8 shows that one extremityof the focal chord will be a point on the curvewhich is 2a or 6 units above the focus, and theother extremity is a second point 2a or 6 unitsbelow the focus. Using this information andre-calling that the vertex is one-half the distance

145

148


Isa4ay

(C )

DIRECTRIX

rso

FOCUS(o,,413

(D)

Figure 10-9,Farabolss corresponding to four forms of the equation.

148

. 149

x 2-4oy

x


(3,0)X

(C)

(B)

F

3,0)

DI

X

Figure 10.10.Sketch of a parabola.

(0)


from the focus to the directrix, plot three morepoints as shown in figure 10-10 (B).

Now a smooth curve through the vertex andthe two points that are the extremities of thefocal chord is a sketch of the parabola in thisproblem. (See fig. 10-10 (C).)

To find the equation of the hyperbola refer tofigure 10-10 (D) and use the procedure usedearlier. We know by definition that any pointP(x, y) on the parabola is equidistant from thefocus and directrix.Thus we equate these two distances and

a)2+ y2 = x + a

However, we have found the distance a to beequal to 3 so we substitute and

(x - 3)2 + y2 x + 3

Square both sides

(x - 3)2 + y2 = (x + 3)2

Expand

x2 - 6x + 9 + y2 = x2 + 6x + 9

Cancel and combine terms to obtain the equa-tion of the parabola

y2 = 12x

If we check the consistency of our findings,we see that the form of the equation and thesketch agree with figure 10-9 (A). Also, the 12in the right side of the equation corresponds tothe 4a in the general form and is correct sincewe determined that the value of a was 3.

NOTE: When the focus of a parabola lies onthe Y axis, the equated distance equation is

(iryF 2)2 + x2 = y + a

PRACTICE PROBLEMS: Give the equation,the value of a, and the length of the focal chordfor the parabola which is the locus of all pointsequidistant from the point and line given in thefollowing problems.

1. The point (-2,0) and the line x = 22. The point (0,4) and the line 3? = -4

148

3. The point (0,-1) and the line y = 14. The point (1,0) and the line x = -1

ANSWERS:

1. y2 1111. -8x, a = -2, f.c. = 8

2. x2 = 16y, a = 4, f.c. = 16

3. x2 = -4y, a = -1, f.c. = 4

4. y2 = 4x, a = 1, Lc. = 4

FORMULA GENERALIZATION

All of the parabolas in the preceding sectionhad the vertex at the origin and the correspond-ing equations were in one of four forms asfollows:

1. y2 = 4ax

2. y2 = -4ax

3. x2 = 4ay

4. x2 = -4ay

In this section we will present four moreforms of the equation of a parabola, generalizedto consider a parabola with a vertex at point V(h,k). When the vertex is moved from the originto a point V(h,k) the x and y terms of the equationare replaced by (x h) and (y - k). Then thegeneral equation for the parabola which opens tothe right (fig. 10-to (A)) is

(y - k)2 = 4a(x - h)

The four general forms of the equations forparabolas with vertex at the point V(h,k) are asfollows:

1. (y - k)2 = 4a (x - h), corresponding toy2 = 4ax, parabola opens to the right

2. (y - k)2 = -4a (x h), corresponding toy2 = -4ax, parabola opens to the left

151

c hapter 10CONIC SECTIONS

3. (x - h)2 4a (y - k), corresponding tox2 = 4ay, parabola opens upward

4. (x - 102 = -4a (y - k), corresponding to

x2 = -4ay, parabola opens downward.

The method for reducing an equation to oneof these standard forms is similar to the methodsused for reducing the equation of a circle.

EXAMPLE: Reduce the equation

y2 - 6y - 8x + 1 = 0to standard form.

SOLUTION: Rearrange the equation so thatthe second degree term and any first degreeterms of the same unknown are on the left side.Then group the unknown term which appears onlyin the first degree and all constants on the right.

y2 - 6y = 8x -1

Then complete the square in y

y2 - 6y + 9 = 8x - 1 + 9

(y - 3)2 = 8x + 8

To get the equation in the form

(y - k)2 = 4a(x h)

factor an 8 out of the right side. Thus

(y - 3)2 z 8(x + 1)

is the equation of the parabola.PRACTICE PROBLEMS: Reduce the equa-

tions given in the following problems to stan-dard form.

1. + 4 ag 4y

2. y2 - 4x = 6y + 9

3. 4x + 8y + y2 + 20 = 0

4. 4x - 12y + 40 + x2 = 0

ANSWERS:

1. x2 = 4(y - 1)

2. (y - 3)2 = 4x

3. (y + 4)2 = -4(x + 1)

4. (x + 2)2 = 12(y - 3)

THE ELLIPSE

An ellipse is a conic section with an eccen-tricity less than one.

Referring to figure 10-11, let

PO = a

FO = c

OM = d

where F is the focus, 0 center,-and P and13' are points on the ellive. Then from the de-finition of eccentricity,

149

- a-c=ed- ead a

a + cd+ a =e a+ c=ed+ ea

Addition and subtraction of the two equationsgive:

2c = 2ae or c = ae

2a = 2de or d =

Place the center of the ellipse at the originso that one focus lies at (-ae,0) and one directrixis the line x = -a/e.

Referring to figure 10-12, there will be a pointon the Y axis which will satisfy the conditionsfor an ellipse. Let

(14)

2

P"O = b

FO = c


Then

Figure 10-11.Development of focusand directrix.

P"F = 1b2 + c2

and the ratio of the distance of P" from thefocus and the directrix is e so that

2+ c2

= ea

Multiplying both aides by ale gives

jb2 c2

b2 + c2 = a2

Iia2 c2so that

(15)

Now combining equations (14) and (15) gives

b = ia2 a2e2

and the distance from (x, y) to the directrixax = is

The ratio of these two distances is equal to e'so that

or

j(x + ae)2+ y2

a

+ ae)2+ y2 = e +

= ex + a

Squaring both sides gives

x2 + 2aex + a2e2+ y2 = e 2x2 + 2aex a2

Canceling like terms and transposing terms in xto the left-hand side of the equation gives

x2 - e2 x2 + y2 = a2 - a2e2

Removing a common factor,

x2(1 - e2) +y2 a2(1 e2)

(17)

LI =MO MO .M.M.IM 1M ON. =0 ANO .= IMO 41.1MO gol

or /.b = ±a117.72 (16) e/

Refer to figure 10-13. If the point (x, y) is i1 eon the ellipse, the ratio of its distance from F 6 co f .1to its distance from the directrix will be e: The

i .distance from (x, y) to the focus (-ae,0) will be MI P Fft

cI

(x + ae)2 + y2

160

'163

Figure 10-12.Focus directrix,and point P".

PIO

0 X


Dividing equation (17) through by the right-hand member

'2 2

+ 1a a2(1 - e2

)

From equation (16) we obtain

b = ±a ii-=?2(1 - e2)

= b2

a

so that the equation becomes

(18)

This is the equation of an ellipse in standardform. In figure 10-14, a is the length of thesemimajor axis and b is the length of the semi-minor axis.

The curve is symmetrical with respect to thex and y axes, so that it is easily seen that ithas another focus at (ae,0) and a correspondingdirectrix x = a/e.

The distance from the center through thefocus to the curve is always designated a and iscalled the semimajor axis. This axis maybe ineither the x or y direction. When it is in the ydirection, the directrix is a line with the equa-tion

y = k

Figure 10-I3.The ellipse.

X aae

-0IRECTRICESI

X s a-r

MAJORAXIS

Figure 10-14. Ellipse showing axes.

In the case we have studied, the directrixwas denoted by the formula

x = k

where k is a constant equal to - a/e.The perpendicular distance from the midpoint

of the major axis to the curve is called the semi-minor axis and is always signified by b.

The distance from the center of the ellipse tothe focus is called c and in any ellipse the follow-ing relations hold for a, b, and c

c = ja2 - b2

b = ia2 - c2

1)32 c2

Whenever the directrix is a line with the equa-tion y =k the major axiswillbe in the y directionand the equation of the ellipse will be as follows:

(19)

Otherwise everything remains as before andthe equation is given by (18).


In an ellipse the position of the a2 and b2terms indicate the orientation of the ellipseaxis. As shown in figure 10-14 the value isthe semimajor or longer axis.

In the previous paragraphs formulas weregiven which related a, b, and c and, in the firstportion of this discussion, a formula relating a,co and the eccentricitywas given. These relation-ships will be used to find the equation of anellipse in the following example.

EXAMPLE: Find the equation of the ellipsewith center at the origin and having foci

2 VA6t

(±2 .4,0) and an eccentricity equal to 7

SOLUTION: With the focal points on the xaxis the ellipse is oriented as in figure 10-14and the standard form of the equation is

x2 2

-aT Il32= 1

With the center at origin the numeratorsof the fractions on the left are y2 and y2 sothe problem is to find the values of a and b.

The distance from the center to either ofthe foci is the value c (fig. 10-14) so in thisproblem

c = ±2

from the given coordinates of the foci.The values of a, c, and e (eccentricity) are

related by

or

c = ae

From the known information, substitute thevalues of c and e

a = *2 473

2E3

a = *2 4.rg x 7

2 laand a = *7

a2 = 49

Then, using the formula

b = a -TT- c9-

or

b2 = a 2 - c2

and substituting for a2 and c2

b2 = 49 - (±2 1.)2

b2 = 49 - (4 x 6)

b2 = 49 - 24

gives the final required value of

b2 = 25

Then, the equation of the ellipse is2 2x

49 25

PRACTICE PROBLEMS: Find the equationof the ellipse with center at the origin and forwhich the-following properties are given.

1. Foci at (± 117/, 0) and an eccentricity

of 4

152

2. b = 5, e

3. a=7,eANSWERS:

1

2

3.

2IL42

2

+

+

+

.2=

=

=

1

1

1

32

.,2

25

2

36

x2

72 2

ELLIPSE AS A LOCUS OF POINTS

An ellipse may be defined as the locus ofa point which moves so that the sum of its


distances from two fixed points is a constantequal to 2a.

Let the foci be F1 and F2at (±ae,0), as shownin figure 10-15, and let the directrices be

e

Then

F, P = e a - xj = a - ex

F2P = e +

so that

= a + ex

F1P + F2P = a - ex + a + ex

F1P + F2P = 2a

Whenever the center of the ellipse is at somepoint other than (0,0), say at the point (h,k),figure 10-16, the equation of the ellipse must bemodified to the following form

- h)2 - k)2

a2 b2 1 (20)

Subtracting h from the value of x reduces thevalue of the term (x - h) to the value which xwould have if the center were at the origin. Theterm (y - k) is identical in value to the valuecf y if the center were at the origin.

REDUCTION TO STANDARD FORM

Whenever we have an equation in the form

Ax2 + Cy2 + Dx + Ey + F = 0 (21)

where the capital letters refer to independentconstants and A and C have the same sign, wecan reduce the equation to the standard form foran ellipse. Completing the squares in x and yand performing a few simple algebraic trans-formations will change the form to that of equa-tion (20).

THEOREM: An equation of the second degree,in which the.xy terAn is missing and the co-efficients of ic4 and yh are different but have thesame sign, represents an ellipse with axesparallel to the coordinate axes.

Figure 10-15.Ellipse, center at origin.


4x2 + 9y2 - 40x - 54y + -145 = 0

to the standard form of an ellipse.SOLUTION: Collect terms in x and y and re-

move the common factors of these terms.

4x2 - 40x + 9y

2 - 54y + 145 = 0

4(x2 - 10x) + 9(y2 - 6y) + 145 = 0

Transpose the constant terms and completethe squares in x and y. When there are factoredterms involved in completing the square, as inthis example, an error is frequently made. The

153

0

Figure 10-Ie.Ellipse, center at (h,k).

X


factored value operates on the term added in-side the parentheses as well as the originalterms. Therefore, the values added to the rightside of the equation will be the product of thefactored value and the term added to completethe square.

Complete the square in x noting that there willbe a product added to the right side

32 20x 100) 2 (100)+ + + y - -Oh t

3 9

4(x2 - 10x + 25) 9(y2 - 6y + 9)

= -145 + 4(25) + 9(9)

= -145 + 100 + 81

= 36

4(x - 5)2 + 9(y - 3)2 = 36

3lo 2

300x + y2 =+

2 y2

-32 +

-288 +9 300

Divide through by the right-hand (constant)term. This reduces the right member to 1 asrequired by the standard form.

4(x - 5)2+

90 - 3)2 1

- 5)2 (y - 3)29 4

This reduces to the standard form

(x - 5)2 ( - 3)220)2 (2)

1

Corresponding to equation (20) and represents anellipse with the center at (5,3), its semimajoraxis (a) equal to 3, and its semiminor axis (b)equal to 2.


3x2+ y2 + 20x + 32 = 0

to the standard form of an ellipse.SOLUTION: .First, collect terms in x and y.

Ai in tht previous example, the coefficients ofe and yzmust be reduced to 1 in order to facili-tate completing the square. Thus the coefficientof the xz term is divided out of the two termscontaining x, as follows:

3x2 + 20x + y2 4- 32 = 0

2 20x3 ( )x + + y2 = -32

3

154

157

10 2 12-02 +

0 2 43 (x + 1 -Tr + y =

Divide through by the right-hand term.

+3

10)23/

43

9 \(x 10)

2

4

2Z_ =4

2( 10) 23 + 1=

(1)


This reduces to the standard form

= 1

= 1

PRACTICE PROBLEMS: Express the fol-lowing equations in the standard form for anellipse.

1. 5x2 - 110 x + 4y2 + 425 ?, 0

2. x2 - 14x + 36y2 - 216y + 337 = 0

3. 9x2 - 54x + 4y2 + 16y + 61 = 0

4. 3x2 - 14x + 4y2 + 11 = 0

ANSWERS:

(X - 11)2 y2 9 = 1

(6)2

2 2

2. 1(6)

(x - 3)2 4. (y+ 2)2 1

3. (3)2

2+ y--- =

/2 .5\\ 3 )

1

THE HYPERBOLA

A hyperbola is a conic section with an ec-centricity greater than one.

The formulas

c = ae

and

d =

developed in the section concerning the ellipsewere derived so that they hold true for any valueof eccentricity. Thus, they hold true for thehyperbola as well as for an ellipse. Since e isgreater than one for a hyperbola, then

c ae and c > a

d and d < a

Therefore c >a >d.According to this analysis, if the center of

symmetry of a hyperbola is the origin, the fociwill lie farther from the origin than the direc-trices. An inspection of figure 10-17 shows thatthe curve will never cross the Y axis. Thus,the solution for the value of b, the semiminoraxis of the ellipse will yield no real value forb. In other words, b will be an imaginerynumber. This can easily be seen from theequation

b = 1a2 - c2

since c > a for a hyperbola.However, we can square both sides of the

above equation, and since the square of animaginary number is a negative real mimber wewrite

or

155

-b2 = a2 - c2

and, since c = ae,

58

.o2= c2 - a2

b2 = a2e2 - a2 = a2 (e2 - 1)


X

0

Figure 10-17.The hyperbola.

Now we can use this equation to obtain the equa-tion of a hyperbola from the following equationwhich was developed in the sectionon the ellipse.

2 2x 4. - 1a2 a2( 21 - e)

and since

a2(1 - e2) = -a(e2 - 1) = -b2

we have

2 2x

a2b2

This is a standard form for the equation ofa hyperbola. The solution of this equation fory gives

by ±- X - aa

wilich shows that y is imaginary only whenx4 < a2. The curve, therefore, lies entirelybeyond the two lines x = *a and crosses the xaxis at x = *a.

The two straight lines

bx+ay2=Oandbx-ay= (22)

can be used to illustrate an interestingpropertyof a hyperbola. The distance from the linebx - ay = 0 to a point (x1, yi) on the curve isgiven by

d =bx

1- ay

1

a + b(23)

Since (xi , ) is on the curve, its coordinatessatisfy the Ligation

b2x12 - a2y12 = a2b2

which may be written

or

(bx1

- ay 1) (bx1

+ ay 1) = a2b2

bx - ay -1 1 bx1 + ay1

Now substituting this value into equation (23),gives us

a2b2

d = a2b2

1a2 b2 (13x1 ay))1

As the point (x1, yi) is chosen farther andfarther from the center of the hyperbola, theabsolute values for xi and yi will increase andthe distance d will approach zero. A similarresult can easily be derived for the linebx + ay = 0.

The lines of equation (22) which are usuallywritten

y = x and y = +xa a

are called the asymptotes of the hyperbola. Theyare very important in tracing a curve and study-ing its properties. The asymptotes of a hyper-bola, figure 10-18, are the diagonals of the rect-angle whose center is the center of the curve andwhose sides are parallel and equalto the axes of

156

159


the curve. The latus rectum of a hyperbola is

equal to 2b2

Another definition of a hyperbola is the locusof a point that moves so thatthe difference of itsdistances from two fixed points is constant, Thefixed points are the foci and the constantdifference is 2a.

The nomenclature of the hyperbola is slightlydifferent from that of an ellipse. The trans-verse axis is 2a or the distance between the in-tersections of the hyperbola with its focal axis.The conjugate axis is 2b and is perpendicular tothe transverse axis.

Whenever the foci are on the Y axis and thedireetrices are lines of the form y=*.k, where kis a constant, the equation of the hyperbola willread

x21

a2

b

This equation represents a hyperbolawith itstransverse axis on the Y axis, Its asymptotesare the lines by - ax = 0 and by + ax = 0.

ANALYSIS OF THE EQUATION

The properties of the hyperbola most oftenused in analysis of the curve are the foci, direc-trices, length of the latus rectum, and the equa-tions of the asymptotes.

Reference to figure 10-17 shows thatthefociare given by the points F1 (c,0) and F2 (-c,0)when the equation of the hyperbola is inthe form

2 2

a2 b2

If the equation were

2 2

b2 a2the foci would be the points (0,c) and (0,-c).The value of c is either determined from the for-mula

1

or the formula

c22

= a2 + b

c = ae

ay cbx

0

Figure 10-18.Using asymptotes to sketch a hyperbola.

157

160

X


Figure 10-17 also shows that the directricesare the lines x = ± or, in the case where thehyperbolas open upward and downward, yThis is also given earlier in this discussion

aas d =

The equations of the asymptotes are givenearlier as

bx + ay = 0 and bx - ay = 0Or

y = --a x and y = +xa

It was also pointed out that the length of the latus2b 2rectum is equal to

The properties of a hyperbola can be deter-mined from the equation of a hyperbola or theequation can be written given certainproperties,as shown in the following examples. In theseexamples and in the practice problems immedi-ately following, all of the hyperbolas consideredhave their centers at the origin.

EXAMPLE: Find the equation of thehyperbola with an eccentricity of 3/2, directricesx = *4/3, and foci at (±3,0).

SOLUTION: The foci lie on the X axis at thepoints (3,0) and (-3,0) so the equation is of theform

2 2

'`a2 b2

This fact is also shown by the equation of thedirectrices.

Before proceeding with the problem one pointshould be emphasized: in the basic formula forthe hyperbola the a2 term will always-be the de-nominator for the x2 term and the b2 term thedenominator for the y2 term. The orientation ofthe axis of symmetry is not dependent onthe sizeof a2 and bz as in the ellipse; it lies along orparallel to the axis of the positive x4 or y2term.Since we have determined the form of the equa-tion and since the. center of the curve in thissection is restricted to the origintheproblem isreduced to finding the values of a2 and b2

First, the foci are given as (±3,0) and sincethe foci are also the points (±c,0) it follows that

c = ±3

The eccentricity is given and the value of a2can be determined from the formula

c = ae

a

a ±63

a = ±2

a2 = 4

The relationship of a, b, and c for thehyperbola is

and

b2 = c2 - a2

b2 = (±3)2 - (±2)2

b2 = 9 - 4

b2= 5

When these values are substituted in the equa-tion

the equation

2 2x _ =

a2b2

2 2L.4 5

results and is the equation of the hyperbola.The equation could also be found by the use

of other relationships which utilize the giveninformation.

The directrices are given as

158

X =3


and, since and, when a > 0

or

d = ae

a = de

Substituting the values given for d and e resultsin For a < 0

a = *1 (1)

a = 1 (-0

a = 2 Then

When d > 0

c = 2 (2)2

c = 3

c = -2 (1)2

c = -3

When d < 0 c = *3

therefore

and

4a3

a = -2

(-0

a = *2

a2 = 4

While the value of c can be determined by thegiven information in this problem, it could alsobe computed since

With values for a and c computed, the valueof b is found as before and the equation can bewritten.

EXAMPLE: Find the foci, directrices,eccentricity, length of the latus rectum, andequations of the asymptoten of the hyperboladescribed by the equation

2 2x y_ i9 16 a

SOLUTION: This equation is of the form

2 225__ y_a2 b2

and the values for a and bare determined by in-c = ae spection to be

and a has been found to equal *2 and e is given as3 then2 '

3c =*2 ()2

and

159

a2= 9

a = *3

b2

= 16

b =*4


With a and b known, find c by using theformula

b2 = c2 - a2

c 2 = a2 + b2

c = ± Ia2 + b2

c = ± VT) -Th

c = ±

c = ±5From the form of the equation we know that thefoci are at the points

F1(c' 0)

and

F (-c 'so the foci = (±5,0).

When a > 0

and when a < 0

x = -3 (1)5

so the directrices are the lines

The eccentricity is found by the formula ±9x =c

e - -a-

±5e = 1-3

5e =

Reference to figure 10-17 shows that with thecenter at the origin, c and a will have the samesign.

The directrix is found by the formula

d

or, since this equation will have directricesparallel to the Y axis, use the formula

ax e Finally, the equations of the asymptotes are theequation of the two straight lines

Then

The latus rectum (1. r. ) is found by

2b21. r. = To-

1. r.

1. r.

2(16)3

32=3

x = ±3 (1)5

and

160

163

bx + ay = 0

bx - ay = 0

Chapter 10-CON/C SECTTONS

In this problem, substituting the values of aand d in the equation gives

4x + 3y = 0

4x - 3y = 0

4x ± 3y = 0

The equa!dons of the lines asymptotic to the

curve can also be written in the form

and

or

and

aIn this form the lines are

4Y = -3 x

and4y = - ix

or4y = ±

If we think of this equation as a form of theslope intercept formula

y = mx + b

from chapter 9, the lines would have slopes of

1± -a and each would have its Y intercept at the

origin as shown in figure 10-18.PRACTICE PROBLEMS:1. Fine the equation of the hyperbolawith an

eccentricity of 4, directrices x = ± r, and fociat (±111, 0).

2. Find the equation of the hyperbola with aneccentricity of 5/3, foci at (±510), and directricesx = ± 9/5.

Find the foci, directrices, eccentricity, equa-tions of the asymptotes, and length of the latusrectum of the hyperbolas given in problems 3and -4.

3.

4.

2x 21

= 1

9

2

9

2-

9 4

161

ANSWERS:

1. x2 - y2 = 1

,2 ..22. -a- 1

9 16

3. foci = (*3 V-2,0);eccentricity = VI; 1.r. = 6;

4. foci = (* Nir.310);

eccentricity = r.* 3y = 0.

directrices, x =S-3y

asymptotes x* y = 0.

directrices x;

8'=

3- asymptotes 2x

The hyperbola can be represented byanequation in the form

Ax2 + Cy2 + Dx + Ey + F = 0

where the capital letters refer to independentconstants and A and C hav e different signs. Theseequations can be reduced to standard form inthesame manner in which similar equations for theellipse were reduced to standard form. Thegeneral forms of these standard equations aregiven by

and

(x - h)2 (y - k)2

1a

(y - k)2 (x - h)2_ L

a

POLAR COORDINATES

So far we have located a point in a plane bygiving the distances of the point from two per-pendicular lines. The location of a point can bedefined equally well by noting its distance andbearing. This method is commonly used aboardship to show the position of another ship ortarget. Thus, 3 miles at 350 locates the posi-tion of a ship relative to the course of the shipmaking the reading. We can use this methodto develop curves and bring out their properties.Assume a fixed direction OX and a fixed point0 on the line in figure 10-19. The position ofany point P is fully determined, if we know thedirected distance from 0 to P and the angle thatthe line OP makes with reference line OX. Theline OP is called the radius vector and the angle

1.64:


PDX is the polar angle. The radius vector is EXAMPLE: Change the equationdesigned P while 0 is the angle designation.

2Point 0 is the pole or origin. As in conven- y =tional trigonometry, the polar angle is positivewhen measured counterclockwise and negative from rectangular to polar coordinates.when measured clockwise. However, unlike the SOLUTION: Substitute p cos 0 for xand P sinconvention established in trigonometry, the 0 for y so that we haveradius vector for polar coordinates is positiveonly when it is laid off on the terminal sideof the angle. When the radius vector is laidoff on the terminal side of the ray produced be-yond the pole (the given angle plus 180°) anegative value is assigned the radius vector.For this reason, there may be more than oneequation in polar coordinates to describe agiven locus. The concept of a negative radiusvector is utilized in some advanced mathematics.For purposes of this course the concept is notexplained or used. It is sufficient that thereader remember that the convention of analways positive radius vector is not followedin someloranches of mathematics.TRANSFORMATION FROM CARTESIANTO POLAR COORDINATES

At times it will be simpler to work with theequation of a curve in polar coordinates than incartesian coordinates. Therefore, it is im-portant to know how to change from one systemto the other. Sometimes both forms are useful,for 3ome properties of the curve may be moreapparent from one form of the equation and otherproperties more evident from the other.

Transformations are made by applying thefollowing equations which can be derived fromfigure 10-20.

x = p cos 0

Y = P sin 0

p2 = X2 + y2

tan =

(24)

(25)

(26)

(27)

Figure 10-19. Defining the polar coordinates.

162

165

or

p sine = p2 cos2a

sine = p cos20

sin 0P =

cos20

p = tan 6 sec@

EXAMPLE: Express the equation of thecircle with its center at (a,0) and with a radiusa, as shown in figure 10-21.

(x - a)2+ y2 = a2

in polar coordinates.

Figure 10-20.Cartesian and polarrelationship.


SOLUTION: First, expanding this equationgives us

x2 2ax + a2+ y2 = a2

Rearranging terms we have

x2 + y2 = 2ax

The use of equation (26) gives us

p2 = 2ax

and applying the value of x given by equation(24), results in

p2 = 2ap cos

Dividing through by p we have the equation of acircle with its center at (a,0) and radius a inpolar coordinates

p = 2a cos 9

TRANSFORMATION FROM POLARTO CARTESIAN COORDINATES

In order to transform to an equation incartesian or rectangular coordinates from anequation in polar coordinates use the followingequations which can be derived from figure10-22.

cos =

sin 0 =

tan 9 =

sec 9

csc 8

cot 0

(28)1/ x2 y2

(29)

lef X y

(30)

ix2 + y2

(31)

+ v2 (32)

slx2 + y2(33)

x(34)

EXAMPLE: Change the equation

p = sec 9 tan 8

to an equation in rectangular coordinates

SOLUTION: Applying relations (28), (31),and (32) to the above equation gives

2 21/7-4. y2k x/

Dividing both sides by 11-c-2-7/4y , we obtain

or2y = x

which is the equation we set out to find.

Chinge the following equationto an equation in rectangular coordinates.

Figure 10-21.--Circle with center (a,0).

163

3P sin - 3 cos 9


( x y )

X

Figure 10-22.--Polar to cartesian relationship.

SOLUTION: Written without a denominatorthe polar equation is

p sin 0 -3 p cos e = 3

Using the transformations

p sin 0 = y

p cos 0 = x

we havey - 3x = 3

as the equation in rectangular coordinates.PRACTICE PROBLEMS: Change the equa-

tion in problems 1 through 4 to equations havingpolar coordinates.

1. x2 + y2 = 4

2) a3 x22. (x2 +

3. 3y - 7x = 10

4. y = 2x - 3Change the equations in the following problemsto equations having Cartesian coordinates.

5. p = 4 sin 9

164

6. p = sin A + cos A

7. p = a2

ANSWERS:

1. P = ±2

2. P = a 1.1c',"(TI

103. p = 3 sin 8 - 7 cos 0

-34. P = sine - 2 cost)

5. x2 + y2 - 4y = 0

6. x2 + y2 = y + x

2 27. x + y = a4

CHAPTER 11

TANGENTS, NORMALS, AND SLOPES OF CURVES

AyIn chapter 9, the notation Erc was introducedto represent the slope of a line. The straightline discussed has a constant slope and thesymbol Ay was defined as (y2 - y1) and Axwas defined as (x2 - x1). In this chapter wewill discuss the slope of curves at specificpoints on the curves. We will do this with aslittle calculus as possible but our discussionwill be directed toward the study of calculus.

SLOPE OF A CURVE AT A POINT

In figure 11-1, the slope of the curve isArepresented at two different places by y

TheAxAyvalue of taken on the lower part of the curve

AXwill be extremely close to the actual slope atP1 because P1 lies on a nearly straight portionof the curve. The value of the slope at P2 willbe less accurate than the slope near P1 becauseP2 lies on a portion of the curve which hasmore curvature than the portion of the curvenear 131. In order to obtain an accurate mea-sure of the slope of the curve at each point, assmall a portion of the curve as possible shouldbe used. When the curve is nearly a straightline a very small error will occur when findingthe slope regardless of the value of the incre-ments Ay and.Ax. If the curvature is great andlarge increments are used when finding theslope of a curve, the error will become verylarge.

Thus, it follows that the error can be re-duced to an infinitesimal if the increments arechosen infinitely small. Whenever the slope ofa curve at a given point is desired, the incre-ments Ay and Ax should be extremely small_Consequently, the arc of the curve can be re-placed by a straight line, which determines theslope of the tangent at that point.

It must be understood that when we speakof the tangent to a curve at a specific point weare really considering the secant line, whichcuts a curve in at least two points. This secantline is to be decreased in length, keeping theend points on the curve, to such a small valuethat it may be considered to be a point. Thispoint is then extended to form the tangent tothe curve at that specific point. Figure 11-2shows this concept.

DIRECTION OF A CURVE

If we allow

y = f(x)

to represent the equation of a curve, then

It!xis the slope of the line tangent to the curve

at P (x, y).The direction of a curve is defined as the

direction of the tangent line at any point on thecurve. Let 0 equal the inclination of the tan-gent line; then the slope equals tan 0 and

165

AX

is the slope of the curve at any point P (x, y).The angle 0 1 is the inclination of the tangentto the curve at P1 in figure 11-3. This angleis acute and the value of tan 0 1 is positive.Hence the slope is positive at point P1. Theangle 02 is an obtuse angle and tan 8 2 is neg-ative and the slope at point P2 is negative. Alllines which lean to the right have positiveslopes and all lines which lean to the left havenegative slopes. At point P3 the tangent to thecurve is horizontal and 8 equals 0. Thismeans that

= tan 0

168

4-E = tan 00 = 0ax


Figure 11-1.Curve with incrementsAy and AX.

St

0 tx

Figure 11-2.Curve with secant lineand tangent line.

X

The fact that the slope of a curve is zerowhen the tangent to the curve at that point ishoriwntal is of great importance in calculuswhen determining the maximum or minimumpoints of a curve. Whenever the slope of acurve is zero, the curve may be at either amaximum or a minimum.

Whenever the inclination of the tangent to acurve at a point is 900, the tangent line is

Figure 11-3.Curve with tangent lines.

vertical and parallel to the Y axis. This re-sulta in an infinitely large slope

61Y = tan 90° = coax

TANGENT AT A GIVEN POINTON THE STANDARD PARABOLA

166

The standard parabola is represented by theequation

y2 = 4ax

Let Pi with coordinates (xi, yi) be a point onthe curve. Choose 13' on the curve,figure 11-4,near the given point so that the coordinates of13' are

(x1

+ ax y1 + ay)

Since P' is a point on the curve

y2 = 4ax

the values of its coordinates may be substi-tuted for x and y. This gives

or

169

(Yi + ay) 2 = 4a(x1 + x)

2 + 2y1 ay + (ay) 2 = 4ax1

+ 4a ax (I)

Chapter 11TANGENTS, NORMALS, AND SLOPES OF CURVES

Figure 11-4.--Parabola.

AySolving for we findAx

(Ay)2

4a AXAX 2y1 2yi

(Ay)2

2a Ax

yl 2y 1

(3)

Before proceeding, a discussion of the term

y)2

AX2y1

in equation (3) is in order. If we solve equation(2) for Ay we find

The point P1(x1, yi) also lies on the curve and thenwe have

2yl = 4axl

2Substituting this value for yi into equation (I)transforms it into

4ax1

+ 2y1

Ay + (Ay)2 = 4ax1

+ 4a Ax

Simplifying we obtain

2y1Ay + (Ay)2 = 4aAx

Divide through by Ax, obtaining

which gives

2y14

4y thol2 4aAx+Ax Ax Ax

2y a 4a (AO2

1 Ax Ax

(2)

and

2y1

Ay + (Ay)2 = 4aAx

Ay(2y1 + Ay) = 4aAx

4aAx"Y 2y1 + ay

Since the denominator -contains a term not de-pendent upon Ay or Ax, as we let Ax approachzero Ay will also approach zero.

NOTE: We may find a value for Ax that willmake Ay luss than 1 and then when y issquared it will approach zero at least as rapidlyas Ax does.

We now refer to equation.(3) again and make2

Azthe statement that we may disregard

since it approaches zero when Ax approacheszero.

Then

167

Ay 2aAx y1

(4)


The quantityzi is the slope of the line con-

necting Pi and V. From figure 11-4, it isobvious that the slope of the curve at Pi isdifferent from the slope of the line connectingPi and V. y

As Ax and Ay approach zero, the ratio Ax

will approach more and more closely the trueslope of the curve at Pi. We designate theslope by (m). Thus, as Ax approaches zero,equation (4) becomes

ylThe equation for a straight line in the

point slope form is

2a

yl m(x

2aSubstituting for m gives

Yi

Y -2a

y = (x - X )1 Y1

1

Clearing fractions we have

2yyi yi = 2ax - 2ax1 (5)

but2

yl 4axl

Adding equations (5) and (6) yields

yyi = 2ax + 2ax1

Dividing by y1 gives

(6)

2ax 2ax1

Yyl Y1

which is an equation of a straight line in theslope intercept form. This is the equation of

the tangent line to the parabola

y2 = 4ax

at the point (x1, y1).

EXAMPLE: Given the equation

y2 = 8x

find the slope of the curve and the equation ofthe tangent line at the point (2, 4).

SOLUTION: Put the equation in standardform as follows: Solve for (a) by letting

y2 = 8x

have the form

T hen

168

171

and

y2 = 4ax

4a = 8

a = 2

2a = 4

The slope m at point (2, 4) becomes2a

yl2(2)

4

The slope of the line is 1 and the equation ofthe tangent to the curve at the point (2, 4) is

1

2ax2ax

1

yl Yl

(2) (2) (x) (2) (2) (2)+

= x + 2

This method, used to find the slope andequation of the tangent for a standard parabola,can be used to find the slope and equation ofthe tangent to a curve at any point regardlessof the type of curve. The method can be usedto find these relationships for circles, hyper-bolas, ellipses, and general algebraic curves.

This general method is outlined Is follows:To find the slope (m) of a givell curve at thepoint Pi (xi, yi) choose a second point r onthe curve so that it has coordinates (xi + Ax,

yl 4- Ay) and substitute the coordinates of Irin the equation of the curve and simplify.Divide through by Ax and eliminate termswhichcontain powers of Ay higher than the first power,as previously discussed. Let Ax approach zero


butand -62-- y will approach the absolute value for the

A xslope (m) at point Pl. Finally solve for (m).

When the slope and coordinates of a pointon the curve are known, the equation of thetangent line can be found by using the point andslope method.

EXAMPLE: Using the method outlined, findthe slope and equation of the tangent line to thecurve

x2 + y2 = r2 at (x1, y1) (7)

SOLUTION: Choose a second point P1 suchthat it has coordinates

(x1 + Ax, yi + Ay)

Substitute into equation (7) and

(x1 + Ax)2 + (yi + Ay)2 = r2

thus

2x

1+ 2x

1Ax + (Ax)2

+ y2 + 2y1Ay + (Ay) 2 = r2

1

then

2 22x

1Ax + (Ax)

2 + 2y1Ay + (Ay)2 = r

2 - x1

- y1

2 2 2=r

=0

Divide through by Ax

2x1Ax (442 2371 Ay (A02.

AX "7-6,-7 Ax

and eliminating (Ay)2 results in

2x1 + ax + 2y1 -bati = 0

Solve for as follows:

-2x1 - Ax

AX 2y1

111 =

AX

-2x1

- bix

2y1

Let Ax approach zero and

Now use the point slope form of a straight line-xl

with the slope equal to and find at point(x1 yi)

Rearranging

and

but

y - yl 121(x -xl)

-xi= (x - x )

1

2 2yy1 - y1 = -xix +

2 2YY. = -2{ x + x +

1 1 -v

1

2 2 2x1 + y1 = r

Then, by substitution

= 0= -xix r2

169

and

-x1x r2

+

Y1 Y1

which is the general equation of the tangentline to the curve

r '17 2

x2 + y2 = r2 at (x1, y1)


EXAMPLE: Using the given method, withminor changes) find the slope and equation ofthe tangent line to the curve

- y2 = k2 at (xi, y1) (8)

SOLUTION: Choose a second point Pi suchthat it has coordinates

(x1 + ax, y1 + py)

Substitute into equation (8) and

(x1

+ ax)2 - (y1

+ ay)2 = k2

and

2 2Xi+ 2xiax+ (Ix)2 - y - 2y1ay- (ay)2k

2 In%

then subtract equation (8) from (9) and obtain

2x1Ax + (ax) 2 - 2y1ay - (ay)2= 0

then divide by Ax and we have

ay2x1

+ - 2y1 -x-

Let Ax approach zero and

(4,y)2

2x1 - 2y1 gc = 0

Solving for zfci results in

ax ylUse the point slope form of a straight line tofind the equation of the tangent line at point(xi, y1) as shown in the following:

y - y1 = m(x - x1)

= m =x

1

= 0

x1Substitute for In

yl

Y1- x 1)

Multiply through by yi and

2 2yy1 - yi = xix xl

Rearrange to obtain

yyl xlx2 2- xi + y1

= x x L2 2)1 19. Y1

Substitute x1

- y1

for k 2 and2 2

yyi = xix - k2

Divide through by y1 to obtain

xlx k2

=

which is the equation and slope desired.PROBLEMS: Find the slope (in) and equa-

tion of the tangent line, in problems 1 through6, at the given points.

2 41. y =

2. y2 = 12x

3. x2 + y2 = 25

4. x2 + y2 = 100

5. x2 - y2 = 9

6. x2 - y2 = 3

7. Find the slope of y = x8. Find the slope of

y = 2x2- 3x + 2ANSWERS:

170

6_1'3

X 4 11. y = + =

2. y = x + 3, m = 13x 25 33.

-3x 254. 4 + -3in

5x 9 5Y t

in T+ 3, tn = 2

7. m = 48. in = 5

2

at (3, 2)

at (3, 8)

al (-3, 4)

al (6, 8)

al (5, 4)

at (2, 1)

at (2, 4)

al (2, 4)


EQUATIONS OF TANGENTSAND NORMALS

In figure 11-5, the coordinates of point P1on the curve are (xi, yi). Let the slope of thetangent to the curve at point Pi be denoted by

ml. Knowing the slope and a point throughwhich the tangent line passes, the equation ofthat tangent line can be determined by usingthe point slope form.

Thus, the equation of the tangent line (MP1)is

y - y1 = m(x - x1)

The normal to a curve at a point (xi, yi) isthe line which is perpendicular to the tangentline at that point. The slope of the normal

1line is then where, as before, the slope

of the tangent line is m1' This is shown in thefollowing:

If

then

Figure 11-5.Curve with tangent andnormal lines.

Notice that if the slope of the tangent ismi and the slope of the normal to the tangentis rn2 and

m1=tan6 1in2nil

m2 = tan (8 + 900)

= -tan [180° - (0 + 90°)]

= -tan (900 - 9)

= -cot 0

1tan 0

1

ml

then the product of the slopes of the tangentand normal equals -1. The relationship be-tween the slopes of the tangent and normalstated more formally is: The slope of thenormal is the negative reciprocal of the slopeof the tangent.

Another approach iu show the relationshipbetween the slopes of the tangent and normalfollows: The inclination of one line must be900 greater than the other. Then

02 = 01 + 900

therefore If

1

2 - tan 82 = m2m1

The equation of the normal through P1 is and

1

yl xl)

171

a174

tan 82 = tan (81 + 900)


then

sin (01 + 900)tan (01+ 90°)= go )

sin 01 cos 90° + cos 8

1sin 90°

therefore

cos 01 cos 90° - sin el sin90°

cos 01=

sin 01

= - cot 01

1tan 91

1tan 02 -O1

LENGTHS OF SUBTANGENTAND SUBNORMAL

The length of the tangent is defined as thatportion of the tangent line between the pointP1 (xi, yi) and the point where the tangentline crosses the X axis. In figure 11-5, thelength of the tangent is (MP1).

The length of the normal is defined as thatportion of the normal line between the point P1and the X axis, That is (P1R) which is perpen-dicular to the tangent.

The projections of these lines on the Xaxis are known as the length of the subtangent(MN) and the length of the subnormal (NR).

The relationships between the slope of thetangent and the lengths of the subtangent andsubnormal follows:

From the triangle MP1N, in figure 11-5,

and

P N1tan 8 = m =

1 MN

MN = P1Nml

The line segment (MN) is the length of thesubtangent said (PIN) is equal to the verticalcoordinate yi. Therefore, the length of the

ylsubtangent is .

but

and

In the triangle NP1R,

tan 0 = NP1

NR

tan 0 = m1

NR = m1NP

1

1111y1

Therefore, the length of the subnormal is miyi.

From this, as shown in figure 11-5, thelength of the tangent and normal may be foundby using the Pythagorean theorem.

NOTE: If the subtangent lies to the right ofpoint Mt it is considered positive, if to theleft it is negative. Likewise, if the subnormalextends to the right of N it is positive, to theleft it is negative.

EXAMPLE: Find the equation of the tan-gent, the equation of the normal, the length ofthe subtangent, the length of the subnormal,and the lengths of the tangent and normal of

2 4y = -5 x at (3, 2)

SOLUTION: Find the value of 2a from

172

Since

then

y2 = 4ax

2 4x3

44a

1a =

2a 2=.1-

4


Thc slope is2

y2a "Y_ 1

m 7- 31

Using the point slope form

5' yl 111(x x1)

then, at point (3, 2)

y - 2 = -13-(x - 3)

x-5

andxy = -5- + 1i

which is the equation of the tangent line.Use the negative reciprocal of the slope

find the equation of the normal as follows:

Y -

then

2 = -3(x - 3)

= -3x + 9

y = -3x + 11

The length of the normal is equal to

hylin1)2 + (3r1)2

=A2

1) + (2)23

9

6.323

to PROBLEMS: Find the equation of the tan-gent and normal, and the lengths of the sub-tangent and subnormal in the following:

1. y2 = 12x,

2. x2 + y2 = 25,

3. x2 - y2 = 9,

4. y = 2x2 - 3x + 2,

ANSWERS:

1. Equation of tangentEquation of normal

Length of subtangent 6

Length Of subnormal

The length of the subtangent is

Y1 2ii-i 1- = 6

1

and the length of the subnormal is

y1m1 = 2(5 = i1) 2

To find the length of the tangent we use thePythagorean theorem. Thus, the length of thetangent is

671)2

1(6)2 (2)2

= V-476

= 6.32

173

at (3, 6)

at (-3, 4)

at (5, 4)

at (1, 1)

y = x + 3y = -x + 9

2. Equation of tangent

Equation of normal

Length of subtangent

6

Y =

Y

163

Length of subnormal 5


Equation of normal


Length of subnormal




Length of subnormal

PARAMETRIC EQUATIONS

Equations used previously have been func-tions involving two unknowns such as x and y.The functions have been in either Cartesian orpolar coordinates and have been defined by oneequation. If a third variable is introduced itis called a parameter. When two equationsare used, each containing the parameter, theequations are called parametric equations.

MOTION IN A STRAIGHT LINE

To illustrate the application of a parameterwe will assume that an aircraft takes off froma field which we will call the origin. Figure11-6 shows the diagram we will use. The air-craft is flying on a compass heading of duenorth. There is a wind blowing from the westat 20 miles per hour and the airspeed of theaircraft is 40C miles per hour. Let the direc-tion of the positive Y axis be due north and thepositive X axis be due east as shown in figure11-6. Use the scales as shown.

MILES

500

400

300

200

100

10 20 30 40 50

MILES

Figure II-ELAircraft position.

One hour after takeoff the position of theaircraft, represented by point P, is 400 milesnorth and 20 miles east of the origin, If weuse t as the parameter, then at any time t theaircrafts position (x, y) will be given by xequals 20t and y equals 400t.The equations are

and

x = 20t

y = 400 t

and are called parametric equations. Noticethat time is not plotted on the graph of figure11-6. The parameter t is used only to plot theposition (x, y) of the aircraft.

We may eliminate the parameter t to obtaina direct relationship between x and y as follows:

If

then

= 400 61-0

y = 20x

and we find the graph to be a straight line.When we eliminated the parameter the resultwas the rectangular coordinate equation of theline.

MOTION IN A CIRCLE

Consider the parametric equations

x = r cos t

and

y r sin t

These equations describe the position of apoint (x, y) at any time t. They can be trans-posed into a single equation by squaring bothsides of each equation to obtain

174

2 2x = r cos2,

y2 = r2 sin2t


and adding

x2 + y2 = r2 cos2t + r2 sin2t

Rearranging we have

x2 + y2 = r2 (cos2t + sin2t)

cos2t + sin2t = 1

X2 + y2 = r2

but

then

which is the equation of a circle.This means that if various values were

assigned to t and the corresponding values ofx and y were calculated and plotted, the re-sult would be a circle. In other words, thepoint (x, y) moves in a circular path.

Using this example again, that is

By substituting we findr cos t

Ax -r sin t cot t

Comparing this result with equation (7) of theprevious section, we find that in rectangularcoordinates the slope is given as

x1

yl

while in terms of a parameter it is

m = - cot t

OTHER PARAMETRIC EQUATIONS

EXAMPLE: Find the equation of the tangentand the normal and the length of the subtangentand the subnormal for the curve representedby

x = t2

and

and given that

and

x r cos t

y = r sin t

Ax=

and

at

given that

and

y = 2t + 1

t = 1

-421 = 2 tA t

m = -r sin t2 At

m = = r cos t1 At

we are able to express the slope at any pointon the circle in terms of t.

AxNOTE: These expressions for AT and lt,Y

may be found by using calculus, but we willaccept them for the present without proof.

If we know -Al and --1-CA we may find AY-At At Ax

leich is the slope of a curve at any point.

That is,

Ay Tt-m = Ax Axat

2At

SOLUTION: Since t equals 1 we write

and

and

then

x = 1

Y = 3

_A/ 2 1

Ax 2t I

=


The equation of the tangent line when t is equal we haveto 1 is

y - 3 = 1(x - 1)

y = x + 2

The equation of the normal line is

y - 3 = -1(x - 1)

y = -x + 4


ylm

The length of the subnormal

yim = (1) (3) = 3

EXAMPLE: Find the equation of the tangentand normal and the lengths of the subtangentand subnormal to the curve represented by theparametric equations

and

at the point where

given that

and

x = 2 cos

y = 2 sin

8 = 45°

= -2 sin

m = L = cot 45° = -1Ax

In order to find (x1' y1) we substitute

0 = 45°

in the parametric equations and

x1

= 2 cos 45° = 2 (-I) =2

y1 = 2 sin 450 = 2(4) =

The equation of the tangent is

y = m(x - xi)

Substituting we have

y = -1(x -

or

x + y = 2 Nri

The equation of the normal is

y - = 1(x -

orx y 0


Y1 r--m 71. v 2

The length of the subnormal isA v = 2 cos 0 yim = (a) (-1) = r2AO

SOLUTION: We know that The horizontal and vertical tangents of acurve can be found very easily when the curve

02. is represented by parametric equations. TheAO 2 cos 0 - cot 0 slope of a curve at any point equals zero when

AX AX -2 sin 8 the tangent is parallel to the x axis. In para-Ao metric equations the horizontal and vertical

tangents can be found easily by settingThen at the point where

0 = 45°

176

179

= 0A t


andAx = 0A t

For the horizontal tangent solve -41f equals

zero for t and for the vertical tangent solveAx equals zero for t.At

EXAMPLE: Find the points of contact of thehorizontal and vertical tangents to the curverepresented by the parametric equations

and

x = 3 - 4 sin 0

y = 4 + 3 cos 0

Plot the graph of the function by taking 0

from 00 to 3600 in increments of 30°.Given that

AX = - 4 cos 0AO

and

= - 3 sin 0Ao

SOLUTION: The graph of the function showsthat the figure is an ellipse, figure 11-7, andconsequently there will be two horizontal andtwo vertical tangents. The coordinates of thehorizontal tangent points are found by firstsetting

This gives

Then

and

- 3 sin 9 = 0

sin 0 = 0

= 0° or 180°

Substituting 0° we have

x = 3 - 4 sin 00

= 3 - 0

= 3

and

y = 4 + 3 cos 0°

= 4 + 3

= 7

Substituting 1800 we obtain

x = 3 - 4 sin 180°

= 3

= 3

and

- 0

y = 4 + 3 cos 180°

= 4 - 3

= 1

The coordinates of the points of contact ofthe horizontal tangents to the ellipse are (3, 1)and (3, 7).

The coordinates of the vertical tangentpoints of contact are found by setting

= 0AO

177

(-1, 4)

( 3 7)

(3,1)

(7,4)

180

Figure 11-7.Ellipse.

X


We find

from which

- 4 cos 0 = 0

0 = 90° or 270°

Substituting 900 we obtain

x = 3 - 4 sin 90°

= 3 - 4

= 4

and

y = 4 + 3 cos 900

= 4 + 0

= 4

Substituting 270° gives

x = 3 - 4 sin 270°

= 3 + 4

and

= 7

y = 4 + 3 cos 270°

= 4 + 0

= 4

The coordinates of the points of contact ofthe vertical tangents to the ellipse are (-1, 4)and (7, 4).

PROBLEMS: Find the equations of the tan-gent and the normal and the lengths of the sub-tangents and the subnormal for each of thefollowing curves at the point indicated.

1. x = t3

y = 3tat t = -1

Ax Aygiven At = 3t2 and

2. x = t2 + 8

y = t2 + 1

at t = 2

3.

Ax 162./given = 2t and = 2tat At

x = t2y = t

at t = 141x 1 and

biy = 2tgiven =at t4. Find the points of contact of the hori-

zontal and vertical tangents to the curvex = 2 cos 0

given

178

y = 3 sin 0

-42-c = -2 sin 0A0

3 cos 0

ANSWERS:



Length of subnormal

y = x - 2y = -x - 43

3

2. Equation of tangent y = x - 7Equation of normal y = -x + 17Length of subtangent 5



Equation of normal



4. Coordinates of the points of contact ofthe horizontal tangent to the ellipse are(0, 3) and (0, -3) and the vertical tangentto the ellipse are (2, 0) and (-2, 0).

y = 2x - 1-x 3-2- 4- -ff

CHAPTER 12

LIMITS AND DIFFERENTIATION

Limits and differentiation are the beginning ofthe study of calculus, which is an important andpowerful method of computation.

LIMIT CONCEPT

The study of the limit concept is very im-portant as it is the very heart of the theory andoperation of calculus. We will include in thissection the definition of limit, some of the in-determinate forms of limits, and some limitformulas, along with example problems,

We will start with values lying between

and

x = 2

x = 6

indicated by line A in figure 12-1. This intervalmay be written as

0 < I x - 4 1 < 2

DEFINITION OF LIMIT The Corresponding interval for y is between

Before we start differentiation ther e are c er- y = 4tain concepts which we must understand. One ofthese concepts deals with the limit of a function. andMany times it will be necessary to find the valueof the limit of a function. y = 36

The discussion of limits will begin with anintuitive point of view. We now take a smaller interval aboutWe will work with the equation

y = f(x) = x2 x = 4 (line B)

which is shown in figure 12-1. The point P by using values ofrepresents the point corresponding to

and

y = 16

x = 4

and

x = 3

x = 5

The behavior of y for given values of x near and find the corresponding interval for y to bebetweenthe point

x = 4

is the center of the discussion. For the present andwe will exclude the point P which is encircledon the graph.

179

= 9

y = 25


Interval of

s

Interval of

f(x)

2 - 6 (A) 4 363 - 5 (9) 9 25

3.5 4.5 t3) 12.25 20.25

3.9 4.1 (D) 15.21 16.91

(8)

Figure 12-1.(A) Graph of y = x2;(B) value chart.

These intervals for x and y are written as

0 < lx - 4 1 < 1

and

9 < y <25

As we diminish the interval of x around

x = 4 (line C and line D)

we find the values of

2y = x

to be grouped more and more closely around

y = 16

This is shown by the chart in figure 12-1.Although we have used only afew intervals of

x in the discussion, it should be apparent thatwe can make the values about y group as closelyas we desire by merely limiting the valuesassigned to x about

x = 4

Because the foregoing is true, we may now saythat the limit of x2, as x approaches 4, resultsin the value 16 for y and we write

lim x2 = 16x-4

In the general form we may write

urn f(x) = Lxa

and we mean that as x approaches a, the limit off(x) will approach L and L is called the limit off(x) as x approaches a. No statement is madeabout f(a) for it may or may not exist although thelimit of f(x), as x approaches a, is defined.

If f(x) is defined atx = a

and for all values of x near a, and if the functionis continuous, then

lim f(x) = f(a)xa

We are now ready to define a limit.Let f(x) be defined for all x in the interval

near

x = a

bu . f necessarily at

x = a

180

'1133

Chapter 12LIMITS AND DIFFERENTMTION

Then there exists a nuinber L such that for everypositive number C

11(x) - LI<E

provided that we may find the number 8 such that

0<ix-al< 6Then we say L is the limit of f (x) as x approachesa and we write

lim f(x) = Lx a

This means that for every challenge numberE We must find a number 6 in the interval

0 < Ix- al<6

such that the difference between f(x) and L issmaller than the number e.

EXAMPLE: Suppose we are given

lim x2 + x - 2x 1 x - 1 3

and the challenge number is 0.1.SOLUTION: We must find a number 6 such

that in the neighborhood of

for all points except

x = 1

x = 1

we have the difference betweenf(x) and 3 smallerthan 0.1.We write

and

x2 + x - 2 +0.1x - 1

x2 + x - 2 - 3x - 1

(x1-.21(21.11)x -

and we consider only values where

x 1

Simplifying the first term, we have

181

(x + 2) (x - 1)

Finally, combine terms as follows:

and

or

then

x + 2 - 3 = x - 1

Ix - 11 < 0.1

-0.1 <X 1 < 0.1

0.9 <x<1.1

and we have fulfilled the definition of the limit.If the limit of a function exists and the function

is continuous then

lirn f(x) = f(a)x a

For instance, in order to find thelimit of thefunction x2 - 3x + 2 as x approaches 3, we sub-stitute 3 for x in the function. Then

f(3) = 32 - 3(3) + 2

= 9 - 9 + 2

=2

Since x is a variable it may assume a valueas close to 3 as we wish, and the closer wechoose the value of x to 3, the closer f(x) willapproach the value 2. Therefore, 2 is called thelimit of f(x) as x approaches 3 and we write

lirn (x2 - 3x + 2) = 2x-3

PROBLEMS: Find the limit of each of thefollowing functions.

urn 2x2 - 11. x-1 2x- 1

Lim-2

22. (x - 2x + 3)x

'184


3.lim

a a

4. tlim (5t2 - 3f + 2)0

5. Urn E3 - EE 6 E - 1

lim Z2 - 3Z + 26.0 Z - 4

ANSWERS:

1. 12. 33. a - 14. 25. 42

16 -

2

INDETERMINATE FORMS

Whenever the answer obtained by substitu-tion, in searching for the value of a limit,assumes any of the following forms, anothermethod for finding the correct limit must be used.

51 (c4) Of 0°, 0043f 100

These are called indeterminate forms.The proper method for evaluating the limit

depends on the problem and sometimes callsfora high degree of ingenuity. We will restrict themethods of solution of indeterminate forms tofactoring and division of the aimerator and de-nominator by powers of the variable. Later inthe study of limits, L'Hospital's rulewill be usedas a method of solving indeterminate forms.

Sometimes factoring will resolve an indeter-minate form.

EXAMPLE: Find the limit of

x2 - 9- as x approaches 3x 3

SOLUTION: By substitution we find

lim x2 - 9 0x 3 x - 3 0

182

165

which is an indeterminate form and is thereforeexcluded as a possible limit. We must nowsearch for a method to find the limit. Factoringis attempted and results in

x2 - 9 (x + 3) (x -3)x - 3 x - 3

then

= x + 3

urn (x + 3) = 6x-3

and we have a determinate limit of 6.Another indeterminate form is often met when

we try to find the limit of a function as the inde-pendent variable becomes infinite.


x4 + 2x3 - 3x2 + 2x3x4 - 2x2 + 1

as x becomes infinite.SOLUTION: If we let x become infinite in the

original expression the result will be

lim x4 + 2x3 - 3x2 + 2xx

3x4 - 2x2 + 1

co

at)

which must be excluded as an indeterminateform. However, if we divide both mnneratorand denominator by x4 we obtain

2 3 2+ -

lim xX 41. 2 1

1" 4x x

1 + 0 - 0 + 03 - + 0

13

and we have a determinate limit of 13'PROBLEMS: Find the limit of the following:

lim x2 - 41. x 2 Tc7":2

2 lim 2x + 3x IT 71

Chapter 12LIMITS AND DIFFERENTIATION

lim 2a2b - 3ab2 + 2ab3. a 0 5ab - a3b2

lim x2 - x - 64. x-3 x - 3

5lim x4 - a4

x a x - a-

lira (x - a)2 - x26. a 0 aANSWERS:

1. 4

2 17

2 - 3b36

5

4. 5

5. 4a3

6. -2x

LIMIT FORMULAS

To obtain results in calculus we will fre-quently operate with limits. The proofs oftheorems shown in this section will not be givenas they are quite long and demand considerablediscussion.

The theorems will be stated and examples willbe given. Assume that we have three simplefunctions of x.

Further, let these functions have separate limitssuch that

lim u = Ax alirnv=B

a

lim w= Cx a

183

Theorem 1. The limit of the sam of twofunctions is equal to the sum of the limits.

lira [ f(x) + g(x)] = A + Bx a

Um f(x) + lim g(x)

This theorem may be extended to include anynumber of functions such as

lira [f(x) + g(x) + h(x)1 = A B + Cx a

lim f(x) + lira g(x) + llm h(x)xa x aEXAMPLE: Find the limit of

(x - 3)2

SOLUTION:

Um (x - 3)2 =x 3

= 9

= 0

as x-3

Urn (x2 - 6x + 9)x 3

ijin x2- Um 6x+ lira 9x 3 x-3 x 3

- 18 + 9

Theorem 2. The limit of a constant c timesa function f(x) is equal to the constant c timesthe limit of the function.

Um cf(x) = cA = c urn f(x)a x a


2x2 as x-3SOLUTION: Um 2x2 = 2 Um x 2

x-3 x 3= (2) (9)

= 18

Theorem 3. The limit of the product of twofunctions is equal to the product of their limits,

llm f(x) g(x) = ABxa

= f(x) (xlim g(x))\x a a



(x2 - (1-23) as x--2

SOLUTION:

urn (x2 - x) (/2i) = ABX 42

= (x2 - x)) (Umx-2 x--2

= (4 - 2) (11)

= 4

3.

4.

ANSWERS:

1.

2.

3.

4.

5x4 as x-2 (Theorem 3)

2x2 + x - 4 4)3x - as x-3 (Theorem7

4

21

80

172

Theorem 4. The limit of the quotient of twofunctions is equal to the quotient of their limits,provided the limit of the divisor is not equal tozero.

lim f(x) Axa g(x) B

lim f(x)xalim g(x)xa

if B


3x2 + x - R as x-3

SOLUTION:

3x2 + x - 6x-3 2x - 5

lim 3x2 + x - 6

lim 2 x - 5x-3

= 24

PROBLEMS: Find the limits of the following,using the theorem indicated.

1. x2 + x + 2 as x--1 (Theorem 1)

2. 7(x2 - 13) as x-4 (Theo.am 2)

184

INFINITESIMALS

In chapter 11, the slope of a curve at a givenpoint was found by taking very small incrementsof ay and Ax and the slope was said to be equalto This section will be a continuation ofthis concept.

DEFINITIONS

A variable that approaches .0 as a limit iscalled an infinitesimal. This may be written as

orlim V = 0

V-0and means, as recalled from a previous sectionof this chapter, that the numericalvalue of V be-comes and remains less than any positive chal-lenge number e .

If the

then

lim V = L

lim V - L = 0

which indicates that the difference between avariable and its limit is an infinitegimal. Con-versely, if the difference between a variable anda constant is an infinitesimal, then the variableapproaches the constant as a limit.

EXAMPLE: As x becomes increaSingly1large, is the term an infinitesimal?


SOLUTION: By the definition of itginitesi-1mai, 6 approaches 0 as x increases in value,x"

12 is an infinitesimal. It does this, and is there-

fore an infinitesimal.EXAMPLE: As x approaches 2, is the ex-

pression x2 - 4 - 4 an infinitesimal?x - 2SOLUTION: By the converse of the definition

2 - 4of infinitesimal, if the differencebetween

x

and 4 approaches 0, as x approaches 2, the ex-

pression x2 -24 - 4 is an infinitesimal. Byx-

direct substitution we find an indeterminateform; therefore we make use of our knowledgeof indeterminates, and write

x2 - 4 (x + 2) (x - 2) x + 2x - 2 x - 2

andUrn (x + 2) = 4x--2

The difference between 4 and 4 is 0 and the ex-

x2-- 4pression x 2

- 4 is an infinitesimal, as xapproaches 2.

SUMS

An infinitesimal is a variable that approaches0 as a limit. We state that and 6 , in figure12-2, are infinitesimals because they bothapproach 0 as shown.

Theorem 1. The algebraic sum of any numberof infinitesimals is an infinitesimal.

In figure 12-2, as e and 8 approach 0, noticethat their sum approaches 0 and by definitionthissum is an infinitesimal and the truth oftheorem 1 has been shown. This approach may beused for the sum of any number of infinitesimals.

PRODUCTS

Theorem 2. The product of any number ofinfinitesimals is an infinitesimal.

In figure 12-3, the product of two infinitesi-tnals, and6, is an infinitesimal as shown. The

product of any number of infinitesimals is alsoan infinitesimal by the same approach as shownfor two numbers.

Theorem 3. The product of a constantand aninfinitesimal is an infinitesimal.

This may be shown, in figure 12-3, by holdingeither e or 6 constant and noticing their productas the variable approaches 0.

185

/ /-- / /44

/4,......

Z56........ 0

/4

/ 2 44 C :#5 G/ ...1 ._,/_ _,E.

/6/7 as

/16

/7Ti.

f / S /7 '

256/6 9 i#/ 65

rir/I .5 / 5

af. 6* 6 f 32 256

/ 257 45 ii $ /_

256 256 256 256 256 /28

i'0

\0

Figure 12-2.Sums of infinitesiinals.

Eef I

/44

2/6

/6-71

_Z...

256I

-.1.0

/ / i .?-a tgii di/ / / / / /

:ir # TC 6f 256 /02*

,..i. / / / / /256 /0244 140%/4 n-- Z-F

/ / / / / /644 Pr X /024 Of% low/ / / / /

256 256....4.,

tail 1094 /650 iiis

0 1 NFigure 12-3.--Products of infinitesimals.

MATHEMATICS, VOLUME 2'

CONCLUSIONS then it is said to be discontinuous at

The term irdinitesimal vas used to describe x = athe term As as it approaches zero. The quantityAx was called in increment of : where incrementwas used to imply that we have added a small We will use examples to show the above state-amount to x. Thus x + Ax indicates thatwe are ments,holdi: g x constant and adding a small but varial:Ileamount to which we will call Ax. EXAMPLE: In figure 12-4, is the function

A very small increment is sometimes called adifferential. A small Ax is indicated by dx. The f(x) = x2 + x - 4differential of 0 is dO and that of y is dy. Thelimit of Ax as it approaches zero is of coursezero, bt this does not meanthat the ratio of two continuous at f(2)?uinfinitesimals cannot be a real number or a realfunction of x. For instance, no matter how small SOLUTION:

dxAxis chosen, the ratio will still be equal to 1. f(2) = 4 + 2 - 4dxIn the section on indeterminate forms, a = 2

0method for evaluating the form -6 was shown.This form results whenever the limit takes theform of one infinitesimal over another. In everycase the limit was a real number.

DISCONTINUITIES

The discussion of discontinuities will bebased upon a comparison to continuity which isdefined by:

A function f(x) is continuous at

x = a

if f(x) is defined at

x = a

and has a limit as x--a, as follows:

lim f(x) = f(a)

Notice that for continuity a function must fulfilthe following thre 3 conditions:

1. f(x) is defined at x = a.2. The limit of f(x) exists as x approaches a.

The value of f(x) at x = a is equal to thelimit of f(x) at x = a.

If a function f(x) is not continuous at

and

lim x2 + x - 4 = 2x-2

and

Um f(x) = f(2)x-2

Therefore the curve is continuous at

x = 2

EXAMPLE: In figure 12-5, is the function

x - 2

continuous at f (2) ?

SOLUTION:

f(2) is undefined at

x = 2

and the function is therefore discontinuous at

x = 2

However, by extending the original definitionx = a of f(x) to read

186

,169

Chapter 12.LIMITS AND DIFFERENTIATION

Figure 12-4. Function f(x) = x2 + x -4.

x2 - 4Figure 12-5.Function f(x) =

11. It should be obvious thatthevalue of the tan-2gent at is undefined. Thus thefunctionis saidto be discontinuous at 11.

PROBLEMS: In the following definitions ofthe functions find where the functions are dis-continuous and then extend the definitions so thatthe functions are continuous.

f14) =

24

x - 2' 1. x - 2

4, x = 2

we will have a continuous function at

x = 2

NOTE: The value of 4 at x = 2 was foundby factoring the mnnerator of f(x) and thensimplifying.

A common kind of discontinuity occurs whendealing with the tangent function of an angle.Figure 12-6 is the graph of the tangent as theangle varies from 0° to 90°; that is, from 0 to

lol

2. f(x) 22,2

gm' 3x + 3

3. f(x) = x2 + x - 129x - 9

ANSWERS:

1. x 2, f(2) 3

2. x -3, f(-3) -4

73. x = 3, f(3) am T

490


1

1

1

11

I 9,9

1(111 2

0 ) y = f(x)

( 0) As recalled from the discussion of limits,;1 11as Ax is made smaller, Ay gets smaller also,

1

but the radio approaches 20. This is shown1 Ax

and

= = 22Ax 2

AyWe are interested in the ratio because thelimit of this ratio as Ax approaches zero is thederivative of

1

1

1

1

Figure 12-6.Graph of tangent function.

INCREMENTS AND DIFFERENTIATION

In this section we will extend our discus-sion of limits and examine the idea of the deriva-tive, the heart of differential calculus.

We will assume we have a particular func-tion of x, such that

y = x2

If x is assigned the value 10, the correspondingvalue of y will be (10)2 or 100. Now, if we in-crease the value of x by 2, making it 12, we maycall this increase of 2 an increment or Ax. Thisresults in an increase in the value of y and wemay call this increase an increment or Ay. Fromthis we write

y + Ay = (x + Ax)2

= (10 + 2)2

= 144

As x increases from 10 to 12, y increases from100 to 144 so that

Ax = 2

Ay = 44

in table 12-1.

Table 12-1. Slope values.

Variable Values of tl'e variable

ax 2 1 0. 5 0. 2 0.1 0. 01 0.0001

Ay 44 21 10. 25 4.04 2.01 0. 2001 0. On)0001

Ay_ax 22 21 20. 5 20. 2 20.1 20. 01 20.0001

There is a much simpler way to find that thelimit of Atit as Ax approaches zero is, in thisAxcase, equal to 20. We have two equations

and

y + Ay = (x + Ax)2

y = x2

By expanding the first equation so that

y + Ay = x2 + 2x Ax + (4x)2

and subtracting the second from this, we have

= 2x Ax + (4)2

Dividing both sides of the equation by ax gives

= 2x + AxAx

188

isl.


Now, taking the limit as Ax approaches zero

Thus,

lim =Ax--0 AX

(2)

NOTE: Equation (2) is onewayof expressingthe derivative of y with respecttox. Other waysare

= y' = V (x) = D(x) =lim Ay.

Ax--0 Ax

Equation (2) has the advantage that itis exactand true for all values of xt Thus if

then

and if

then

x = 10

= 2 (10) = 20dx

x = 3

= 2(3) = 6dx

This method for obtaining the derivative of ywith respect to x is general and may be formu-lated as follows:

1. Set up the function of x as a function of(x + Ax) and expand this function.

2. Subtract the or iginal function of x from thenew function (x + Ax).

3. Divide both sides of the equation of Ax.4. Take the limit of all the terms in the equa-

tion as Ax approaches zero. The resulting equa-tion is the derivative of f(x) with respect to x.

GENERAL FORMULA

In order to obtain a formula for the deriva-tive of any expression in x, assume the function

y = f(x) (3)

189

1S2

so that

y + Ay + Ax) (4)

Subtracting equation (3) from equation (4) gives

Ay = f(x + Ax) f(x)

and dividing both sides of the equation by Ax wehave

y. f(x + Ax) - f(x)Ax Ax

The desired formula is obtained by taking thelimit of both sides as Ax approaches zero, sothat

or

urn f(x + Ax) - f(x)Ax--0 Ax Ax

lim f(x + Ax) - f(x)dx Ax-,0 Ax

NOTE: The notation 51 is not to be con-dx

sidered as a fraction which has dy for the mimer-ator and dx for the denominator. The expression

-41 is a fraction with Ay as its numeratorand AxAx

as its denominator and gr is a symbol repre-Ay

senting the limit approached by as Ax ap-Ax

proaches zero.

EXAMPLES OF DIFFERENTIATION

In this last section of the chapterwewill useseveral examples of differentiation to obtain afirm understanding of the general formula.

dyEXAMPLE: Find the derivative for theax

functiony = 5x3 3x + 2

and determine the slope of its graph at

X = -1, 41 04 1Draw the graph of the function, as shown infigure 12-7


( A )

Expand equation (5), then subtract equation (6)from equation (5) and simplify to obtain

f(x + Ax) - f(x)

= 5 [ 3x2 px + 3x(Ax)2 + (Ax)3] - 3Ax

Divide through by Ax and we have

f(x + Ax) - f(x)Ax

1 = 5 [3x2 + 3xAx + (Ax)2] - 3t

Take the limit of both sides as Ax-- 0 and

2 I1

-2

(B)

lim + Ax)Ax--0 Ax,

X1 2

x - II

-Irr 0I

qr. 1

dy

dx12 0 -3 0 12

Y 0 2.89 2 1.1 4

Figure 12-7.(A) Graph of f(x) = 5x3 - 3x+ 2; (B) chart of values.

SOLUTION: Finding the derivative by for-mula, we have

f(x + Ax) = 5(x + Ax)3 - 3(x + Ax) + 2 (5)

And

f(x) = 5x3 -3x + 2 (6)

then

= 152 - 3dx

Using this derivative let us find the slope ofthe curve at the points given.

Thus we have a new method of graphing anequation. By substituting different values ofx inequation (7) we can find the slope of the curve atthe point corresponding to the value of x.

EXAMPLE: Differentiate the function, that

is' find of

and then find the slope of the curve at

x = 2

SOLUTION: Apply the formula for the de-rivative, and simplify as follows:

190

133

1 1f(x + Ax) - f(x) x + Ax

Ax Ax

Ax

- 1x x + Ax)


Now take the limit of both sides as Ax-0 and

dx x2

In order to find the slope of the curve at thepoint where x has the value 2, substitute 2 for x

dyin the expression for --

1dx 2

1= _

EXAMPLE: Find the slope of the tangent lineon the curve

f(x) = x 4- 4

x = 3

at

SOLUTION: We need to find a which is thedxslope of the tangent line at a given point. Applythe formula for the derivative; then,

f(x + Ax) = (x + Ax)2 + 4

and

f(x) = x2 + 4

Expand equation (8) so that

f(x + Ax) = x2 + 2xAx + (Ax)2 + 4

then subtract equation (9) from equation (8) and

(8)

(9)

f(x + Ax) - f(x) = 2x Ax + (Ax)2

Now, divide through by Ax arid

f(x + Ax) - f(x)Ax

then take the limit of both sides as Ax--0 and

Substitute 3 for x in the expression for the de-rivative to find the slope of the function at

so that

x = 3

slope = 6

In this last example we willsetthe derivativeof the function f(x) equal to zero to find a maxi-mum or minhmun point on the curve. By maxi-mum or minim* _ of a curve we mean the pointor points througn which the slope of the curvechanges from positive to negative orfromnega-tive or positive.

NOTE: When the derivative of a function isset equal to zero this does not mean that in allcases we will have found a maximum or minimumpoint on the curve. A complete discussion ofmaxima or minima may be found in most calcu-lus texts.

We will require that the following conditionsare met:

1. We have a maximum or minimum point.2. The derivative exists.3. We are dealing with an interior point on

the curve.When these conditions are met the derivative

of the function will be equal to zero.EXAMPLE: Find the derivative of the func-

tion

y = 5x3 - 6x2 - 3x + 3

and set the derivative equal to zero and find thepoints of maximum and minimum on the curve,then verify this by drawing the graph of thecurve.

SOLUTION: Apply the formula for 3!as fol-lows:

and

f(x + Ax) =(10)

5(x + Ax)3 - 6(x + Ax)2 - 3(x + Ax) + 3

f(x) = 5x3 - ex2 - 3x + 3 (11)

Expand equation (10) and subtract equation (11),obtaining

191

184

MATBEMATICS, VOLUME 2

f(x + Ax) - f(x) = 5(3x2Ax 3xAx2 + Ax3) -6(2xAx + Ax2) - 3Ax

Now, divide through by Ax and take the limit asAx--0, so that

= 5(3x2) 6(2x) - 3cbc

= 15x2 - 12x - 3

Set dx equal to zero, thus

then

and

15x2 - 12x - 3 = 0

3(5x2 - 4x - 1) = 0

(5x + 1)(x - 1) = 0

Set each factor equal to zero and find the pointsof maximum or minimum are

and

5x = - 1

x = 1

The graph of the function is shown infigure 12-8.PROBLEMS: Differentiate the functions in

problem 1 through 3.

1. f(x) = x2 - 3

2. f(x) = x2 - 5x

3. f(x) = 3x2 - 2x + 34. Findthe slope of the curve

at the points

= x3 - 3x + 2

x = -2, 0,. and 3

192

Figure 12-8.Graph of 54(3 - 6x2 -3x + 3.

5. Find the values of x where the function

f(x) = 2x3 - 9x2 - 60x + 12

has a maximum or a minimum.

ANSWERS:

1. 2x2. 2x - 53. 6x - 24. m = 9, -3, and 245. x = -2, x = 5

CHAPTER 13

DERIVATIVES

In the previous chapter on limits, we usedthe delta process to find the limit of a functionas Ax approached zero. We called the resultof this tedious and in some cases lengthy pro-cess the derivative. In this chapter we willexamine some rules used to find the derivativeof a function without using the delta process.

To find how y changes as x changes, we take

the limit of as AX- 0 and writeAx

limAx - 0 Ax

which is called the derivative of y with re-

spect to x, and we use the symbol to indi-

cate the derivative and write

urnAx Ax dx

In this section we will take up a number of

rules which will enable us to e4bily obtain thederivative of many algebraic functions. In thederivation of these rules, which will be calledtheorems, we will assume that

lim Ex_± Ax) - f(x)Ax

ordy. lim Aydx Ax--0 Ax

exists and is finite.

DERIVATIVE OF A CONSTANT

The method used to find the derivative of aconstant will be similar to the mita processused in the previous chapter but will include ananalytical proof. A diagram is used to give ageometrical meaning of the function.

FORMULA

Theorem 1. The derivative of a constant iszero. Expressed as a formula, this may bewritten as

lim ndx Ax--0 Ax

when y is parallel to the x axis.

PROOF

In figure 13-1, the graph of

y = e (a constant)

the value of y is the same for all values of

x, and any change in x (that is, Ax) does notaffect y, then

and

and

Ay = 0

Another way of stating this is that when x isequal to x1 and when x is equal to x1 + Ax,y has the same value. Therefore,

193

and

115(P::

y = c

y + Ay =


11110MIIMIIMION A X IMIIMIIM100IP

xi xit ax

Figure 13-1.--Graph of y = c (a constant).

so that

and

then

f(x + Ax) - f(x) c - cAx Ax Ax

lim Ay nAxO Ax

dy lim f(x±,Ax) - fdx AxO Ax

lira c - c-AxO Ax

The equation

y = c

therefore

lim f(x + Ax) - f(x)dx Ax--0 Ax

= 0

VARIABLES

In this section on variables, we will extendthe theorems of limits covered previously. Re-call that a derivative is actually a limit. Theproof of the theorems presented here involve thedelta process, and only a few of these pro'Aswill be offered.

POWER FORM

Theorem 2. The derivative of the function

y = xn

where n is any number is given by

nxn-1dx

Proof: By definition

dy lim (x + Ax)n - (x)nAx-0 Ax

The expression (x + Ax)n may be expandedrepresents a straight line parallel to the x by the binomial theorem intoaxis. The slope of this line will be zero forx -

all values of x. Therefore, the derivative is n + nxn-1 Ax + n(n-1) x112Ax2 + . . . + Axilzero for all values of x.2

EXAMPLE: Find the derivative a of theSubstituting in the expression for the deriva-function tive, we have

SOLUTION:

and

Y = 6

Y = 6

y + Ay = 6

nxn- l+ xn(n-1) n-2 2Ax +. .dx x-0

Ax

Simplifying, this becomes

dy lim [ix11-1 + + . . .a"-= Ax-0

194

Chapter 13DERIVATIVES

Letting ax approach zero, we have

Thus, the proof is complete.EXAMPLE: Find the derivative of

f(x) = x5

cnT3TTION: Apply Theorem 2 and Mid

5 nx = x

The previous example is a special case ofthe power form and indicates that the deriva-tive of a function with respect to itself is 1.

EXAMPLE: Find the derivative of

and

f(x) = ax, a = constant

SOLUTION:

f(x) = ax

...:tore f(x + Ax) = a(x + Ax)

and

so that given

n = 5

n - 1 = 4

a nxn-1dx

and substituting values for n find that

= 5x4dx


f(x) = x

so that

Therefore

= ax + aAx

Ay = f(x + Ax) f(x)

= (ax + aAx) - ax

= aAx

lim aAxdx Ax

= a

The previous example is a continuation ofSOLUTION: Apply Theorem 2 and find the derivative of a function with respect to

itself and indicates that the derivative of aX

n = X function with respect to itself, times a con-stant, is that constant.

therefore EXAMPLE: Find the derivative ofn = 1

f(x) = 6xand

SOLUTION:n - 1 = 0

so that = 6dx

= 1

A study of the functions and their deriva-tives in table 13-1 should further the under-standing of this section.

195


Table 13-1. Derivatives of functions.

f(x) 3 x x2 x3 x4 3x2 9x3 x-1 x-2 9x-4

clz0 1 2x 3x2 4x3 6x 27x2 -x-2 -2x-3 -12x-5

then

PROBLEMS: Find the derivatives of the andfollowing:

1. f(x) = 21

2. f(x) = x

3. f(x) = 21x

4. f(x) = 7x3

5. f(x) = 4x2

6. f(x) = 31-2

ANSWERS:

1. 0

2. 1

3. 21

4. 21x2

5. 8x

6. -6x-3

SUMS

Proof:

du dvdx dx dx

y = g(x) + h(x) (1)

y + Ay = g(x + Ax) + h(x + Ax) (2)

Subtract equation (1) from equation (2) and

Ay = ex + Ax) + h(x + Ax) - g(x) - h(x)

Rearrange this equation such that

Ay = g(x + Ax) - g(x) + h(x + Ax) - h(x)

Divide both sides of the equation by Ax andthen take the limit as AxO and

lim Ay lim g(x + Ax) - g(x)Ax-0 Ax AxO AX

lim h(x + Ax) - h(x)

but, by definition

lim ex + Ax) - g(x) duAxO Ax dx

and

lim h(x + Ax) - h(x) dv

Theorem 3. The derivative of the sum of AxO Ax dx

two or more functions of x is equal to the sumof their derivatives, then by substitution

Assume two functions of x which we willcall u and v, such that du dv

dx dx dxu = g(x)

EXAMPLE: Find the derivative of the func-and tion

and also

v = h(x)

y u +

= g(x) + h(x)

196

y = x3 - 8x2 + 7x - 5

SOLUTION: Theorem 3 indicates that weshould find the derivative of each term andthen show them as a sum; that is, if


then, if

then

y =

y = 7x,

Y = -5,city. ndx

y = x3 - 8x2 + 7x - 5

ca.= 33E2 16x + 7 + 0dx

= 3x2 - 16x +

PROBLEMS: Find the derivative of the fol-lowing:

1. f(x) = x2 + x - 1

2. f(x) = 2x4 + 3x + 16

3. f(x) = 2x3 + 3x2 + x - S

4. f(x) = 3x3 + 2x2 - 4x + 2 + 2x-1 - 3x-3

ANSWERS:

1. 2x + 1

2. 8x3 + 3

3. 6x2 + 6x + 1

4. 9x2 + 4x - 4 - 2x -2 + 9x -4

PRODUCTS

Theorem 4. The derivative of the productof two functions of x is equal to the first func-tion multiplied by the derivative of the secondfunction, plus the second function multiplied bythe derivative of the first function.

If

y = uv

197

then

dv du= vdx d dx

This theorem may be extended to includethe product of three functions. The result willbe as follows:

If

theny = uvw

= uv + vwd--1-1/(1u + uwdx dx dx dx


f(x) = (x2 2)(x4 + 5)

SOLUTION: The derivative of the first fac-tor is 21, and the derivative of the second fac-tor is 4x4. Therefore

(x2 2)(4x3) t,x4 + 5)(2x)

= 4x5 - 8x3 + 2x5 + 10x

= ex5 - 8x3 + 10x


f(x) = (x3 - 3)(x2 + 2)(x4 - 5)

SOLUTION: The derivatives of the threefactors, in the order given, are 3x2, 2x, and3x2.

Therefoi e

fy (x) = (x3 - 3)(x2 + 2)(4x3)

(x2 2)(x4 5)(3x2)

+ (x3 - 3)(x4 - 5)(2x)

then

(x) = 4x8 + 8x6 - 12x5 -24x3

+ 3x8 + 6x6 - 15x4 - 30x2

+ 2x8 - 6x5 - 10x4 + 30x

= 9x8 + 14x -18x5- 25x4 - 24x3 -30x2+30x

gigc

4


PROBLEMS: Find the derivatives of thefollowing:

1. f(x) = x3(x2 -

2. Mc) = (x3 - 3)(x2 + 2x)

3. f(x) = (x2 - 7x)(x5 - 4x2)

4. f(x) = 2)(x2 3)(x3 - 4)

ANSWERS

1. 5x4 - 12x2

2. 5x4 + 8x3 - 6x-6

3. 7x6 - 42x5 - 16x3 + 84x2

4. 6x5 - 10x4 - 12x3 + 6x2 + 16x + 12

QUOTIENTS

Theorem 5. The derivative of the quotientof two functions of x is equal to the denomina-tor times the derivative of the numerator ministhe numerator times the derivative of the de-nominator, all divided by the square of the de-nominator,

If

then

du dvvdx

dx v2

EXAMPLE: Find the derivative of the func-tion

x2 - 7gx) 2x + 8

SOLUTION: The derivative of the mimeratoris 2x, and the derivative of the denominator is2. Therefore

(x)(2x + 8)(2x) - (x2 - 'n(2)

(2x + 8)2

4x2 + 16x - 2x2 + 14

(2x + 8)2

2x2 + 16x + 144(x + 4)2

x2 + 8x + 72(x + 4)2


x41. Alt) =

2. f(x) = x + 7

x2 + 3x + 53. gx) = 3x - 4

ANSWERS:

1.

2.(x + 7)2

3+ 620 + 15x2 + 8x + 12)

.(20 - 4)2

POWERS OF FUNCTIONS

Theorem 6. The derivative of any func-tion of x raised to the power n, whe .e n isany number, is equal to a times the po?inomialfunction of x to the (n - 1) power times thederivative of the polynomial itself.

2x5 - 8x3

(x2 - 2)2

x2 + 14x 4. 3

If

y = uwhere u is any function of x then

dudx dx

F.

198


EXAMPLE: Find the derivative of the func- RADICALStion

y = (x3 - 3x2 + 7

SOLUTION: Apply Theorem 6 and find

7(x3 3x2 + 2x)6 (3x2 - ex + 2)aEXAMPLE: Find the derivative of the func-tion

(x2 + 2)3f(x) =

SOLUTION: This problem involvesTheorem5 and Theorem 6. Theorem 6 is used to findthe derivative of the numerator, then Theorem5 is used to find the derivative of the resultingquotient.The derivative of the numerator is

3(x2+ 2)2 (2x)

and the derivative of the denominator ii 1.Then, by Theorem 5

(x - 1) [3(x2 + 2)2 (2x)1 - (1)(x2 +dx(x - 1)2

6x(x2 + 2)2 (x - 1) - (x2 + 2)3(x - 1)2

(x2 + 2)2 [6x(x - 1) - (x2 + 2)](x - 1)

)3

(x2 + 2)2 (5x2 - 6x - 2)(x - 1)2


1. f(x) = (x3 + 2x - 6)22. f(x) = 5(x2 + x + 7)4

3. f(x)=

2(x + 3)3

ANSWERS:

1. 2(x3 + 2x - 6)(3x2+ 2)

2. 20(x2 + x + 7)3 (2x + 1)3. 18x(x + 3)2 - 6(x + 3)3

9x2

199

To differentiate a function containing a radi-cal, replace the radical by a fractional expo-nent then fiLd the derivative by applying theappropriate theorems.EXAMPLE: Find the derivative ofnu-f(x) = v 2x - 5SOLUTION: Replace the radical by theproper fractional exponent, then

f(x) = (2x2 - 5)1/2and by Theorem 6

1/2x 5) -1(4x)dx 2

(2x2 - 5)4/2 (4x)

= 2(2x2 - 5)-1/2

2x= / n

N/2x` -5

2x 077i2x

2 - 5EXAMPLE: Find the derivative of

f(x) - 2x + 1/377i

SOLUTION: Replace the radicalproper fractional exponent, thus

2x + 1f(x)(3x2 + 2)1/2

by the

At this point a decision is in order. Thisproblem may be solved by either writing

f(x) -(3x2 2+ 2)1/

2x + 1(3)

and applying Theorem 6 in the denominatorthen applying Theorem 5 for the quotient, orwriting

f(x) =I (2x + 1)(3x2 + 2)-1/2 (4)


and applying Theorem 6 for the second factor Find the derivative of each factorthen applying Theorem 4 for the products

The two methods of solution will be com-pleted individually as follows:

Use equation (3)

f(x)(3x2

+ 2)1/2

2x + 1

Find the derivative of the denominator

d 2 1/2(3x + 2)dx Now apply Theorem 4 and

by applying the power theorem and 21-3/21+ (3x2+ 2)-1/2r(x) = (2x+ 1) [lx(3x2

dx + = 2d

and

d 2 -1/2 1 2 -1/2-1(3x + 2) (3x + 2) (6x)

-3x(3x2 + 2r3/2

d 2 1/2 1 2 1/2-1 (6x)(3x + 2) = -- (3x + 2)dx 2

= 3x(3x2+ 2)-1/2

Multiply both numerator and denominator by

and

(3x2 2)41'2

The derivative of the numerator is-3x(2x + 1) + 2(3x2 + 2)ft(x) -

d (3x2+ 2)3/2(2x + 1) = 2dx

Now apply Theorem 5 and

r(c) (3x2

+ 2)1/2(2) -(2x + 1) [3x(3x2 + 2)4/2]

(3x2 + 2)

-6x2 - 3x + 6x2 + 4(3x2 + 2)3/2

4 - 3x(3x2

+ 2)3/2Multiply both numerator and denominator by which agrees with the solution of the first

method used.(3x2 + 2)1/2

PROBLEMS: Find the derivatives of theand simplify, then following:

r(x) = 2(3x2 + 2) - 3x(2x + 1)(3x2

+ 2)31'2

6x2 + 4 - Ox2 - 3x(3x2 + 2)3/2

=4 - 3x(3x +

1. f(x) = NrZ

2. f(x) = 4-317

3. f(x) = acc - 4

4. f(x) =31r

- 3x + 2

ANSWERS:

To find the same solution, by a different1. ormethod, use equation (4) 2frc 2x Ng

1f(x) = (2x + 1)(3x2

+ 2)-1/2 2. - or - le-2-12?-

200

2103


3 3 3x - 43' 2 V-57:7 or 2(3x - 4)

but

8x - 34 r---0

or- 3x + 2)1 therefore,

LIE = 2(x + x2)(1 + 2x)(8x - 3) 3I4x2 - 3x + 2 dx

3(4x2 - 3x + 2)

CHAIN RULE

A frequently used rule in differential calculusis the chain rule. This rule links togetherderi-vatives which have related variables. The chain andrule is

dy sly dudx du dx

when the variable y depends on u and u inturadepends on x.


y = (X + X2)2

SOLUTION: Let

and

Then

and

Now,

u = (x + x2)

y = u2 then

dy = 2udu and by substitution

and

and

EXAMPLE: Find ca wheredx

y = 12t4 + It

t = x2 + 4

SOLUTION: By the chain rule

sly dtdx dt dx

48t3 + 7dt

dtdx= 2x

= (48t3 + 7)(2x)dx

dx du dx dx

du = 1 + 2xdx

and substituting into equation (5) gives

az du= + 2x)

dx du dx

(5)

201

20f4y,;

E.IY. = [48(x2 + 4)3 + (2x)dx

dyPROBLEMS: Find in the following:aTc

1, y = 3t3 + 8t and

t = x3 + 2

2. y = 7n2 + 8n + 3 and

n = 2x3 + 4x2 + x


ANSWERS: dxThe reciprocal of Ty- is the derivative of1. [9(x3 + 2)2 + 8] (3x2) the direct function, and we find

2. [14(2x3 + 4x2 + x) + 8] (6x2 + 8x + 1) 1 y3dx dx y - 2

INVERSE FUNCTIONS dy

Theorem 7. The derivative of an inverse dvEXAMPLE: Find the derivative of thefunction is equal to the reciprocal of the deriva- drtive of the direct function. function

In the equations to this point, x has been2the independent variable and y has been the x = y

dependent variable. The equations have beenin a form such as dx d4SOLUTION: Find to be ai = 2ydy

= x2 + 3x + 2Suppose that we have a function like

then

1 1

dx dxdy

dy_dy. PROBLEMS: Find the dertwative of theand we wish to find the derivative No- dxdx 6 following functions:tice that if we solve for y in terms of x, usingthe quadratic formula, we get the more cora- 1, x = 4 - y2plicated function

2. x = 9 + y2Y = 2x

If we call this function the direct function,then

ANSWERS:

11. - 2y

1 1 1

Y Y

is the inverse function. It is easy to deter- IMPLICIT FUNCTIONS

mine a- from the inverse function.dx In equations containing x and y, it is notEXAMPLE: Find the derivative a of the always easy to separate the variables. If we

function do not solve an equation for y, we call y an1 1 implicit function of x. In the equation

x = -Y Y x2 - 4y = 0

dxSOLUTION: Find the derivative thus y is an implicit function of x, and x is alsocalled an implicit function of y. If we solveddx

+1 this equation for y, that iscTI 4-

Y Y

-2 1x2y =4

Y Ythen y would be called an explicit function ofx. In many cases such a solution would befar too complicated to handle conveniently.

Chapter 13-DERIVATIVES

When y is given by an equation such as ANSWERS:

x2 + xy2 = 0

y is an implicit function of x.Whenever we have an equation of this type

in which y is a function of x, we can dif-ferentiate the function in a straightforwardmanner. The derivative of each term con-taining y will be followed by cd4c. Refer totheorem 8.

EXAMPLE: Obtain the derivative 5-17- of thefollowing

dx.

X2 +2 = 0

SOLUTION: Find the derivative

d 2Tx-

2x

and the derivative

2dx

.(xy ) = xtcy; + (y2)(1)

Therefore,

d (x2 +2) = 2x + 2xy 2

dx dx

Solving for we find that

-2xy 2x + y2

and

1-5x4 - 4y3

12xy2 - 15y4

2.

3.

IL- 2xy

x2 + 3y2

TRIGONOMETRIC FUNCTIONS

If we are given

y = sin U

we may state that, from the general formula,

lim sin(u + Au) - sin uAu-0 Au

sin u cosAu + cos u sthAu - sin uAu-.0 Au

Urn sin u(cosAu - 1) lim sinAu cos uAu-0 Au Au-0 Au

It can be shown that

and

(8)

Um cosAu - 1 0Au-0 Au (7)

lim sinAulu-.0 Au

= 1 (8)

Therefor, by substituting equations (7) and(8) into equation (8)

Thus, whenever we differentiate an implicit = COB U (9)

function, the derivative will usually Isontainterms in both x and y.

PROBLEMS: Find the derivative of theNow, we are interested in finding the deriva-2-

folbwing: tive of the function sin u so we apply thechain rule

1. x5 + 4xy3 - 3y5

2: 2

2. x3y2 = 3

1=crahli3. x2y + y3 = 4

203

!ge6


and from the chain rule and equation (9) we and substituting into equation (10) find thatfind

cos2 u + sin2 udu(sin u) = cos u duax- cos2 udx

In words, this states that to find the deriva-tive of the sine of a function, we use the cosineof the function times the derivative of thefunction,

By a similar process we find the deriva-tive of the cosine function to be

dx (cos = - sin u aidu-

1

cos u

sec2 u

Now, using the chain rule and equation (11)we find

dx du dx

The derivatives of the other trigonometric 2 du= sec ufunctions may be found by expressing them in dxterms of the sine and cosine. That is

dx (tan u) = ci (sin u\dx \cos AV

By stating the other trigonometric func-tions in terms of the sine and cosine andusing similar processes, the following deri-vatives may be found to be

and by substituting sin u for u, cos u for v, duand du for dx in the expression of the quotient cv u) = cos u(sin -c-ii

theorem

we have

du dvv uc_(1) dx _arlvz

(sin u)du du cos u

(cos u) = - sin u aidudx -

dTX- (tan u) = sec2 dx

2 du(cot u) - cec udx dx

zr (sec u) = sec u tan u

(10) du(clic u) - cec u cot ucos u E (sin u) - sin u E (cos u) dx

Taking

and

EXAMPLE: Find the derivative of the func-tion

= sin 3xati (sin u) = cos u SOLUTION:

= cos 3x (3x)dx dx(cos u) - sin udu SI 3 cos 3x


EXAMPLE: Find the derivative of the func- Combining all of these, we find thattion

y = tan2 3x= (2 tan 3x)(sec2 3xX3)dx

SOLUTION: Use the power theorem and = 6 tan 3x sec2 3x

= 2 tan 3x d (tan 3x)dx cTi

PROBLEMS: Find the derivative of the fol-lowing:

then find 1. y = sin 2x

2. y = (cos x2 )2(tan 3x) = sec2 3x (3x)

dx dxANSWERS:

and1. 2 cos 2x

mdiax, =dx 2. - 4x cos x2 sin x2

205

CHAPTER 14

INTEGRATION

The two main branches of calculus are dif-ferential calculus and integral calculus. Havinginvestigated differential calculus in previouschapters, we now turn our attention to integralcalculus. Basically, integration is the inverseof differentiation just as division is the inverseof multiplication, and as subtraction is the in-verse of addition.

DEFINITIONS

Integration is defined as the inverse et dif-ferentiation. When we were dealing with dif-ferentiation, we were given a function F(x) andwere required to find the derivative of thisfunction. In integration we will be given thederivative of a function and will be requiredto find the function. That is, when we are giventhe function f(x), we will find another functionF(x) such that

dit1.1cdx " (1)

In words, when we have the function f(x), wemust find the function F(x) whose derivative isthe function f(x).

If we change equation (1) to read

dr(x) = f(x) dx (2)

we have used dx as a differential. An equiva-lent statement for equation (2) is

F(x) f f(x) dx

We call f(x) the integrand, and we say F(x) isequal to the indefinite integral of f(x). Theelongated S, that is,f, is the Litegral sign. Thissymbol is used because integration may beshown to be the limit of a sum.

INTERPRETATION OF AN INTEGRAL

We will use the area under a curve for theinterpretation of an integral. It should berealized, however, that an integral may repre-sent many things, and it may be real or abstract.It may represent plane area, volume, or surfacearea of some figure.

AREA UNDER A CURVE

In order to find the area under a curve, wemust agree on what is desired. In figure 14-1,where f(x) is equal to the constant 4, andthe °curve° is the straight line

y = 4

The area of the rectangle is found by multiplyingthe height times the width. Thus, the area underthe curve is

A =4(b a)

Tha next problem will be to find a methodfor determining the area under any curve, pro-vided that the curve is continuous. in figure14-2, the area under the curve

y = f(x)

between points x and x Ax is approximatelyf(x)Ax. We consider that Ax is small and the areais given to be .A A. This area under the curveis nearly a rectangle. The area AA, under thecurve, would differ from the area of the rec-tangle by the areaof the triangle ABC if AC werea straight line.

When Ax becomes smaller and smaller, thearea of ABC becomes smaller at a faster rate,and ABC finally becomes indistinguishable froma triangle. The area of this triangle becomesnegligible when A x is sufficiently small.

206

12(9

4

Chapter 14INTEGRATION

Figure 14-1.Area of a rectangle.

X x+ax

Figure 14-2.Area AA.

Therefore, for sufficiently small values of ax we

can say that

X

AA As f(x)Ax

Now, if we have the curve in figure 14-3, thesum of all the rectangles will be approximatelyequal to the area under the curve andbounded by

the lines at a and b. The difference between theactual area under the curve and the sum of the

areas of the rectangles will be the sum of the

areas of the triangles above ekich rectangle.As Ax is made smaller and smaller, the sum

of the rectangular areas will approach thevalue

Figure 14-3.Area of strips.

of the area under the curve. The sum of thoareas of the rectangles may be indicated by

A = urn f(x) Ax (3)n.4 k=1

where Z (sigma) is the symbol for sum, n is thenumber of rectangles, f(x)Ax is the area of eachrectangle, and k is the designation number ofeach

rectangle. In the particular example just dis-cussed, where we have four rectangles, we would

write

4AL f(xu.) Ax

k=1 "

and we would have only the sum of four rec-tangles and not the limiting area under the curve.

When using the limit of a sum, as in equa-tion (3), we are required to use extensive alge-braic techniques to find the actual areaunder the

curve.To this point we have been given a choice of

using arithmetic and finding only anapproxima-tion of the area under a curve, or we could use

207

210


extensive algebraic preliminaries and find theactual area.

We will now use calculus to find the areaunder a curve fairly easily.

In figure 14-4, the area under the curve, froma to by is shown as the sum of the areas of

A and A . The notation A means thea c c b a carea under the curve from a to c.

The Intermediate Value Theorem statesthat

aAb f(c) a)where f(c) in figure 14-4 is the function at an in-termediate point between a and b.

We now modify figure 14-4as shown infigure14-5.

When

x = a

aAa = 0

It is seen in figure 14-5 that

aAx + xA (x+ Ax) = aA(x+Ax)

Figure 14-4.Designation of limits.

Figure 14-5.Increments of area at f(c).

therefore, the increase in area, as shown, is

A A =xA(x+ Ax)

but reference to figure 14-5 shows

xA(x+ Ax) = f(c)Ax

where c is a point between a and b.Then, by substitution

= f(c) Ax

or

AA = f(c)Ax

and as PX approaches zero we have

208

AA_

AV..°

= lim f(c)cx

= 1(4


Now, from the definition of integration

aAx = f f(x) dx

= F(x) + C

and

aAa = F(a) + C

but

aAa = 0

therefore

(4)

F(a) + C = 0

and solving for C we have

C = -F(a)

By substituting -F(a) into equation (4) we find

If we let

then

aAx = F(x) - F(a)

x = b

aAb F(b) -F(a)where F(b) and F(a) are the integrals of thefunction of the curve at the values b and a.

The constant of integration C is omitted inequation (5) because when the function of thecurve at b and a is integrated C will occur withboth F(a) and F(b) and will therefore be sub-tracted from itself.

EXAMPLE: Find the area under the curve

(5)

y = 2x - 1

in figure 14-13, bounded by the vertical lines ata and b, and the x-axis.

SOLUTION: We know that

aAb F(a)

9

7

6

5

4

3

2

1

a2 3 4 5 6

X

Figure 14-6.Area of triangle and rectangle.

and find

F(x) = f f(x)clx

= j (2x - 1)dx

= x2 - x (this step will bejustified later)

Than, substituting the values for a and b intoF(x) (that is, xz x) find that when

and when

209

x b= 5

F(b) = 25 - 5= 20


Then by substittu'-g these values in

find thataAb "3) F(a)

aAb = 20 - 0

= 20

We may verify this by considering figure 14-6 tobe a triangle with base 4 and height 8 sitting ona rectangle of height 1 and base 4. 13:1 knownformulas, we find the area under the curve to be20.

CONSTANT OF INTEGRATION

A number which is independent of the variableof integration is called a constant of integration.This is to say that two integrals of the samefunction may differ by the constant of integration.

1NTEGRAND

When we are given a differential (or de-rivative) and we are to find the function whosederivative is the differential we were given,we call the operation integration.

If we have

_2dx x

and are asked to find the function whose de-rivative is this value, x2, we write

dy x2cbc

then

or we write

y = f x2dx

x3 + C

The symbol is the iategral sign, f x2dx is theintegral of x dx, and xh is called the integrand.The C is called the constant of integration.

INDEFINITE INTEGRALS

When we were finding the derivative of a func-tion, we wrote

or

dx

whe:e we say the derivative of F(x) isf(x). Ourproblem is to find F(x) when Nye are given f(x).

We know that the symbol J ...dx is the in-verse of ccly2c- , or when dealing with differentials,the operator symbols d and f are the inverse ofeach other.

That is

F(4 f f(x)dx

and when the derivative of each side is taken,d annulling ft we have

dF(x) = f(x)dx

or where f ...dx annuls Tcx we have

df(x) r itlxdx dx N= JOE)

From this, we find that

d(x3) = 3x2dx

then

f 3x2dx = x3 + C

Also we find ths.t

d(x3 + 3) = 3x2dxthen

f 3x2dx = x3 + 2Again, we find that

d(x3 - 9) 123x2dx

210

-0.1:213


then

f 3x2clx = x3 - 9

This is to say that

and

d(x3 + C) = 2dx

3x2dx = x3 + C

where C is any constant of integration. SinceC may have infinitely many values, then a dif-ferential expression may have infinitely manyintegrals which differ only by the constant.We assume the differential expression has atleast one integral.

Because the integral contains C and C is in-definite, we call

F(x) + C

an indefinite integral of f(x)dx. In the generalform we say

f f(x)clx = F(x) + C

With regard to the constant of integration,a theorem and its converse state:

If two functions differ by a constant, theyhave the same derivative.

If two functions have the same derivative,their difference is a constant.

EVALUATING THE CONSTANT

To evaluate the constant ui integration we willuse the following approach.

If we are to find the equation of a curvewhose first derivative is 2 times the independentvariable x, we may write

= 2xdx

ordy = 2xdx (6)

We may obtain the desired equation for the curveby integrating the expression for dy. That is,integrate both sides of equation (6).If

dy = 2xdx

then

but

and also

therefore

f dY = f 2xdx

f dY =

f 2xclx = x2 + c

y = x2 + C

We have obtained only a general equation of thecurve because a different curve resultsfor eachvalue we assign to C. This is shown in figure14-7. If we specified that

and

x = 0

Y = 6

we may obtain a specific value for C and hencea particular curve.

Suppose that

then

or

y = x2 + CI x = 01 and y = 6

6 = 02 + C

C = 6

By substituting the value 6 into the generalequation, the equation for the particular curveis

y = x2 + 6

which is curve C of figure 14-7.

The values for x and y will determine thevalue for C and also determine the particularcurve.of the family of curves.

In figure 14-7, curve A has a constant equalto -4, curve B has a constant equal to 0, andcurve C has a constant equal to 6.

211

214


Figure 14-7.Family of curves.

EXAMPLE: Find the equation of the curveif its first derivative is 6 times the independentvariable, y equals 2, and x equals 0.

SOLUTION: We may write

and

therefore

Solving for C when

and

.-- = 6xdx

fdy = f 6xclx

y = 3x2 + C

x = 0

y = 2

we have

or

2 = 3(02) + C

C = 2

y = 32 + 2

RULES FOR INTEGRATION

and the equation is

Although integration is the inverse of dif-ferentiation, and we were given rules for dif-ferentiation, we are required to determine theanswers in integration by trial and error. Thereare some formulas which aid us in the determi-nation of the answer.

In this section we will discuss four of therules and hnw they are used to integrate standardelementary forms. In the rules we willlet u andv denote a differentiable function of a variablesuch as x. We will let C, n, and a denote con-stants.

Our proofs will involve remembering thatwe are searching for a function F(x) whose de-rivative is f(x)dx.

Rule 1. f du = u + C

The integral of a differential of a function isthe function plus a constant.

Proof: If

then

and

212

d(u + C)du

d(u + C) = du

ft:111=u+ c

EXAMPLE: Evaluate the integral

fdxSOLUTION: By Rule 1 we have

fdx=x+ C

Chapter 14I4TEGRATION

Rule 2. fadu = a fclu = au + C EXAMPLE: Evaluate the integral

The integral of the product of a constant and f (2x - 5x + 4)dxa variable is equal to the product of the constantand the integral of the variable. That is, a con- SOLUTION: We will not combine 2x and -5x.stant may be moved across the integral sign. Then, by Rule 3NOTE: A variable may NOT be moved acrossthe integral sign. f (2ic - 5x + 4)dx

Proof: If= f 2xdx - f+ 5xdx + f 4dx

d(au + C) = (a) d(u + C)= 2 f xdx - 5 f xdx + 4 f dx

= au2 5x2

then =2 1

C - C2 + 4x + C3

f adu = a f du = au + C2 5 2= x -

2x + 4x + C

EXAMPLE: Evaluate the integralwhere C is the sum of C1, C2, and C3.

f 4dx This solution requires knowledge of Rule 4 whichfollows.

SOLUTION: By Rule 2 n+ 1r uf 4dx = 4 f dx

Rule 4. u uu n + 1

and by Rule 1 The integral of undu may be obtained by add-ing 1 to the exponent, then dividing by this new

f dx = x + C exponent. NOTE: If n is minus 1, this rule isnot valid and another method must be used.

therefore Proof: If

f4dx = 4x + C

Rule 3. f (du+ dv+ dw) = fdu+ fdv+ fdw

=u+ v+ w+ C

The integral of a sum is equal to the sum of thenthe integrals.

Proof: If

then

d(u + v + w + C) = du + dv + dw

f (du + dv + dw) = (u + C1) + + C2)

+ + C3)

=u+v+w+Cwhere

C C1

+ C2 + C3

213

4+ 11n-- + C) =

(n.n + 1 n + 1

= un

un+1f undu = + Cn + 1


f x3 dx

SOLUTION: By Rule 4

3+ 1f x3 (ix + c

x4"T c

)31.6


EXAMPLE: Evaluate the integral 1. f x2dx

dx 2. f 4xdx

SOLUTION: First write the integral 3. f (x3 + x2 + x)dx

dx 4. f 6dx5

5. dxas

f 7x-3dx ANSWERS:

then, by Rule 2 write x31. + C3

'7 f x-3dx2. 2x2 + C

and by Rule 44 3 2x x x

7 f x 3.3dx 4 4' 3 2

4. 6x + C

-2

f 73

fls

= - 72

+ C2x


1 2 \

x2 + x )dx

SOLUTION:

1 2

f + x3)dx

1 2

= f x2dx + f x3dx

3 5T Ix x

= + c + C3 5 +2 3

3 5

2x2 3x3

3 5+ C

is

55.

DEFINITE INTEGRALS

The general form of the indefinite integral

f f(x)dx = F(x) + C

and has two identifying characteristics. First,the constant of integration was required to beadded to each integration. Second, the result ofintegration is a function of a variable and has nodefinite value, even after the constant of inte-gration is determined, until the variable is as-signed a numerical value.

The definite integral eliminates these twocharacteristics. The form of the definite inte-gral is

fbf(x)thc = F(b) + C - [F(a) + C ]

= F(b) - F(0(7)

where a and b are given value.... : ,tice that thePRACTICE PROBLEMS: Evaluate the fol- constant of integration does not appear in the

lowing integrals: final expression of equation (7). In words,

214

Chapte. 4.4.4NTEGRATION

this equation states that the difference of thevalues of

for

and

f f(x)dxa

x = a

x = b

gives the area under the curve defined by f(x),the x axis, and the ordinates where

and

x = a

x = b

Figure 14-8.Upper and lower limits.

where b is the upper limit and a is the lowerlimit.

EXAMPLE: Find the area bounota.d by thecurve

2y = x

UPPER AND LOWER LIMITS the x axis, and the ordinates

In figure 14-8, the value of b is the upper x = 2limit and the value at aisthe lower limit. These andupper and lower limits may be any assignedvalues in the range of the curve. The upper x = 3limit is positive with respect to the lower limitin that it is located to the right (positive in our as shown in figure 14-9.case) of the lower limit. SOLUTION: Substituting into equation (8)

Equation (7) is the limit of the sum of all thestrips between a and b, having areas of f(x)Ax.That is f f(x)dx = F (x) 1 = F(b) - F (a)

a ax=b

urn f(x)Ax = f f(x)dx 3 ax=a a f x`cbc

2

To evaluate the definite integral x3 I 3

3 I 2

fa f(x)dx.2 23

find the function F(x) whose derivative is f(x)dx 3at the value of b and subtract the function at thevalue of a. That is 27 8

bf(x)dx F(x)

a I a

F(b) - F(a)

(8)

,

215

--3- -3-

_ 193 ;

1= 63


Figure 14-9.Area from x = 2 to x= 3.

We may make an estimate of this solution 1:15considering the area desired in figure 14-9 atbeing a right triangle resting on a rectangleThe triangle has an approximate area of

A bh2

(1)(5)2

5

and the area of the rectangle is

A = bh

= (1)(4)

= 4

and

-4 -3 3 4

Figure 14-10.Area under a curve.

SOLUTION: Substituting into equation (8)

f f(x)dx = F(x) I 2a I -2

= F(2) - F(-2)

=2 nf e dx

which is a close approximation of the area -2found by the process of integration.

EXAMPLE: Find the area bounded by the3 2curve

y = X23 -2

the x axis, and the ordinates

x = -2and

x = 2as shown in figure 1440.

216

..219

= [163

Chapter 14--/NTEGRATION

The area above a curve and below thex axis,as shown in figure 14-11, will through integr, -tion furnish a negative answer.

Then when dealing with area as shown infigure 14-12, each of the areas shown must befound separately. The areas thus foundare thenadded together, with area considered as the ab-solute value.

Figure 14-11.Area above a curve.

Figure 14-12.Negative and positive valueareas

EXAMPLE: Find the area between the curve

* x

and the x axis, bounded by the lines

x = -2

and

x = 2

SOLUTION: These areas must be computedseparately; therefore we write

217

0Area A 32 f f(x)dx

-2

0f xdx

-2

x2 0

2 -2

0 - [1]32

-2

and the absolute value of -2 is

Then

1-21 = 2

2Area B = f f(x)dx

0

x2I 22 0

= 2

and adding the two areas A and B we find

A + B Is 2 + 2

21 4

7


NOTE: INCORRECT SOLUTION: If the func-tion is integrated from -2 to 2 the following in-correct result will occur

This is o14-12. Suof makinglution. Athis common

EXAMPLthe curve

2Area = f f(x)dx

-2

2f xdx-2

x2 2

2 -2

1 [Inam.

= 0 (INCORRECTSOLUTION)

bviously not the area shown in figurech an example emphasizes the valuea commonsense check on every so-

sketch of the function will aidsense Judgment.E: Find the total area bounded by

the x axis, and t

and

y = x3 - 9xhe lines

x = -3

as shown in figure 1SOLUTION: The

and below the x axis,the areas separately,using their absolute valTherefore

x = 34-13.area desired is both above

therefore we need to findthen add them togetherues.

0A = f (x3 - 9x)dx1 -3

_x4 9 2102 x I -3

181 11211

81T

+3

fivre 14-13.Positive and negative vahleareas.

The area

and

Then

218

221

3A2 = f

3- 9x)dx

0

3x4 9 2=2 x4 2 1 0

_81 81 ro4 2 "

I 11 II 4 I 4

81 81Al +2

-4

+4

_ 1624

1


PROBLEMS: and

1. Find, by integration, the area under the x = 2curve

y = x + 4

bounded by the x axis and the lines

x = 2

and

x = 7

Verify this by a geometric process.

3. Find the area between the curve

y = x3 - 12x

and the x axis, from

to

x = 3

2. rind the area under the curveANSWERS:

y = 3x2 + 2 1. 42 1-2-

bounded by the x axis and the lines 2. 12

x = 0 3. 39 1/4

2194 "1,

f222

CHAPTER 15

INTEGRATION FORMULAS

In this chapter several of the integrationformulas and proofs are discussed and ex-amples are given. Some of the formulas fromthe previous chapter are repeated because theyare considered essential for the understandingof integration. The formulas in this chapterare basic and should not be considered a com-plete collection of integration formulas. Inte-gration is so complex that tables of integralshave been published for use as referencesources.

In the following formulas and proofs, u, v,and w are considered functions of a singlevariable.

POWER OF A VARIABLE

The integral of a variable to a power isthe variable to a power increased by one anddivided by the new power.

Formula:

f xn dx

Proof:1d - +n + 1

therefore

n+ 1

xn + 1 + n 7i -1

(n + 1)xn+(n + 1)

(n + 1)xn(n + 1)

= xn

_II+ 1f xn dx = + c, n -1

EXAMPLE: Evaluate

f x5 dx

SOLUTION:

f x5 dx x5+ 1=5 + 1

= + C

EXAMPLE: Evaluate

f x-5 dx

SOLUTION:

fx-5 dx =-4

+ c

PRODUCT OF CONSTANTAND VARIABLE

When the variable is multiplied by a con-stant, the constant may be written either be-fore or after the integral sign.

Formula:

faciu=a fdu = au + C

Proof:

d(au + C) = a d + g-a)

= a du

therefore

faduma f du mg au + C

EXAMPLE: Evaluate

f 17 dx

Chapter 15INTEGRATION FORMULAS

SOLUTION:

fl7clx=l7fcIx

= 17x + C

EXAMPLE: Evaluate

f3x4dx

SOLUTION:

f 3x4 dx =3 f x4 dx

SUMS

The integral of an algebraic sum of differen-tiable functions is the same as the algebraicsum of the integrals of these functions takenseparately.

Formula:

f (du + dv dw) = f du + f dv + f dw

Proof:

d(u + v + w + C) = du + dv + dw

therefore

f du +

where

Then

and

f dv + f dw =u+ C1+v+C 2+w+C3

C1

+ C2 + C3 = C

f du+ fdv+ fdw=u+v+w+C

f (du + dv + dw) =fdu+ fdv + fdw

=u+ v+w+ C

221

EXAMPLE: Evaluate

f (3x2 + x) dx

SOLUTION:

f (3x2 + dx = f 3x2 dx + f x dx

= x3 + C1 + z21

3 x2= x2

EXAMPLE: Evaluate

f (x6 + x-3) dx

SOLUTION:

+ C2

f (x5 + x-3) dx = f x5 dx + fx3 dx

6 -2x

X6 X-2 + C

X6 1 f4+

2x

PROBLEMS: Evaluate the following inte-grals:

'224

1. f x6 dx

2. f x-4 dx

3. f 17x2 dx

4. fvr drf7x1/2

6. f (x7 + x6 + 3x3) dx

7. f (6- x3) dxANSWERS:

1.

12.


POWER OF A FUNCTION OF x

The integral of a function raised to a powerand multiplied by the derivative of that functionis found by the following steps:

1. Increase the power of the function by 1.2. Divide the result of step 1 by this in-

creased power.3. Add the constant of integration.Formula:

a+ 1f un du = " + 1 + C n -1n

Proof:fun+ ) n

d c n + 1 u"

= un du

therefore

and

and

Then

du = 2 dx

n = 2

f un du

and by substitution

f (2x - 3)2 (2)dx = (2x - 3)3

When using this formula the integral mustcontain precisely du. If du is not present itmust be placed in the integral and then com-pensation must be made.

EXAMPLE: Evaluate

and

f (3x + 5)2 dx

SOLUTION: Let

u = (3x + 5)

du = 3 dx

un+1

f un du = n + 1 + C We find dx in the integral but not 3 dx. A 3 mustbe included in the integral in order to fulfill therequirements of du.

NOTE: Recall that In words, this means the integral

f (3x + 5)2 dxcbdc {-1 (2x - 3)1 = (3) (1) (2x - 3)2

needs du in order that the formula may be used.

= 2(2x - 3)2Therefore, we write

EXAMPLE: Evaluate f (3x + 5)2 dx3

f (2x - 3)2 (2) dx

SOLUTION: Let

u (2x - 3)-

and recalling that a constant may be carriedacross the integral sign, we write

f (3x 5)2 .x 1a f (3x + 5)2 3 dx

222

;1%.5

4


Notice that we needed 3 in the integral for duand we included 3 in the integral, then com-pensated for the 3 by multiplying the integralby 1/3.Then

3_Is) (3x-,.3- 5)3 + C

= 1(3x + 5)3 + C9

EXAMPLE: Evaluate

f x(2 + x2)2 dx

SOLUTION: Let

u = (2 + x2)

and

Then

du = 2x dx

2 rf x(2 + x2)2 jdx =-2- x(2 + x2)2 dx

.1 r2

2x(2 + x2)2

1 /2 4. 2,3

(2 + x2)3

6 + C

PROBLEMS: Evaluate the followinggrals:

1. f (x2 + 6) (2x) dx

2. fx2 (7 + x3)2 dx

3. f (3x2 + 2x)2 (6x + 2) dx4. f (6x3 + 2x)1/2 (9x2 + 1) dx

5. f (x2 + 712 x dx

ANSWERS:

(x2 + 6)21. + C

(7 + x3)32. + t.;

inte-

223

(3x2 + 2x)33. + C

3

(6x3+ 2x)3/24. + C

15. 4- C2(x2 + 7)

QUOTIENT

In this section three methods of integratingquotients are discussed but only the secondmethod will be proved.

The first method is to put the quotient intothe form of the power of a function. The secondmethod results in operations with logarithms.The third method is a special case in hich thequotient must be simplified in order to use thesum rule.

METHOD 1

If we are given the integral

2x

(9 - 4x2)1/2 cbc

we observe that this integral may be written as

-1/2f 2x(9 - 4x2) dx

and by letting

u = (9 - 4x2)

and

du = -8x dx

the only requirement for this to fit the form

f un du

is the factor for du of -4. We accomplish thisby multiplying 2x dx by -4, giving -8x dx whichis du. We then compensate for the factor -4by multiplying the integral by - 1/4.


Then PROBLEMS: Evaluate the following inte-grals:

2-1/2

22x, dx =-- f 2x(9 - 4x ) dx

r=

1j (-4) (2x)(9 4x2)-1/2

9 A.,-1/2- I f-8x(9 - 4f) `44.'4

21/2

(4) (9÷c )2

2 1/2(9 - 4x ) + C2

EXAMPLE: Evaluate

I.

dx(3 + x-)

SOLUTION: Write

1. f x 1/2 dx+ x2)

2 f dx

3. f dx(3x + 2)5

ANSWERS:

9 1/21. (2 + + C

Ar,

(3x + 1)2/3.

-13. + C12(3x + 2)4

METHOD 2

In the previous examples, if the exponent of9 -1/2 u was -1, that is

f 1-7172 dx = f x(3 + dx

(3 +

Then, let where

fun du

u = (3 + x2)

we would have applied the following.and

du 2x dx Formula:= fc-14)The factor 2 is used in the integral to give du =lnu+C,u 0and is compensated for by multiplying the inte-gral by 1/2, Proof:Therefore

d(ln u + C) = du9 -1/2 -1/2f x(3 + xm) dx =I f 2x(3 + x2) dx therefore

2 1/2(i) amc.

1/2= 13 + x ) + C

224

Plnu+ CEXAMPLE: Evaluate the integral

f.dx


SOLUTION: If we write

fdx = f x-1 dx

we find we are unable to evaluate

f x-1 dx

by use of the power of a variable rule so wewrite

j2idx= hix+

because the 1 dx in the numerator is preciselydu and we have fulfilled the requirements for

= u + C

EXAMPLE: Evaluate

r 2-I 2x + 1 w'

SOLUTION: Let

and

u = 2x + 1

du = 2 dx

then we have the form

therefore

f = ln u + C

r 2 dx = in (2x + 1) + CJ 2x + 1

EXAMPLE: Evaluater 23x + 1 dx

SOLUTION: Letu=3x+ 1

and

du = 3 dx

We find we need 3 dx but we have 2 dx, There-fore, we need to change 2 dx to 3 dx, We do thisby writing

dx = (A) (i) f 23x + 1 2 3 3x + 1

2f 2 3x + 1 dx

3 f 3x 3+ 1 dx

3ln (3x + 1) + C

where the 2/3 is used to compensate for the 3/2used in the integral.Therefore

225

r 2 Air =_2_ ln(3x + 1) + C3x + 1 w' 3


r x3. j dx2 - 3x2

n 3

3 + 2xANSWERS:

11. T1g3x+ 2) + C

12. - ln(5 - 2x) + C

3. - ln(2 - 3x2) + C

4. I ln(3 + 2x4) + C

METHOD 3

In the third method that we will discuss, forsolving integrals of quotients, we find that to

2 28:.


integrate an algebraic function which has a Integrating separately,numerator which is not of lower degree than thedenominator we proceed as follows1 f dx = x + c

1Change the integrand into a polynomial plusa fraction by dividing the denominator into the andnumerator. After this is accomplished, applythe rules available. f dx = -1n(x + 1) + C2

EXAMPLE: Evaluate

r16x2 - 4x - 8 dxJ 2x + 1

SOLUTION: Divide the denominator into thenumerator, then

(16x2 - 4x - 82x + 1

therefore

r xx + 1 dx = x - ln(x + 1) + C

where

dx = f (8x - 6 - dx2 C = C

1+ C2

PROBLEMS: Evaluate the following inte-dx grals:2f 8x dx f 6 dx

r2x+ 1

and, integrating each separately, we have

f 8x dx = 4x2 + C1 3x - 82. j dx

and

and

1. r2x2 + 6x + 5dxx + 1

. f 6x3 + 13x2 + 20x + 23dx- f 6 dx = -6x + C2 3

2x + 3

f 2x +2 1 dx = -1n(2x + 1) + C34. r2lx + 16x + 4

dx.1 3x + 1

ANSWERS:

x2 + 4x + ln(x + 1) + CThen, by substitution, find that

r16x2 - 4x - 8 21.

2.

3.

dx = 4x - 6x - ln(2x + 1) + C2x + 1

where3x - 81n(x) + C

x3 + x2 + 7x + ln(2x + 3) + CC = Cl + C2 + C3

7 2 14. -I x + 3x + ln(3x + 1) + CEXAMPLE: Evaluate

dxCONSTANT TO A VARIABLE POWER

In this section a discussion of twoform.s of aSOLUTION: The numerator is not of lower constant to a variable power is presented. Thedegree than the denominator; therefore we di- two forms are au and ell where u is the variablevide and finq that and a and e are the constants.x A.p f 1

x + 1 `11' x + 1 dx

1=fdx-fi dx

Formula:


Proof: SOLUTION: Let

d(au + C1) = au hi a du u = 2x

then and

fa In a du = a + C1

but ln a is a constant, then

and

faulnadu=lna fall&

ln a f au du = + C1

Then, by dividing both sides by ln a, we have

Ina r 4,,_ au C1

Ina Jau

'""6-. Ina lna

du = 2 dx

The integral should contain a factor of 2 in orderthat

du = 2dx

Thus we add a factor of 2 in the integral and com-pensate by multiplying the integral by 1/2.

Then

and letting therefore

we have

and

1C = 1 32xI f 321c 2 dx = (i) +2

In a

32x

f au du a= + C 2 ln 3 + CIn

f 32x dx = f (2) 32x dx2

= f 3 2x2 dx

EXAMPLE: Evaluate

SOLUTION: Let

f 3x

U = X

du = 1 dx

therefore, by knowing that

f au"

aIn a

and using substitution, we find that3x

f 3x dx = + C

EXAMPLE: Evaluate but we have

f 32x dx

EXAMPLE: Evaluate

f7x1r dx

SOLUTION: Let

and

In order to use

U = X2

du = 2x dx

auf au du = lna 4- C

the integral must be in the form of

227

21/4do

f bx2

f bx2 7x dx


therefore we remove the 7 and insert a 2 by thereforewriting

f 7xbx2 dx = f 4)* 7xbx2

= 1 fi7xbx2dx2 7

= 2xbx2 dx2

2bx2

7 bx2TInb C


1. f 1o2x cbc

2. f 73x

3. f 02 x dx

4. f (3x2 + 1)cbc

ANSWERS:

102x1 2 ln 10 +

73x2. + C

9x2u° 2 ln 9 +

2(3x2 + 1)4' 6 ln 2 + C

We will now discuss the second form of theintegral of a constant to a variable power.

Formula:

Proof:

f eu clu = eu + C

d(eu eu

228

and

fe du =e + C

EXAMPLE: Evaluate

feXdx

SOLUTION: Let

=

du = 1 dx

The integral is in the correct form to use;f e du = e + C

therefore, using substitution, we find

f ex cbc = ex + C

EXAMPLE: Evaluate

f e2x dx

SOLUTION: Let

and

u = 2x

du = 2 dx

We need a factor of 2 in the integral and write

f e2x dx = f e2x dx2

= 1 f e2x 2clx2

and

231

2

EXAMPLE: Evaluate

2fxe`" dx

SOLUTION: Let

u = 2x2

du =4x dx


Here a factor of 4 is needed in the integral,therefore

fxe`d4 dx = r 4 las.j -dr xe dx

1=

4r 4xe

2x2dx

EXAMPLE: Evaluate

x2

ex3

+ C

dx

SOLUTION: Write the integral

2r xj dx = fx2 e-x3 dx

exq"

and let3

U = -X

and

du = -3x2 dx

therefore

3f x2 e-x dx - f -3x2 e3-x

1 -x3= + C

dx


21. f -2xe-x dx

2. fe4x dx

3.fe(2x - 1) dx

4. fex2

229

ANSWERS:

1. e-x2 + C

12. -4- e

4x + C

1 (2x - 1)e

2+ C

4. -e-x2 + C

TRIGONOMETRIC FUNCTIONS

Trigonometric functions, whichcomprise onegroup of transcendental functions, may be dif-ferentiated and integrated in the same fashionas the other functions. We will limit our proofsto the sine, cosine, and secant functions,butwilllist several others.

Formula:

Proof:

and

therefore

f sin u = -cos u + C

d(cos u + C) = -sin u du

d(-cos u + C) = sin u du

f sin u du = -cos u + C

Foftnula:

f cos u du = sin u + C

Proof:

d(sin u + C) = cos u du

therefore


Formula:f sec2u du= tan u + C

Proof:/sin u c)

d(tan u + C)\COB U


and by the quotient rule

d(sin u)+ cos u(cos u) - sin u(-sin u) ducos AV 2cos u but we do not have 3 dx. Therefore, we multiplythe integral by 3/3 and rearrange as follows:

and

du = 3 dx

therefore

=6os2u + sin2u)cos2u

du3 rf sin 3x dx = 3- i sin 3x dx

(---cosi2u) du

= sec 2udu

f sec2u du = tan u + C

To this point we have considered integralsof trigonometric functions which result in func-tions of the sine, cosine, and tangent. Thoseintegrals which result in functions of the co-tangent, secant, and cosecant are included inthe following list of elementary integrals,



f sec2u du= tan u + C

f csc2u du = -cot u + C

f sec u tan u du = sec u + C

f csc u cot u du .-csc u + C

EXAMPLE: Evaluate

f ain 3x dx

SOLUTION: We need the integral in the formof f (3 sin 2x + 4 cos 3x) dx

then

and

1 r3 j sin 3x 3 dx

1 f sin 3x 3 dx = 1 (-cos 3x) + C3 3

1= - cos 3x + C

EXAMPLE: Evaluate

f cos(2x + 4) dx

SOLUTION: Let

u = (2x + 4)

du = 2 dx

Therefore,

f cos(2x + 4) dx = -?2- f cos(2x + 4) dx

1 f cos(2x + 4) 2 dx2

1 .- sm(2x + 4) + C

EXAMPLE: Evaluate

f (3 sin 2x + 4 cos 3x) dx

SOLUTION: We use the rule for sums andwrite


therefore, we let

u = 3x

= f 3 sin 2x dx + f 4 cos 3x cix

Then, in the integral

f 3 sin 2x dx

letu = 2x


and

but we have

du = 2 dx

3 dx

To change 3 cbc to 2 dx we divide by 3 and multiplyby 2, with proper compensation, as follows:

f 3 sin 2x dx = (-1-) (1) f 3 sin 2x dx

= 1 f 1 (3 sin 2x dx)2 3

=1f2sin2xdx2

3= (-cos 2x) + C

2 1

3= - 2 cos 2x + C1

The second integral

f 4 cos 3x dx

with

u = 3x

and

du = 3 dx

is evaluated as follows:

f 4 cos 3x dx = (1)(i) f 4 cos 3x dx

f3 cos 3x dx

4= 3 (sin 3x) + C2

Then, by combining the two solutions, we have

f (3 sir. 2x + 4 cos 3x) dx

= - 4 cos 2x + C1 + sin 3x + C2

3- cos 2x + 4 sin 3x + C2 3

where

and

C1 + C2 = C

EXAMPLE: Evaluate

f sec2 3x dx

SOLUTION: Let

u = 3x

du = 3 dx

We need 3 dx so we write

and

f sec2 3x dx = 1 fsec2 3x dx3

= I f sec2 3x 3 cbc3

1= (tan 3x) + C

3

EXAMPLE: Evaluate

csc 2x cot 2x dx

SOLUTION: Let

u = 2x

du = 2 dx

We require du equal to 2 dx so we write

f csc 2x cot 2x dx = f csc 2x cot 2x six2

f 2 csc 2x cot 2x dx

EXAMPLE:

SOLUTION:

1-2

csc 2x + C

Evaluatef csc2 3x dx

Letu = 3x


and

then

du = 3 dx

f csc2 3x dx = f 3 csc2 3x dx3

1= - cot 3x + C3

EXAMPLE: Evaluate

f sec tan

SOLUTION: Let

-2-

and

then

dx

1du = dx2

f sec 21 tan 3- cix = 2 f 1 sec2

2-E- tan dx2 2 2 2

= 2 sec f + c


1. fcos 4 xdx

2. f sin 5x dx

3. f sec2 6x dx

3. I tan 6x C6

1 .4. -2- sin (6x + 2) + C

15. - cos(2x2) + C

26. - 1- cot 5x + C

7. 9 sec -a- c

TRIGONOMETRIC FUNCTIONSOF THE FORM fun du

The integrals of powers of trigonometricfunctions will be limited to those which may,by substitution, be written in the form

and

fun du

EXAMPLE: Evaluate__

f sing x cos x dx

SOLUTION: Let

u = sin x

du = cos x dx

By substitution

f sin4 x cos x dx = fu4 du

115

5

4. f 3 cos(6x + 2) dx Then, by substitution again, find that

5. f x sin (2x2) 5

5sin5 x + C5

6. f 2 esc2 5x dxtherefore

x x7. f3 sec tan dx f sin4 x cos x dx = sth5 x5

+ CANSWERS:

1 . s in 4x + C

2. - 1 cos 5x + C

232

EXAMPLE: Evaluate

f cos3 x(-sin x) dx


SOLUTION: Let

and

Write

U = COS X

du = -sin x dx

r 3 u4j u du = + C4

and by substitution

u4COS

4 x+ C =4 4 + C

f cos 2x dxsin3 2x

5. f cos3 x sin x clx

6. f sin x cos x (sin x + cos x) dx

ANSWERS:

1 31. sin x + C3

1 52. sin x + C

5

3. sin2 x + C

-1PROBLEMS: Evaluate the following inte- 4. + Cgrals: 4 sin2

2x

1. f sin2 x cos x clx

2. f sin4 x cos x dx

3. f 2 sin x cos x cix

\

233

1 45. - 7-1- cos x + C

6. sin3x - cos3 x + C3

-

4

4

I

1

CHAPTER 16

COMBINATIONS AND PERMUTATIONS

This chapter deals with concepts required forthe study of probability and statistics. Statisticsis a branch of science which is an outgrowth ofthe theory of probability. Combinations andpermutations are used in both statistics andprobability, and they, in turn, involve opera-tions with factorial notation. Therefore, combi-nations, permutations, and factorial notationare discussed in this chapter.

DEFINITIONS

A combination is defined as a possible se-lection of a certain number of objects takenfrom a group with no regard given to order.For instance, suppose we were to choose twoletters from a group of three letters. If thegroup of three letters were A, B, and C, wecould choose the letters in combinations of twoas follows:

AB, AC, BC

The order in which we wrote the letters is ofno concern. That is, AB could be written BAbut we, would still have only one combination ofthe letters A and B. ,

If order were considered, we would referto the letters as permutations and make a dis-tinction between AB and BA. The permutationsof two letters from the group of three letterswould be as follows:

AB, AC, BC, BA, CA, CB

The symbol used to indicate the foregoingcombination will be 3C2 meaning a group ofthree objects taken two at a time. For theprevious permutation we will use 3P2, meaninga group of three objects taken two at a timeand ordered.

An understanding of factorial notation isrequired prior to a detailed discussion of com-binations and permutations. We define theproduct of the integers 1 through n as n frac-torial and use the symbol n! to denote this. Thatis,

31 = 1.2.3

6 I = 123.4.5.6n1 = 1.2.3. (n - 1).n

EXAMPLE: Find the value of 5!SOLUTION: Write

5! = 1.2.3.4.5

= 120

EXAMPLE: Find the value of

5!3 I

SOLUTION: Write

and

then

5! = 5.4.3.2.1

31 = 3.2.1

5! 5442-131 3.21

and by simplification

234

1237

5.4.3.2.1 5.43* 2* 1

= 20

Chapter 16--COMBINATIONS AND PERMUTATIONS

The previous example could have been solved by This theorem allows us to simplify an expres-writing sion as follows:

51 31 4.53! 3!

= 54 = 3! 4.5

Notice that we wrote = 21 3-4.5

5! = 5.4.3.2.1 =

and combined the factors Another example is

3.2.1 (n + 2)1 = (n + 1)1 + 2)

as = n 1 (n + 1)(n + 2)

31 = (n - 1)1n(n + 1)(n + 2)

5! = 4! 5

then

and

51 = 31 4.5

EXAMPLE: Simplify

(n + 3)1n!

EXAMPLE: Find the value of SOLUTION: Write

61 - 4!4!

SOLUTION: Write

then

61 = 41 5-6

then

(n + 3)! = n! (n + 1)(n + 2)(n + 3)

(n. 4_2_1)1 n! (n+ 1)(n + 2)(n + 3)n! n!

= (n + 1)(n + 2) (n + 3)

4! = 411 PROBLEMS: Find the value of problems1-4 and simplify proslems 5 and 6.

6! - 4! 41(5.6 - 1)4! 4!

= (5.6 - 1)

= 29

Notice that 4! was factored from the expression

6! - 4!

THEOREM

If n and r are positive integers, with n grzaterthan r, then

n! = r 1 (r + 1)(r + 2) .n

235

1. 6!

2. 3! 4!

3. 8!11!

5! - 3!4.31

nl5.

6. n!ANSWERS:

1. 7202. 144

2381


3.

4.5.6.

1 n(n - 1) ... (n - r + 1)99019n(n + 1) (n + 2)

n r 1. 2- 3. - rThe denominator may be written as

COMBINATIONS

As indicated previously, a combination is theselection of a certain number of objects takenfrom a group of objects without regard to order.We use the symbol 5C3 to indicate that we havefive objects taken three at a time, without re-gard to order. Using the letters A, B, C, D, andE, to designate the five objects, we list thecombinations as follows:

ABC ABD ABE ACD ACE

ADE BCD BCE BDE CDE

We find there are ten combinations of five ob-jects taken three at a time. We made the se-lection of three objects, as shown, but we calledthese selections combinations. The word com-binations infers that order is not considered.

EXAMPLE: Suppose we wish to know lowmany color combinations can be made fromfour different colored marbles, if we use onlythree marbles at a time. The marbles arecolored red, green, white, and blue.

SOLUTION: We let the first letter in eachword indicate the color, then we list the pos-sible combinations as follows:

RGW RGY RWY GWY

If we rearrange the first group, RGW, to formGWR or RWG we still have the same colorcombination; therefore order is not important.

The previous examples are completely withinour capabilities, but suppose we have 20 boysand wish to know how many different basketballteams we could form, one at a time, from theseboys. Our listing would be quite lengthy and wewould have a difficult task to determine that wehad all of the possible combinations. In fact,there would be over 15,000 combinations wewould have to list. This indicates the need fora formula for combinations.

FORMULA

The general formula for possible combina-tions of r objects from a group of n objects is

1.2.3...r =r!

and if we multiply both numerator and denomi-nator by

(n - r). 2.1

which is

- r)!

we have

n(n - 1) (n - r + 1)(n - r). 2.1nCr r! (n - r). .2.1

The numerator

is

Then

n(n - 1)... (n - r + 1)(n - r). 2.1

n1

n!Cn r - r! (n - r)!

This formula is read: The number of combi-nations of n objects taken r at a time is equalto n factorial divided by r factorial times nminus r factorial.

EXAMPLE: In the previous problem where20 boys were available, how many differentbasketball teams could be formed?

SOLUTION: If the choice of which boy playedcenter, guard, or forward is not considered,we find we desire the number of combinationsof 20 boys taken five at a time and write

236

where

and

n!C* =n r r 1 (n- r)!

n = 20

r 5_

Chapter 16COMBINATIONS AND PERMUTATIONS

Then, by substitution we have

20!ncr 20

c5 5! (20 - 5)1

20!51 15!

15! 16.17.18.19.20151 51

16.17.18.19.205. 4. 3. 2. 1

16.17.3.19.11

= 15,504

EXAMPLE: A man has, in his pocket, asil-ver dollar, a half-dollar, a quarter, a dime, anickel, and a penny. If he reaches into hispocket and pulls out three coins, how many dif-ferent sums may he have?

SOLUTION: The order in not important,therefore the number of combinations of coinspossible is

6C =1

6 3 31 (6 - 3)1

6!31 31

31 4.5.631 3!

4.5.63. 2. 1

4. 51

= 20

This seems to violate the rule, °division byzero is not allowed,' but we define 0! as equal 1.Then

311

31 01 3!

which is obvious if we list the combinations ofthree things taken three at a time.

PROBLEMS: Find the value of problems 1-6and solve problems 7, 8, and 9.

1. 6C2

2.6

C4

3. 15C5

4. 7C7

6C3 + 7C35.

13C6

6. (7C3) (6C3)14C4

7. We want to paint three rooms in a house,each a different color and we may choose fromseven different colors of paint. How many colorcombinations are possible, for the three rooms?

8. If 20 boys go out for the football team,how many different teams may be formed, one ata time?

9. Two boys and their dates go to the drive-in and each wants a different flavor ice creamcone. The drive-in has 24 flavors of ice cream.How many combinations of flavors may theychoose?

ANSWERS:

EXAMPLE: Find the value of

3C3

SOLUTION: We use the formula given andfind that

1.

2.

3.

4.

5'

A

15

15

3,003

1

5C3 3 31 (3 - 3)1

31

156

10031 01 143


7. 35

8. 167,960

9. 10, 626

PRINCIPLE OF CHOICE

EXAMPLE: A man ordering dinner has achoice of one meat dish from four, a choice ofthree vegetables from seven, one salad fromthree, and one dessert from four. How manydifferent menus are possible?

SOLUTION: The individual combinations areas follows:

meat . 4The principle of choice is discussed in re-lation to combinations although it is also, later vegetable 7C4in this chapter, discussed in relation to permu-tations. It is stated as follows: salad

3C1If a selection can be made in n1 ways, andafter this selection is made, a second selection dessert ... C.. 4 1can be made in n2 ways, and after this selec-tion is made, a third selection can be made in The value ofn3 ways, and so forth for r ways, then the r

41selections can be made together in C =4 1 1 ! (4 - 1)!

nln2n3nr ways 4!31

EXAMPLE: In how many ways can a coachchoose first a football team and then a basket-ball team if 18 boys go out for either team?

SOLUTION: First let the coach choose a andfootball team. That is

18!18c11 11! (18 - 11)1

181111 7!

11! 121314.15.16.17.18111 7.6.5.4.3.2.1

= 31, 824

The coach now must choose a basketball teamfrom the remaining seven boys. That is

= 4

7!C7 4 = 41 (7- 4)!7!

41 31

5.6.72.3

= 35

and

-3 1 1! (3 - 1)!

7!7c

5 51 (7 - 5)1 3121

7151 21 = 351 6.7 therefore, there are5! 21

67 (4)(35) (3) (4) = 1680.2 different menus available to the man.

PROBLEMS: Solve the following problems.= 21 1. A man has 12 different colored shirtsand 20 different ties. How many shirt and tieThen, together, the two teams can be chosen in combinations can he select to take on a trip, if

(31, 824) (21) = 668,304 ways he takes three shirts and five ties?

238

.1241


2. A petty officer, in charge of posting thewatch, has in the duty section 12 men. He mustpost three different fire watches, then post fouraircraft guards on different aircraft. How manydifferent assignments of men can he make?

3. If there are 10 third class and 14 secondclass petty officers in a division which mustfurnish two second class and six third classpetty officers for shore patrol, how many dif-ferent shore patrol parties can be made?

ANSWERS:1. 3,410,8802. 27,7203. 19,110

PERMUTATIONS

Permutations are similar to combinations butextend the requirements ofcombinations by con-sidering order.

Suppose we have two letters, A and B, andwish to know how many arrangements of theseletters can be made. It is obvious that the answeris two. That is

AB and BA

If we extend this to the three letters A, B, andC, we find the answer to be

ABC, ACB, MC, BCA, CAB, CM

We had three choices for the first letter, andafter we chose the first letter, we had only twochoices for the second letter, and after thesecond letter, we had only one choice. This isshown in the °tree" diagram in figure16-1. No-tice that there is a total of six different pathsto the ends of the "branches" of the "tree"diagram,

STARTINGPOINT

FIRSTCHOICE

SECONDCHOICE

THIRDCHOICE

0

Figure 16-1."Tree" diagram.

If the number of objects is large, the treewould become very complicated; therefore, weapproach the problem in another manner, usingparentheses to show the possible choices. Sup-pose we were to arrange five objects inas manydifferent orders as possible. We have for thefirst choice six objects.

(6) ( ) ( ) ( ) ( ) ( )

For the second choice we have only five choices.

(6) (6) ( ) ( ) ( ) ( )

For the third choice we have only four choices.

(6) (5) (4) ( ) ( ( )

This may be continued as follows:

(6) (5) (4) (3) (2) (1)

By applying the principle of choice we find thetotal possible ways of arranging the objects tobe the product of the individual choices. That is

6.5.4.3.2.1

and this may be written as

61

This leads to the statement: The number ofpermutations of n objects, taken all together, isequal to nl.

EXAMPLE: How many permutations ofseven different letters may be made?

SOLUTION: We could use the "tree" but thiswould become complicated. (Try it to seewhy.)We could use the parentheses as follows:

(7) (6) (5) (4) (3) (2) (1) = 5040

The easiest solution is to use the previousstatement and write

7P7 = 71

That is, the number of possible arrangementsof seven objects, taken seven at a time, is 7NOTE: Compare this with the number of COM-BINATIONS of seven objects, taken seven at atime.

239

'242


If we are faced with finding the number ofpermutations of seven objects taken three at atime, we use three parentheses as follows:

In the first position we have a choice ofseven objects.

(7) ( ) ( )

In the second position we have a choice of sixobjects

(7) (6) ( )

In the last position we have a choice of five ob-jects,

(7) (6) (5)and by principle of choice, the solution is

7.6.5 = 210

FORMULA

At this point we will use our knowledge ofcombinations to develop a formula for the num-ber of permutations of n objects taken r at atime.

Suppose we wish to find the number of per-mutations of five things taken three at a time.We first determine the number of groups of three,as follows:

5C3

= 10

Thus, there are 10 groups of three objects each.We are now asked to arrange each of these

ten groups in as many orders as possible. Weknow that the number of permutations of threeobjects, taken together, is 31. We may arrangeeach of the 10 groups in 31 or six ways. Thetotal number of possible permutations of 5C3then is

5C3 3! = 10.6

which is written

5C3 31 =5P3

Put into the general form, then= PnCr r! n r

240

and knowing that

then

but

therefore

nCr =

n!nCr r! = r!r! - r)!

n!(n - r)!

nCr r! = nPr

nlnPr - (n r)!

EXAMPLE: How many permutations of sixobjects, taken two at a time, can be made?

SOLUTION: The number of permutations ofsix objects, taken two at a time, is written

6!t6-r2ff6!4!

4 ! 5. 64!

= 5. 6

= 30

EXAMPLE: In how many ways can eightpeople be arranged in a row?

SOLUTION: All eight people must be in therow; therefore, we want the number of permu-tations of eight people, taken eight at a time,which is

8!P8 8 (8 - 8)!810!

(Remember that 0! was defined as equal to 1)then

243

8! 875 5. 4.3.2.10! 1

= 40,320


Problems dealing with combinations and per-mutations often require the use of both formulasto solve one problem.

EXAMPLE: There are eight first class andsix second class petty officers on the board ofthe fifty-six club. In how many ways can theyelect, from the board, a president, a vice-president, a secretary, and a treasurer if thepresident and secretary must be first classpetty officers and the vice-president and treas-urer must be second class petty officers?

SOLUTION: Since two of the eight first classpetty officers are to fill two different offices, wewrite

8P2 8!

(8 - 2)!

8!6!

= 7.8= 56

Then, two of the six second class petty officersare to fill two different offices; thus we write

6P2

= 5.6

= 30

The principle of choice holis inthis case; there-fore, there are

56.30 = 1680

ways to select the required office holdees. Theproblem, thus far, is apermutationproblem, butsuppose we are asked the following: In howmany ways can they elect the office holders fromthe board, if two of the office holders must befirst class petty officers and two of the officeholders must be second class petty officers?

SOLUTION: We have already determinedhow many ways eight things may be takentwo ata time and how many ways six may be takentwo at a time, and also, how many ways theymay be taken together. That is

8P2 = 50

and

then

6P2 = 30

P6P

2= 16808 2

Our problem now is to find how many ways wecan combine the four offices, two at a time.Therefore, we write

241

4C24!

= 2! (4 - 2)!4!

= 2! 2!4.3.2.1_

2. 2

= 6

Then, in answer to the problem, we write= 10,080

8P2 6P2 4C2

In words, if there are 4C2 ways of combiningthe four offices, and then, for everyone of theseways there are 8P2 6P2 ways of arrangingtheoffice holders, then there are

8 2 6 2 4 2ways of electing the petty officers.

PROBLEMS: Find the answers to the fol-lowing.

1. 6P32. 4P3

3. 72 524. In how many ways can six people be seated

in a row?5. There are seven boys and nine girls in

a club. In how many ways can they elect fourdifferent officers designated byA,B,C, and D if:

(a) A and B must be boys and C and Dmust be girls?

(b) two of the officers must be boys andtwo of the officers must be girls?

ANSWERS:1. 1202. 243. 8404. 7205. (a) 3,024

(b) 18,144

.244


In the question "How many different arrange-ments of the letters in the word STOP can bemade?" were asked, we would write

4P4 41

= 24We would be correct since all letters are dif-ferent. If some of the letters were the same,we would reason as given in the following prob-lem.

EXAMPLE: How many different arrange-ments of the letters in the word ROOM can bemade?

SOLUTION: We have two letters alike. Ifwe list the possible arrangements, using sub-scripts to make a distinction between the O's,we have

R 0102M 0102M R 01M 02R M 0102R

R 0201M 0201M R 02M 01R M 0201R

R 01M 02 0102R M 01R M 02 M 0

1R 02

R 02M 01 0201R M 02R MO1

M02R 01

R M 0102 01M R 02 01R 02M M R 0102

R M 0201 02M R 01 02R 01M M R 0201

but we cannot distinguish between the O's andR 0102M and R 0201M would he the same ar-rangement without the subscript. Notice in thelist that there are only half as many arrange-ments without the use of subscripts or a totalof twelve arrangements. This leads to the state-ment: The number of arrangements of n items,where ri, r2, and rk are alike, is given by

n!r 11 r

21 rk I

In the previous example n was equal to fGur andthere were two letters alike; therefore,wewouldwrite

4! 4-3.2.12! 2.1

= 12

EXAMPLE: How many arrangements canbemade using the letters in the word ADAPTA-TION?

SOLUTION: We use

where

and

and

Then

n Ir 1! r 2! . . rk!

n = 10

r1

= 2 (two T's)

r2

= 3 (three A's)

n!rl! r2i rk!

10!2! 3!

2.5.6.7.8.9.101

= 302, 400

PROBLEMS: Find the number of possiblearrangements of the letters in the followingwo rds.

1. DOWN2. STRUCTURE3. BOOK4. MILLIAMPERE5. TENNESSEEANSWERS:1. 242. 45,3603. 124. 2,494,8005. 3,780Although the previous discussions have been

associated with formulas, problems dealingwithcombinations and permucations may be analyzedand solved in a more meaningful way withoutcomplete reliance upon the formulas.

EXAMPLE: How many four-digit numberscan be formed from the digits 2, 3,4, 5, 6, and 7

(a) without repetitions?(b) with repetitions?

SOLUTION: The (a) part of the question is astraight forward permutation problem and we

Chapter 16-COMBINATIONS AND PERMUTATIONS

reason that we want the number of permutationsof six items taken four at a time.

Therefore

p = 6!6 4 (6 - 4)!

=6- 5.4-3.2.1

2. 1

= 360

The (b) part of the question would become quitecomplicated if we tried to use the formulas;therefore, we reason as follows:

We desire a four digit number and find wehave six choices for the first digit. That is,we may use any of the digits 2, 3, 4, 5, 6, or 7in the thousands column which gives us sixchoices for the digit to be placed in the thousandscolumn. If we select the digit 4 for the thousandscolumn we still have a choice of any of the digits2, 3, 4, 5, 6, or 7 for the hundreds column.This is because we are allowed repetition andmay select the digit 4 for the hundreds columnas we did for the thousands column, This givesus six choices for the hundreds column.

Continuing this reasoning, we couldwrite thenumber of choices for each place value columnas shown in table 16-1.

Table 16-1. -Place value choices.

thousands

column

hundreds

column

tens

column

units

column

six

choices

six

choices

six

choices

six

choices

In table 16-1, observe that the total numberof choices for the four digit number, by theprinciple of choice, is

tition. We would be required to start in the unitscolumn because ai. odd number is determined bythe units column digit. Therefore, we haveonlythree choices. That is, either the 3, 5, or 7.For the tens column we have five choices andfor the hundreds column we have four choices.This is shown in table 16-2,

Table 16-2. -Place value choices.

hundreds

column

tens

column

units

column

four

choices

five

choices

three

choices

In table 16-2, observe that there are

4.5.3 = 60

three-digit odd numbers that can be formedfrom the digits 1, 3, 4, 5, 6, and 7, withoutrepetition.

PROBLEMS: Solve the following problems.1. Using the digits 4, 5, 6, and 7, how many

two-digit numbers can be formed:(a) without repetitions?(b) with repetitions?

2. Using the digits 4, 5, 6, 7, 8, and 9, howmany five-digit numbers can be formed:

(a) without repetitions?(b) with repetitions?

3. Using the digits of problem 2, how manyfour-digit odd numbers can be formed, withoutrepetitions?

ANSWERS

6.6.6.6 = 1,296 1. (a) 12(b) 16

Suppose, in the previous problem, we were 2. (a) 720to find how many three-digit odd numbers could (b) 7,776be formed from the given digits, without repe- 3. 180

243

246

CHAPTER 17

PROBABILITY

The history of probability theory dates backto the 17th century and at that time was relatedto games of chance. In the 18th century it wasseen that probability theory had applicationsbeyond the scope of games of chance. Some ofthe applications in which probability theory isapplied are situations with outcomes such as lifeor death and boy or girl. In the present century,statistics and probability are applied to insur-ance, annuities, biology, and social investiga-tions.

The treatment of probability in this chapter islimited to simple applications. These applica-tions will be, to a large extent, based On gamesof chance which lend themselves to an under-standing of basic ideas of probability.

BASIC CONCEPTS

If a coin were tossed, the chance that it wouldland heads up is just as likely as the chance itwould land tails up. That is, the coin has nomore reason to land heads up than it has to landtails up. Every toss of the coin is called a trial.

We define probability as the ratio of the dif-ferent number of ways a trial can succeed (orfail) to the total numbers of ways in which it mayresult. We will let p represent the probabilityof success and q represent the probability offailure.

One commonly misunderstood concept ofprobability is the effect prior trials have on asingle trial. That is, after a coin has beentossed many times and every trial resulted inthe coin falling heads up, will the next toss of thecoin result in tails up? The answer is "notnecessarily," and will be explained later in thischapter.

PROBABILITY OF SUCCESS

If a trial must result in any of n equallylikely ways, and if s is the number of successful

ways and f is the number of failing ways, thenthe probability of success is

where

244

41"'s

s + f = n

EXAMPLE: What is the probability that acoin will land heads up?

SOLUTION: There is only one way the coincan land heads up, therefore s equals one.There is also only one way the coin can landother than heads up; therefore, f equals one.Then

and

s = 1

f = 1

Thus the probability of success is

P = s + f1

1 + 1

1

TThis, then, is the ratio of successful ways in

which the trial can succeed to the total numberof ways the trial can result.

EXAMPLE: What is the probability that adie (singular of dice) willland withathree show-ing on the upper face.

SOLUTION: There is only one favorableway the die may land and there are a total offive ways it can land without the three face up.

Chapter 17PROBABILITY

and

and

s = 1

= 5

P = s + f

1

EXAMPLE: What is the probability of draw-ing a black marble from a box of marbles if allsix of the marbles in the box are white?

SOLUTION: There are no favorable waysof success and there are six total ways, there-fores

and

then

s = 0

f = 6

013 0 + 6

0

= 0

EXAMPLE: What is the probagility ofdrawing a black marble from a box of sixblack marbles?

SOLUTION: There are six successful waysand no unsuccessful ways of drawingthe marble,therefore

and

then

s = 6

f = 0

66 + 0

e"6"

= 1

The previous two examples are the extremesof probabilities and intuitively demonstrate that

the probability of an event ranges from zero toone inclusive.

EXAMPLE: A box contains six hard leadpencils and twelve soft lead pencils. What isthe probability of drawing a soft leadpencil fromthe box?

SOLUTION: We are given

and

then

245

s = 12

f = 6

12P

1182

2=

PROBLEMS:

1. What is the probability of drawing anace from a standard deck of fifty-two playingcards?

2. What is the probability of drawing a blackace from a standard deck of playing cards?

3. If a die is rolled, what is the probabilityof an odd number showing on the upper face?

4. A man has three nickels, two dimes, andfour quarters in his pocket. If he draws a singlecoin from his pocket, what is the probabilitythat:

(a) he draws a nickel?(b) he draws a half-dollar?(c) he draws a quarter?

ANSWERS:

1.

2.

3.

113

1

1

(b) 04

(c) -(4


PROBABILITY OF FAILURE

If. a trial results in any of n equally likelyways, and s is the number of successful waysand f is the number of failures then, as before,

ors + f n

n - s = f

The probability of failure is given by

q =S + f

n - s

A trial must result in either success or fail-ure. If success is certain then p equals one andq equals zero. If success is impossible thenp equals zero and q equals one. By combiningboth eventsthat is, in either casethe proba-bility of success plus the probability of failureis equal to one as shown by

and

then

If, in any ev

then

In the caseof success is

P = s + f

q =S + f

P+cl=s+f+s+f= 1

ent

p + q = 1

q = 1 P

of tossing a coin, the probability

and the probability of failure is

q = 1 P

1

2

EXAMPLE: What is the probability of notdrawing a black marble from a box containingsix white, three red, and two black marbles froma box containing six white, three red, and twoblack marbles?

SOLUTION: The probability of drawing ablack marble from the box is

211

Since the probability of drawing a marble isone, then the probability of not drawing a blackmarble is

q = 1 - p

= 1 -

9IT

PROBLEMS: Compare the following prob-lems and answers with the preceding problemsdealing with the probability of success.

1. What is the probability of not drawing anace from a standard deck of fifty-two playingcards?

2. What is the probability of not drawing ablack ace from a standard deck of playing cards?

3. If a die is rolled, what is the probability ofan odd number not showing on the upper face?

4. A man has three nickels, two dimes, andfour quarters in his pocket. If he draws a singlecoin from his pocket, what is the probability that:

(a) he does not draw a nickel?(b) he does not draw a half-dollar?(c) he does not draw a quarter?

ANSWERS:

246

249


, 2526

3.

4. (a) I(b) 1

5(c)

EXPECTATION

In this discussion of expectation we will con-sider two types. One is a numerical expecta-tion and the other is value expectation.

Numerical Expectation

If you tossed a coin fifty times you would ex-pect the coin to fall heads about twenty-fivetimes. Your assumption is explained by thefollowing definition.

If the probability of success in one trial isp, and k is the total number of trials, then pk isthe expected number of successes inthek trials.

In the above example of tossing the coinfiftytimes the expected number of heads (successes)is

En = pk

where

En = expected number

p = probability of heads (successes)

k = number of tosses

Substituting values in the equation, we find that

1E = () 50n 2

= 25

EXAMPLE: A die is rolled by a player. Whatis the expectation of rolling a six in 30 trials?

SOLUTION: The probability of rolling a sixin one trial is

and the number of rolls is

therefore

k = 30

En = pk

1= (30)

= 5

In words, the player would expect to roll asix five times in thirty rolls.

EXAMPLE: If a box contained seven num-bered slips of paper, eachnumbered differently,how many times would a man expect to draw asingle selected number slip, if he returned thedrawn slip after each draw and he made a totalof seventy draws?

SOLUTION: The probability of drawing theselected number slip in one drawing is

1= "Ff

247

and the number of draws is

therefore

k = 70

En = pk

= 10

Note: When the product of pk is not an inte-ger, we will use the nearest integer to plc.

Value Expectation

We will define value expectation as follows:If, in the event of a successful result, a personis to receive m value and p is the probability ofsuccess of that event, then nip is his value ex-pectation.

If you attended a house party where a doorprize of $5.00 was given and ten people attendedthe party, what would be your expectation? Inthis case, instead of using k for expected num-ber we use m for expected value. That is

250

Eir = pm


where

p = probability of success

m -T. value of prize

and

Ev = expected value

Then, by substitution

= pm

= (A-) $5-00

= $.50

EXAMPLE: In a game, a wheel is spun andwhen the wheel stops a pointer indicates one ofthe digits 1, 2, 3, 4, 5, 6, 7, or 8. The prizefor winning is $16.00. If a person needed a 6 towin, calculate the following:

(a) What is his probability of winning?(b) What is his value expectation?SOLUTION

(a) p 1

1(b) p = and m = $16.00

therefore

Ev = pm = $16.00

= $2.00

PROBLEMS:1. When a store opened, each person who

made a purchase was given one ticket on a chancefor a door prize of $400. At the close of theday 2,000 people had registered.

(a) If you made one purchase, what isyour expectation?

(b) If you made 5 purchases, what isyourexpectation?

2. Each person at a Bingo game purchaseda fifty-cent chance for the jackpot of twentydollars. If fifty people purchased chances,what is each person's

(a) probability of winning?(b) probability of not winning?(c) expectation?

ANSWERS:

1. (a) $.20(b) $1.00

2. (a) 1)-49

(b) 50(c) $.40

COMPOUND PROBABILITIES

The probabilities to this point have beensingle events. In the discussion on compoundprobabilities, events which may affect otherswill be covered. The word °may* is used be-cause included with dependent events and mu-tually exclusive events is the independent event.

INDEPENDENT EVENTS

Two events are said to be independent if theoccurrence of one has no effect on the occur-rence of the other.

When two coins are tossed at the sametime or one after the other, whether one fallsheads or tails has no effect on the way thesecond coin falls. Suppose we call the coinsA and B. There are four ways in which thecoins may fall, as follows:

1. A and B fall heads.2. A and B fall tails.3. A falls heads and B falls tails.4. A falls tails and B falls heads.The probability of any one way for the coins

to fall is calculated as follows:

= 1

and

n = 4

therefore

1p 74-

This probability may be determined by con-sidering the product of the separate probabil-ities; that is

248

251

1p that A falls heads is y1p that B falls heads is -2-

4


and the probability that both fall heads is

1.1_i-4

In other words, when two events are inde-pendent, the probability that one and then theother will occur is the product of their sepa-rate probabilities.

EXAMPLE: A box contains three red mar-bles and seven green marbles. If a marble isdrawn, then replaced and another marble isdrawn, what is the probability that both mar-bles are red?

SOLUTION: Two solutions are offered.First, there are, by the principle of choice,10 10 ways in which two marbles can be se-lected. There are three ways the red marblemay be selected on the first draw and threeways on the second draw and by the principleof choice there are 3 3 ways in which a redmarble may be drawn on both trials. Then therequired probability is

9= 100

The second solution, using the product ofindependent events, follows: The probability of

3drawing a red ball on the first draw is yr) and

the probability of drawing a red ball on the3

second draw is Therefore, the probability

of drawing a red ball on both draws is theproduct of the separate probabilities

3 3 9'-i rrO 0PROBLEMS:1. If a die is tossed twice, what is the prob-

ability of a two up followed by a three up?2. A box contains two white, three red, and

four blue marbles. If after each selection themarble is replaced, what is the probability ofdrawing, in order:

(a) a white then a blue marble?(b) a blue then a red marble?(c) a white, a red, then a blue marble?

ANSWERS:

1

82. (a) 81

DEPENDENT EVENTS

In some cases one event is dependent onanother. That is, two or more events are saidto be dependent if the occurrence or nonoccur-rence of one of the events affects the probabil-ities of occurrence of any of the others.

Consider that two or more events are de-pendent. If pi is the probabilityof a first event,p2 the probability that after the first happensthe second will occur, p3 the probability thatafter the first and second have happened thethird will occur, etc., then the probability thatpalrlodeuveetpntsl pii2.1 happ3pen in the given order is the

EXAMPLE A box contains three white mar-bles and two black marbles. What is theprobability that in two drawsboth marbles drawnwill be black. The first marble drawn is notr eplaced.

SOLUTION: On the first draw the probabilityof drawing a black marble is

2131 5

and on the second draw the probability of draw-ing a black marble is

1

P2 71.

then the probability of drawing both black mar-bles is

249

P P1 P2

2 1

1

10

EXAMPLE: Slips numbered one throughnineare placed in a box. N two slips are drawn,what is the probability that

(a) both are odd?(b) both are even?SOLUTION:(a) The probability that the first is odd is

5P1

252


and the probability that the second is odd is4

P2

therefore, the probability that both are odd is

P = P1 P2

5 4

518

(b) The probability that the first is even is4

131

and the probability that the second is even is3

P2 =

therefore, the probability that both are even is

PROBLEMS: In the following problems as-sume that no replacement is made after eachselection.

1. A box contains five white and six redmarbles. What is the probability of success-fully drawing, in order, a red marble then awhite marble?

2. A bag contains three red, two white, andsix blue marbles. What is the probability ofdrawing, in order, two red, one blue and twowhite marbles?

3. There are fifteen airmen inthe line crew.They must take care of the coffee mess and lineshack cleanup. They put slips numbered 1through 15 in a hat and decide that any who drawsa number divisible by 5 will be assigned thecoffee mess and any who draws a number divis-ible by 4 will be assigned cleanup. The firstperson draws a 4, the second a 3, and the thirdan 11. As fourth person to draw, what is theprobability that you will:

(a) be assigned the coffee mess?

4 3 (b) be assigned the cleanup?ti ANSWERS:

1 i 3-6 1 II-----

A second method of solution involves the use 12of combinations. 770

(a) There are a total of nine slips taken two1at a time and there are five odd slips taken two 3. (a) i

at a time, therefore

5C2 (b)

1

P =9C2

MUTUALLY EXCLUSIVE EVENTS51

21 (31) Two or more events are called mutually ex-= clusive if the occurrence of any one of them

21 (71) excludes the occurrence of the others. Theprobability of occurrence of some one of two

5 or more mutually exclusive events is the sumro of the probabilities of the individual events.It sometimes happens that when one event

(b) There are a total of 9C2 choices andfour has occurred, the probability of another eventeven slips taken two at a time, therefore is excluded, it being understood that we are re-

ferring to the same given occasion or trial.4C2

P For example, throwing a die once can yield: 6, but not both, in the same toss. The

1 7.Dility that either a 5 or a 6 occurs is the6 6- . cf their individual probabilities.

250

253

Chapter 17PROBABILI:TY

EXAMPLE: From a bag containing five whiteballs, two black balls, and eleven red balls, oneball is drawn. What is the probability that it iseither black or red?

SOLUTION: There are eighteen ways inwhich the draw can be made. There are twoblack ball choices and eleven red ball choiceswhich are favorable, or a total of thirteen fav-orable choices. Then, the probability of suc-cess is

13

Since drawing a red ball excludes the drawingof a black ball, and vice versa, the two eventsare mutually exclusive. Then, the probabilityof drawing a black ball is

2131 =

and the probability of drawing a red ball is

Therefore the probability of success is

P 131 + P2

2 11 13

EXAMPLE: What is the probability of draw-ing either a king, a queen, or ajack from a deckof playing cards?

SOLUTION: The individual probabilities are

4king =

queen = 452

4jack =

251

254

Therefore the probability of success is4 4 4

P 52 4.

12

3

EXAMPLE: What is the probability of roll-ing a die twice and having a 5 and then a 3 showor having a 2 and then a 4 show?

SOLUTION: The probability of havinga 5 andthen a 3 show ir,

1 1

p1 = -6

1

and the probability of having a 2 and then a 4 showis

P2 (1)1

Then, the probability of either p1 or p2 is

P =131 +132

1 1

+

1

18

PROBLEMS:1. When tossing a coin, whatisthe probabil-

ity of getting either a head or a tail?2. A bag contains twelve blue, three red, and

four white marbles. What is the probability ofdrawing:

(a) in one draw, either a red or a whitemarble?

(b) in one draw, either a red, white, orblue marble?

(c) in two draws, either a red marblefollowed by a blue marble or a red marble fol-lowed by a red marble?

3. What is the probability of getting a total ofat least 10 points in rolling two dice?

(HINT: You want either a total of 10, 11, or12.)


ANSWERS:

1. 1

2. (a) 11

(b) 1

(c) -54

1

EMPIRICAL PROBABILITIES

Among the most important applications ofprobability are those in situations where wecannot list all possible outcomes. To this pointwe have considered problems in which theprobabilities could be obtained from situationsin terms of equally likely results..

Because some problems are so complicatedfor analysis we can only estimate probabilitiesfrom experience and observation. This is em-pirical probability.

In modern industry probability now plays animportant role in many activities. Qualitycontrol and reliability of a manufactured articlehave become extremely important considera-tions in which probability is used.

RELATIVE FREQUENCY OF SUCCESS

We define relative frequency of success asfollows. After N trials of an event have beenmade, of which S trials are success, then therelative frequency of success is

STs1-

Experience has shown that empirical proba-bilities, if carefully determined on the basis ofadequate statistical samples, can be applied tolarge groups with the result that probability andrelative frequency are approximately equal. Byadequate samples we meana large enough sampleso that accidental runs of "luck," both good andbad, cancel each other. Withenoughtrials, pre-dicted results and actual results agree quiteclosely. On the other hand, applying a probabil-ity ratio to a single individual event is virtuallymeaningless.

For example, table 17-1 shows a small num-ber of weather forecasts from April 1 to April 10.The actual weather on the dates is also given.

Observe that the forecasts on April 1, 3, 4,6, 7, 8, and 10 were correct. We have observedten outcomes. The event of a correct forecasthas occurred seven times. Based on this infor-mation we might say that the probability for

7future forecasts being true is . This num-ber is the best estimate that we can make fromthe given information. In this case, since wehave observed such a small number of outcomes,it would not be correct to say that the estimateof P is dependable. A great many more casesshould be used if we expect to make a good esti-mate of the probability that a weather forecastwill be accurate. There are a great many fac-tors which affect the accuracy of a weatherforecast. This example merely indicates some-thing about how successful a particular weatheroffice has been in making weather forecasts.

Another example may be drawn from in-dustry. Many thousands of articles of a certaintype are manufactured. The company selects100 of these articles at random and subjectsthem to very careful tests. In these tests it isfound that 98 of the articles meet all measure-ment requirements and perform satisfactorily.

98This suggests that iifjo- s a measure of therelizbility of the article.

One might expect that about 98% of all of thearticles manufactured by this process will besatisfactory. The probability (measure ofchance) that one of these articles will be satis-factory might be said to be 0.98.

This second example of empirical probabilityis different from the first example in one veryimportant respect. In the first example all ofthe possibilities could be listed and in the secondexample we could not do so. The selection of asample and its size is a problem of statistics.

Considered from another point of view, sta-tistical probability can be regarded as relativefrequency.

EXAMPLE: In a dart game, a player hitthe bull's eye 3 times out of 25 trials. What isthe statistical probability that he will hit thebull's eye on the next throw?

SOLUTION:

and

hence

2524..

255

N = 25

S = 3

3I- =23


Table 17-1. Weather forecast.

Date Forecast Actual weatherDid the actualforecasted eventoccur ?

1 Rain Rain yes

2 Light showers Sunny No

3 Cloudy Cloudy Ye s

4 Clear Clear Yes

5 Scatteredshowers

Warm and sunny No

6 Scatteredshowers

Scattered showers Yes

7 Windy andcloudy

Overcast and windy Yes

8 Thundershowers Thundershowers Yes

9 Clear Cloudy and rain No

10 Clear Clear Yes

EXAMPLE: Using table 17-2, what is the thereforeprobability that a person 20 years old will liveto be 50 years old?

SOLUTION: Of 95,148 persons at age 20, where81,090 survived to age 50. Hence

P 81'09095,148

= 0.852

EXAMPLE: How many times would a die beexpected to land with 5 or6 showing in 20 trials?

SOLUTION: The probability of a 5 or 6 show-ing is

1p = -s

The relative frequency is approximately equalto the probability

P p

253

-11'56

and

S=

N = 20

Rearranging and substituting, find that

SN

1 S3 20

203

= 7


Table 17-2. Mortality table (based on 100,000 indi-viduals 1 year of age).

Age Number of people

5 98, 382

10 97, 180

15 96, 227

20 95, 148

25 93, 920

30 92, 461

35 90, 655

40 88, 334

45 85,255

50 81, 090

55 75, 419

60 67, 777

65 57, 778

70 45,455

75 31,598

80 18, 177

85 7, 822

90 2,158

(NOTE: The number observed in an experi- 3. Using table 17-2, find the probability thatment may differ from that predicted; therefore, a person whose age is 30 will live to age 60.the results may be taken to the nearest integer.) ANSWERS:

1. 1PROBLEMS:2. 15

1. In a construction crew there are six elec- 3. 0.733tricians and 38 other workers. How many elec-tricians would you expect to choose if you chose DATA PROCESSINGone man each day of a week for your helper?

2. How many times would a tossed die be Data processing is an extremely large andexpected to turn up 3 or less in thirty tosses? complex field and applications are usually made

254

72' 5 7


for individual situations. For a general under-standing of how data processing can be relatedto probability and statistics, the functional op-eration of the computer is needed.

COMPUTER OPERATIONS

High speed computers are used in dataprocessing because they are able to solve prob-lems in seconds where humans may requiremonths and even years to solve the sameproblem.

A problem which arises, when using a com-puter, is how the human can communicate withthe computer. This communication is a functionof mathematics. We refer to this form of mathe-matics as computer-oriented mathematics.

Digital computers are high speed adding ma-chines. To perform multiplication they makerepeated additions. To perform subtraction theaddition sequence may be reversed, and.divisionis the process of repeated subtraction.

An example of an operation of a digitalcomputer is finding the square root of 36. Weknow that the sum of the first n odd integersis equal to n2. The computer is programed tosubtract successive odd integers from 36 untilzero is reached. That is,

36 - 1 - 3 - 5 - - 9 - 11 0

When zero is reached, the computer then countsthe number of odd integers it has subtracted andthis sum is the square root of 36. For a furtherdiscussion on computers and number systemsrefer to Mathematics, Volume 3, NavPers 10073.

APPLICATION TO PROBABILITY

Many problems may be solved on computerswith the use of the proper mathematical model.The mathematical model may include any of thevariables and the probability with which theyoccur. The collection of statistical data playsan important part in building a mathematicalmodel for the computer.

The mathematical model may be used to de-termine probabilities by the use of computers.That is, the computer will 'play a game° manytimes and give a result comparable to manytrials for determining probabilities.

To understand how a game of chance may beused to produce a useful result, consider the

problem of determining the product of 3/8 and2/3. Place eight ping-pong balls of which threeare coated with a conducting material in onecontainer. Place three other ping-porg ballsof which two are coated with the conducting ma-terial in a second container. A trial consists ofa detecting head from the computer touching aball in each container. If both balls are coateda point is registered; if not, azero is registered.The number of points registered divided by thetotal trials, if the number of trials is large, willclosely approximate the fraction 6/24.

The reasoning for the result is that theprobability of touching a coated ball in thefirst container is 3/8 and the probability oftouching a coated ball in the second containeris 2/3. The events of touching a coated ball inthe first container and one in the second con-tainer are independent, and the probability of bothballs being coated is the product of the individualprobabilities. This is exactly what we set out todetermine.

The preceding example is extremely simpli-fied, in comparison with the complexity of theactual statistical problems solved by computers.However, it does serve to indicate some of thepossibilities of computer-oriented mathematics.

USAGE OF STATISTICAL DATA

Suppose a squadron, through years of opera-tion, has accumulated statistical data on theoperation of an aircraft. By using a computer,the probability of the failure of an engine can bedetermined from the many bits of informationregarding individual parts of the engine. A re-lated problem is to determine how the enginefailure probability can be decreased.

While changiag all of the components of anaircraft engine to try to improve efficiency isunsound, the mathematical model, from a se-lected group of experiments, may be used topredict the change in efficiency for any combi-nation of component changes.

One solution is by trial and error, but thiswould take years of time. Another solution is tomake a few flights using various configurationsof the engine and then use the computerto simu-late years of operation. Researchers would thendetermine the probability of failure, based on thenew 3tatistical data obtained from the few flights.By this high speed determination of the failureof the engine with different combinations of na

255

258


parts, the optimum design of the modified enginemay be determined.

The field of data processing with computersis just beginning and many new techniques are

being developed. It is hoped that this briefintroduction will stimulate the reader tosearch further for a better understanding ofthis new field.

256

259

APPENDIX I

LOGARITHMS OF TRIGONOMETRIC FUNCTIONS

3804 sin Diff.1' csc tan Diff.

1' cot sec Diff.1 , cos 4-14 i. °

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46421

0t

12804 cos Diff.1 '

sec cot Diff.I' tan cso Diff.

i ,t

sin .51o

257

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Abscissas, 122Acute angles, ratio for, 26Addition of vectors, 78Algebraic operatims, logarithms,Angle between two lines, 127Angles:

300 - 60°, 4245°, 43cofunctions, 53cosine of, 30, 47coterminal, 32degree system, 17greater than 90°, 47quadrantal, 45radians, 18reference, 36sine of, 47special, 42standard position, 31tangent of, 48

Angular velocity, 21Antilogrithms, 6Applications:

equilthrium, 100logarithms, 15probability, 255

Area:formula, 74of a sector, 22under a curve, 206

Basic concepts of probability, 244

INDEX

Common logarithms:characteristics, 3fractions, 4

14 integers, 3mantissa, 3

Complementary angles, 53Composite circle diagram, 50Compound probabilities, 248Computer operations, 255Constant:

derivation of, 193of integration, 210to a variable power, 226

Constant and variable, product of, 220Coordinates:

converting Cartesian to polar, 162polar, 83rectangular, 31

Cosine:graph of, 51law of, 56of an angle, 30, 47

Coterminal angles, 32Curve:

direction of, 165slope of, 165

Cartesian to polar coordinates, 162Center of gravity, 99Chain rule of differentiation, 201Changing bases, logarithms, 7Circle:

defined by three points, 140diagrams, composite of, 50general, 138motion in, 174

Cofunctions and complementary angles,Co logarithms, 12Combinations, 236Combining vectors, 78

Definite integrals, 214Definition of:

combinations and permutations, 234limit, 179

Derivative of a constant, 193Differentiation:

chain rule, 201general, 188implicit functions, 202inverse functions, 201powers of functions, 195products, 197quotients, 198radicals, 199sums, 196

53 trigonometric functions, 203variables, 194

Direction of a curve, 165Discontinuities, 186

270

273

INDEX

Distance:between two points, 122of a point from a line, 132

Division:logarithms, 11of a line segment, 123of vectors, 89

Ellipse:equation, 152general, 149locus of points, 152standard form, 153

Empirical probabilities, 252Equation:

ellipse, 152hyperbola, 157parametric, 174straight line, 128tangent and normal, 171trigonometric, 118

Equilibrium:applications, 100first condition, 92rotational, 98second condition, 98translational, 92

Events:dependent, 249independent, 248mutually exclusive, 250

Expectation:numerical, 247value, 247

First condition of equilibrium, 92Formula:

area, 74combinations, 236limit, 183parabola, 148permutations, 240reduction, 107

Free body diagrams, 95Frequency of success, relative, 252Functions, transcendental, 117Fundamental identities, 103

General differentiation, 188Graph of the:

cosine, 51sine, 51tangent, 51

Gravity, center of, 99

Hyperbola, 155

Identities, 103Implicit functions, differentiation, 202Inclination and slope, 124Increment and differentiation, 188Indefinite integrals, 210Independent events, 248Indeterminate forms, 182Infinitesimals:

conclusions, 186products, 185sums, 185

Integrand, 210Integration:

area und3r a curve, 206constant, 210definite integrals, 214definitions, 206evaluation of constant, 211general, 206indefinite, 210integrand, 210interpretation, 206limits, 215rules, 212

Integration of:constant to a variable power, 226power of a function of x, 222power of a variable, 220product of constant and variable, 220quotient, 223sums, 221trigonometric functions, 229, 232

Interpolation, logarithms, 9Inverse functions:

differentiation, 201relations, 111trigonometric, 108

Laws of:cosines, 56sines, 55tangents, 72

Length of subtangents and subnormals, 172Limit:

concept, 179definition of, 179formulas, 183lower, 215upper, 215

Limited solutions, trigonometric, 120Line:

equations:normal form, 130


Line: (continued)equations: (continued)

parallel and perpendicular, 130point-slope form, 128slope-intercept form, 129

inclination and slope, 124midpoint, 124segment, division of, 123

Lines, parallel and perpendicular, 126Logarithmic solutions, triangles, 68Logarithms:

definitions, 1division, 2forms, 1multiplication, 1powers, 2rules, 1

Mantissa, 4Measures with radians, 20Measuring angles, 17Midpoint of a line, 124Mils, 19Motion in a:

circle, 174straight line, 174

Multiplication:logarithms, 9vectors, 88

Mutually exclusive events, 250

Natural logarithms, 7Negative characteristics, logarithms, 14Numerical expectation, 247Oblique triangles, solutions:

three sides, 61two angles and one side, 59two sides and an opposite angle, 64

sides and included angle, 63Operations, computer, 255Ordinate, 122

Parabola, 144Parallel and perpendicular lines, 126, 130Parametric equations, 174, 175Periodicity of functions, 51Permutations, 239Point-slope form, 128Polar coordinates, 83, 161Polar to Cartesian coordinates, 163Power of a:

function of x, 222variable, 220

Powers of functions, differentiation, 198Principal values, 108

Principle of choice, 238Probability:

application of, 255basic concepts, 244compound, 248of failure, 246of success, 244

Product of constant and variable, 220Products, differentiation, 197Properties of triangles, 23Pythagorean Theorem, 23

Quadrantal angles, 45Quadrant system, 35Quotients, differentiation, 196

Radians, 18, 20Radicals, differentiation, 199Raising to a power, logarithms, 13Ratios for acute angles, 26Reciprocals and ratios, 104Rectangular coordinates, 31Reduction formulas, 107Reference angle, 36Relations among inverse functions, 111Relative frequency of success, 252Right triangles, 25Rotation, 92Rotational equilibrium, 98Rules of integration, 212

Scalars, 77Second condition of equilibrium, 98Sector, area of, 22Similar triangles, 24Sine:

graph of, 55law of, 55of an angle, 47

SI, pe-intercept form, 129Slope of a curve at a given point, 165Special angles, 42Squared relationships, 105Standard position, angles, 31Statistical data, usage of, 255Straight line, motion of, 174Subnormal, length of, 172Subtangent, length of, 172Subtraction of vectors, 79Sums, difierentiation, 196

Tangent:graph of, 51law of, 72of an angle, 48

2'175

INDEX

Tangents:and normals, 171at a given point, 166

Transcendental functions, 117Translation, 92Translational equilibrium, 92Triangles:

logarithmic solutions, 68properties of, 23similar, 24

Trigonometric:equations, 118formulas:

addition and subtraction,double angle, 115halt angle, 116

functions:definition, 33differentiation, 203inverse, 108logarithms, 6

limited solutions, 120

ratios, 29reciprocals and ratios, 104square relationships, 105tables, 26, 39

Usage of statistical data, 255Use of:

tables, trigonometric, 39trigonometric ratios, 29

Value expectation, 247Variable, power of, 220Variables, differentiation, 194

112 Vectors:addition, 78combining, 78division of, 89multiplication, 88solutions, 80, 81subtraction, 79symbols, 77

Velocity, angular, 21

2

27673