Engineering Mathematics: Loolsolveo Problems

lOOlsoLvEo PROBLEMs IN -

ENGINEERING MATHEMATICS _____ .....

,

.._.....,.,.

Second Edition

JAIME R. TIONG •1

BSCE,.UV 1985 (Summa Cum Laude) College of Engineering First Summa Cum Laude , 1st Placer, PICE National Students' Quiz, 1985 .. Awardee, Outstanding Alumni, UV . ·' Prestaent, Excel First Review and Training Center, Inc. ·· · Past President, Rotary Club of Cebu North Pr~sident, UV Engineering & Arch. Alumi Association ··Del~ate, Rotary International Convention, Chicago, USA r(~mt(l~,. Ramon Aboitiz Foundation Inc. Triennial Awards ' . . ' Former Plant Engineer, University of the Visayas ·· Fo~mE!r Reviewer, Besaviiia Engineering Review Center f,ormer Reviewer, Salazar Institute of Technology ·· Former Faculty, UV College of .Engineering Author, Various Engineering Reviewers

.,..·. UY

·~

ROMEO A. ROJAS Jl-i.

"·'BSEE, CIT 1991 (Cum Laude), BSECE, OT 1996 1st Placer, RME Licensure Examinations, Oc;:tober 1997 8th Placer, REE Licensure Examinations, April 1999 Former Faculty, Cebu Institute of Technology Former Technical Assistant, CIT Automation Center Reviewer, Excel Review Center Author, Various Electrical Engineering Rexiewers

IMPORTANT: Any copy of this book not bearing the signature ofany one ofthe authors or ofthe publisher on this page shall be considered as comingfrom an illegal source.

AUTHOR I PUBLISHER

TABLE OF CONTENTS

•ndamenta~~~

Preface to the First Edition Preface to the Second Edition Dedication

DAY

of Numbers Conversion

1

THEORY: Number 1 Types of Numbers 1 Numerals 1 Digits 2 Real Numbers 2 Imaginary number 2 Complex number 3 System of Numbers 3 Fractions 3 Types of Fractions 3 Composite numbers 4 Prime numbers 4 Fundamental Theorem of Arithmetic 4 Types of Prime Numbers 5 Perfect Number 5 Abundant Number 6 Deficient Number 6 Perfect Number 6 Amicable Number 6 Friendly Number 6 Factorial 6 Significant Figures & Digits 7 Forms of Approximation 7 Conversion 7 Celsius Scale 7 Fahrenheit Scale 7 Kelvin Scale 8 Rankine Scale 8 Degrees, Radians, Grads & Mils 9 Trivia 9 Quote 9 T"EST (50 Problems for 2 hours) SOLUTIONS

15

Noles

20

10

Algebra

2

THEORY: Properties of Integers Properties of Addition Properties of Multiplication Additive Identity Additive inverse Multiplicative Identity Multiplicative Inverse Properties of Equality Properties of Zero Exponents Radical Surd Types of Surds Special Products Proportion Properties of Proportion Least Common Denominator Least Common Multiple Greatest Common Factor Remainder Theorem Factor Theorem Trivia Quote

23 23 23 23 23 24 24 24 24 24 25 25 25 26 26 26 26 27

27 27 27 27 27

TEST (50 Problems for 3.75 hours) SOLUTIONS Notes

•

~@tJc mfal If

DAY Equation,

Theorem & Logarithms

THEORY: Quadratic Equation Nature of Roots 5 Properties of Roots Binomial Theorem Properties of Expansion Pascal's Triangle Coefficient of any term Formula for rth term

28 34 43

3 45 45 46 46 46 46 47 47

Stjm of Coefficients 47 Sum of Exponents 47 Degree of Polynomial/Equation 47 logarithm 47 Common & Natural Logarithms 48 ·Euler's Number 48 Binary Logarithm 48 Properties of Logarithms 48 Trivia 48 48 Quote TEST (40 Problems for 3 hours) SOLUTIONS Notes

49 54 60

DAY4

Mixture, Digit, Motion Problems THEORY: Age Problems Work Problems Mixture Problems Digit Problems Motion Problems Coin Problems Trivia Quote

TEST (65 Problems for4.5 hours) SOLUTIONS Notes 63 63 64

64 64 64

65 65

TEST (40 Problems for 4 hours) SOLUTIONS Notes

66 73 84

'~'f

DAY

II

~

,,' '

l'i.!'' ·' I.

Variation, Problems & Progression THEORY: Clock Problems Variation Problems Diophantine Equations Sequence Series Progression Arithmetic Progression Geometric Progression

Infinite Geometric Progression Harmonic Progression Other related sequences Fibonacci Numbers Lucas Numbers Figurate Numbers Triangular numbers Square numbers Gnomons Oblong numbers Pentagonal numbers Cubic numbers Tetrahedral numbers Cubic numl;>ers Square pyramidal numbers Supertetrahedral numbers Trivia Quote

5 87

88 88

88 88

89 89

89

DAY n Diagram, trmutation, Combination & Probability THEORY: Venn Diagram Combinatorics Fundamental Principle of Counting Permutation Inversion Cyclic Permutation Permutation with Identical Elements Assortment Combination Relation between Permutation And Combination Probability Principles of Probability Mutually Exclusive Events Independent Events Binomial Distribution Odd

89 89 90 90 90 90 90 90 90 90 90 90 90 90 90 90 9t 91 92 100 114

6 117 118 118

118 118 . 119 119 ' 119

119 119 120 120 120 121

121 121

Odd For Odd Against Mathematical Expectation Card Games Probability with Dice Trivia Quote

122 122 122 122 123 124 124


125 131 141

DAY Geometry

1

THEORY: 143 Definition of Geometry 143 Branches of Geometry 144 Basic Postulates of Euclid Basic Geometry Elements and 144 Figures 144 Types of Angles 145 Bisector 146 Units of Angles ·Polygons 146 147 Triangles 147 Other Types of Triangles 148 Quadrilaterals 148 Types of Quadrilaterals 150 Bramaguptha's Theorem 150 Ptolemy's Theorem Areas and Perimeters of 150 Regular Polygons 151 Perimeter 151 Circles Useful Theorems involving . 152 Circles 153 Ellipses 154 Trivia 154 Quote

TEST (50 Problems for 3.75 hours) 155 161 SOLUTIONS Notes 176

DAY Geometry THEORY: Polyhedrons Five Regular Polyhedrons Platonic Solids Prisms Cylinders Pyramids Cones Frustum of Pyramid Frustum of Cone Prismatoid Prismoidal Formula Sphere Zone Spherical Segment Spherical Sector Spherical Pyramid Spherical Wedge Torus Ellipsoid Spheroid Trivia Quote TEST (30 Problems for 2.5 hours) SOLUTIONS Notes

DAY THEORY: Definition of Trigonometry Branches of Trigonometry Classification of Triangles Solution to Right Triangles Pythagorean Theorem Solution to Oblique Triangl(!s Law of Sines Law of Cosines Law of Tangents Trigonometric Identities

8 179 179 179 181 182 182 182 183 183 183 183 184

184 184 184 184

185 185 185 185 186

186 187 191 200

9 203 203 203 204 204 204 204 205 205 205

Exponential Form of the . Fundamental Tngonometnc Function Amb1guous Case Other Parts of Triangle Radius of Inscribed Circle and Circumscribing Circle Plane Areas (Triangles) Plane Areas (Quadrilaterals) Ptolemy's Theorem Important Properties of Triangles Important Points in Triangles Conditions for Congruency Conditions for Similarity Tnv1a Quote TEST (50 Problems for 4 hours) SOLUTIONS Notes

DAY •

Sphencal onometry

=EST (15 Problems ~OLUTION 206 206 207 207 208 209 209 209 209 210 210 210 210 211 217 230

·o

1

THEORY:

I

·. •

~ '"' ·

1

Definition of Spherical Tngonometry Great Circle Small Circle Pole Polar Distance

233 233 234 234 234

Spherical Wedge Spherical Triangle

234 234

Propositions Tnangle of Spherical Solutions to Right Triangle Nap1er's Rules Quadrantal Spherical Triangle Solution to Oblique Triangles Area of Spherical Triangle Terrestrial Sphere Pnme Meridian International Date Line Greenwich Mean Time Coordinated Universal Time Latitude i':lnd Longitude Terrestrial Sphere Constants Trivia Quote

235 235 235 236 236 237 237 237 237 238 238 238 239 239 239

Notes . ~~,: ·

'fi:;;

for 1.5 hours)

..·

DAY

Geometry ~;~~ \~0 ' Points . Lines & ' c· I Ire es

240 242 247

11

/'''i,X;AJtalytic

THEORY Rectangular Coordinates System Distance Formula Distance Between Two Points in space Slope of a Line Angle Between Two Lines Distance Between a Point and a line Distance Between Two Lines Division of Line Segment Area by coordinates Lines Conic sections General Equation of Conics Circles Trivia Quote TEST (50 Problems for 4 .hours) SOLUTIONS Notes

~··· ~:'{t:

DAY

249 250 250 250 251 251 251 252 252 252 253 253 254 255 255 256 261 275

12

1\a.lytic Geometry '" '"• . . Parabola, Ellipse & Hyperbola THEORY: Parabola Ellipse Hyperbola Polar coordinates Trivia Quote TEST (55 Problems for 4 hours) SOLUTIONS Notes

277 279 281 284 285 285 286 292 :l(fT

.j,·'·:·• . r'

.

DAY .

Dlff~re_ntlal

':

13

ulus .(l1m1ts & D · f ) enva IVeS THEORY: Definition of Calculus Limits Theorems of Limits One-Sided Limits Continuity Special Limits Derivatives Algebraic Functions Exponential Functions Logarithmic Functions Trigonometric Functions Inverse Trigonometric Functions Hyperbolic Functions Inverse Hyperbolic Functions Trivia Quote

309 309 309 310 311 311 311 311 312 312 312 312 31.2 312 313 313


314 319 329

.~;;

:.*. 1t<~..

DAY

14

0~,.al Calculus rna/Minima &

Max~

DAY

a~~

338 343 359

15

THEORY: Definition of Integral Calculus Definite and Indefinite Integrals Fundamental Theorem of Calculus Basic Integrals Exponential & Logarithmic Functions Trigonometric Functions Inverse Trigonometric Functions Hyperbolic Functions Trigonometric Substitution Integration by Parts Plane Areas Centroid Length of Arc First Proposition of Pappus Volume Second Proposition of Pappus Work Hooke's Law Moment of inertia Ml:lltiple Integrals Trivia Quote

361 361 361 362 362 362 362 362 363 363 363 364 364 364 365 365 366 366 366 366 366 366

.

mum Minimum Values 331 Max1ma I M1mma 332 Time :Rates . 332 Relat1on between the vanables & maxima I minima values 332 Trivia 337 Quote 337 TEST (35 Problems for 3 hours) SOLUTIONS Notes

.

u.ral Calculus 0':


Time Rates) THEORY:

. ·· .

.~ :·~

• . • .{•.•.·.· ..·.· . . ·. .

DAY

1.£)ifferential .

367 373 391

16

>iEouations

THEORY: Types of DE Order of DE Degree of DE Types of Solutions of DE Applications of DE Trivia Quote

393 394 394 394 395 396 396


DAY

397 401 410

11

THEORY: Complex Numbers 413 Different Forms of Complex Numbers 413 Mathematical Operation of Complex Numbers 414 Matrices 415 Sum 6( two matrices 416 Difference of two matrices 416 Product of two matrices 416 Division of matrices 417 Transpose matrix 417 Cofactor of an entry of a matrix 417 Cofactor matrix . • 417 Inverse matrix 417 Determinants 418 Properties of Determinants 418 Laplace transform 419 Laplace transforms of elementary functions 419 Trivia 420

r"'f"''t"' ~

'

'II


421 428 438

DAY

18

-\{• '!:"''',

~it.'.

Physics

THEORY: Vector & Scalar Quantities Classifications of Vectors Speed and Velocity Distance and Displacement Acceleration Laws of Motion Force Frictional Force

441 441 442 442 442 442 442 443

Centripetal Force Law of Universal Gravitation Work Energy Law of Conservation of Energy Power Momentum Law of Conservation of Momentum Impulse Types of Collisions Coefficient of Restitution Gas laws Properties of Fluids Archimedes Principle Trivia Quote TEST (40 Problems for 3 hours) SOLUTIONS Notes

·'"t'W

·~-1-B~%\,

DAY

·!~lJtingineering . h~rdcs (Statics)

443 443 444 444 445 445 445 445 445 445 445 446 446 446 447 447 448 453 462

19

THEORY: Definition of Terms Branches of Mechanics Conditions for Equilibrium Friction Parabolic Cable Catenary Moment of inertia Mass moment of inertia Trivia Quote

465 465 465 466 466 467 467 468 469 469


470 475 484

;,.:. 5 '~ ',., ~'3',,

-"

DAY

., "::<'•···

~~~t~,£" E~~~~=~ii~~ (Dynamics)

20

THEORY: Types of Rectilinear Translation Horizontal Translation Vertical Translation Free Falling Body Curvilinear Translation Projectile or Trajectory Rotation D'Aiembert's Principle Centrifugal force Banking of Highway Curve Trivia Quote

487 487 488 488 489 489 490 490 491 491 492 492


493 499 511

DAY

21

Strength of Materials THEORY: Definition of Terms Simple St~ess Types of Normal Stress Simple Strain Hooke's Law Stress-Strain Diagram Thermal Stress Thin-Walled Cylinder Torsion Helical springs Trivia Quote

513 513 514 514 514 515 515 516 516 517 517 517


518 522 528

~~~> ,,;~;;:r:>,;: .. ~~~>l: 'tf;~·.·:·~: :. ;~r1:

DAY

< >«;

Engineering mfi!fiebhomy (Simple & Compound Interest

22

THEORY: Definition of Terms Consumers & Producers Goods and Serv1ces Necessity and Luxury Market Situations Demand Supply Law of Supply and Demand Interest Simple Interest Discount Compound Interest Continuous Compounding Nominal & effective rates of interest Trivia Quote

531 531 .532 533 534 534 535 535 536 536' 537


539 545 551

'iJ'''l[''''''

.

DAY

·,#z~!;< ·~-il;;\ '

1Z. ·.'

•

•

::~'~l:t Engmeermg

"foilomy (Annuity, Depreciation, Bonds, Breakeven analysis, etc. THEORY: Annuity Capitalized Cost Annual Cost Bonds. Depreciation Break Even Analysis Legal Forms of Business Organizations Trivia Quote

531

538 538 538

23 553 555 555 555 557 558 558 559 559


I

I :RJ·;~i;t~t£

A. B.

c.

D. E.

F. G.

560 568 580

1ces

GLOSSARY UNITS & CONVERSION PHYSICAL CONSTANTS NUMERATION MATH NOTATION GREEK ALPHABETS DIVISIBILITY RULES

583 625 633 634 634 635 636

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Topics

~ Mon

D D Tue

~

_j D ,-Cl LJ D [_] Theory

Wed

Problems

Thu

Solut1ons

Fri

Notes

Sat

What is a number? A number is an item that describes a magnitude or a position. ~-'

What are the types of numbers? Numbers are classified into two types, namely cardinal numbers and ordinal numbers.

il ~'

,,./

Cardinal numbers are numbers which allow us to count the objects or ideas in a given collection. Example, 1 ,2,3 ... , 1000, 100000 while ordinal numbers state the position of the individual objects in a sequence. Example, First, second, third ..

IIYJ:l.!lt are

numerals?

Numerals CJre symbols, or combination of ·;ymb<)ls wt1ich describe i1 number.

Cardinal and Ordinal Numbers Numerals and Digits System of Numbers -Natural numbers, Integers, Rational numbers, lrrationC!I numbers & imaginary numbers Complex numbers Types of Fractions Composite Numbers Prime Numbers Defective and Abundant Numbers Amicable Numbers Significant Figures and Digits Forms of Approximation Conversion

The most widely used numerals are the Arabic numerals and the Roman numerals. Arabic numerals were simply the modification of the Hindu-Arabic number signs and are written in Arabic digits. Taken singly, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 and in combination 20, 21, 22, ... 1999, ... The Roman numerals are numbers which are written in Latin alphabet. Example MCMXCIV. The following are Roman numerals and their equivalent Arabic numbers: I 1 C = 100 v =5 0=500 X = 10 M "' 1000 L :: 50

=

The Romuns used the following to indicate large nurnbers:

2 l 00 l s21v~d Problems ·in Engineering Mathematics (2"d Edition) by Tiong & Rojas 1.

Bracket - to (11ultiply it by 100 times.

Examples: - 4, -1, 0, 3, 8

3. Rational numbers - are numbers

lVI = 500 2.

which can be expressed as a quotient (ratio) of two integers. The term "rational" comes from the word "ratio".

Vinculum (bar above the humber) - to multiply the number 1000 times.

Examples: 0.5,

v= 3.

5,000

Doorframe- to multiply the number by 1000000 times

fVl = 5, 000,000 What i2J! digit? A digit is a specific symbol or symbols used alone or in combination to denote a number. For example, the number 21 has two digits, namely 2 and 1. In Roman numerals, the number 9 is denoted as IX. So the digits I and X were used together to denote one number and that is the number 9. In· mathematical computations or in some engineering applications, a system of numbers using cardinal numbers was established and widely used. ~re

real numbers?

The number system is divided into two categories namely, real numbers and imaginary number. Real numbers are classified as follows: 1.

, -3, 0.333 ...

In the above example, 0.5 can be

I

What

2

3

Natural numbers - numbers which are considered as the "counting numbers". Examples: 1 ,2, 3 ...

2. Integers- are all the natural number, the negative of the natural numbers and the number zero.

1 expressed as - and -3 can be 2

-6

Day l - Systems of Numbers and Conversion 3 Imaginary number and its equivalent:.

i =~ i2 = -1

The absolute value of a real number is the numerical value of the number neglecting the sign.

i =-i =- ~ i4 = 1 3

For example, the absolute value of- 5 is 5

What is a complex number?

while of -x is x. The absolute value

A complex number is an expression of both real and imaginary number combined. It takes the form of a + bi, where "a" and "b" are real numbers.

examples are rational numbers.

If a = 0, then pure imaginary number is produced while real number is obtained when b = 0.

The number 0.333 ... can also be

What is a system of numbers?

expressed as -,.hence the two

2

express·ed as __!_ and therefore a

3

A system of numbers is a diagram or chart

rational number.

which shows the two sub-classifications of the two basic classifications of numbers, namely real numbers and imaginary numbers.

The number 0.333 ... is a repeating and non-terminating decimal. As a rule, a non-terminating but repeating (or periodic) decimal is always a rational number. Also, ajl integers are rational numbers. 4. Irrational numbers - are numbers which cannot be expressed as a quotient of two integers. Examples:

System of Numbers

/Real Numbers

Imaginary Number

The numbers in the examples above can never be expressed exactly as a quotient of two integers. They are in fact, a non-terminating number with non-terminating decimal.

Ia Iis

either positive or zero but can never be negative. What are fractions? Fractions are numbers which are in the form of

~or a/b,

where a is called!he

numerator which may be any integer while b is called the denominator which may be any integer greater than zero. Fraction is also defined as a part of a whole. What are the types of fractions? 1. Simple fraction - a fraction in which the numerator and denominator are both integers. This is also known as a common fraction. Examples:

..J2, n, e, ...

2

3

6 7

.

2. Proper fraction - is one where the numerator is smaller thai! the denominator.

Examples:

5

2

7' 3

3. Improper fraction - is one where the numerator is greater than the denominator.

What is an imaginary number? An imaginary number is denoted as "i" which is equal to the square root of negative one. In some other areas in mathematical computation, especially in electronics and electrical engineering it is denoted as "j".

What is an absolute value?

'

Examples:

Pi~l
5

12

2' 7

4. Unit fraction - is a fraction with unity for its numerator and positive integer for its denominator.

Day 1 - Systems of Numbers and Conversion 5

!_lt:'O 1 Solved Problems in Engineering Mathematics (2"ct Edition) by Tiong & Rojas

11. Undefined fraction- a fraction with a

1

Examples:

4·

25

5. Simplified fraction - a fraction whose numerator and denominator are 1ntegers and their greatest common factor is 1.

Examples:

"

1

denom1nator of zero. The example below means that 8 is divided by 0, which is an impossibility because nothing can be divided by zero.

8 0

8

Examples: -

2' -11

6. An lnteJer Represented as fraction - a fraction in which the denominator is 1. 2 3 E xamples: -1 , -1 7. Reciprocal- a fraction that results from interchanging the numerator and der.1ominator.

· 12. Indeterminate fraction - a fraction which has no quantitative meaning. Examples:

0

0

13. Mixed number- a number that is a combination of an integer and a proper fraction.

Examples: 4 is the 4 8. Complex fraction -a fraction in which the numerator or denominator, or both are fractions.

5

3

-

Examples: __4_

7'

8

•r""''"''

II

1 2

8 11

Examples: 5-. 9-

-~

reciprocal of

-~1' -13 4

2

9. Similar fractions -two or more simple fraction that have the same denominator.

What is a composite number? Composite numbers are positive integers that have more than two positive whole number factors. It can be written as product of two or more integers, each greater than 1. It is observed that most integers are composite numbers. The number 6 is a composite number because its factors are 1, 2, 3 and 6. The number 1 is the only natural nu"'lber ti1at is neither composite nor prime. What is a prime nu111ber?

Examples:

1

4

g-· ·g-·

!_

9

10. Zero fraction - a fraction in which the numerator is zero. A zero fraction is equal to zero.

Examples:

0

0

2 11

A prime number is an integer qreater than 1 that is divisible only by 1 and it~elf.

2,3, 5, 7, 11, 13, 17, 19,23,29,31, 37, 41,43,47,53, 59,61,67, 71, 73, 79,83, 89, 97,101, 103,107,109, 113,127,131, 137. 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199,211,223, 227,229,233,239,241,251,257,263, 269,271,277,281, 283, 293,307, 311, 3.13, 317,331, 337, 347, 349,353, 359, 367,373,379,383,389,397,401,409, 419,421,431,433,439,443,449,457, 461,463,467,479,487,491,499,503, 509,521,523,541,547,557,563,569, 571,577,587,593,599,601,607,613, 617,619,631,641,643,647,653,659, 661,673,677,683,691,701,709,719, 727,733, 739, 743, 751, 757, 761, 769, 773,787,797,809,811,821,823,827, 829,839,853,857,859,863,877,881, 883,887,907,911,919,929,937,941, 947,953,967,971,977,983,991,997, The number 2 is the only prime number which is an even number. What are the types of prime numbers? Na.tural prime numbers are those that have only two factors; 1 and the number. Twin primes are a set of two consecutive odd primes, which differ by two. The following are twin primes less than 100.

11' 13 17, 19

Symmetric primes are a pair of prime numbers that are the same distance from a given number in a number line. Symmetric primes are also called Euler primes. The following are symmetric primes for the number 1 through 25. Number

According to the fundamental theorem of arithmetic, " Every positive integer greater !h.an 1 is a prime or can be expressed as a unique product of primes and powers of primes". The following 1s a list of the prime numbers less than 1,000.

29, 31 41,43 59, 61 71, 73

3, 5 5, 7

1

2 3 4 5

6 7 8

Symmetric prime None None None

3,5 3, 5, 3, 5,

7 7

11 11; "3, 13

9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

7,11; 5,13 1. 1a a 11

5, 17; 3, 19 11, 13; 7, 15; 5,19 7, 19; 3, 23 11,17.5 ;;_· .. 13, 17, ''i' 9; 7, 23 15, 17; '!,., 19; 3, 29 11,23;6.2~ 3,31 17, 19; 1.l, 23; 7, 29; 5, 31 9,29; 7, 31 17, 23; 11, 29; 3, 37 19, 23; 13, 29; 11,31; 5, 37 13, 31; 7, 37; 3,41 17,29; 13, 33; 5,41; 3,43 19, 29; 17, 31; 11, 37; 7,41; 5, 43 19, 31; 13, 37, 7,43; 3,47

25

Emirp (prime spelled backwards} is a prime number that remains a prime when its digits are reversed. The following are emirps less than 500:

11, 13, 17, 31, 71, 73, 79, 97, 101, 107, 113,131, 149,151, 157, 167, 179,181, 191,199,311,313,337,347,353,359, 373,383,389 Relatively prime numbers are numbers whose greatest common factor is 1. Unique product of power of primes is a number whose factors are prime numbers raised to a certain power. Example of unique product of pow~r of primes:

360

=23 . 32 . 51

What is a perfect number? A perfect number is an integer that is ~ equal to the sL m of all its possible divisors, except the number itself. Example:

6, 28, 496 ...

In the case of 6, the factors or diviscrs .-e 1, 2 and 3. When the factors are added the sum is ~qual to the number itself and shown in the following equation.

1+2+3=6

6 1001 Solved Problems in Engineering Mathematics (2"d Edition) by Tidng & Rojas What are an abundant numbers and deficient numbers?

2P-1 (2P

If the sum of the possible divisors is greater than the number, it is referred to as abundant number.

where:

A defective number is an integer'with the sum of all its possible divisor is less than the number itself. It is also called deficient number.

p and ( 2P

Number

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

Factors Excluding Itself

1 1 1, 2 1 1, 2, 3 1 1, 2, 4 1, 3 1,2, 5 1 1,2,3,4,6 1 1, 2, 7 1,3,5 1, 2, 4, 8 1 1,2,3,6,9

1 1,2,4,5, 10 1, 3, 7 1' 2, 11 1 1,2,3,4,6,8, 12 1, 5

Example: n! = n(n- 1) ... 3, 2, 1

-1) are prime numbers

Tvoe

0 1 1 3 1 6 1 7 4 8 1 16 1 10

D D D D D

9 15 1 21 1 22 11 14 1 36 6

Perfect

D D D D D A D D D D D A D A D D D A D

What is a perfect number? Perfect number is a number that is equal to the sum of its factors excluding itself. They are mathematical rarities that have no practical use. The formula to find a perfect number is a follows:

~

(0 !)(0 + 1) = (0 + 1)!

i (2 -1)

6

3

2 (2 -1)

28_

5

2 (2 -1)

496

The factorial symbol ( ! ) was introduced by Christian Kramp in 1808.

7

8128

What are significant figures or digits?

2

26 (2 -1)

What are amicable numbers? Sum

(n!)(n+1) = (n+1)!

Perfect number

2

integer number, fewer value-carrying (non" zero) digits.

If n = 0, by definition:

Formula

4

The following is a list of the first 25 numbers with its corresponding type, D for deficient and A for abundant.

-1)

Day I - Systems of Numbers and Conversion_ 7

Amicable numbers or friendly numbers refer to two integers where each is the sum of all the possible divisors of the other. The smallest known amicable numbers are 220 and 284. The number 220 has the following factors/divisors: 1, 2, 4, 5, 10, 11, 20, 22, 44, 55, & 110 which when added sums up to 284, while the number 284 has the following divisors 1, 2, 4, 71, and 142 which adds up to 220. There a.re more than 1000 pairs of amicable numbers have been found. Th~ following are the ten smallest pairs of amicable numbers.

220 and 284 1184 and 1210 2620 and 2924 5020 and 5564 6232 and 6368 10744 and 10856 12285 and 14595 17296 and 18416 63020 and 76084 66928 and 66992 What is a factorial? A factorial denoted as n!, represents the product of all positive integers from· t to n, inclusive.

0!(1)=1! 0! =1

Significant figures or digits are· digits that define the numerical value of a number.

Examples: 3.14159 shall be rounded up to 3.1416 3.12354 shall be rounded down to 3.1235 Truncation refers to the dropping of the next digits in order to obtain the degree of accuracy beyond the need of practical calculations.This is just the same as rounding down and truncated values will always have values lower than the exact values. Example: 3.1415 is truncated to 3 decimal as 3. 141 What is

A digit is considered significant unless it is used to place a decimal point. The significant digit of a number begins with the first non-zero digit and ends with the final digit, whether zerc, or non-zero. Examples:

16.72 . 1.672 X 10 0.0016

3

4 significant figures 4 significant figures 2 significant figures

Example 2 is expressed in scientific notation and figures considered significant are 1, 6, 7 and 2 excluding 103 Example 3 has 2 significant figures only because the 3 zeros are used only to place a deciMal. The number of significant digit is considered the place of accuracy. Hence, a number with 3 significant digits is said to have a three place accuracy and a number with 4 significant figures is said to have a four place accuracy.

a conversion?

Conversion is the process of getting the equivalent value in another unit of measure of a certain value with a different given unit of measure. Most conversions can be done conveniently by using a prepared conversion table while other conversions can be done through mathematical computations using formulas. The authors suggest to the user of this book to familiarize the values in the conversion table which is found at the last part of this book labeled as "Appendix A".

to

How convert a temperature in degree Celsius to degree Fahrenheit and vice versa? The unit "Celsius" was named after the Swedish astronomer, Anders Celsius (1701 -1744). In this unit of temperature, the boiling point and freezing point are 100 degrees and 0 degree, respectively.

What are the forms of approximations? There are two forms of approximations, namely rounding and truncation. Hounding of a number means replacing IIi" n11rnber with diH >flier JHirnber having I• ·w• ., •;1qnlfle
The unit "Fahrenheit" was named after the German physicist, Gabriel Daniel Fahrenheit (1686- 1736). In this unit of temperature, the boiling point and freezing point are 212 degrees and 32 degrees, respectively.

8 10.01 Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas Solution: For same reading, F

=C

Day .1 - Systems of Numbers and Conversion · 9 How to convert one unit of an angle to another unit?

Problema There are how many grads in 1200 mils?

9

F=-C+32 'iling point 80

F=-F+32 5 O.BF = -32

70

F = -40°

90

There are four units commonly used to measure an angle. They are degrees, radians, grads and mils.

5 9

60

1:30

100- 0~

_ _ tj·: ''· __ )::"'

oc-

·

20

212-32

0

~--li'oF- 32

-10

Centigrade

Fahrenheit

Scale

Scale

By ratio and proportion: F-32 C·-0 --=-212-32 100-0

c

F-32

-=-100

180 100 C=-(F-32) 180

Absolute temperature may be expressed in Kelvin or in Rankine. Kelvin was named after British physicist, William Thompson (1824-1902) the First Baron, Kelvin.

The formulas for conversion to absolute temperature are as follows:

Solution:

°K= °C+273

0

R=°F+490

Problem: Express the temperature of 60°C to absolute temperature.

9

5

Solution: °K = °C+273 °K=60+273

Problem: Convert 45°C to °F.

Unit Degree Radian Grad Mil

Problem&

5

F=-C+32

or

The following is a tabulation of the unit of angle measurement and the corresponding value in one revolution:

Rankine was named after Scottish engineer and physicist, William John Macquom Rankine (1820-1872).

9

C =~(F-32)

II

How to convert temperature in degree Celsius or degree Fahrenheit to absolute temperature?

The degree is the mostpommonly used measure of an angle. The radian is an angle subtended by an arc equal to the length of the radius of the circle.

°K =333°

5

9 5

F=-(45)+32 F = 113°F

Problema At what temperature will the Centigrade scale and the Fahrenheit scale will have the same reading?

Express the temperature of 150°F to absolute temperature. Solution: 0 ' R=°F+490 0

R = 150+490

0

R = 640°

1200 mils 6400 mils

x=75 grads. Study Appendix B - Prefixes which is found in the last part of this book. Proceed to the next page for your first test. Detach and use the answer sheet provided at the last part of this book. Use pencil number 2 in shading your answer. GOOD LUCK!

'Ol:ribia:

How many mils are there in 90 degrees?

x 90° 6400 mils 360° x =1600 mils

Did you know that... the symbol 1t (pi), which is the ratio of the circumference of a circle to its diameter was introduced by William Jones in 1706 after the initial letter of the Greek word meaning "periphery".

~uote:

Problema How many radians is equivalent to 320 grads? · Solution: _x__ = 320 grads 2Tt radians 400 grads · x = 5.03 radians

9 F=-C+32

x 400 grads

1 Revolution 360 21t 400 6400

Problem:

Solution:

Solution:

Problema 145 degrees is equivalent to how many grads? Solution: X 145° 400 grads = 360"

x = 161.11 grads

"I could hardly ever known a mathematician who was capable of reasoning." -Plato

Dayl -Systems of Numbers and Conversion 11 B.

C. D.

8

2.37 X 10" 0.2371 X 10"9 o.oo2371 x 1o· 11

7. EE Board October 1994

Topics

~ Mon

D Tue

D D Theory

Wed

~

D

Problems

Thu

D D D D

Solutions

Notes

Fri

Cardinal and Ordinal Numbers Numerals and Digits System of Numbers -Natural numbers, Integers, Rational numbers, Irrational numbers & imaginary numbers Complex numbers Types of Fractions Composite Numbers Prime Numbers Defective and Abundant Numbers Amicable Numbers Significant Figures and Digits Forms of Approximation Conversion

Sat

7 + Oi is A.

B. C.

D.

irrational number real number imaginary number a variable

13. EE Board April1993 Express decimally: Seven hundred twentyfive hundred thousandths A.

B.

c. D.

14. EE Board April1993 Express decimally: Four and two tenth

8. ECE Board Marc:h 1996

A.

The number 0.123123123123 ........ is

B. C.

A. B. C. D.

irrational surd rational transcendental

/«"ECE Board April1991 Round off 6785768.342 to the nearest one-tenth.

D.

6785768 6785768.4 6785768.3 None of these

Express 45" in mils A.

B. C. ~.fi:

•• ME Board October 1996

A. B.

c. D.

il

2 3

4 5

,£icE Board April1991 Round off 0.003086 to three significant figures.

D.

34.0

4: ME Board April 199& ·'which number has three significant figures?

A. B. C. D.

0.0014 1.4141 0.01414 0.0141

~1!:CE

Board April1991

'Round off 149.691 to the nP.arest integer A. B.

C.

D.

0.003 0.00309 0.0031 0.00308

A. B. C. D.

Y,ECE Board April1991 Round off 34.2814 to four significant figures. A.

B. C.

34.2814 34.281 34.28

149.69 149.7 150 149

90 57.3 100 45

B.

Express decimally: Fourteen Ten thousandths

C.

A.

Jl'/• CE Board May 1993

B. C. D.

0.0014 0.00014 0.014 0.14

u. ECE Board Marc:h 1996 MCM~CIV is equivalent to what number?

A: B.

C. D.

1964 1994 1984 1974

12. EE Board April 1993 Express decimally: Fourty-Sevenmillionth

&~CECE Board April1991 Round off 2.371 x 10"8 to two significant

A.

figures.

c

2.4 X 10·8

ME Board April1997

10. EE Board April1993

__.x.

A.

80 mils 800 mils 8000mils 80000 mils

What is the value in degrees of 1 radian?

A. "ifow many significant digits do 10.097 have?

0.042 4.02 4.2 0.42

.A$~/ECE Board November 1995

D. A. B. C. D.

0.000725 0.00725 0.0725 0.725

B.

D

0.00000047 0.0000047 0 000047 0. 00000004 7

D.

3200 mils is equal to how many degrees? A.

B. C. D.

45" 90" 180" 270"

18. ECE ~rd November 1995 An angular unit equivalent to 1/400 of the circumference of a circle is called A. B. C. D.

mil degree radian grad

12 lOOi Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas .19. EC:E Board April1999 4800 mils is equivalent to __ degrees.

A. B. C.

D.

135 270 235 142

,.a( ME Board April 199ft

A.

B. C. D.

micro femto tera atto

,:Ji:· If a foot has 12 inches, then how A. B.

c. D.

3 4 6 8

A.

10"2

B.

10-e

C. D.

10"3

D.

2.667° C 1.334°C 13.34° c 37.8° C

u.

EE Board October 1990

:&7: RME Board October 1994

:J~ In a hydrographic survey, a certain

The prefix pico means

point below the surface of the water measures 12 fathoms. It is equivalent to a deep of how many feet?

What is the absolute temperature of the freezing point of water in degree Rankine?

A. B.

c. D.

492

0 460 273

-~ "8 Board October 1994

·What is the Fahrenheit equivalent of 100 degrees Celsius?

A.. B.

c. D.

200 180 212 100

Z31EE Board Aprii199:J The temperature 45° C is equal to

A.

45° F 113° F

B. c. 5rF D.

81° F

Z4;ME Board October 1994 · How many degrees Celsius is 80 degrees Fahrenheit? A. B.

c. D.

13.34 1.334 26.67 2.667

A.· B. C. D.

10"

10"12 of a unit 10-e of a unit 10"15 of a unit 10·9 of a unit

A.

100 feet

B. C. D.

200 feet 400 feet 800 feet

A. B. C.

:&8. ME Bo8rd April1999

The prefix nano is opposite to

A. B. C. D.

mega giga tera hexa

:&9•1 foot is to 12 inches as 1 yard is to

_·_spans.

D.

4 6

c. 9 D.

24

/ :J{). EE Board .Juae 1990 A one-inch diameter conduit is equivalent to

A.

254mm

B. C.

25.4 mm 100mm 2.54mm

D.

57· EE Board October 1994

A. B. C. D.

8x 8x 8X 8x

10"2 102 10"3 10_.

:JS(vVhich of the following is equivalent to hectare?

·'1

A.

a.

C. D.

100 ares 2 acres 1000 square meters 50000 square feet

:J9. Ten square statute miles is equivalent

· to

A. B. C.

w;

sections.

100 5 10 20

The legendary ship, Titanic that sunk in 1912 was estimated to be at the sea bottom at a deep of 18 cables. How deep it is in feet?

40. The land area of the province of Cebu 'is 5088.39 sq. km. This is equivalent to

A. B.

A. B.

c. D.

A. B.

72 60 48 36

441 statute mile 414 nautical mile

' Carry out the following multiplication and express your answer in cubic meter: 8 em x 5 mm x 2in.

:JZ~ How many feet difference is 1 nautical ' mile and 1 statute mile?

A.

12

C. D.

many hands are there in one foot?

:&Cn EE Board October 1994 The micro or ~ means

c.

I)

10 to the 12 h power is the value of the prefix

Day 1 - Systems of Numbers and Conversion 13

How many degrees Celsius is 100 deg~::ees Fahrenheit? B.

II

zs. ME Board October 1996 1

D.

12,000 12,343 12,633 12,960

C.

D.

.u~"'u Board October 1994

. :~s: ME Board October 1994

How many square feet is 100 square meters?

A

B. C.

D.

5088.39 hectares 1964.64 sq. miles 2257907.2 acres 5.08839 acres

,/How many cubic feet is 100 gallons of liquid?

328.1 929 32.81 1076

A. B. C. D.

74.80 1.337 13.37 133.7 /

,.u;·io ~rd Octo.,...1994

:56.

A certain luxury ship cruises Cebu to Manila at 21 knots. If it will take 21 hours to reach Manila from Cebu, the distance traveled by the ship is nearly A. B

847.5km 507 15 statute mile

'

ME Board April1998 How many cubic meters is 100 gallons of liquid? A. B.

1.638 3.785

14 .1001 Solved Problems in Engineering- Mathematics (2nd Edition) by Tiong & Rojas C. D.

0.164 0.378

4~.-ME Board October 1994 .·How many cubic meters is 100 cubic feet of liquid?

C.

D.

49_.ME Board April ~:99s c5'ne horsepower is equivalent to

A. A. B.

C. D.

3.785 28.31 37.85 2.831

4.4•-'f~n (10) cubic meters is equivalent to !l'>

5

10 104

B. C.

D.

Topics

746 watts 7460 watts 74.6 watts 7.46 watts

~ Mon

$0(ME Board October 1994 "'1iow many horsepower is 746 kilowatts?

A. B.

A.

5

B.

10 20 100

C. D.

c. D.

500 74.6 100 1000

0 0

Theory

Problems

45. ME Board Aprii199S The standard acceleration due to· gravity is ·:;..:.

A.

B. C. D.

0 0 0 0 D Tue

r'llow many stere?

32.2 ft/s 2 980 fUs 2 58.3 fUs 2 35.3 fUs 2

Solutions

0

Notes

VVed Thu Fri

I

Cardinal and Ordinal Numbers Numerals and Digits System of Numbers - Natural numbers, Integers, Rational numbers, Irrational numbers & imaginary numbers Complex numbers Types of Fractions Composite Numbers Prime Numbers Defective and Abundant Numbers Amicable Numbers Significant Figures and Digits Forms of Approximation Conversion

Sat

46. ME Board October 1996 A ?kg mass is suspended in a rope. What is the tension in the rope in Sl?

ANSWER KEY

A. B. C. D.

68:67 N 70 N 71 N 72 N

47. A 10-liter pail is full of water. Neglecting the weight of the pail, how heavy is its water content?

A. B.

c. D.

5kg 6.67 kg 10 kg 12.5 kg

2. B 3. c 4. D 5. c 6.A 7. B 8. c 9. c . 10. A 11. 8 12. c 13. 8 ,.,._

48;;'The unit of work in the mks system is .
A.

B.

6

10 7 10

14. c 15. B 16. B 17. c 18. D 19. 8 20. D 21. A 22. c 23. 8 24. c 25. c 26. 8

1. D

l<

27. A 28. B 29. A 30. B 31. A 32. D 33. A 34. D 35. D 36. 8 37. D 38. A 39. c

RATING

40. B 41. c 42. D 43. D 44. B 45. A 46.A 47. c 48. B 49.A 50. D

~

'-~

'

c:J c:J c:J 0

43-50 Topnotcher 30-42 Passer 25-29 Conditional 0-24 Failed IfF AILED, repeat the test.

16 .1061 Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas

Ill

a

The number 10.097 has 5 significant figures.

MCMXCIV

II

El

The number 0.003086 when rounded off to three significant digitsbecomes 0.00309.

II :t·

The number 34.2814when rounded off to four significant digitsbecomes 34.28.

II

0.0014 has two significant figures 1.4141 has five significant figures 0.01414 has four significant figures 0.0141 has three significant figures m-Ans

II

=

M CM . XC IV = 1000 + 900 + 90 + 4 = 1994

Using the formula,

·c=~(F-32) 9

·c=%(100-32)

=

-~ 1000000

Fourty-seven millionth = 0.000047

lEI Seven hundred twenty-five hundred 725 = 0.00725 thousandths = 100000

Ill

Four and two tenth

=4.2

°C=37.8°C

Ell The freezing point of water is equal to 32•F oro•c.

8

The number 2.371 x 10' when rounded off to two significant digitsbecomes 2.4 x 10'8 .

II 7 + Oi = 7 thus, the answer is, "real number".

II Repeating decimal number is a "rational number".

II

II Fourteen ten thousandths=

14

10000 Fourteen ten thousandths= 0.0014

R= "F+460 R=32°+460

0

R =492°R •

Using the formula,

•F =~(•c)+32

45° 6400 mils= 360° x =800 mils X

lrl By ratio and proportion: x 1 rad 360° .= 2Jt rad

.

By ratio and proportion: x 3200 mils 360° = 6400 mils

1 span is equivalent to 9 inches 1 yard = 3 ft = 36 inches, thus . 1 span 36 mches x - - - - :::: 4 spans 9inches

El .

10mm

1 mch = 2.54 em x--- =·25.4 mm 1 em

hand

1 foot= 12 mches x - .-h4 1nc es 1 foot = 3 hands

m

°F=113°F

Let: x = the difference between a nautical mile and a statute mile 6080 - 5280 x = 800 feet X ::

=~(a0-32) 9

oc = 26.67•c

X =270°

m

1 statute miie = 5280 ft. 1 nautical mile = 6080 ft

oc

By ratio and proportion: x 4800 mils 360° = 6400 mils

to o·

The prefix nano is equivalent 1 9 of a unit while the prefix giga is equivalent to 9 10 of a unit . ·

°F=~(45)+32

9

Ill

El

.

·c=~(F-32)

Grad

The prefix pico is equivalent to 10'12 of a unit

1 hand is equivalent to 4 inches, thus

Using the formula,

m

El

°F = 212°F

°F=*(•c)+32

m

10·6 means micro

II

Using the formula,

Ill

m

°F=~(100)+32 5

m

X= 57.3°

X= 180°

The number 6785768.342 when rounded off to the nearest one-tenth becomes 6785768.3.

0 0

El

By ratio and proportion:

The number 149.691 when rounded off to the nearest integer becomes 150.

II

II

Fourty-seven millionth

•

________________D_a...y_l_-_S_y._s_t..,..e_ms __o_f_N_u_m __,bers and Conversion 17

m I hu prefix tera is equivalent to 1012 of a unit

El 1 fathom is equivalent to 6 feet, thus 12 fathoms= 6(12) = 72 feet

Ell 1 cable is squivalent to 120 fathoms, thus: 18 cables= 120(18) = 2160 fathoms 6feet 18 cables= 2160 fathoms x--fathom

18 100 1'Solved Problems in Engineering Mathematics (2"d Edition) by Tiong & Rojas ft3 3

-

18 cables= 12 960 feet '

100 gallons x _ = 13.37 ft 7.48 gallons

1 meter is equivalent to 3.281 ft, thus 2 1 m 2 = (3.281 )2 ft 2 1 m 2 = 10.76 ft 2 100m2 = 100(10.76) ft 2 100m2 = 1076 ft

El

3

3.79 li m 100 gallons x - - x - - gallon 1000 li

=

Iii

x-~-=

m

3.281 ft

m m

m

1 hectare = 100 ares &Core: 1 are= 100 sq. meters 1 1 hectare = 100 ares x - 0.:..0.::_::_sq.:.o.·-.-m.:.. 1 are 1 hectare= 10,000 sq. meters

The unit of force (tension) in the Sl system is newtons (N). 9 Tension= 7 kg m) = 68.67 N

x(

m 1 square statute mile = 1 section 10 square statute mile = 10 sections

m

-~~

El Density of water ( p ) = 1000

_k~ mo

,,~ kg Density of water ( p ) = 1 -.liter

1 square km = 0.386102 sq. miles

1 cubic ft. = 7.48 gallons

3

The following are the standard gravitational acceleration: 2 32.2 ft/s 981 cm/s 2 9.81. m/s 2

o·

A= 5088 _39 km 2 x 0.3861 02" miles km A = 1964.64 sq. miles

x(-m-)

= 100 ft 3 3 V = 2.831 m

V

2

II 1

hp = 746 watts 1 hp = 0.746 kilowatts

746 kW x.

Given volume is 100 cu. ft.

1 cubic meter = .1 stere, thus, 10m 3 = 10 steres

1m 0.8 m 100 em 1 m = 0.005 m 5 mm x 1000 mm 4 3 0.08(0.005)(2) = 8 x 1 m 8 em

3

m

V = 21 knots = 21 nautical miles hour D=21(21) D = nm x 1.15 statute miles 441 nautical mile D 507.15 statute mile

II 1 horsepower = 746 watts

1 gallon = 3.79 liters 1000 liters = 1 cubic meters

100 gallons= 0.379 m

Solving for distance, D = Vt

-

m

Day 1 -Systems of Numbers and Conversion 19

W= p·V

W=1

~ x 10 liters = 10 kg liter

m 7

1 joule= 10 ergs

hp

=1000 hp

22 100 1 Solved Problems in Engineering Mathematics ~2~d Editionl by Tiong &: Roiaa

Topics

0

Mon

Tue

I"

0

Wed

Theory

0 0 0

Problems

Solutions

0

Thu

0 Fri

LJ Sat

Notes

What are properties of integers? Integers have special properties. Computations of integers will become easier by understandir;g these special properti~s. The commutative property, for instance, allows you to change the order of adding or multiplying while the associative property allows you to change grouping. The properties of adpitiori of integers: \.

2. Commutative property

5. Inverse property a +(-a)= 0

The properties of multiplication of integers:

:\ Associative proper1y 1 <

The num 2r 0 is called the additive identiy

a(b+c) = ab:t-ac

a+b=b+a

ll)

a+O=a

6. Distributive property

a + b = integer

1

4. Identity property

The number -a is called the additive inverse

Closure property

(a

Properties of Addition of 1nteger Properties of Multiplication of Integers Properties uf Equality Properties of Zero Properties of Exponents Properties of Radicals Surds Special Products Properties of Proportion Least Common Denominator Least Common Multiple Greatest Common Factor Remainder Theorem Factor Theorem

"

1

II'

Closure properly 1 '

)

o~h

illl••
24

~ 00 I· Solved

Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas

2. Commutative· property

4. Substitution property If a =b, then a can be replaced by b in any expression inl'olving .

ab=ba

3. Associative property

a 5. Addition I Subtraction property

(ab)c=a(bc)

a+1=a ~r

would say that 3 is the power and that "power" and "exponent" mean the same thing. The misuse has probably come from a misunderstanding of statements such "nine is the second power of three". The exponential notation states that if a is a real number, variable or algebraic expression and n is a positive number, then:

If a = b, then a + c = b + c If a = b, then a - c = b - c

4. Identity property

Day 2 -Fundamentals in Algebra 25

5. Inverse property

!

= ~ with c

c c

~0

lfa+c=b+c,thena=b If ac = be and c 0 , then a

*

The number ~a is called the . multiplicative inverse

6. Distributive property

=b

The properties of zero: Consider a, b and c as integers or real numbers or variables of an algebraic expression.

a(b+c)=ab+ac

a+O=a and a-O=a

2.

a(O)=O

3.

~ == 0 ,

4.

~

5.

If ab = 0, then a = 0 or b

7. Multiplication property of zero a(O) =0

The properties ofequality of integers: Consider a, b and c as integers or real numbers or variables of an algebraic expression.

with

2. Symmetric property

=a

3. Transitive property

If a = b and b = c, then a = c

cya ·cyb = 'fab

~ =(Wf =22 ~=4

¥5 ·~675 = ~(5)(675) = ~3375 = 15

3.

- = n - b;t:O % b'

cya~

we= 3rso =¥5 V10 \/10

Property

Example

4.

'ifiFa = mzya

~~=1tfl5

1. am +a"= am+n

x2 + xa = x2+3 = xs

5.

(cya)" =a

(~f =2x

?t

!fa"= lal

~(-12) 4 =l-121

2.

am m-n -=a a"

X

8

-=X x3

a-3

s =X

2 (y6) = y12

3. (am)" =amn

:::12 (For n =even no.)

5.

a ;t 0

4

(2x) =2 4 x 4 =16x 4

(abt =ambm

(~r=24=~ 4 4

(~r = : :

x

m

6.

is undefined

=0. This is

known as Zero-Factor property

Exponent is a number that gives the power to which a base is raised. For example, in 32 , the base is 3 and the exponent is 2.

a=a

cya)

!fa"= lal

What Is an exponent?

·1. Reflexive property

If a = b, then b

a

m

=(

The properties of exponents with corresponding examples:

4. 1. ·.

r,;;

vam

n factors

7. Cancellation property a(;)=1

2.

0

"---y---J

If a = b, then ac = be If a = b, then

1.

Example

a" = a · a · a · a · · · ·

6. Multiplication I Division property

The number 1 is called the multiplicative identiy

Property

Exponent should not be misunderstood as "power" Power is a word that is almost never used in its correct, original sense any more. Strictly speaking, if we write 32 = 9, then 3 is the base, 2 is the exponent and 9 is the power. But almost everyone, including most mathematicians,

7.

8.

a

5 (4x)3 = 3 (4x)

1 =am

(x

2

+2t

index of the radical. For example,

=1

What is a radical? Radical refers to the symbol that indicates

F.

It was first used in 1525 by a root, Christoff Rudolff in his Die Coss.

cya ,

In the expression, n is called the Index, a (the expression inside the symbol) is called the radicand while the symbol

J

is called radical

What is a surd? Surd is a radical expressing an irrational number. The surd is described after the

-5 =1I( x5

a0 = 1 (a;;, 0)

= -12

(For n =odd no.)

x

((44

a"=~ -m

x

~(-12)3

.J3 is a

quadratic surd, ~ is a cubic surd, ~ is a quartic surd and so on. Different types of surds: Pure surd, sometimes called an entire surd contairls no rational number and all its terms are surds. Example:

.J3 + J2 .

Mixed surd is a surd that contains at least

s.J3 is a mixed surd because 5 is a rational number while .J3 is one rational number.

a surd.

26. 1001 Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas Binomial surd is an expression of two ter111s with at least one term a surd. Example: 5 +

Trinomial surd is an expression of three terms with at least two or. them are surds and cannot be expressed as a single surd, otherwise it will become a binomial surd. Example: 5 +

Solution:

What is a proportion? Proportion is a statement that two ratios are equal.

F2

What is a special product?

d

2

2

( x + y) = x + 2xy + y (x -

2

y) = x 2

-

2xy +

2

l

3

3

2

2

=x

3

2

3x y + 3xy2- y

-

3

3

4. Difference of two cubes 3

3

x -y =(x-y)(x

2

+xy+l)

5. Sum of two cubes 3

x +y

3

=(x- y)(x

2

-

xy +

l)

6, Square of a trinomial 2

2

2

2

( x + y + z ) = x + y + z + 2xy :t 2xz + 2yz

GCF = 14 What is a Remainder Theorem?

4.

LCD= 360 What is a least common multiple (LCMI?

5.

6

If~=~ then a- b = c- d b

.

b

d

If~ = ~ then a + b = c + d b

If ·

d. d.

b

d

~ = ~ then a + b = c + d b

d'

~a:x=y:dl

u

means

23 (32 )(5)

d

a c b d 3. If - =- , then - =b d a c

extremes

(x + y) = x + 3x y + 3xy + y

(x- yf

c

d.

a-b

c-'d

antecedent

3. Cube of a binomial

= 32

LCD=

In number ( 1), quantities a and d are called extremes while x and y are called means. If x = y, then its value is known as mean proportional. In the ratio xly, the first term x is called the antecedent while the second term y is called the consequent.

2. Square of a binomial

9

a x 1. If - = - . then a : x = y : d

b

(X + y){X - y) = x2 - y2

=2 3

12=3·2

a c a b 2. If - =- then - =-

1. Sum and difference of same terms or Difference of two squares

8

15=3·5

y

With x, y and z as real numbers or variables or algebraic expression, the following are the special products.

GCF = 2(7)

2

Properties of proportion

F2 + J3

Special products are the expressions where the values can be obtained without execution of long multiplication.

Day 2 - Fundamentals in Algebra 27

r: ·o

consequent

What is a least common denominator (LCD)? Least common denominator (LCD) refers .to the product of several prime numbers occurring in the denominators, each taken with its greatest multiplicity.

A common multiple is a number that two other numbers will divide into evenly. The least common multiple (LCM) is the lowest multiple of two numbers.

Remainder Theorem states that if a polynomial in an unknown quantity x is divided by a first degree expression in the same variable, (x- k), where k may be any real number or complex number, the remainder to be expected will be equal to the sum obtained when the numerical . value. of k is substituted for x in the polynomial. Thus,

remainder = f(X) x->k

What is a Factor Theorem?

Problema What is the least common multiple of 15 and 18? Solution: 15=3·5 18

= 3 .2 2

LCM = 32 (5}(2) LCM

= 90

What is a greatest common factor (GCFI? A factor is a number that divides into a larger number evenly. The greatest common factor (GCF) is the largest number that divides into two or more numbers evenly.

Problema What Is the greatest common factor of 70 and 112? Solution: 70

=2. 5. 7

Problem:

112=2 4 -7

What is the least common denominator of 8,9,12and15? • '

Common factors aro 2 and 7.

Factor theorem states that if a polynomial is divided by (x - k) will result to a remainder of zero, then the value (x- k) is a factor of the polynomial. Both remainder theorem and factor theorem were suggested by a French mathematician, Etienne Bezout (17301783).

Proceed to the next page for your second test. Detach and use the answer sheet provided at the last part of this book. Use pencil number 2 in shading your answer. GOOD LUCK I

'Orribia: Did you know that. .. the two long parallel lin;)S (=)as a symbol for equality was introduced by Robert Recorde in 1557!

~note: "Among the great things which are found among us, the existence of Nothing is the greatest," - Leonardo da Vinci


,.,,}

SS'~CME Board October :1995

D.

Solve for the value of x and y. 4x + 2y = 5 13x -3y = 2

c.o. EE Board Apri11997

A.

Topics

0

Properties of Addition of Integer Properties of Multiplication of Integers Properties of Equality Properties of Zero Properties of Exponents Properties of Radicals Surds Special Products Properties of Proportion Least Common Denominator Least Common Multiple Greatest Common Factor Remainder Theorem Factor Theorem

Mon

:t-

u

Theory

~

Problems

0 0

Solutions

Notes

~ Tue

0 0 0 0

Wed

Thu

Fri

Sat

<,sj~
10[~+~ ]=A

If 16 is 4 more than 4x, find 5x- 1.

A. B. C.

3

D.

5

2[3~-4~]=A

14 12 A.

si~ EE Board October :199:l

x

F .md t h e va Iue o f x .m -+-1 + -2x = 4 7 - 2 x .

3

A. B. C. D.

4

16.47 12.87 18.27 20.17

n: EE Board October :199:1 '"Find the value of x in the equations:

50/9

B. 80/9 C.· 70/9 D.

60/9

54. EE Board October :1997 ' Find the values of x and y from the equations: X - 4y + 2 = 0 2x + y -4 = 0

A. B. C. D.

11/7, -5/7 11/9, 8/9 4/9, 8/9 3/2, 5/3

B·: C. D.

y = 1/2, X = 3/2 y =3/2, x =112 y = 2, X= 1 y = 3, X = 1

x=-2,y=-3,z=-1

Multiply the following: (2x + 5y)(5x- 2y)

A.

B.

C. D.

2

10x - 21xy + 10/ 2 -10x + 21xy + 10x2 + 21xy- 10y -10~- 21xy -10y2

10t

Sb. ME Board October :199&

61. EE Board March 1998

Solve the simultaneous equations:

Determine the sum of the positive valued solution to the simultaneous equations: xy = 15, yz = 35,.zx = 21.

2x 2 -3y2 = 6 3x 2 +2y 2 = 35

A. A. B.

C. D.

x = 3 or -3; x = 3 or -3, x = 3 or -3, x 3 or -3,

=

y = 2 or -2 y = -2 or 1 y = -2 or -1 y = 2 or -3

B.

c: D.

15 13 17 19

c.z. ECE Board April1991 57. ECE Board May :1997 Find the value of w in the following equations: 3x- 2y +w = 11 x + 5y -2w = -9 2x+y-3w = -6

A. B. C. D.

B.

Solve for the value of x. 2x-y+z=6 x-3y- 2z = 13 2x- 3y- 3z = 16

c. D.

6~.

4 3 2 1

x2y1zs

2Y Xy 1

~ Xy 1

~ Xy ECE Board November 199~

5x

-=---- 2x2 + 7x + 3

Solve the simultaneous equations: x+y=-4 x+z-1=0 y+z+1=0

A

y = -5, Z = 3 y =2, Z =-3 ·1, y =: -3, Z = 2 '

C.

X "- -1, X ::: 1,

X ~

t (x-3 yz3 f~

Simplify the following equation

S91 ME Board October 199&

A II t;

2

( xyz-3 )-~

A.

3 2 4 -2

ss. EE Board October 199~

A. B. C .. D.

2 3

Simplify: (x y z-

B.

4 X+3 2

-

x-3

_j_ x-3

x +3 2x 2

-

3x - 2

2x+1

+ -:::--2 x +x- 6


30 1001 Solved Problems in Engineering Mathematics (2"d Edition) by Tiong & Rojas D.

D.

2 X+3

&4, ECE Board April1.99:l s;mplify

~,l [,-!,-l (,,,-Tl

rr

&8. ME Board October :199& X+2 .mto part1a 'Ifract1on. . Reso Ive

A.

5

A. B.

C.

X 3 y2

D.

~ y2

&s. ECE Board April :199:1 Simplify: 7

8

+

2

-

8(7) "" 1 + 5(7)8 + 49(7)8 - 2 8

a

c. ~

A. C.

-5a -3a -7a

D.

-4a

B.

X

B.

2 2 ( b -4b+16) ( b -16) = ~-----:::-'--'-----< b3 +64

A. B. C. D.

b+4 b-4

"'!1/1.

b+2

A.

B.

2

b -4 c. -

D.

C. D.

b+2 b-4

&7. ECE Board April

y = 5" y =9 Y = 52n y = 18

:199~

x-z x+z a+ b

C. D.

3x ( X - 3 )( X + 2 ) 3x ( X + 3 )( X + 2 ) 3x ( X + 3 )( X - 2 ) 3x (X- 3 )(X- 2)

so.

If p - q = 5 and pq =

81 Evaluate: 64x . 4Y. 256"Y 4x+3y 64x+ 3y 43x+y

A. B. C. D.

= 9Y

A. B.

G.

81Y 3-x = 243

-

x2 .

k 25k 25 + k k 25

81.. ME Board Apri11.995

2a -2x

Simplify bm/n

( a + X )( a -X ) ( a + x )( a + x ) 2x-2a

A.

Jbffi

7&. ME Board April1.99&

B.

n bm+n

Factor the expression x 2 + 6x + 8 as completely as possible.

c.

ifbiTi

D.

-

A. B. C.

{ X+ 4 ( X- 4 ( X- 4 0 .. ( X + 6

)( X + 2 ) )( X + 2 ) )( X- 2 ) )( X + 2 )

bm n

82. ME Board Aprill998 Find the value of x which will satisfy the following expression:

77. ECE Board November 1990 (a-b) 3 =? A. LJ.

a

3

a

3

-

3a 2 b + 3ab 2 + b 3

--

3a 2 b- 3ab 2 2

-

b3

<:

a:I + 3a b + 3ab 2

-

b3

Il

;~·

-·

b3

1

~ , then p2 + q2

equals

Give the factors of a 2

729 140 243

1.5

B. C. D.

D.

2

1

A.

12 9 11 10

Solve for x in the following equations.

A. B.

2.25

75. ECE Board November 1.991.

-1~

27x

9

D.

36 2.5

Factor the expression 3x 3 - 3x 2 -18x

Find a: A. B.

A. B. C.

79· ME Board Aprill.995

~=10 am

/(2. ECE Board Aprlll99~

Solve for : _x_ = _Y_ = _z_ Y (b-e) (a-c) (a-b)

A. B. C.

~ 1~

11o. ME Board October 199& The value of (3 to 2.5 power) square is equal to:

&&. Solve for x:

A.

4( 52n+1) _ 10( 52n-1) ,Evaluate: y = 2(5 2n)

Given: (an)( am)= 100,000

+ x 2 +2x+5

A

is a perfect square.

74. ECE Board April :1990

c

~

78. Find the value of k so that 4~ + 6x + k

. 7~d:ECE Board April :199~

A. B. C. D.

&9. CE Board May :199& Find the value of A in the equation: A B(2x + 2) x 2 + 4x + 10 --=--~- = - + -::-'------"x3 + 2x 2 + 5x x x 2 + 2x + 5

y2 3

D.

6 2 x-4 x-3 3 - - -5x-4 x-3 -6- - -5x-4 x-3 7 - - -5x-4 x-3

-----

2 2.5

X 5

c.

x2 - 7x + 12

·

B. y2

C. D.

a- b

2

3a b + 3ab

2

A. B. C. D.

3/2 9/4 18/6 None of these

,Jx- 2 = JX + 2.

3Z 1001 Solved Problems in Engine~ring Mathematics (2"d Edition) by Tiong & Rojas B. C.

~

8~. Simplify V~

A.

D.

B.

89. What is the least common multiple of 15 and 18? "'

~

Jab ab c. 'Jab B.

D.

(n -1)! n! (n- 1)"

ab

~

A. B.

3 5

C.

90

D.

270

I~"

,. If x to the 3/4 power.equals·8, x equals

A. B.

A.

c.

c.

B.

320 2 180

D.

16

D.

90 ·

8S. Solve for x:

.Jx + 2·J2x + 3 -3 = 0

A.

3

B. C. D.

23 3 and 23

20

2 4 6

D.

192

.c,!S;EcE Board April 1999 "'t>iven: f(x) = (x + 3)(x- 4) + 4. When f(x) is divided by (x - k), the remainder is k. Find

A.

2

B. C.

4 6

D.

8

Solve for X from the given equation:

A. B.

C. D.

a= 5, b = 7

=

a= -5, b 7 a=-5,b=-7 a= 5, b = -7

What is the mean of x, y and z?

92. EE Board April 1996

A.

EE Board March 1998 The polynomial x 3 + 4x 2 - 3x + 8 is divided by x - 5, then the remainder is,

B.

c.

4 2 3

D.

5

A. B. C. D.

175 140 218

B.

C. D.

200

9~· Find the quotient of 3x5 - 4x + 2>t? + 3 36x + 48 divided by x -2>t? + 6.

a+b+c 3 a+b+c

2 a+b+c abc abc

--·

a+b+c

3

..,..u

.

Board October 1991

Hf(x) = 2>t? + 2x + 4, what is f(2)?

C.

4x + 2 16 ,(l +X+ 2

D.

8

A. B.

A. B. C. D.

88. EE Board Aprll199'7

3>t? - 4x - 8 3>t? + 4x + 8 3>t? - 6x - 8 3>t? + 6x + 8

94~
If ri is any positive integer, when (n-1)(n2)(n-3) ... (3)(2)(1)

4y 3 + 18y 2 +By- 4 by (2y + 3).

A.

A.

=

e
10

A. B.

42.31 50 38.62 5'7.12

:lOO.

97. The mean of x and y is a, the mean of y and z is b and the mean of x and z is c. ·

~8 ~2.J8X =2

r

If two numbers namely 250 and 850 are removed, what is the arithmetic mean of the remaining numbers? ·.

D.

96. ·The expression x4 + ax3 + 5~ + bx + 6

/

.s'- CE Board November 1991

A.

13

when divided by (x - 2) leaves a remainder of 16 and when divided by (x + 1) leaves a remainder of 10. Find a and b.

9:1. The numbers 12 and 16 has the greatest common divisor of

A. B. C.

D.

99· ECE Board April :1998 The arithmetic mean of 80 numbers is 55.

c.

What is the lowest common factor of 10 and 32?

-9 6 9

15

k..

90. ECE Board April 1998

,)14. ME Board April 1996

11.

c.

98. · ECE Board April 1999 Find the mean proportional of 4 and 36. A. B. C. D.

72 24 12

20

ECE Board Aprill.998

The arithmetic mean of6 numbers is 17. If two numbers are added to the progression, the new set of numbers will have an arithmetic mean.of 19. What are the two numbers if their difference is 4? A.

21, 25

B. C. D.

23,27 8, 12 16,20


II

14

18 10 140 X=18 70 X=-

X

16=4X+4 x=3 5x-1=5(3)-1 5x -1= 14

Topics

D Mon

.__

!"

Tue

D D D D ~l D D D Wed

Theory

Problems

Solutions

Thu

Fri

froperties ~~Addition of Integer Properties of Multiplication of Integers Properties of Equality Properties of Zero Properties of Exponents Properties of Radicals Surds Special Products Properties of Proportion Least Common Denominator Least Common Multiple Greatest Common Factor Remainder Theorem Factor Theorem

Bl

II [ x; 1 +

2:

= 47- 2x} 2

4x+4+6x = 564-24x 34x=560 X= 16.47

{

10 [A + A]=A}-1 X y 10A

' ' '

A

-~

"

'

-•o

"'"''"''

0 0 D D

43-50 Topnotcher

Substitute equation 1 to equation 2:

2(4y-2)+y-4=0 8y-4+y-4=0 9y=8 --+ Equation 3


9

--+ Equation 1

{2[3~-4~]=A}1

30-42 Passer 25-29 Conditional

--+ Equation 2

32 X=--2

RATING

40. 8 41. c 42. D 43. D 44. 8 45.A 46.A 47.C 48.8 49.A 50. D

2x+y-4,;0

X=4(%)~2

Y

ANSWER KEY 27.A 14. c 28. 8 15. 8 16. 8 29. A 30. 8 17. c 31. A 18. D 32. D 19. 8 33.A 20. D 21. A 34. D 35. D 22.C 36. 8 23. 8 37. D 24. c 38.A 25. c 39.C 26. 8

--+ Equation 1

y =%

.!.--101 x 1

1. D 2. 8 3. c 4.0 5. c 6.A 7.8 8. c 9. c 10.A 11. 8 12. c 13. 8

x-4y+2 =0 x=4y-2

Ill 1 1 1 -+-=X Y 10

Sat

. Notes

9

14 X=9

m

4x+2y = 5

~-.!!. =1 y

X

--+ Equation 2

0-24 Failed

y=

%- 2x

13x- 3y = 2

--+ Equation 1 --+ Equation 2

Substitute equation 1 in equation 2:

If FAILED, repeat the test.


~-a(_!__.!.)= 1 10

X

6

8

X

13x-3(%-2x) = 2

8

---+-=1 X 10 X

.!i= 1+~ X

10

18x=2+~ 2

18x=~ 2

X=.!_

2


36 100 !··solved Problems in Engineering Mathematics (2nd Edition)' by Tiong & Rojas Substitute equation 5 in equation 7:

y

=%-2(~)=%

~~

( 2x

2

-

9[7w1;38]-w=-12 y+(5+y)+1=0

63w- 342 -17w = -204 w=3

2

3y = 6 ) 3

6x 2 -9y 2

=18-7 Eq.

1

m

)fl

-7 Eq. 1

= 13

-7 Eq. 2

+ 4y = 10 -7 Eq. 2

6x + 4y

- (

6x 2 - 9y2 ) = 70-18

Z=2 -=X

X= -1

Consider above equation as Eq. 4 -

(2x + 5y){5x- 2y) = 10x2

2

9(2) = 18 6x 2 =54 X=±3

Iii

5x 2x 2 + 7x +3

ID

x-z=3 x=3+z

3x - 2y + w = 11

-7 Eq. 1

3x+15y-6w=-27 ·

-7Eq.4

-10y 2

Subtract equation 2 from equation 5:

Ill

5x + 5z = 5 z = 1-x

-7 Eq. 3

Multiply equation 2 by equation 3:

4XY+ 25xy

(2x + 5y)(5x- 2y) = 10x2 +21xy -10y 2

x + 5y - 2w = - 9 -7 E;q. 2 2x + y- 3w = - 6

-

Multiply equation 1 by equation 2: 6x - 3y + 3z = 18 -7 Eq. 5

(5x- 3y + 3z)- (x- 3y- 2z) =18-13

(xy)(yz)(zx) = 15(35}(21)

Substitute

xy = 15, in equation 4:

17y -7w = -38 7w-3 y=-17 Value of y above is considered as Eq. 5

2x =4 X=2

II

-7 Eq. 6

Subtract equation 6 by equation 3:

(2x + 10y -4w) -(2x + y- 3wl= -18- (-6) 9y-@= -12 Consider the above equation as ;q. 7

4(2x + 1)(x- 2) (2x + 1)(X+ 3)(x- 2)

4

+ y =-4

X

+ z - 1 = 0 -7 Eq. 2

-7 Eq. 1

X=3 Substitute zx = 21, in equation 4:

Subtract equation 1 from equation

2:

-:+

{

x~ [ x·sy·J (x y·2 r~ 2

21y = 105

Eq. 4

Thus, x + y + z = 3 + 5 + 7

Jr

=

Y=5

xtz-(x+y)=1-(-4)

z =5 + y

X+3

m

35x = 105

y + z + 1 = 0 -7 Eq. 3 Multiply equation 2 by 2: 2x + 10y- 4w = -18

2x+1 (x + 3)(x- 2)

- 3x -.2) - (2x + 1)(x + 3)(x ~ 2)

Substitute yz = 35, in equation4:

X

z3

+~~-

4(2x 2

_

15z =105 Z=7

x=3+(1-x)

(3x + 15y -6w) -(3x -2y + w) = -27-11

x +3 (2X+ 1)(x- 21

8x 2 -12x-·8 (2x + 1)(x + 3)(x- 2)

-:+ Eq. 4

X=3+Z Subtract equation 4 by equatiqn 1:

7

5x 2 -10x-x 2 -6x-9+4x2 +4x+1 (2x + 1)(x + 3)(x- 2)

2

xyz = 105


1

2x +1 x +3 2. +-2-·2x - 3x - 2 x + x - 6

5x • (2x + 1)(x + 3)

(xyz) . = 11025

Consider above equation as Eq. 6

2-

X.Y

5x(x- 2)- (x + 3)(x + 3) + (2x + 1)(2x + 1) (2x + 1)(x + 3)(x- 2)

Multiply the three given equations:

.\1~

y· 7 z -3 =

-2

m

X= -4- (-3)

( 2x - 3 y - 3z) - ( x - 3 y - 2z) = 16 -13

y =±2 6x

x·~y-~zlf

-9-1+~ 6-1-U =x -6+1+~ 22y 22Z22

Solving for x:

13y2 =52

2

(xyz· 3 r~

2x- 3y- 3z = 16 -7 Eq. 3

2

Subtract equation 2 from equation 3: 2

(x...sy-9zs){x~y--lz-!)

z = 5 + (-3)

Subtract equation1 from equation 2: 2

(x2y3z·2t (x-3yzT!

=6

y = -3

2x- y + z = 6 x- 3y- 2z

2

t>x

2y

m

Solving for z:

( 3x 2 + 2y = 35 ) 2 2


= 15

4

=x

[x-~yi

r

= x4 [

x4 x· 3 y- 2 x- 3~ y3 1

[

x· 5 y~ J

1

•

]3

38 100 i Solved Problems in Engineering Mathematics (2"d Edition) by Tiong & Rojas ~ Eq.1 A=1-B Equate constants: 2 = - 4A - 3B ~ Eq. 2

~

=

x-1y~ = '£_

• r· 2

-

X

6

(3) '(3t' ':' (3) 6x- x = 5

5

5x =5 8(7)8 +1 + 5(7)8 + 49(7)8 -

2

= r7 2 - (8)7.7 1 + (5)r +

49 7 < "> 72

Substitute equation 1 in equation 2: 2 ::; - 4(1 -B) - 3B 2 =- 4 + 4B- 3B B=6 A=1-6=-5

=r(49-56+5+1 >

i"

Day_g_- f'll.rldamentals in Algebra 39

m

=-7"

2 ( b2 - 4b + 16 ) ( b -.16 ) x-b3 + 43 2 ( b - 4b + 16 ) ( b- 4 )( b +.4 ) x-~----~~~~~ ( b + 4 )( b2 - 4b + 42 ) X=b-4

Thus,

X+2 6 5 =----x2 - 7x + 12 x- 4 x- 3

(b-e) by-cy x=y--=-.-a-c a-c (a-b) ay-by z=y--=--a-c a-c by -cy ay -by X+Z=--+-a-c a-c ay-cy y(a-c) X+Z=---=--a-C a-c X+Z=Y

ID

m

X+2 A B =--. +-x2-7x+12 x-3 x-4 x+2 =A(x-4)+B(x-3) (x- 3)(x- 4) (x- 3)(x- 4) x+2=A(x-4)+B(x-3) x + 2 = Ax - A4 + Bx - B3 Equate coefficients of x: 1=A+B

4( 52n+1) -1

c

+ 2 x +2x + 5

x 3 +2x 2 + 5x By equating constants:

3 (a-b) =a3 -3a 2b+3ab 2 -b 3 2 4x + 6x + k = 0

(X+

52" 2

"

5

[\1

(a"){am) = 100,000

~

Eq.1

a" = 10 ; a" = 10 am a

~

Eq. 2

m

J=243 Substitute am

ED p-q=5 By squaring both sides:

(P- q)2

p

=100, in equation 1:

1000m = 100000

m=2 ~

Eq.2

Substitute equation 2 in equation 1: (3)4Y(3t' = (3)5

Substitute

m = 2, in equation 3: 2

a = 100 a= 10

-

=52 2 2pq + q = 25 p 2 + q2 = 25 + 2pq

=25+2(~)

p2 +q 2 = 25 + k

~ Eq. 1

(a")( am)= (a")"'= 100000

2

2 2 p +q

a" = 1000 (a")(am) =am"= 100000

Squaring both sides: (3)ox = ( 3)4y

r=0.~5k

3x 3 - 3x 2 - 18x = 3x ( x 2 -- x - 6 } 3x 3 - 3x 2 -18x = 3x ( x-3 )( x + 2 )

am= 100

= {9)Y

=0

II

Ill

(amt = 10000

(3)3x = {3)2y

\n2

k = 2.25

1oamam = 100000

{81)Y(3tx = 243 (3) 4Y(3tx = {3)5

l±

Since it is a perfect square, then

a" (100) = 100000

{27)x

[

2(52n+1) _ 5( 5 2n-1}

A =2

Iii

Iii

x 2 + 1.5x + 0.25k = 0

Substitute (2) in (1):

10=5A

25 (3) .

2 x +6X+8 = (x+4)(x+2)

y=9

Ax 2 + 2Ax + 5A + 2Bx 2 + 2B)\. + Cx

11 [ II

IZI

a

o( 52n-1)

1 2 1 -5 )-5(5 " ·5- ) y= 52" 1 1 y=2(5 )-5(5- )

A B(2x+2) . = ~ + -=-'"----'x3 + 2x 2 + 5x x x 2 + 2x + 5

2 2 a -· x =(a+ x)(a- x)

2(5 2")

y=

2(5

x2 +4x+10

64x4y = (4)3x(4)Y = (4}3x+y

X+2 X+2 x 2 -7x+12- (x-3)(x-4)

Ill y=

A(x 2 +2x+5}+Bx(2x+2)+Cx 2 x('x + 2x + 5}

m

X= 1

~~~

Ell

bm/n = (bm)~ =

Ill

Jx- 2 = JX + 2 By squaring both sides: ( Jx- 2

r

~

=(

JX + 2

r

x-2=x+4JX+4 4JX =-6

40 __! 00 i~olved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas

2

x = 23

-+ Absurd

X=3

~8 ~2$x

Ell

=2

3

x +2x 2 +6}3x5

8

=

~2$x

= (2)4

~2../8X

=2

=16

.J8X=4

By raising both sides to exponent 2:

v~=~

•

X= (8)3

By squaring both sides: x +2J2x +3 = 9 2../2X+ 3 = 9 -

X

El

m

By squaring both sides: 4(2x + 3) = (9- x)

2

8x + 12 =81-18x + x 2

x2 -· 26x = -69

""' ' \

f(x) = 2Y! + 2x + 4 f(2) = 2(2)2 + 2(2).+ 4 = 16

II

f(r) =Sa+ 2b + 42 16 =Sa +2b+42 -26 =Sa +2b b=-13-4a When divisor is (x + 1), r

=-1 & f(r) = 10

3

f(r)=(-1t +a(-1) +5(-1t +b(-1)+6

Remainder= f [-%]

10=-a-b+12 2=a+b Substitute equation 1 in equation 2:

liiilll

= 3

3

4

f(r) = 2 + a(2} + 5(2t + b(2) + 6

0 . f(y) =4y 3 + 1Sy2 + Sy _ 4

=2 & f(r) = 16

f(r)=-a-b+12

n! = n(n-1)(n-2)(n-3) ... (3)(2)(1) (n -1)! = (n-1)(n-2)(n-3) ... (3)(2)(1)

.Jx + 2~2x + 3

When divisor is (x- 2), r

(-) 8x 3 -16x 2 +48

m

4

f(x) = x + axl + 5Y! + bx + 6

Note: Remainder = f(r).

12x3 + 36x

Note: Using remainder theorem,

11!9!11

X= 16

-

divisot= 2y + 3 = y- [ -%]

X=2

EFI

B

8x 3 -16x 2 + 48

Sx =(4)2 =16

[x~ =8J

3x 2 +6x+8 4x 3 + 2x 2 + 36x +48

(-) 6x 4

2$x = (2)3 = 8

/ ab

3(5) + 8 = 218

k=4 k=-2

6x 4 -4x 3 -16x 2 +36x


fab' v7a--' b =(ab)'

-

-

(-) 3x 5 -6x 4 +18x~


lab r> vif.ili "(ab)' lab [ Vifab = (ab)"']1'

ID

Remainder= (5)3 + 4(5) 2

lei)

VVab- ~(ab)~

IIJ

k -2k-8 =0 (k-4)(k+2)=0

3x + 8 ; divisor = x- 5

Note: Using remainder theorem, remainder= f (5).

x-13=±10

Etl jab-

-

2

2

(x -13) = 100

Note: Since x = 9/4 will not satisfy to the given general equation when substituted, this equation is classified as defective and thus, the a~swer is "None of these".

~

k =k2 -k-8

f(x) = x3 + 4x 2

2

(x-13) =-69+{13t

9 4

X=-

a

1!1

By completing the square:

JX =-~


15 =5·3 18 =6·3 LCM = 5 · 6 · 3 = 90 10 =5·2 32 = 2 . 2 . 2 . 2 . 2 Lowest common factor

t[-%]=4[~%J +18[-%J +8[-~]-4

a =-5

ID

b=-13-4(-5}

f(k) =(x + 3)(x- 4) + 4 f (X) =

12=4.3=4·3 16=4.4 =4·4 Greatest common divisor = 4

=15

t[-%]=11

f(x)=x 2 -4x+3x-12+4

=2

2=a+(-13-4a) -3a

X2 - X- 8

Remainder = f(k)

f(k) = k2 - k - 8 Substitute the given remainder = k in equation 1:

b=7

Iii X+Y=a·

2

'

y+z=b·

2

X+Z=C

. '

2

By adding a, b and c: a+b+c= X+y +Y+Z+X+Z

'

2 1

2

a+ b + c = -[2x + 2y + 2z]

2

2

4~ -1001 Sb1ved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas

.

'~

'.

~~

a+b+c=x+y+z Mean = x + Y + z _ a + b + c

-

3

-

3

Topies

D

Let: x = the mean proportion of 4 and 36

t

= __>5_

Jx

2

x

Mon

36 = 144

X=

12

Tue

D D D D D D [I] D

m Let: x = the arithmetic sum of 80 numbers,

II

Arithmetic mean = __>5_ = 55 80 X 80(55) 4400

=

=

y = new Arithmetic mean _ x-(250+850) y. 80-2

Theory

Wed

Problems

Thu

Solutions

Fri

Notes

y = 42.31

111

'~~

Let:

l

x = the first number x + 4 = the second number y = sum of the original 6 numbers.

=

y+x+(X+4) 19 6+2 102 + 2x + 4 = 19

8

2x

= 46

X=

23

'---~ ~-- ---~..,.-,- ~~---r.

-

T [

6

y = 17(6) = 102

=19(8)

--

Sat

.•

Arithmetic mean = 'j_ = 17

106+2X

-~

Properties of Addition of Integer Properties of Multiplication of Integers Properties of Equality Properties of Zero Properties of Exponents Properties of Radicals Surds Special Products Properties of Proportion Least Common Denominator Least Common Multiple Greatest Common Factor Remainder Theorem Factor Theorem

1

I

·.

;\.:;\; ~--i-i· il'l·'·r

I

li !~ \

1

'J

X+ 4 = 27 l

44

r--.

Dj I D Mon

I"

'II

Tue

I>JJ ·II

~

Theory

~ Wed

D D D D D D

Problems

Thu

Solutions

Fri

Notes

Sat

What is a Quadratic Equation?

'

Topics -~-·-

Quadratic Equation Quadratic Formula Properties of Roots Discriminant and Nature of Roots Binomial Theorem Binomial Expansion Properties of Binomial Expansion Pascal's Triangle Degree of Polynomial or Equation Logarithms Properties of Logarithms Modulus of Logarithms

The following is the quadratic formula:

Quadratic is an expression or an equation that contains the variable squared, but not raised to any higher power. Quadratic equation in x contains x2 but not x3 . I he general quadratic equati'm is ~~xpressed

as: Ax 2 +Bx+C = 0

where, A, Band Care real numbers and wrth A ±0. When B = 0, quadratic equation is known o~s a pure quadratic equation. 1\ quadratic equation in x is also known as .r ~u~cond-degree

polynomial equation.

1111 • :;olution to a quadratic equation is •·rill<~~ by factorin~J or by th(~ liSe of the
-B±JB2 -4AC

X=-----

2A

The quantity Js 2 - 4AC in the above equation is known as the discriminant. The discriminant will determine the nature of the roots of the quadratic equation. The table below shows the value of the discriminant and its corresponding nature of roots.

JB

2

-4AC

0 >0 <0

Nature of roots Only one root . . (Real and ~qual) Real and unequal Imaginary and unequal

The sum and product of the roots of a quadratic equation can be solved even without using factoring or quadratic

Day 3 ~Quadratic Equation, Binomial Theorem, Logarithms 47

4~ 100 i Solved Problems in Engineering Mathematics (2"d Edition) by Tiong & Rojas

formula as long as the general equation is given.

The following are just a few examples of a binomial expansion:

The following are the properties of the roots of a quadratic equation:

(X+Y)

0

Sum of the roots: When the two roots are added, the result is:

B

r1+r2=-;;.

"

Product of the roots: When the two roots are multiplied, the result is:

Binomial (x + y)o (X+ y)1

2

What is a Binomial Theorem? Binomial is an expression containing two terms joined by either+ or-. Binomial theorem gives the result of raising a binomial expression to a certain power. The expansion and the series it leads to are called the binomial expansion and the binomial series, respectively.

il

The binomial theorem is expressed as follows:

(x'f:y)" =x~ +•nx"-!;y +·p(n-: )xn-~y~· + ·•·

1

. ·.·'f'

. "

.

.

•·· ..

·21

·.

... +nxy't!~1 + yn Binomial coefficient is a coefficient of x in the expansion of ( x + y)", The binomial coefficient

3

2

2

2

(x + y) = x + 3x y + 3xy + y

3

As observed in the binomial expansions above, some properties were established and are enumerated as follows: Properties of Binomial Expansion of (x + y)": 1. The number of terms in the resulting expansion is equal to n + 1.

0 ri·r2""A

ncm gives the number of ways

of picking m outcomes {not in any particular order) from n possible outcomes. The binomial coefficient forms the rows· of the Pascal's triangle.

(x+y)": Sum= n(n + 1)

( x + y ) = x 2 + 2xy + y 3

Sum of expon~nts of the expansion of

=1

(x+yf=x+y Let r1 and r2 be the roots of a quadratic !3quation:

In Italy, this triangular pattern is known as Tartaglia's triangle while in many parts of Asia, it is referred to as Yang Hui's triangle.

2. The exponent of x decreases by 1 in succeeding terms, while that· exponent of y increases by 1 in succeeding terms. 3. The sum of the exponents of each term· is equal ton. 4. The first term is x" arid the last term is y" and each of the terms has a coefficient of 1

Pascal's Tiangle 1 1 1

1 2

{x+d {x+d {X+ y)4 (X+ y)s (X+ y)6

What is a Degree of a Polynomial or Equation?

1

1 3 3 1 1 4 6 4 1 15101051 1 6 15 20 15 6 1

Another way to determine the coefficient of any term in the binomial expansion is to use the following formula:

Example: C =(Coeff. ofPT)(Exponent of x of PT) E;xponent ·of y of PT + 1 where: C = coefficient of any term PT = preceding term The r1h term of the binomial expansion of ( x + y )" may be calculated using the following formulas:

r

2)- .. (n- r + 2) x(n-r+1)

Each number in the triangle is equal to the sum of the two numbers immediately above it.

2. What is the degree of the polynomial3x 4 y- 2x 3 z4 + 7yz 5 ? . Answer: 7. 7 is the sum of 3 and 4 in the second term.

(r -1)!

=

H

Y

rtt\ = nCr-1X(n-r+1)yr-1

The coefficients of a binomial expansion can also be conveniently obtained by arranging them in a triangular array or pattern. This is known as Pascal's Triangle named after the famous French mathematician Blaise Pascal (16231662).

1. What is the degree of the monomial 7x5 .. Answer: 5

What is a Logarithm? n(n -1)(n-

th

5. The coefficient increases and then decreases in a symmetric pattern. The Pascal's Triangle:

Degree of a polynomial or equation with only one variable refers to the exponent of the variable. For a polynomial or equation that contains two or more variables, the degree is the maximum sum of the exponents of the variables in a single term.

A term involving a variable with a specific exponent is obtained by using the following formula:

y'

t-

needed to give x.

-J- - -..1

Logr6=4 maybewOttena•

----n

r ~

=16

_ =-n(n-1)(n-2)···(n-r+1) x" 'y' r!

Sum of the coefficients of 'the expansion of

(x

The logarithm of a number or variable x to base b, Iogb x , is the exponent of b

y)": Sum = ( Coeff. of x + coeff. of y )"

The term "logarithm" comes from Greek words, "logus" meaning "ratio" and "arithmus" meaning "number". John Napier (1550 -1617) invented logarithm in 1614 using e = 2.718 ... for its base. Logarithm with base e {loge or In) is called the natural logarithm or the Napierean logarithm. In 1616, through the suggestion of John Napier, Henry Briggs (1561 0 1630), a professor of Geometry at

48 ·1001 Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas Gresham College in London, improved the logarithm using 10 as the base. The logarithm with base 10 is known as common logarithm or the Brlggsian logarithm. The number ·e· which is the base of the natural logarithm is known as Euler's number, named after the Swiss mathematician, Leonhard Euler (17071783) and is defined as,

e

= um(1+!)n ·-

n

Binary logarithm (denoted as lb) is a logarithm with a base value of 2. Relation between natural logarithm and common logarithm: The natural logarithm can be converted into a common logarithm and vice versa. To obtain this, a factor known as the modulus of logarithm is necessary, such as: logx = 0.43431nx

Proceed to the next page for your third test. Detach and use the answer sheet provided at the last part of this book. Use pencil number 2 in shading your answer. GOOD LUCK!

Topics

ll

~ribia:

Mon

D

Did you know that ... Isaac Newton while a student at age 22, invented differential and integral calculus, discovered the law of universal gravitation, formulated the three laws of motion, developed the new theory o flight in just 18 months and set a record of the most productive periods of achievement by an individual in the history of science!

Tue

D ·'

Problems

What are the Properties of Logarithms?

=logx +logy

1.

log(xy)

2.

log(~)= logx -logy =nlogx

3.

logx"

4'.

logx logbx=Iogb

5. 6.

logbx 1og.x= logba log.a=1

Wed

0

Solutions

Notes

I

'~o

~-

~-

Thu

D [_] D D

"The art of asking the right questions in mathematics is more important than the art of solving them." - Georg Cantor

lnx = 2.30261ogx

The coefficients 0.4343 and 2.3026 are the referred, to as the modulus of logarithm.

Quadratic Equation Quadratic Formula Properties of Roots Discriminant and Nature of Roots Binomial Theorem Binomial Expansion Properties of Binomial Expansion Pascal's Triangle Degree of Polynomial or Equation Logarithms Properties of Logarithms Modulus of Logarithms

'o.,.,,...,._.--~

J11 ~~~:.

1,,

.~

tJ

'

l 1

i ~

Fri '

Sat

1011 ECE Board March 1996 !"he equation of whose roots are the reciprocal of the roots of 2x2 -- 3x - 5 = 0 IS, 2

5x + 3x - 2 2x2 + 3x - 5 3Jt - 3x + 2 2x-2 + 5x- 3

1\ ll. C. U.

=0 =0 ::: 0 =0

8.

c. D.

±3 ±4 ±5

J:04: ME Board October :1.996 Solve for x that satisfies the equation 6x2

...:7x-5 = 0. A.

5 -1 -or-

lOZI

3

•·qual to

B.

3 3 -or-

EE Board October 199~ In the equation Jt + x = 0, one root is x f\ 1!.

I:. I>

1 5 1/4 none of these

( O:Jt ECE Board Aprilt990 .• nlw for the value of "a" in the equation ·•" I ,,,·• I 1G '~ 0. I\

t .'

C. D.

2

2 8 7 -7 -or5 15 3 3 -or5 4

lOS: EE BoCJ~rd Oetober 1:997 Find the values of x in the equation 24~ + 5x- 1 "= 0

SO

A.

100 1 ·solved Problems in Engineering Mathematics (2"d Edition) by Tiong & Rojas 1.1.1.: Given the equation 3x + Bx + 12 = 0. What is the value of B so that the roots of the equation are equal?

(~.1)

t~ ~)

A.

4

(~

~) 2'5

B. C.

8

c. D.

(

~8' -~) 3

J...,;;

B.

6'5

1.06: EE Board October 1.990

i"

2

D.

Determine k so that tl.e equation 4x + kx + 1 = 0 will have just one real solution.

A. B.

3 4

C. D.

5 6

A. B.

C. D.

10

8064 8046 8046 4680

x l 5 x 10 x x5 y

t

Find the 5th term of the expansion of

/ ( x2 +1 )10

B.

A.

- 0.113,- 0.887

C.

B.

- 0.331;- 0.788 -0.113,-0.788 -0.311,- 0.887

D.

D.

J.08: If 1/3 and -3/2 are the roots of a

8

B.

I

C. D.

6~ + 7x - 3 = 0 6x2 - 7x + 3 = 0 6~ - 7x- 3 = 0 6~ - 7x + 1 = 0

0? A.

B.

0.6 -0.6

D.

0.75

c. -o.8

A.

8

C.

-16 16

D.

-8

B.

D.

D.

-39396 128a11

1.24: ME ~ard April 1.997 What is the value of log tci base 10 of

D.

B. C.

1.181 What is the coefficient of the term 'free of x of the expansion of (2x- 5y) 4 ?

256 526 265 625

-548~l

103 1650 x 103 161700 X 103 167100 x 100 167100 x

l

-154,288 xV - 1,548,288 xV

0

A. B. C. D.

1 2 3

1.21.:

ECE Board April 1.995

c. D.

What is the value of (log 5 to the base 2) +(log 5 to the base 3)?

A. B. C. D.

7.39 3.79 3.97 9.37

1.26: Find the value of log4 (log3 5).

A. B.

C. D.

1.460 0.275 1.273 0.165

:l%7: Given: log4 7 = n. Find log4! 7

4 /the term next to 495x y ? 660 792 990 1100

10.9 99.9 9.9 9.5

1.25: ECE Board April 1998

What is the sum of the coefficients in the expansion of (x + y- z) 8 ?

A. B.

A 13. C. D.

D

.1.-1.6: What is the numerical coefficient of A. B.

A. B. C.

D.

-148,288 x3

A. B. C.

3.76 5.84 4.48 2.98

1000 33 ?·

What is the sum of the coefficients of the expansion of (2x -1 )20 ?

2 100 'expansion of (x + x ) ? A. B. C.

(A.

tzo:ECE Board November 1.995

8

1.1.0: What is the discriminant of the equation 4~ = 8x- 5?

-33669 256a1t

D. 63,360 126,720 506,880 253,440

usf What is the fourth term of the

1.09: Which of the following is a root of this quadratic equation, 30~ + 49x + 20 =

Evaluate the log6 845 = x:

d:;~·Find the 6th term of (3x- 4y~~)

'In the expression of ( x + 4y ) , the numerical coefficient of the 5th term is,

D.

1.98

C.

D.

12

B. C.

178

-66339 128att

C.

260 x 8 5040 x 8 210 x 8 420 x

quadratic equation, then the equation is

A.

C. D.

B.

A.

ulli'ECE Board April 1.998

A.

.

1.86 1.68

-66939 256att

B. A.

)t6

B.

A.

·l

x

C.

1

A.

1.23: CE Board November 1.997

1.07: ME Board April 1.996

2 Solve for x: 10x + 10x + 1 = 0

I

\_2a

10 -12

.l·J:il

".n7: CE Board November 1.996 · Find the 6th term of the expansion of 1--3

Find the term involving y5 in the .;expansion of (2x 2 + y) 10 2

Day 3 - Quadratic Equation, Binomial Theorem, Logarithm 51

0 1 2 3

t:u:r CE Board November 1.99~ ECE Board Nov. 199:J I 1nd the valut! of loq., 'Ill •

C.

1/n

n -1/n - n

1.28: CE Board November :199:2 CE Board May 1.994 If log a 10 = 0.25, what is the value of log 1oa?

A.

n

2 11

I .::_f ,.~

52 l 00 l·.Solved Problems in Engineering Mathematics (2"d Edition) by Tiong & Rojas

C. D.

:134: ECE Board November :199:1

6 8

Given: Iogb 1024

:l:t9: ECE Board November :1995 Given: log b y = 2x + log b x. Which of the following is true?

A. B.

Y = b2x y = 2xb y=-

D.

y = xb2x

.

b

-7x 10 to the -7x power

D.

2

2

:1401 Solve for the value of x:

6

Find b.

log2x 3 +log-= 6.278 X

.~

·B. C.

i:l

D.

-1

B ..

9

C.

-1 and

D.

1 and- 9

-7 log to the base 10

9~

:13&: ECE Board April :1993 Solve for the value of x in the following equation: x 3109 x = 1OOx .

A. B.

c. D.

12

8 30 10

it ,··~

'l

'" --~

:137: EE Board October :l99:t Given: log 6 + x log 4 = log 4 + log (32 + 4•). Find x.

logx

n

n log x C. log (x to the 1/n power)

B.

n D.

(n -1)1ogx

:l:J:tl ECE Board November :1990 Log (MN) is equal to:

A. B.

C. D.

Log M- N Log M + N N log M Log M + Log N

:l:J:JI ME Board April :1997 What expression is equivalent to log ( x ) -'log ( y + z )? A.

B. C.

D.

log log log log

x + log y + log z [ XI ( y + z )] x -log y - log z y + log ( x + z )

A. B.

2 3

D.

6

c. 4 ECE November :1998 If log of 2 to the base 2 plus log of x to the

:1:581

base 2 is equal to 2, then the value of x is,

A. B.

4 -2

D.

-1

c. 2

:1391 ME Board October :1997 Find the value of x if log12 x = 2.

.~1

·{

A .. 144

B.

414

D.

524 425

c.

A.

•'>'·~· .· ·,'j~t

7

:l:J:l: ME Board April :1996 Log of the n1h root of x equals log of x to 1/n power and also equal to:

A.

c.

2560 16 4

A.

:l:JO: ME Board October :1996 Which value is equal to log to tlte base e of e to the -7x power?

B. C. E.

A. B.

=~

:l:JS: Given: log3 (~- 8x) = 2. Find x.

2x

c.

A.

Day 3 - Quadratic Equa!1e_n, Bin_o!'lial Theorem, Logarithm 53

,f-

379.65 365.97 397.56 356.79

~

Day 3- Quadratic Equation, Binomial Theorem, Logarithm 55

ml

2x 2

-

2

7 ± ~(7) -_ 4(6)(-5) __ X=--'--'-,.:__,_ 2(6)

3x- 5 = 0

(2x+2)(x-25)=0

7 ± 13 X=-12

X= 2.5

Topics

D tv! on

D Tue

[QJ

D D D D D D Theory

Problems

Wed

Thu

Solutions

Fri

Notes

Sat

RATING

ANSWER KEY 101. A 102. D 103. A 104.A 105. D 106. 8 107. A 108. A 109. 110. 8

c

111. D 112. A 113. 114. B 115. 8 116. 8 1·17. 8 118. D 119. D 120.A

c

121. 8 122.A 123.A 124.C 125. 8 126. 8 12"1. D 128.8 129. D 130. A

Quadratic Equation Quadratic Formula Properties of Roots Discriminant and Nature gt;Roots Binomial Theorem Binomial Expansion Properties.of Binomial Expansion Pascal's Triangle Degree of Polynomial or Equation Logarithms Properties of Logarithms Modulus of Logarithms

131. A 132. D 133.8 134. 8 135. 136. D 137. 8 138. 139. A 140.C

c c

c:J

34-40 Topnotcher

c:J 26-33

o o

Passer

X= -1 Thus, Thus. the roots of the second equation are:

1

X1 = - = 0.4 2.5 1 x2 = - = -1 -1

•

5 x1 =3

1 and x 2 = - 2

24x 2 + 5x -1 = 0

Using the quadratic formula;

-5±~(5) 2

Solving for the second eqution: X

2(24)

(x-0.4)(x+1)=0

-5 ± 11

X=-48

2

x + x- 0.4x- 0.4 = 0 { x 2 + 0.6x- 0.4 = 0 ). 5 5x 2 + 3x -2 = 0

• •

x(x+1)=0

•

1

3

X=O where: A = 4; 8 = k & C = 1

X= -1

8 2 -4AC = 0

a8 -17a 4 +16 = 0 2 8 Let: x = a4 and x = a x2 -17,+16=0

Using the quadratic formula;

17±~(17) -4(1)(16) _.!....:__:__ __ X= _ 2

k 2 -4(4)(1)=0 k

17 ±15 X=--

X=

Thus,

a= ±1

= 16

k = ±4

•

2

x1 = 1 a4 = 1

2

10x2 + 10x + 1 = 0 Using the quadratic formula:

2


1

Thus, x1 = B and x2 = -

Note: There is only one solution to the equation (4x2 + kx + 1 = 0), if the discriminant (8 2 - 4AC) is equal to zero.

20-25- Conditional 0-19 Failed

-4(24)(-1)

= ----L'----'--,-::-c---

2(10) -10 ± 7.746 X=---20

X2 = 16

a4

= 16

a= ±2 Thus, x1

ml 6x 2 - 7x- 5 = 0 1J:;lll~J the quadratic formtila;

-10±~(10) 2 -4(10)(1)

lml

=- 0.887 and x2 =- 0.113 x1=

i

and

x 2 =-~

S6 ..100 I Solved Problems in Engineering Mathematics (2"d Edition) by Tiong & Rojas

-B

x1 +x2 = -;;;.

where: A = 4; 8 = - 8 & C = 5

Day 3 - Quadratic Equation, Binomial Theorem, Logarithm 57

a

(X+X2)100

Substitute:

.c A

82

i+(-%)=~~

8

-

-

il'

=0 2 8 = 144

4{3){12)

B=~A

liB

C =_:!A

is the 61h term Note: The term involving 10 of the expansion (2~ + y)

i

Substitute to the general quadratic equation: 2

Ax + Bx + C

=0

lx +(-=~At,, 0 2 )

6x?. + 7x- 3 = 0

0!- - ( 32x 10) y5 = - -1-

(10-5)!5! 61h term= 8064 x 10y5

~~~

2

-49 ± J(49)2 -=4(30).(20) X=-------·---2(30)

5th term =

X=---

10

60

Thus, x, X2

11.1

-49±1

= - - = -0.8 60 -49-1

(__!__)n-r+ (-3)'-1 2a where: n = 16; r = 6

-

1 6th term = 1s C 5 - a (2

C

4

(~}. 6 (lx..!.)

4

)11 (-3)

5

)( 243) 16! ( 1 - (16-5)!5! 2048 a 11 -

12 (

1) x4

4368(-243) 16 = 2048 a 11 + 16

5th term= 210 x8

2

4x - 8x + 5 = 0 where: A = 4; B = - 8 & C = 5 Discriminant = B2 - 4AC = (- 8) 2 - 4(4)(5) = -16

Noie: The roots of the equation (3x2 + Bx + 12 0) are equal, if the discnminant (82 - 4AC) is equal to zero.

1111

a

5th term= nc,_, (x)n-r+ 1(4y)'" 1

Note: The term in the expansion (2x sd which is free of x is the last term or the 5th term.

where: n = 12; r = 5 5th term= 12 C 4 (x) 8(4y}

4

i2!

8

4

"" - - - - - - (x )(256y ) (12-4)!4! 5th term = 126,720

x8 l

(3x-4d

6th term= ncr-1 (3x)n-r+ \-4y)'" 1

sth term= ncr-1 (2x)"·r+ 1(-5y)'" 1 where: n

= 8; r = 6

where: n 8 C5

(3x) 3(-4y) 5

81 3 5 ) (27x )(-1024y ) 8-5 !5! 6th term = - 1,548,288 x 3y 5

El Note: To solve the sum of the coefficients 20 of (2x -1 ) , substitute one ( 1) to x, calculate, then subtract a value of (-1 )20 from the result Sum of coefficients= [(2)(1) -1t- (-1)20 =0

IBI Note: To solve the sum of the coefficients 8 of (x + y -z) • substitute one(1) to all the variables and calculate. Sum of coefficients = ( 1 + 1 - 1 )8 = 1

lfD

lED

6th term = 66339 128 a11

(x + 4y)t2

~~~

(1)(625 4 ) . y

= 625 l

r-1

60

IDI =

1

10!

= - - = -0.833

)16

n

= (10-4)!4lx

and

1 -3 (2a

6th term = C

X

5th term

!4!

= (

. 495 (8) Coefficient of next term "" - - 4 +1 = 792

( <, )r-1

(4-4)

6th term=

Coefficient of next term =

1 )10 ( X2+;:

where: n = 10; r = 5

-49 ± 1

a

1&1

1h 5 term = n c r-1 (x2)n-r +1

30x + 49x + 20 = 0 Using the quadratic fon'ru.l)a;

I

10! 97 6 x (x ) (100-3)!3! 4th term= 161,700 x103

{Exponent of y) + 1

6th term= 1oCs (2x2)s Ys

' 7 1 "J'6 2 [ Ax + 6Ax- 2.A = 0 A

= 100 c3 (x)97(x2)3

(Coefficient of PT)(Exponent of x}

61h term= ncr-1 (2x2 )n-r+ 1 (yf 1 where: n ::: 10; r = 6

A:x +(!_A 5 )

•

4th term

8 = ±12

2

2

where: n = 100; r = 4

4AC = 0

~(-%)=~ 6

41

4th term= ncr-1 (x2tr+ 1(~)'"1

x1·x2 = -

2

5th term = 4 C 4 (2x) 0 (-5y) 4

=4; r = 5

log8 48

= 10log 9to 48 6 = 1.86 10

logs 845 = log1o 845 = 3. 76 log 10 6

IBI

log10 1000

ml

log 2 5 + log 3 5 =

33

= 3.3 log10 1000 = 9.9

~91o ~ + log10 5 log1o 2' log1o 3

= 3.79 ---.._....

.JII

1

58. 1001 Solved Problems in Engineering Mathematics (2"d Edition) by Tiong & Rojas .

•

IBI

log (1og 5) = log1o(log3 5) 3 4 log1o4

Iogb 1024 =

~

2 log1o 1024 5 log1ob = 2 log.10 b -_ ~..:.:!_--=. log1o 1024 . 2.5

!.,

Day 3 --Quadratic Equation, Binomial Theorem, Lo$J!rithm 59

'~

II log6 + xlog4 = log4 + log(32 + 4") log6 + log4" = log4 + log(32 +4") log(6){4"} = log(4)(32+4")

log1o b = 1.204

1&1 ,

log 4 ~=log

4 1-log 4 7

(6)( 4"} = ( 4}(32) ~ (4)( 4•)

Note: Logarithm of 1 to any base is eq~Jal to zero. 1 log 4 - = 0 - log 4 7 = - n

IB1

•

le~g 3 ( x2

-

8x)

=2

log10 (x -8x) = 2 'log 10 3 log10 (x 2 - 8x) = 2 log 10 3 = log 10 (3) 2 log10 (x 2 - 8x) = log10 9 x 2 - 8x = 9

log.10=0.25 log1o 10 = 0.25 log 10 a log 10. Iog1oa = - 10 --= 4 0.25

x2

-

x=9

Iogb y -·Iogb x = 2x

ml

log1o Y- log1o x = 2x log 10 b log10 b log10 y -log 10 x

=2x log10b

x31ogx

Take logarithm on both sides: log4" =log64 xlog4 = log64 log64. X=--

log4

X=3

ml

=1OOx

Take logarithm on both sides: log 3109 x = log 1OOx

log2 2 + log2 x = 2 log10 2 + log10 x = 2 log10 2 log10 2

x

1+ log1o x = 2 log10 2

log1o 'j_ = log1o b2x X

• •

1!1

Ill!.

y

= (-'7X)(1)

Equating factors to zero:

=-7x log

vx

1 =log(x)' =-log(x) 1

n

.

31ogx +2 = 0 31ogx = -2

-2

log MN

X=2

(3(1ogx) + 2)(1ogx -1} = 0

log. e-7• = (-7x)log. e

nC

log1o x = 1 log10 2 · log10 x = log10 2

3(1ogx)(logx) = lcg1 00 + logx 3(1ogx) 2 -logx- 2 = 0

= xb2x

= log M + log N

logx=3 x = 0.215 (absurd) logx -1 = 0

(x) -log (y + z) =log(, :z]

logx = 1

3

log2x -logx = 6.278 -log6 2x 3 log-= 5.49984 X

2 2

(x +1)(x -9) = 0 X= -1

X

·iog2x3 + log6.: logx = 6.278

2x

4" = 64

8x -9 = 0

Iogb y = 2x + Iogb x

log2x 3 +log§.= 6.278

log2x 2 = 5.49984

2(-4") = 128 2

7

ml

(6}(4"} = (4)(~2+4x)

b = 16

•

El

log12 x = 2 X:::

122

X= 144

=antilog5.4998

x ""158055.6425 X= ±397.56

62 100 1 Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas

Topics

D D D Mon

Tue

I~

~

Theory

Age Problems Work Problems Mixture Problems Digit Problems Motion Problems Coin Problems

Wed

D ~ D D D D

Problems

Thu

Solutions

Fri

Notes

Sat

Age Problem

1/5 -?This is what the person One of the most common problems in Algebra is the age problem. This type of problems must be solved meticulously by qiving more emphasis to the tenses (i.e. past, present or future) of the statement.

~

Example: fhe ages of a certain person in the past, present and future in terms of x are as lui lows:

II

This is the work For a complete job, rc;tte

Work Problem · ;11ppuse that a person can do a certain w< '' k in 5 days. This means that the said 1"'"·"'1 c.1n finish 1/5 of the work in one ""Y lllu~;. lli~; rah~ IS 1/~i of the work P• 'I

d.iy

finished in 1 day

x time,= 1 .

When there is a specific work and specific time and manpower, the rate of doing the work may be computed using the number of man-hours.

64 100 I Solved Problems in Engineering Mathematics (2"d Edition) by Tiong & Rojas .. ,..,

Example: If 20 bakers can bake 40 pizzas in 8 hoars, how many bakers can bake 10 pizzas in 2 hours? Solution: Get the rate (in man-hour) of baking a pizza. Rate=

,

(20bakers )(8hours)

Day 4- Age, Work, Mixture, Digit, Motion Problems

Digit Problems Let h, t and u be the hundreds, tens, and units digit, respectively. A three-digit number must be represented in the following manner. Number= (h ){100) + (t)(1 0) + (u)

Dime 10 cents

Quarter 25 cents

A two-digit number is represented by;

40pizzas Number= (t)(1 0) + ( u)

Rate= 4baker- hour pizza

Half Dollar 50 cerits

Motion Problems This means that to bake a pizza, you need either 4 bakers to work in 1 hour or 1 baker to work in 4 hours. No. of bakers= ( 4

ba~er-hourJ(10 pizzas) p1zza

In Algebra, the problems pertaining to motion deals only with a uniform velocity, i.e., no acceleration or deceleration in the process. The following is the relationship between the distance, time and velocity.

· 2 hours

= 20 bakers

Time= 0

Time= t

v

Proceed to .the next page for your 4th test. Detach and use the answer sheet provided at the 1ast part of this i:)ook. Use pencil number 2 in shading your answer.

•

GOOD LUCK I

Mixture Problems The easiest way to solve a mixture problem is to draw a rectangle or square which will illustrate the content of the mixture as shown in the following illustration.

I

D=Vt

'ijtribia: ·

V=~ t

t=~ v

Consider a 5 cubic meter mixture containing 65% alcohol and 35% gasoline. Coin Problems

.65% Alcohol

!~

D

The entire mixture

35% Gasoline

Problems in Algebra about coins are more focus on the dollar denomination than local Philippine currency. The following are the equivalent value for each coin.

V= 5m 3 The quantity of alcohol is (0.65)(5) = 3.25 cubic meters while that of gasoline is (0.35)(5) = 1.75 cubic meters. Penny 1 cent

Nickel 5 cents

Did you know that... 161h century italian mathematician and physician Gerolamo Carda no, was the first to introduce the concepts of probability and define it as the number of favorable outcomes divided by the number of possible outcomes. Because of this, he is regarded as the "Father of the Theory of Probability".

~uote: "Where there is matter, there is geometry." - Johannes Kepler

~

Day 4- Age, vvork, Mixture, Digit, Motion Problems 67

uS: A girl is one-third as old as her ,1" / brother and 8 years younger than her sister. The sum of their ages is 38 years. How old is the girl?

"ISO: Debbie is now twice as old as Jerry. Four years ago, Debbie was three times as old as Jerry then. How old is Debbie? A.

D D D D tvlon Tue

Theory

,-------

Topics

...-----~·

! Age Problems j Work Problems 1 Mixture Problems I Digit Problems j Motion ~roblems Coin Problems

Vl/ed

IQJ Problems.

Notes

B. C. D.

4 5

c. 6 D.

lSi: ME Board April1998 ··A pump can pump out water from a tank in 11 hours. Another pump can pump out water from the same tank in 20 hours. How long wiil it take both pumps to pump out the water in the tank?

A. B. C.

5 7 8

A.

D.

10

C.

B. D.

Thu

age now when you were born." If the father is now 38 years old, how old was his son 2 years ago?

Fri

Sat

A. B. C.

19

D.

21

Mary is 24 years old. Mary is twice as old as Ann was when Mary was as old as Ann is now. How old is Ann now? A. B. C. D.

16 18 12 15

.14i: EE Board April1997 The sum of Kim's and Kevin's ages is 18. In 3 years, Kim will be twice as old as ·Kevin. What are their ages now?

A B. C. D.

4,14 5, 13 7, 11 6, 12

14-(:"GE Board February 1994 "Robert is j 5 years older than his brother Stan. However "y" years ago, Robert was twice as old as Stan. If Stan is now "b" years old and b>y, find the value of (by).

A. B.

C. A. B. C.

15 16

D.

18

17

.•144:

JJ is three times as old as Jan-Jan. Three years ago, JJ was four times as old as Jan-Jan. The sum of their ages is A. B. C. D.

20 24 28 36

D.

"A. 400-mm 0

hours. With all the three pipes open, how long will it take to fill the tank?

B. C. D.

hours hours hours hours

A tank is filled with an intake pipe in 2 hours and emptied by an outlet pipe in 4 hours. If both pipes are opened, how long will it take to fill the empty tank? A.

3 hours

B. C.

4 hours 5 hours

D.

6 hours ~":,.,.-4:"" .

1

Js.:i: A tank can be filled in 9 hours by one pipe, 12 hours by a second pipe and can be drained when full by _aJ!l_jrd pipe in 15 hours. How long will it take to fill an empty tank with all pipes in operation?

A

3

B.

4

A.

5

B.

(:)

C.

()

2.00 2.50 2.25 2.75

,63:

s49i

c

pipe can fill the tank alone in

5 hours and another 600-mm 0 pipe can fill the tank alone in 4 hours. A drain pipe 300-mm 0 can empty the tank in 20

17 16 15 14

At present, the sum of the parents' ages is twice the sum of the children's ages. Five years ago, the sum of the parents' ages was 4 times the sum of the children's ages. Fifteen years hence, the sum of the parents' ages will be equal to the sum of the children's ages. How many children are there?

hours hours 1/2 hours 1/2 hours

,,~

15 17

1.!18i""'Six years ago, Nilda was five times as old as Riza. In five years, Nilda will be three times as old as Riza. What is the present age of Riza?

7 6 7 6

J.Si: CE Board November 1993

A.

.w:'ECE Board April1995 ECE Board April1999

14 16" 18 24

7

, ..., J46: Paula is now 18 years old and his /colleague Monica is 14 years old. How many years ago was Paula twice as ol·d as Monica?

14'7: A father tells his son, "I was your

D D D D

Solutions

A. B.

7 hours and 12 minutes 7 hours and 32 minutes 7 hours and 42 minutes

68 100 LSolved Problems in Engineering Mathematics (2"d Edition) by Tiong & Rojas 7 hours and 50 minutes

cr· ,iSS: ME Board April :1995 If A can do the work in "x" days and Bin "y" days, how long will they finish the job working together? x+y xy

A.

~y

B.

2 __!'!_

C.

I'

FY

.s6; ECE Board November :1995 · Pedro can paint a fence 50% faster than Juan and 20% faster than Pilar, and together they can paint a given fence in 4 hours. How long will it take Pedro to paint the same fence if he had to work alone?

A. B. C.

"while Stewart can paint the same house in 16 hours. They work together for 4 hours. After 4 hours, Stewart left and Glenn finished the job alone. How many more days did it take Glenn to finish the job?

A.

2.75 2.50 2.25 3.00

B. C. D.

hours hours hours hours

/

CE Board November :1993 It takes Butch twice as long as it takes Dar'l to do a certain piece of work. Working together they can do tile work in 6 days. How long would it take Dan to do it alone?

.J:S81

.~ ·;

A. B. C. D.

D.

8, 13

A. B. C. D.

10, 15 6, 11 7, 12

12 13

'I'

,Ja,

B. C. D.

9 days

A. B. C. D.

9 hours 18 hours 12 hours 14 hours

.:u;i; ECE Board April :1.999 Mike, Loui and Joy can mow the lawn in 4, 6 and 7 hours respectively. What fraction of the yard can they mow in 1 hour if they work together? A. B. C. D.

47/84 45/84 84/47 39/60

~6:1:

A farmer cari plow the field in 8 days. After working for 3 days, his son joins him and together they plow the field in 3 more days. How many days will it require for the son to plow the field alone?

10 days

11 days 12 days

A. B

10 11

C. D.

(.'''

~~ A goldsmith has two alloys of gold,

1

c: D.

20 days. In how many days can all of them do the work together? 19 17 21 15

20 m of solution with 35% alcohol, 3 40 m of solution with 50% alcohol 3 50 m of solution with 35% alcohol, 3 20 m of solution with 50% alcohol 3 20 m of solution with 35% alcohol, 3 50 m of solution with 50% alcohol 3 40 m of solution with 35% alcohol, 3 20 m of solution with 50% alcohol

B.

:1.64: ECE Board November :1.99:1. Crew No. 1 can finish installation of an antenna tower in 200 man~hour while Crew No. 2 can finish the same job in 300 man-hour. How long will it take both crews to finish the same job, working together? A. B.

A:6o: EE Board April :1.99& '·A and B can do' a piece of work in 42 days, Band C in 31 days and C and A in

A.

3

A.

"""'"'#.[

100 man-hour 120 man-hour 140 man-hour 160 man-hour

the first being 70% pure and the second being 60% pure. How many ounces of the 60% pure gold must be used to make 100 ounces of an alloy which will be 66% gold?

..,_.<"'""

. u)s: ME Board October :1.994

,'jl;h, 'i 'f~

·w'

It takes Myline twice as long as ·Jean a to do a certain piece of work. Working together, they can finish the work in 6 hours. How long would it take Jeana to do it alone?

,_s{:"'Glenn can paint a house in 9 hours

;

C.

J.&"i;

6 8 10 12

D.

Day 4- Age, Work, Mixture, Digit, Motion Problems 69

.<1.59: ME Board April :1.995 A and B working together can finish painting a house in 6 days. A working alone can finish it in 5 days less than B. How long will it take each of them to finish the work alone?

,.,..,,<"~

X+y

D.

,_

-·

On one job, two power shovels excavate 20,000 cubic meters of earth, the larger shovel working 40 hours and the smaller for 35 hours. On another job, they removed 40,000 cubic meters with the larger shovel working 70 hours and the smaller working 90 hours, How much earth can each remove in 1 hour working alone?

A.

il

B. C.

f

D:

169.2, 178.3, 173.9, 2P0.1,

287.3 294.1 347.8 312.4

:lflft: EE Board October :1.997 Ten liters of 25 % salt solution and 15 liters of 35 % salt solution are poured into a drum originally containing 30 liters of 10% salt solution. What is the per cent concentration of salt in the mixture?

A. B. C. D.

19.55% 22.15% 27.05% 25.72%

:1.67: ME Board October :1.99:Z A Chemist of a distillery experimented on two alcohol solutions of different strength, 35% alcohol and 50% alcohol, respectively. How many cubic meters of each strength must he use in order to produce a mixture of 60 cubic meters that contain 40'Vo alcohol?

A. B. C.

40 35 45 38

D. /

/

f

,d9: ME Board October :1.994 Two thousand (2000) kg of steel containing 8% nickel is to be made by mixing a steel containing 14% nickel with another containing 6% nickel. How much of each is needed? A.

B. C. D.

1500 kg of steel with 14% nickel, 500 kg of steel with 6% nickel 750 kg of steel with 14% nickel, 1250 k g of steel with 6% nickel 500 kg of steel with 14% nickel, 1500 k g of steel with 6% nickel 1250 kg of steel with 14% nickel, 750 k g of steel with 6% nickel

,t!fo: How much water must be /evaporated from 10 kg solution which has 4% salt to make a solution of 10% salt? A. B. C. D.

4kg 5 kg 6kg 7 kg

. :1.'f1: EE Bo!li'd October :1.994 If a two digit number has x for its unit's digit and y for its ten's digit, represent the number. A. B

1ox+y 10y +X

·Day 4- Age_,_Wor}S_Mixture, Digit, Motion Problems 71

7 0 100 1 Solved Problems in EnQ'ineering Mathematics (2nd Edition) by Tiong & Rojas

c. D.

yx xy

1?.1-:z: EE Board October 1994

.,;771 GE Board February 199Z The product of

·~

and

i·of

a number is

500. What is the number?

P:.Une number is 5 less than the other. If their sum is 135, what are the numbers?

A. A. B. C. D.

85, 50 80, 55 70, 65 75,60

B.

c. D.

50 75 100 125

.J78: If 3 is subtracted from the ,1?3: ECE Board March 1996 Ten less than four times a certain number is 14. Determine the number. A.

6

B. C. D.

7 8 9

ECE Board March 1996 · ·the sum of two numbers is 21 and one number is twice the other. Find the numbers.

B. C. D.

6, 15 7,1'4 8,13 9,12

J'15: EE Board April1993 lf eight is added to the product of nine and the numerical number, the sum is seventy-one. Find the unknown number.

A. B. C. D.

5 6 7 8

1'76: Find the fraction such that if 2 is (ubtracted from its terms its becomes 1/4, but if 4 is added to its terms it becomes 1/2.

A.

3/5

B.

5/12 5/14 6/13

c. D.

A.

35/55

B.

36/55

c.

~4:

A.

numerator of a certain fraction, the value of the fraction becomes 3/5. If 1 is subtracted from the denominator of the same fraction, it becomes 2/3. Find the original fraction.

D.

3/7 32/41

,,f:t79: ECE Board November 1997 •' The denominator of a certain fraction is three more than twice the numerator. If 7 is added to both terms of the fraction, the resulting fraction is 3/5. Find the original . fraction.

A.

8/5

B.

13/5

c. 5/13 D.

3/5

180: Find the product of two numbers ' such that twice the first added to the second equals 19 and three times the first is 21 rnore than the second.

A. B.

c.

24 32

18

D. 20 j ·'1:81: The tens' digit of a number is 3 less than the units' digit. If the number is divided by the sum of the digits, the quotient is 4 and the remainder is 3. What is the original number?

A. B.

36 47

C.

58

/1,S&: CE Board November 1.994

D.

69

· An airplane flying with the wind, took 2 hours to travel 1000 km and 2.5 hours in flying back. What was the wind velocity in kph?

l8%: The second of the four numbers is three less than the first. the third is four more than the first and the fourth is two more than the third. Find the fourth number if their sum is 35. A. B. C. D.

10 11 12

..,..,.,,.,z,

l83: EE Board April1997 A jogger starts a course at a steady rate of 8 kph. Five minutes later, a second jogger starts the same course at 10 kph. How long will it take the second jogger to catch the first? 20 min 21 min 22 min 18 min ,c''

.1114: EE Board April1997 A boat man rows to a place 4.8 miles with the stream and back in 14 hours, but finds that he can row 14 miles with the stream in the same time as 3 miles against the stream. Find the rate of the stream. A. B. C. D.

1.5 miles per hour 1 mile per hour 0.8 mile per hour 0.6 mile per hour /"

_J.SS: ECE Board November 1998 A man rows downstream at the rate of 5 mph and upstream at the rate of 2 mph. How far downstream should he go if he is to return in 7/4 hours after leaving? A. B. C. D.

2.5 3.3 3.1 2.7

50 60 70 40

,;::187: CE Board May 1.998

13

A. B. C. D.

A. B. C. D.

miles miles miles miles

A boat travels downstream in 2/3 of the time as it goes going upstream. If the velocity of the river's current is 8 kph, determine the velocity of the boat in still water.

A. B.

c. D.

40 kph 50 kph 30 kph 60 kph

/188: Two planes leave Manila for a southern city, a distance of 900 km. Plane A travels at a ground speed of 90 kph faster than the plane B. Plane A arrives in their destination 2 hours and 15 minutes ahead of Plane B. What is the ground speed of plane A?

A. B. C. D.

205 kph 315 kph 240 kph 287 kph

1.89: EE Board April 1.997

#'

A train, an hour after starting, meets with an accident which detains it an hour, after which it proceeds at 3/5 ot its former rate and arrives three hbur after time; but had the accident happened 50 miles farther on the line, it would have arrived one and one-half hour sooner. Find the length of the journey. A. B. C. D.

910/9 800/9 920/9 850/9

miles miles miles miles

72 l 00 I· Solved Problems in Engineering Mathematics (2"d Edition) by Tiong & Rojas · 190: On a certain trip, Edgar drive 231 km in exactly the same time as Erwin drive 308 km. If Erwin's rate exceeded that of Edgar by 13 kph, determine the rate of Erwin. A.

39 kph

B~

44 kph

D.

48 kph 52 kph

c.

·~

')~,

'L •I'!

'\'' :'I:'

,·!

Topics

.,,,

------·-"-'""'""-"".._..._u,_____,,__,

0

( (

Age Problems Work Problems Mixture Problems Digit Problems Motion Problems Coin Problems

Mon

,:

0

1,\\

Tue

0 0

0

[

0 0

Theory

·1v·

nr. :{f i~

Problems

Solutions

0

Notes

Wed

[QJ Thu

Fri

Sat

ANSWER KEY

ill' I

141. B 142.8 143. A 144.0 145.C 146. D 147. B 148.A 149. c 150. B I 151. A 152. B 153. B

154. 0 155. c 156.C 157.A 158.A 159. B 160.A 161. B 162.A 163. c 164. B 165.C 166.A

167. 0 168. A 169. c 170. c 171. B 172. c 173. A 174. B 175.C 176.C 177. c 178. B 179. c

180. A 181. B 182. 0 183. A 184. D 185. A 186.A 187.A 188.C 189. B 190. D

0

RATING

43-5{] Topnotcher

c::J 33-42 Passer c:J 25-32

D

1:11

Conditional

0-25 Failed lfFAILEO, repeat the test.

74

100'1 Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas

Ill

•

x-12=3 X=9

Present 24 X

Nilda Riza

3x = 3(9)

Sum of ages = 9 + 27 =36

X= 18

Ill

Ill Future X +3 v+3

,,. X+ y =18 y=18-x (y + 3) = 2(x + 3)

~Eq.1 ~

Eq. 2

=

Y= 3x

~

Eq.1

z.= x + 8 X + y + z = 38

~

Eq. 2 Eq. 3

~

X

X=6

lEI

X=5

x = 1On- 5

Ill X

Present 38

0

X

Past (b+15-y)=2(b-y) b + 15- y = 2b- 2y

Father Son 38- X=

b-y=15

X-

0

X =19

Past 3x -3 x-3

3x- 3 =4(x- 3) 3x- 3 = 4x -12

Present 3x X

Two years ago, the son was (19- 2) = 17 yea'rs old

n=5

~

1

1

1

1 2

1 4

X

1

-+---=5 4 20 X x = 2.5 hours

---=x = 4 hours

1 1 1 1 -+---=9 12 15 X x = 7.826 hours x = 7 hours & 50 minutes

•

X

Present 2x X

t

~

Let: x = time needed to complete the work 1 1 1 -+-=20 11 X x = 7,.096 hours

Let: n = number of days needed to complete the work 1 1 1 -+-=-

~'

'!:

2b - b- 2y + y = 15

-+ Eq. 1

2x + 30 = x + 15n X = 15n- 30 -+ Eq. 2

•

=10

• • •

liD

15n-30=10n-5

18-x = 28-2x X

X

Equate equations 1 and 2:

(18-x)=2(14-x) Present b + 15 b

Future 2x+30 x+15n

Present 2x

2x = 20n-10

y =13

Past b+ 15- v b-y

Past 2x-10 x-5n

2x -10 = 4x- 20n

·J

Present 18 14

ml

X =17

2x-10=4(x-5n)

i

y=18-5

-2x+45=111

Parents Children

l

x+3x+(x+8)=38

21-x=2x+6

•

Let: x = age of the girl y age of her brother z = age of her sister

(18- x) + 3 = 2x + 6

Ill

Future 3(x+5) x+5

?

3x + 15- 5x + 30 = x + 5- x + 6

Substitute equations 1 & 2 in equation 3:

Substitute y in equation 2:

I

Present

3(x + 5) -5(x -6) = x +5-(x -6)

2x =36

I I

Past 5(x-6) x-6

3x= 27

24- x =x-12

Robert Stan

Day 4- Age, Work, Mixt~e, Digit, Motion Problems 75

2x- 3(x- 4) = x -(x -4) 2x - 3x + 12 = x - x + 4 -x+12=4 X=8 2x=16 Thus, Debbie is now 16 years old.

y n

_!=x+y n xy

•

n=.2L X+Y

Let: A = number of hours, Pedro can paint the house B = number of hours, Juan can paint the house C = number of hours, Pilar can paint the house 1 1 1 1 -+-+-=~Eq.1 A B C 4

_!_ = 1.5(1) · _!_ = 0.666(2-) ~ E . 2 B'B A· q A


76 100 i Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas

~=2(~)

*=1.2(~} ~=0.833(l) ~ Eq. 3 28-5 8 2 -58 128 - 30 = 8 2 - 58

=e-

Substitute equations 2 & 3 in equation 1:

A.

82

1 (1) +0.833 (1) A+0.666 A = 1

4

-

Note: (rate)(time) = 1(complete job) 1 1l 1 =1 ( -+-14+-(x) 9 16; 9 0.6944 + 0.111x = 1 x = 2. 75 hours

ml

x = time for l!utch to finish a certain job working alone y = time for Dan to finish a certain job working alone 1 1 1 -+-=-

~

y 6

X

·~

X= 2y

X

1 1 1 -+-=A B 42 1 1 1 -+-=8 31 1 1 1 -+-=A C 20

~

~

1

1

Add the three equations:

1+ 2

1

+ [ _!_ A B

y = 9 days

•

1

1

-+-=A B 6 A=B-5

~

Eq. 1

~

Eq.2


1 1 1 -+-=B-5 B 6 B+(B-5} 1 B{B-5) =5

2

,. '

-+-+- = 0.106 A B C 1 1 1 1 -+-+- =0.053 =A B C x x = 18.87days

•

Let: x = time for Myline to finish the jo y = time for Jeana to finish the job

1

1

1 -+-=X y 6

~Eq.1

(1)

1 [1 1]

-(3)+ -+- (3)=1 X X y

.~.·.·

2

x~19days

.!)

4 6 7 42+28+24 X=_:_-~-=-.. 168 94 X=168 47 X=84

i

'

Substitute x

--+ Eq. 2

70( 500- 0.875y) + 90y =40000 35000-61.25y+90y = 40000 y = 173.9 m3 /hr X= 500-0.875{173.9)

•c:]+I 10

~

x = 347.8 m3 /hr

35%

I+ c:J

15

=

c:J

30

55

Eq.1

=8 in equation 1:

r~(3)+[i+~ J(3)= 1H

m

70x + 90y ,;, 40000


Let: x = time for the farmer to ~_!}he field y = time for the son to flow the field,

Eq. 3

=0.106

2

•

2

..!] +[..!B + _!_] +[_!_ + ..!] = ~ + __!_ + ~ C A C 42 31 20

2Y=6 1

Eq.t

~·~q.

c

Eq.2


1

c

=(..! + _! +

1

40x + 35y = 20000 X= 500- 0.875y --+ Eq. 1 .

Let: x = fraction of the lawn that can mowed after one hour X

A B

6

y =18 hours

•

=15 - 5 =10 days

Let: x = number of days needed by A, B and C to finished the work working together. 1 1 1 1 -+-+-=-

Eq. 1

-+-=2y y 6

y

•

1

-+-=200 300 X x = 120 man-hours

Let: x = capacity of the larger shovel in m3/hr y = capacity of the smaller shovel in m3/hr

3

Substitute B = 15 in equation 2

•

1

-=-

B = 15 days B = 2 (absurd)

A

1

-+-=y y 6

{B-15}(8-2)=0

El

Let:

2

1

Eq.2


178 + 30 = 0

A= 10 hours

~

1 1 1 1 -+-+-=8 8 y 3 y = 12 days

Let: x =number of man-hours needed by

crew number 1 and number 2 to finish the job.

0.25{10}+0.35{15) + 0.10(30) = x(55) X=19.55%

•

c=]+lso%1=14o%J x

60-x

60

For the 35% solution: 0.35(x) + 0.50(60- x) =0.40{60) 0.35x + 30- 0.5x = 24 X=40 m3

18 1001 Solved Problems in Engineering Mathematics (2"d Edition) by Tiong & Rojas For the 50% solution: .Let:

60-x=20~ m 3

•c=:J c=J c::J +

x

70- 0.7x + 0.6x = 66 x =40 ounces

•

X=5

2x = 140

I~i1i

x

c:J= c:J 2000-x

2000

y = 14

x-5 =65

Ill

L:et:

1111

x = the number

X=6

DD

5x + 35

Thus, the numbers are 7 and 14.

2000 - X = 1500 kg

•c::J- c:J= 6 X

10-x

0.04{10)- O(x) = 0.10(10- x) 0.4 = 1-0.1x

•

~(x) J= 500

X=6 kg

Let: y = tens' digit of the number x = units' digit of the number

The two digit number is represented by: 1Qy +X.

~~~~

Let x = the number

~~l

Let:

X

=100

~ = the fraction y

X -3

3

y

5

3y =

5x -15 ~Eq.1

2

3x=y+21 ~Eq.2

y~2

4


4x-i3 = y -2

,-;,_"

~

y=4x-6

Eq.1

=2[~x -5 ]-2

10 3x=-x-10-2 3 0.333x = 12

X+4

--=y+4 2 2x + 8 = y + 4

3x

~

Eq. 2

X

=36

2x+8=y+4Jar

y

5

3

~

~

Eq. 1 Eq. 2

Substitute equation 1 in equation 2: 3x=(19-2x)+21 5x=40 x=8 y=19-2{8) y=3 :. prqduct of the numbers is 8(3)

(36) 5 -55

13

2x+ y = 19

3

-=-

~.

x = the first num.ber y = the second number

Let:

y = 19- 2x

3x=2y-2

x-2

Therefore, the original fraction is

IB!I

--=-

X y-1

y

y=2(5)+3

y = 13 Let:

3

~ = the fraction

X=5

x 2 = 10000

X=7

Ill

Eq. 2

5x + 35 = 6x + 9 + 21

20

5 y=-x-5

· 9x+8=71

~

5x + 35 = 3(2x + 3) + 21

~=500

•

2x =.14

=3y + 21


·.~

0.14x + 120- 0.06x = 160 X= 500 kg

y+ 7

2

x=7

Eq. 1

Let: x = the number

[ l(x) ][

Let: x =-the first number 2x = the sec;ond number

~

3 5

X+7

5

Therefore, the fraction is - . 14


0.14(x) + 0.06(2000- x) = 0.08(2000)

10

Y = 2x + 3
x + 2x = 21

For 6% substance:

x = numerator of the faction y = denominator of the fraction

y=4(5)-6

1

For 14% substance:

I

Let:

Ax-10=14

~+

ml

10 =2x

X= 70

. . 36 Thus, the fractlon 1s -

55

2x +8 = (4x -6)+4

x+(x-5)=135

. 100

0.70(1 00..., x) + 0.60(x) = 0.66(100)


x = the first number x - 5 = the second number

=

100-x


= 24:

''II

'li

I 80 IOO"l Solved Problems in Engineering Mathematics (2"d Edition) by Tiong & Rojas

Ill Let:

w + (w- 3)+(w +4) +(w + 6) = 35

•=u-~

W=7

'~~

·r.

,''.if

z =13 t+U 10t+u-3

,:\r .I

Ill

t+U

4 t+U 10t+ u -3 = 4t + 4u 6t-3=3u ~· Eq. 2

Let: V 1 = velocity of the first jogger V2 = velocity of the second jogger

t1

. Substitute equation 1 in equation 2:

Point where · jogger 2 · starts running

14 3 ---------·-·

t, =25 min. '·.!··'..

~

Z=7+6

~ Eq.1

at,= 101,-50 V1 +V2

,I

~~ '.':.~

Substitute w = 7 in equation 3:

10t+U= 4 +~

.,

4w=28

t = ten's.digit of the number u = unit's digit of the number Number= 10t + u

--------'-----.:D=ay'-4 -Age, Work, Mixture, Digit, Motion Problems ~J.

14V1 -14V2

t2 = 25-5

Let:

'""

V1 -V2

3V1 + 3V2

11V1 = 1'7V2

t 2 = 20 min.

IIEII

'::':.

v, = velocity of boatman v2 =velocity of stream

V1 = 1.545V2

-~ Eq. 2

Substitute equation 2 in equation 1: 2

49.6(1.545V2 ) = 14(1.545V2 ) -14V/

Total time= 14 hrs

·-.(···· .............................................................

-

i4.832V2

=19.418V/

v2 "'0.76 mph

Direction of the stream current ~~~.%,'{W"''V'

f.t----------------1

!IP.Jl!ll ll&iill

Note: time=

dis tan ce -v,.ei"OcitY

4.8miles

6(u-3)-3=3u fhe two figures below that the same time.

6u-18-3=3u 3u=21 u==7

t =4

/ ................................. ····l·.................................. c...

Number= 47

~~, P

y =w+4

-7 Eq.2

Point where jogger 2 catches up jogger 1

z =w+6

V,

=

->Eq.4

Substitute eq1f3tions 1, 2 and 3 in equation 4: /

I 1\

5 2 4 0.7S =d.75.

--c_~·

S=;2.5 miles

3miles

~~+~=14

Skmx~=_!_km/min 60min

v2 = ~~ km/min

7 Eq. 3 ..----------

w+xr35

. v,_ .. v2

•

hr

z = y + 2 = (w + 4) + 2

.... -

s s 7 -·+-=-

~

~'\WI'

s2

x = second number y = third number z =fourth number

-7 Eq. 1

t, + t2 = ttotol

twtttfl!.t!lklfW*iP!X!lf.. ?'MP.tEt.,. !IC,.i!IM.

w = first number

x=w-;3

14miles

--=-- v2

Number= 10{4)+ 7

•

-v,+V2

t2=t1-5

t=7-3

Let

2__,__ ~----.r

s1

S1 =S 2 = V2t 2

v,t, 8

10

60

60

-t, =-(t,-5)

V1 +V2

60

v,- V2

4 8(V1 +V2 }+4.8(V1 + V2 )

-

(V1 + V2 }(V1 - V2 )

,1 IIV1 ~: 4.8V2 + 4.8V1 + 4.8V2

V/ . . v,v2 + v,V2 - v/ 96V1

=14

=14

14V/-14V/~Eq.1

Let:

=

V, velocity of airplane V2 =velocity of wind

,

11=2 .....................................................................

~ I r

, ! !

+I

s,=1ooo

1000 v, + V2 =----·= soo

'

I

V!+V2-+

2

1

7 Eq. 1

I

...

82 ·100 I Solved Problems in Engineering Mathematics (2nd Edition) by Tibng & Rojas Direction of the wind -

I 2= 2.5

L.,

-<······· ...................................... .

.

-

I·

"-

v.-v, 82=1000

•

v, - v2 = 1ooo =400 2.5

~

1

Eq.2


Let:

v, =ground speed of plane A

V2

Let: t = time needed to travel and reach destination without any delay V = velocity of the train

'= ground speed of plane B

~ ~-v

,

PlaneA.................................. . .............._.............

,, ,,'

:w

s = 900

1

·_.:1-~·. .,

']'

(v, + V2 ) - (v, - V2 ) "' 500 -

400

2V2 = 100

v2 =50 kph

Plane B .

t-2.25

..,cz~ -V+90

Day 4- Age, Work, Mixture: Digit, Motion Problems 83

•

RR

\:

Ill =

-

-

V+8

Direction of stream current

F- :,== DireCtion of -stream current

=0 ·

(V-8)1=(

~V+ 163 ;A3=~

I\

.

3

5

3

3

2

8 800 miles 9

Let: V = rated of Erwin V - 13 = rate of Edgar

~

~v

-V·13

-,~,.,,

s

.

S- 231 km.

b . Su st1tute t = V :

12

s -(S- V)=-+1 5

!~ph

2

11

S-V =t+1

3V

v =2-:r:, )~ph ·E:.·

I9R

5

=0

V

5 250 5 . v 50+V+-S----\i =S+-

lllii6ill

-V

V

~(~)=~

(absurd)

s

V=4

S=308km

t, = h ~

Eq . 1

V=40 kph < :ondition

I

3V

s =

. S-V 1+1+-3- = t+3

(V -240}(V +150) = 0 V

+~(S-50- Vh~+_!] V

S = Vt + 202.5 "" 0

V/

5 50+ V [ V

3

Divide all by 2.25:

S 1 =S 2

50+V + 1+ S-(50+V) =I+~ v ~v 2

=--S

2.25V2 - 81000 + 202.5V = 0 2.25V2 + 202.5V -· 81000 = 0

V 2 +.90V- 36000

"

S =VI

250 -~(§.)=s+.!(~)- 100 50 +§.+~S4 3 3- 3 4 2 4 3

v

. (900) V

~v

Substitute equation 1 in above equation:

Vt = Vt- 2.25V + 90t- 202.5

2.25V- 90

S-V-50

t2

5

I·

~

2.25V- 90t + 202.5 900 But t =

S2=S-V-SO

v +50 v

t, =

Point where the accident happened

Multiply both sides by V:

V-8-

.l

Condition 1: If the accident happened 1 hour after, substitute values to the · general equation:

S1 =S 2 Vt = (V + 90){1- 2.25)

J/ s, = v +50

=

Let: V = velocity of the boat in still water s, = distance traveled upstream s2 distance traveled downstream

Mi£li ,...•. III:J.

General equation: Time consumed by the train traveling, before the accident + Time during which the train was detained + Time needed to continue the course and Time needed reach the destination to travel and reach the destination without any delay + Time of delay

\1

Point where the accident happened

2: If the accident happened 50

111lln'~ fnrthcr. suhstitutP. ve~luP.s to tiH' 'i<'IH!IOIII•qliOIIIOil

308

231 V-13

231 v

v

V

308 v -4004 =52 kph

=

:I·

86 1001 S~lved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas ~

----

-Topics

D D D D 0 D ~ D D tv1on

Tue

Theory

Wed

Problems

Thu

·splutions

Fri

Notes

Sat

~I

What is a clock problem? A clock problem is a mathematical problem which focus on the relationship of tile movements of the hands (hour hand, minute hand, second hand) of the clock. rhis type of problem is for mechanical docks only and never for a digital clocks.

Clock Problems Variation Problems Diophantine Equation Sequence Series Arithmetic Progression Geometric Progression Infinite Geometric Progression Harmonic Progression Figurate Numbers

Most of the problems in clock problem involve only the minute hand and the hour hand. The diagram shows the relation between movement of the minute and hour hands.

rhe longest hand is the second hand while llle shortest hand is the hour hand. By principle, the second hand ($H) always rnoves faster than the minute hand (MH) ;md the minute hand always moves faster lllan the (HH). The relation between the thn~e hands of the clock <:~re as follows: HH"'MH

12

HH=~ 720

MH..,.!t! 12

where : SH is in number of seconds MH is in numb('r of minutes

\

~ 12

The earliest known clock problem was posed in 1694 by Jacques Ozanam in his

~~ 1. Solved P:ro~~ns in Engineering Mathematics {?nd Eqition) by Tiong & Rojas

"Recreations Mathematiques Physiques".

et

What is Variation Problem? Variation problems are problems in Algebra which show the relationship between the variables in terms of expressions such as "directly proportional or inversely proportion or simply proportional". The expression x varies directly as y is expressed as follows: xocy The symbol varies ( oc ) is replaced by an equality symbol and a constant of proportionality, k, hence: X

=ky

The expression x varies inversely as the square of y is expressed as follows:

Day 5- Cl_ock, Variation, Miscellaneous Problems & Progr§lssion 89

Diophantine equation is named after a Greek mathematician, Diophantus of Alexandria (AD c200- c284) who

What is a progression? A progression is simply another term for a sequence.

" developed his own algebraic notation and is sometimes called the "Father of Algebra".

What is are the types of progression?

k_!_ y2

A Diophantine equation is an equation that has integer coefficients and for which integer solutions are required.

1" Arithmetic Progression (AP) 2. Geometric Progression (GP) 3, Harmonic Progression (HP)

A sequence is a set or collection of numbers arran9ed in an orderly manner s"uch that the preceding and the following numbers are completely specified.

A sequence is said to be in arithmetic progression if its succeeding terms have a common difference"

An infinite sequence is a function whose domain is the set of positive integer. If the domain of the function consists of the first n positive integers only, then it is said to be a finite sequence.

The corresponding sum of all the terms in arithmetic progression is called as arithmetic series.

1+3+5+7+9 -

There are only two formulas (i.e. last term and sum) to remember and used in solving a problem in arithmetic sequence.

finite sequence

1,1

\

a0

Elements are the term used to describe the numbers in a given sequence" An element is·sometimes called a term.

S=%(a 1 +a.) or where:

What is an alternating series?

What is the difference between a converging series and a diver..aent series?

=a1 +(n-1)d

,I

I

If an infinite series has a finite sum, it is referred to as convergent sorlos and divergent series if it has no sum nl all

S=%[2a 1 +(n-1)d]

a 1 = first term an= last term (n1h term) n = number of terms d = common difference d a2- a, a3- a2 = ...

=

=

What is a ge.ometric progression? !\ sequence is said to be a geometric

progression if its succeeding terms have common ratio.

.1

IIH~ corresponding sum of all the terms in .,, ·onwtric progression is called as quometric series"

l

S

or

= a1(1-rn) 1-r

r-1

where: a, = first term an = last term (d" term) n = number of terms r= common ratio=

a2 a1

=

a 3. = ... a2

What is an Infinite Geometric Progression? This type of progression is a geometric progression only that the number of terms (n) is extremely large or infinity. If r > 1, sum of all terms is infinite If r < 1, the sum of all terms is

Sum of all terms:

What .is a series?

An alternating series has positive and negative terms arranged alternately.

it,

Last term (n1h term):

What is an element?

Despite its simple appearance, Dir;>phantine equations can be fantastically difficult to solve. A notorious example comes from Fermat's Last T~which is a" = b" + c" , where

Sum of all terms:

S = a.1(rn .:_1) What is an arithmetic progression?

What is the difference between an " infinite and a finite sequence?

Series is the sum of the terms in a sequence.

o)fe

" an == a1rn-1

What i~ a sequence?

The best examples of Diophantine equations are those from Pythagorean Theorem, a 2 + b 2 = c 2 , where a, band c all required to be whole numbers.

Diophantine eqw;rtions may refer to a system of equations where the number of equations is less ihan the number of unknowns. Ttjese equations yield whole a11swers" number for

Last term (n 11' term):

The most common types of progression are:

1 + 3 + 5 + 7 + 9 + ··.-,..infinite sequence X=

Also, there are only two formulas (i.e. last term and sum) to remember and used in solving a problem in geometric sequence.

S=-3_ 1-r where: a, = first term · r = common ratio What is Harmonic Progression? A sequence of numbers whose reciprocals form an arithmetic progression is known as harmonic progression. In solving a problem, it would be wise to convert all given terms into arithmetic sequence by getting its reciprocals. Use the formulas in arithmetic sequence and take the reciprocal of resulting value to obtain the equivalent harmonic term for an answer.

~

Dav 5- Cloc:k, Variation, Miscellaneous Problems & Progression 91

90 1001 Solved Problems in Engineering Mathematics (2"ct Edition) by Tiong & Rojas D.

What are the Fibonacci and Related Sequences? The following are the some of the famous and related sequences:

1.

Fibonacci Numbers - Named after the Italian merchant and mathematician, Leonardo di Pisa or Fibonacci (Figlio dei Bonacci, "Son of the Bonnaccis").

1, 1, 2, 3, 5, 8, •13, 21, 34, 55, 89, 144 ... Each number is equal to the sum of the two preceding numbers.

ObloJOtg numbers: Numbers which can be drawn as dots and arranged in a rectangle shape.

• ••••• ••••• •••••• ••••• •••••• •••• ••• •••• •••••• •• ••• •••• • •••• • ••••• 2 6 12 20 30

......

E.

. (.: 5

Lucas Sequence - Named after Edouard Lucas (1841- 1891). Like the Fibonacci numbers, every term of the Lucas. sequence is the sum of the two preceding numbers.

' fit {jjJ ril .

\~l -:-:11 12

Cubic numbers:

riD

•

35

22

8

.- If

....,__. -· ~re

t=-~ ~:-·

....

~

....

27

64

1' 3, 4, 7, 11' 18, 29, 47, 76, 123 ... Tetrahedral numbers:

Figurate Numbers: Triangular numbers: Numbers which can be drawn as dots and arranged in triangular shape .

A.

. .·.··'·

• •• • •• •• • • •• • ••• •• •••• • •• ••••• • ••• •••••• • •••• • •• 1 3 6 10 15 21 B.

•

•••• ••• •••• •••• •• ••• • ••• •1 ••4 ••• 16 9 C.

25

•••••• •••••• •••••• •~····· ••••• •••••• 36

Gnomons:y.«fmbers which can be drawr;Vas dots on equally long leg7:sa right angle.

•

1

• '• • ••• • 3

5

•• • • • • • 7

•

••• •

• • • • •

9

•• ••

• • • • e •

11

4;. 4

H.

Square numbers: Numbers which can be· drawn as dots and arranged in square shape.

••••• ••••• ••••• ••••• •••••

Did you know that... the eminent German mathematician, Carl Freidrich Gauss' father was an accountant and· young Gauss corrected his father's spreadsheet at the age of 3 !

@uote: "Mathematics is the queen of sciences and arithmetic is the queen of mathematics. She often condescends to render service to astronomy and other natural sciences, but under all circumstances the first place is her due." - Carl Freidrich Gauss

G.

3.

GOOD LUCK I

\!l:ribia:

Pentagonal numbers:

F. 2.

Proceed to the. next page for your 5th test. Detach and use the answer sheet provided at the last part of this book. Use pencil number 2 in shading your answer.

•1 I.

4. ·~ . 10

20

Square pyramidal numbers:

~

•• 14

30

Supertetrahedral numbers:

By piling up tetrahedral numbers, 1, 4, 10, 20, 35, 56, etc, we make four-dimensional numbers: 1, 5, 15, 35, 70, ...

1 1+4 1+4+10 1 + 4 + 10 + 20 1 + 4 + 10 + 20 + 3!1

=1 =5 =15 :15 {()

Day 5 -Clock, Variation, Miscellaneous Problem!J & Progression 93 /

.1.11&: EE Board October 1990

T~pics

Dr Mon

p

j

D

~

j j

Tue

LJ D D lJ ~ D D

Problems

Solutions

Thu

Fri

"'

1gt·; CE Board May 1995

.fu how many minutes after 2 o'clock will the hands of the clock extend in opposite directions for the first time?

C. ·D.

A. . 3:02.30 B. 3:17.37 C. 3:14.32 D 3:16.36

.:144: GE Board February 1997

42.8 minutes 43.2 minutes 43.6 minutes

"'At what time after 12:00 noon will the hour hand and minute hand of the clock first form an angle of 120°?

19ii CE Board November 1995

A. B.

the hands l;le directly opposite each other for the first time?

D.

minutes minutes minutes minut

c.

12:18.818 12:21.818 12:22.818 12:24.818

./1.95: At what time between 8 and 9 o'clock will the minute hand coincide with the hour hand? ~

193:,/'~ CE oard.· May 1997 What ti e after 3 o'clock will the hands of the clo · k be together for the first time?

I

C.

D. /

197: GE Board February 1994 From the time 6:15PM to the time 7:45 PM of the same day, the minute hand of a standard clock describe an arc of A.

60°

B.

go•

D.

540•

c. fso·

A.

B.

3 4

c

5

D.

6

A. R

c. D.

8:42.5 8:43.2 8:43.6 843.9

~oo: CE Board May 1993 ·· Given that "w" varies directly as the product of "x" and "y" and inversely as the square of "z" and that w = 4 when x = 2, y = 6 and z = 3. Find the value of "w" when x = 1, y = 4 and z = 2.

A. B.

199: ECE Board April :1990 lhe resistance of a wire varies directly with its length and inversely with its area. If a certain piece of wire 10 m long and C.10 em in diameter has a resistance of tOO ohms, what will its resistance be if it 1s uniformly stretched so that its length becomes 12 m? /\. ll.

80 90

c

144 120

I)

3 4

C.

5

D.

6

""

~1: ECE Board November 1993 If x varies directly as y and inversely as z, and x = 14 when y = 7 and z = 2, find the value of x when y 16 and z 4.

=

A. B. C. D.

=

14 4 16 8

~: EE Board Marc:h 1998 The electric power which a transmission line can transmit is proportional to the product of its design voltage and current capacity, and inversely to the transmission distance. A 115-kilovolt line rated at 100 amperes can transmit 150 megawatts over 150 km. How much power, in megawatts can a 230 kilovolt line rated at 150 amperes transmit over 100 km?

A.

B.

/

42A minutes

5.22 5.33 5.46 5.54

33147 321.45 346.10 3 36.50

which is proportional to the charge. If the charge is reduced by 50% of its original value at the end of 2 days, how long will it take to reduce the charge to 25% of its original charge?

'1n how many minutes after 7 o'clock will A. B. G. D.

A. B.

198; EE Board Aprill990 A storage battery discharges at a rate

Sat

Notes

A. B.

j

VVed

Theory

il

I

Clock Problems Variation Problems Diophantine Fquation Sequence Series Arithmetic Progression Geometric Progression Infinite Geometric Progression Harmonic Progression Figurate Numbers

A man left his home at past 3:00 o'clock PM as indicated in his wall clock, between 2 to 3 hours after, he returns home and noticed the hands of the clock interchanged. At what time did the man leave his home?

C. D.

785 485 675

595

~--3~ ME Board October 1992

a

'ihe time required for an elevator to lift weight varies directly with the weight and the distance through which it is to be lifted and inversely as the power of the motor. If it takes 30 seconds for a 10 hp motor to lift 100 lbs through 50 feet, what size of motor is required to lift 800 lbs in 40 seconds through 40 feet? A.

B. C. D.

42 44 46 48

Day 5 - Clock, Variation,

94 . 1001 Solved Problems in Engineering Mathematics (2"d Edition) by Tiong & Rojas

Miscellane~us

Problems & Progression 95

r

204: The selling price of a 1V set is double that of its cost If the 1V set was sold to a customer at a profit of 25% of the net cost, how much discount was given to the customer?

A.

a

C. D.

~.7% ~.7% ~.5%

~.j%

205: A group of EE examinees decided to hire a mathematics tutor from Excel Review Center and planned to contribute equal amount for the tutor's fee. If there were 10 more examinees, each would have paid P 2 less. However, if there were 5 less examinees, each would have paid P 2 more. How many examinees are there in the group?

D.

14 16 18 20

A. B. C. D.

B. C.

D.

16 20 18 24

206: EE Board Marc:h J:998

The arithmetic mean of a and b is

A bookstore purchased a best selling price book at P 200.00 per copy. At what price should this book be sold so that, giving a 20% discount, the profit is 30%?

A. B.

JO P 500 P 357 P 400

207: ECE Board November 1993 Jojo bought a second hand Betamax VCR and then sold it to Rudy at a profit of 40%. Rudy then sold the VCR to Noel at a profit of 20%. If Noel paid P 2,856 more than it cost to Jojo, how much did Jojo paid for the unit?

A. B. C. D.

p 4,000 P 4,100 P 4,200 P 4,300

C. D.

a+b 2

Jab ab 2 a-b 2

.zii: The sum of three arithmetic means

B. C. D.

17 18 19 20

:t:t~ECE Board November J:998 ,l:i '~

Find the 30th term of the arithmetic progression 4, 7, 10, ..

A. B. C. D.

75 88 90 91

CE Board May 1994, CE Board November J:994 How many terms of the progression 3, 5, 7, ... must be taken in order that their sum will be 2600?

B. C. D.

A. B. C. D.

114 124 134 144

48 49 50 51

U.J:1!:E Board Marc:h 1998 :t:;ravity causes a body to fall 16.1 ft in the first second, 48.3 in the 2"d second, 80.5 in the 3'd second. How far did the body fall during the 101h second? 248.7 ft

n.

6~

C.

6~ ~

. CE Board May 1998 )

~2·J:'9:

• Determine the sum of the progression if there are 7 arithmetic mean between 3 and 35.

C. D.

11 12 13 14

./'

A besiege fortress is held by 5700 men who have provisions for 66 days. If the garisson losses 20 men each day, for how many days can the provision hold out?

C. D.

In the recent "Gulf War" in the Middle East, the allied forces captured 6400 of Saddam's soldiers and with provisions on hand it will last for 216 meals while feeding 3 meals a day. The provision lasted 9 more days because of daily deaths. At an average, how many died per day?

U'7: CE Board May 1995

A

B.

14; ·19 ... up to the 20th term?

c. D.

A <;

1030 1035 II 140

II

,:145

II

72 74 76 78

.~z.rf'CE Board May 199J:

\:'Jh:1! 1s

tt·"' sum of the progression 4, 9,

171 182 232 216

·2ZO: ECE Board April J:995

B.

In a pile of logs, each layer contains one more log than the layer above and the top contains just one log. If there are 105 logs in the pile, how many layers are there?.

(~.

6~

B.

A.

U6~ME Board Apri11995

A G.

A.

A. B.

:.t~·s: CE Board May 1993

A.

US: EE Board April1997 ,.., A stack of bricks has 61 bricks in the bottom layer, 58 bricks in the second layer, 55 bricks in the third layer, and so on until there are 10 bricks in the last layer. How many bricks are there all together?

D.

between 34 and 42 is

A

.{

A.

209: ECE Board Marc:h 1996 A merchant has three items on sale, namely a radio for P 50, a clock for P 30 and a flashlight for P 1. At the end of the day, he sold a total of 100 of the three items and has taken exactly P 1,000 on the total sales. How many radios did he sale?

308.1 ft 241.5ft 305.9 ft

2~3r-if the first term of an arithmetic 'progression is 25 and the fourth term is 13, what is the third term?

850 500 550 600

,v:z.t-0:~ ME Board October J:996

B. C. D.

C. D.

In a certair.~ community of 1,200 people, 60% are literate. Of the males, 50% are literate and of the females 70% are .literate~ What is the female po'pulation?

A. A. B. C.

B.

208: EE Board Marc:h 1998

15 16 17 18

Day 5 - Clock, Variation, Miscellaneous Prob~J & ProJ.l!ession 92' 96 1001 Solved Problems in Engineering Mathematics (2nd Edition) by T'ong & Rojas

;atZ: GE Board July X99:5 ·"A Geodetic Engineering student got a score of 30% on Test 1 of the five number test in Surveying. On the last number he got 90% in which a constant difference more on each number that he had ·on the immediately preceding one. What was his average score in Surveying?

A. B.

50 55

D.

60 65

c.

/,<' ~~b: ME

Board April 1999 :If the sum is 220 and the first term is 10, find the common difference if the last term is 30.

"pf;blem 226:

C. D.

· When all odd numbers from 1 to 101 are added, the' result is

2~i;' Find the 9 1h term of the harmonic progressio11 3, 2, 3/2 .....

. A. B. C. D.

2500 2601 2501

2 5

c.

3

D.

2/3

2:1'.ci;·'EE Board April 1997 Once a month, a man puts some money into the cookie jar. Each month he puts 50 centavos more into the jar than the month before. After 12 years, he counted his money, he had P 5,436. How much money did he put in the jar in the last month?

A. B.

c. D.

P 73.50 P 75.50 p 74.50 P 72.50

~z71'"How many limes will a grandfather's /clock strikes in one day if it strikes only at the hours and strike once at 1 o'clock, twice at 2 o'clock , thrice at 3 o'clock and so on? A.

210

B.

24 156 300

c.

/If

A girl on a bicycle coasts downhill covering 4 feet the first second, 12 feet the second second, and in general, 8 feet more each second than the previous second. If she reaches the bottom at the end of 14 seconds, how far did she coasts? A.

782 feet

B. C.

780 feet 784 feet 786 feet

4/5 4/9 ..... -~

I

A. B. C. D.

130 140 150 160

233: EE Board October :199:1 The fourth term of a G. P. is 216 and the 61h term is 1944. Find the 81h tem1:

<>"~

"-'. Uth CE Board May :1992 · To conserve energy due to the present energy crisis, the Meralco tried to readjust their charges to electrical energy users who consume more than 2000 kwhrs. For the first 100 kw-hr, they charged 40 centavos and increasing at a constant rate more than the preceding one until the fifth 100 kw-hr, the charge is 76 centavos. How much is the average charge for the electrical energy per 100 kw~hr? A. B. C. D.

3/5 3/8

:a3'i: Find the sum of 4 geometric means between 160 and 5.

c

B.

17649 17496 16749

I>

17964

A.

I

7, 7, 7, 7,

-7/12 -5/6 . -14/5 - 716

The 3'd term of a harmonic progression is 15 and the 91h term is 6. Find the 11th teim. · ·

Z}Sr ECE Board April 1999 II one third of the air in a tank is removed hy each stroke of an air pump, what ''action a I part of the total air is removed in 1; strokes?

A.

4

(\

B. C. D.

5 6 7

II

A..

B.

1/10 1/11

·~

A.

14336

B.

13463 16433 16344

c. D.

,.A:J's: ECE Board April :1998 The sum of the first 10 terms of a geometric progression 2, 4, 8, ... is

:t:J9f'if the first term of a G.P. is 9 and 1he common ratio is -213, find the fifth term.

D

I

The numbers 28, x + 2, 112 form a G. P. What is the 101n term?

he a geometric progression.

('

I

~: CE Board May 1995

1596

60 centavos 62 centavos 64 centavos

1/2' 0.2, 0.125, ~ ..

P 213.23 p 202.'15 P 302.75 P 156.00

D.

II.

.•. ~;;: ECE Board November 1995 Find the .fourth term of the progression

A. B. C. D.

ZJ4: ECE Board April 1999 Determine x so that: x, 2x + 7, 1Ox - 7 will

1\.

·.

A product has a current selling of P 325.00. If its selling price is expected to decline at the rate of 10% per annum because of obsolescence, what w\11 be its selling price four years hence?

2046 225

58 centavos

/"

,:zj"(;~ ME Board October 1996

A B. C.

p

... -2291 CE Board November :199:5

.,&25: EE Board AJiril1997 ·

b.

A. B. C. D.

3500

D.

A. B.

0.102 0.099

,.. ll

0.7122 0.9122 0.6122 0.8122

1023

A. B. C.

8/5 16/9 1517

D.

1314

.,2401 EE Board April 1997 The sev~nth term is 56 and the twelfth term is --1792 of a geometric progression. Find the comro~,..., ral,..; i;jnd the first term. Assume the ratios .are equal.

A. B. C. D.

-2, -1, -1, -2,

5/8 5/8 7/8 7/8

98 .1 00 1. Solved Problems in Engineering Mathematicl:' (2nd Edition) by Tiong & Rojas

,~.u; A person has 2 parents, 4 grandparents, 8 great grandparents and so on. How many ancestors during the 15 generations preceding his own, assuming no duplication?

A. B. C.

131070 65534 32766 16383

D.

.....

-~Zlf$:

EE Board March J:998 Determine the sum of the infinite series:

s= A.

20 19 18 21

B. D. A'

24'3: CE Board November 1:994 /In a benefit show, a number of wealthy men agreed that the first one to arrive would pay 10 centavos to enter and each later arrive would pay twice as much as the preceding man. The total amount collected from all of them wasP 104,857.50 How many wealthy men paid?

A.

18 19 20 21

B.

c. D.

c. D.

z4~: A man mailed 10 chain letters to ten ··of his friends with a request to continue by sending a similar letter tci each of their ten friends. If this continue for 6 sets of letters and if all responded, how much will the Phil. Postal office earn if minimum postage costs P 4 per letter?

B.

c. D.

~ + 2~ + ... + ( i

J

B.

c.

p 6,000,000 P 60,000 p 2,222,220 P 4,444,440

5/2

D.

11/2

A. B. C. D.

i, 2~

....

5/6 2/3 0.84 0.72

Find the ratio of an infinite geometric progression if the sum is 2 and the first term is 1/2.

A. 1/3 B. 1/2

C. 3/4

z.47: EE Board October 1:994

D. 1/4

A rubber ball is made to fall from a height of 50 feet and is observed to rebound 2/3 of the distance it falls. How far will the ball travel before coming to rest if the ball continues to fall in this manner?

A. B. C. D.

200 225 250 275

zs:t(EE Board April 1:997 · If equal spheres are piled in the form of a complete pyramid with an equilateral triangle as base, find the total number of spheres in the pile if each side of the base contains 4 spheres.

feet feet feet feet

A. B. C. D.

. -ziis: EE Board April 1:990 What is the fraction in lowest term equivalent to 0.133133133?

A. B. C.

D.

zs3: Find the 6th term of the sequence 55, 40, 28, 19, 13, ...

133 1\.

10

133

IJ.

c

9 8

()

11

777 133

l:S4: EE Board October 1:997

888

In the series 1, 1, 1/2, 1/6, 1/24, ... , determine the 6th term. ·

133 999

~~ ECE Board April J:998 .i1Find the sum of the infinite geometric progression 6, -2, 2/3, ...

A.

15 20 18 21

666

9/2

l 'p,_:

;

1\

It (: ll

I

1/80 1/74 1/100 1/120

Progression 99

~:

211: ECE Board November :1998

1,073,741 1,730,74 1,073,741,823 1,037,417

Misc~U~n~ous Pro~lems &

r

712

Find the sum of 1. -

Under favorable condition, a single cell bacteria divided into two about every 20 minutes. If the same rate of division is maintained for 10 hours, how many organisms is produced from a single cell?

A.

B. C.

zsol CE Board May 1998

.«:!&:

/

A.

+

4/5 3/4 2/3 1/2

B.

D.

C.

i

/

,..:viz: In the PBA three-point shootout - contest, the committee decided to give a prize in the following manner: A prize of P1 for the first basket made, P 2 for the second, P 4 for the third, P8 for the fourth and so on. If the contestant wants to win a prize of no less than a million pesos, what is the minimum number of baskets to be converted? A.

Day 5 - Cl<>.c:1c, Y:ariation,

ECE Board Aprill998 Find the 198ih digit in the decimal

. Ient to 1785 . from t h e eqUJva - startmg 9999 decimal point.

A. B. C. D.

8 1 7 5

'

' l Day 5- Clock, Variation, Miscellaneous Problems & Progression 101

ml

ml

X

--

12 ---

Topics

D D D D D D

Clock Problems Variation Problems Diophantine Equation Sequence Series Arithmetic Progression Geometric Progression Infinite Geometric Progression Harmonic Progression Figurate Numbers

Mon Tue

Theory

Wed

Problems

Thu

Solutions

X

. t es Note: 1200 ( 30 minutes) = 20 mmu 180°

x=40+~

12 x = 43.6 minutes

X =~+20 12 x=21.818 min

Ill

Therefore, the time is 12:21.818

•

~ Fri

D D Notes

Sat

X

12 RATING

ANSWER KEY 191. D 208. D 192.C 209. A 193. D 210.A 194. B 211. A 195. c 212. D 196. A 213. A 197. D 214. D 198. A 215. c 199. c 216. D 200.A 217. A 201. c 218. c 202.C 219. A 203. D 220.C 204. c 221. D 205. D 222.C 206.C 223.A L'07.C 224.A

225. c 226. B 227.C 228.A 229. B 230. B 231. A 232.C 233. B 234. D 235. B 236.A 237.A 238. B 239. B 240. D 241. B

242.A 243.C 244.0 245. D 246.C 247.C 248. D 249.A 250.A 251. c 252. B 253.A 254. D 255.A

c:J c:J

o

c:J

!

\

12

x=5+~

55-65 Topnotcher

12 x = 5.454 minutes

42-54 Passer 32-42 Conditional

12

mJ

0-31 Failed

x = 43.6 min Therefore, the time is 8:43.6

•

If FAILED, repeat the test. fiWliilk

X=40+~

m&Jiill~1:t!!!JlJiiill!i

:~

i

X= 30+L 12

y=15+~ 12

7 Eq. ~

Eq.2

' Substitute (2) in (1 ):

15 + -~

xc15t-~-

v~,

x

I

12 1G.:lG rnin

llll't"lut<', lilt' lllll<' r•; :1 11·,

X=

:If>

30 + ·--· ! 2._ ~2

!

'

[

t

102 100i Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas

Day 5- Clock, Varia!i
6: 15 = 6(60) + 15 = 375 minutes

Note: When the wire was stretched, the diameter was changed but the volume remains constant assuming there was no losses in the process.

7:45 = 7(60) + 45 =465 minutes

X

X=465-375

ED

When P = 150, V = 115, I = 100 & d = 150

When R = 100 and L = 10

x = 90 minutes

150 = k (115)(100) 150 k = 1.956

2

100=k'(10) 180° x=90min ( -) 30 min

k' = 1

When V = 230, I= 150 and d = 100 When L = 12:

X= 540°

ml

Time when he left his home

R=1(12) D=kC

p = 1.956[(230)(150)] 100

2

R =144 Q

When D

= 2 and C = 0.5 C

ml

y 2 ~ k(0.5C)

;

D=(

r

12 - - : :----Time when he returned home

When w = 4, x = 2, y = 6 and z =

15+~] 12

x=30+~ [

12x = 360 + 15 + ~ 12 x = 31.4it min

So the time when he left home was 3:31.47.

•

Let: x = the number of minute difference between 6:15 and 7:45.

= 0.75 C:

R=k.!:_

A

Let: V

p

30 = k[

(10~~50)]

k = 0.06

When x = 1, y = 4 and z = 2: ~

When t = 40, W = 800 and S = 40 40 = (0.06>[

Eq.1

w=3[(1)(4)]

(2)2

W=3

V=AL

ml

~ Eq.2

p

ml

x=kr

z


Wilen x = 14, y = 7 and z = 2

R"k( ~]

ii .

=48 hp

Let:

x = net col.'t 2x = selling price d =discount 2x (1 -d) =new selling price

2x(1- d)= x + 0.25x 2x - 2xd = 1.25x

2

Wilen y

(800~( 40)]

New selling price = Net cost + Gain

14=k~

R=k'L2

k'=~ v

ws

When t = 30, W = 100, S =50 and P =10

k=3

= volume of the wire

A=Lv

t=k

(3)2

D= 3 days

ml

3

4=k(2)(6)

~ )(o.75C)

Multiple both sides by 12:

mJ

z2

c

When C

P = 675 megawatts·

w =k xy

4 k=-

Y

P=k VI d

k=4

d=0.375

= 16 and z = 4,

d =-S:-7<5%

X X

c

(4}(~6) 16

• x y

r;;

Let: x = number of examinees y = tutor's fee original fee shared per examinee

~

If !here were 10 more examines who wm join,

y =(X+

10{~--2)

m Let: x = number of men in the popuiC!tion y = number of female in the population

10y =2x + 20x 2

y "'0.2x + 2x

X= 1200- y --• Eq. 1

If there were 5 examinees who will backout,

y=(x-5{~+2)

-~ Eq. 2

I

I5IIJI

0.2x 2 + 2x = 0.4x

2

m Let: x = number of radios sold out y = number of clocks sold out z number of flashlight sold out

-

2x

4x =0.2x 2 x = 20 examinees

m x = selling price without t!iscount 0.8x = new selling price (with discount} Profit= Income- Expense!s 0.24x = 0.8x ·- 200 X= 357.14

It!

B!J

1.2(1.4x) =

x + 2856

n2 + 2n - 2600 = 0

~~~~ a,

-a 1

af = 25;

rl~

n

a4= 13

2

(,I

105 = ~(1 + n)

IIi'

ti;'

2

ljl

210=n+n 2

'I' Iilli

n2 +n-210=0

i\)l

~: 'i

(n-14}(n+15)=0

d=-4

i


a3 =a,+ 2d a 3 = 17

mJ

a2

a,= 4;

=7;

a3 = 10

By inspection, d = 3 cL 10

'a 1 ·1·

a,0

4

1

,I

~)

1

111

1!:1

I

a 3 = 25 + 2 ( -4)

1000 = 1000 It checks!

ti~'

105=-(2+n--1)

a 4 =a 1 +Gd

in equation 1:

=~[2a 1 +(n-1)d]

105 =%[2{1)+(n -1){1)]

13=25+3d

50(16) + 30( 4)+ 80 = 1000

x= 16 radios

= 1; d = 1; S = 105 S

a 10 = 305.9

z=80

Therefore,

n=50

d = 32.2

y=4

1 .68x = x + 285 X= 4200

I


d=48.3-16.1

fJEI

=20, y =4 and z = 80 in

j

(n + 51){n- 50)= 0

a,= 16.1; a2 = 48.3; a3·= 80.5

49(16) + 29y = 900

Substitute x equation 2:

(4 + 2n)

n ==-51 (absurd)

16-t-4-t-z = 100

x = price Jojo paid for the VCR 1.4x = price Rudy paid for the VCR 1 .2(14x) =price Noel paid for the VCR

2

.

2

Thus, a2 = 36, a3 = 38 and _a4 == 40 Sum = 36 + 38 + 40 Sur:n = 114

49x + 29y = 900

Lc·

n

. 2600 = 2n ;t n

( 50x + 30y + z)- ( x + y + z) = 1000 -1 00

=4

2

d=2

a10 =16.1+9(32.2)

Substitute x = 20 and y

2600=-(6+2n-2) 2600=

a 10 = a 1 +9d

0.3(0.8x) = 0.8x -200

J

=~[ 2(3) + (n -1){2) J

2600

a5 = a1 + 4d


Assume x = 16:

·

n

d = a2 Z

Let

nr

42 =34+4d

7 Eq. 1 7 Eq. 2

= 100 50x + 30y + z = 1000

y+

X+

a3 =7

a2 = 5;

S='2L2a 1 +(n-1)d

34, a2, a3, a4,42


=

Equate equations 1 and 2:

I

~Eq.2

y = 600 females

5y =2x 2 -10x

=3;

By inspection, d = 2

2

Eq. 1

600- 0.5y + 0.7y = 720

X

y = 0.4x 2 - 2x

0.5X+0.7y=0.6(1200)

~

0.5(1200- y)- 0.7y = 720

5y y = y + 2x - · - -1 0

a,

Thus, the arithmetic mean of a and b is, a+b

X+ y = 1200

X

2

~~~~

Note: Arithmetic mean i? the same term as average.

10y

y "' y - 2x + - - - 20

Day 5- Clocls__'{_aria!i()n, Miscellaneous Problems & Progression lOS

•

l 04 1Ot) 1 S_olved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas

29d

29(3)

Ell

n = -15 (absurd)

i·'l

n = 14 layers of logs

1:::

a,= 4; a2 = 9; a3 = 14;_a4 n = 20 By

ir:~spection,

d=5

S=%[2a 1 +(n-1)d]

1;,1

:=

19;

I

DayS- Clock, Variation, Miscellaneous Problems & Progression 107

106 1001 Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas 20 [2(4)+(20-1)5] 2 =1030

S=

s

37620 = 571n- n2 n2 - 571n + 37620 = 0

•

(n- 76)(n- 495) = 0

S=

n = 495 (absurd)

Ell

By inspection, d = - 3

Ell

an =a1+(n-1)d

Let:

10=61+(n-1)(-3) 10=61-3n+3 n =18

2

the meal can last S = total number of provisions n = number of days the total provisions can last d = number of soldiers died per day

ml

2

d = 0.50; n = 12(12) =144

144

2

[2a 1 +(144 -1)(0,5)]

a 1 =2 a144 =a,+ 143d

n =72+9 n =81 days

fm

8 = %[2a 1 +(n-1)d]

Total provision= 5700 (66) = 376,200

S =%[2a1+(n-1)d]

a1

1 8= ;[2(4)+13(8)]

= 30; as = 90

8= 784 as= a1+4d 90 =30+4d

•

a,=1;an=101;d=2

d =15 376200 = %[ 2(5700) + (n -1)( -20)] 376200 = ~[11400- 20n + 20)

2

376200 = 571 On -1 On

2

S=%[2a 1 +(n-1)d] S=%[2(30)+4(15)] 8=300 300 Average score = - · ". 60 5

a1 = 40; as= 76 a5 =a 1+4d

76 =40+4d d=9 Therefore: a2 =40+9=49 a3 = 49+9 =58 a4 = 58+9 = 67

a1 = 4; d = 8; n = 14

d= -18

&D

•

a,44 = 73.50

Therefore, 18 soldiers died everyday

Note: a1 = 5700; d = - 20

Therefore, total = 2(78) = 156 times

a,4 4 =2+143(0.5)

S = %[2a 1+(n -1)d]

ml

8=78

5436 = 144a1+5148

S = 460800 meals

~1 [2(6400)+80d]

12 . S=-(1+12)

S=%[2a 1 +(n-1)d]

s = 6400(72)

460800 =

2

Note: One day is equivalent to 24 hours.

216 3 x = 72 days

By inspection: a1 = 3: an = 35; n='7+2=9

a1 = 1; a2 = 2; a3 = 3; ... a12 = 12 n 8=-(a,+an)

d=2

5436 =

n S=-(a 1+aJ 2 9 s =-(3+35) 2 s = 171

ml

30=10+10d

X=-

3, a2, a3, a4, as, a6, a7, aa, 35

2

8=2601

a11 = a1 +10d

x = number of days,

1 S= :[2(6.1)+(18-1)(-3)]

a

2

51 8=-(1+101)

(a 1 +a 2 )

n = 11

S = %[2a 1+ (n -1)d]

S=639

n

220 = ~(10 +30)

n = 76 days

a1 = 61; a2·= 58; a3 =55; an= 10

n 8=2(a1 +an)

8 = 220; a1 = 10; an= 30

an= a1 +(n-1)d

40 + 49 + 58 + 67 + 76 A verage = - - - -- - - -

5

Average = 58 centavos

•

1 h3 = 15· a3 = '

15

hs = 6; as=

6'1

a 3 =a, +2d

1

= a 1 + 2d

101=1+(n-1)(2)

15

101=1+2n-2

1 a1=--2d 15

n =51

~Eq.1

a 9 =a 1 + 8d 1 6

-- =a, + 8d -> Eq.2

Day 6- Clock, Variation, Miscellaneous Problems& Progression 109 108 100 l Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas Substitute equation 1 in equation 2:

1

1

6=15-2d+8d

•

h 1 = 3; a, =

=~x-i(~x )=~x

r2 = 9

Volume left after 3'd stroke

6

=1~-2(6~)

216=a,(3)

Note: By inspection, the volume left after each stroke forms a GP whose common ratio is:

as= al7

ag=i+8(i)

a8 =8(3)

5 ag =3

a8

m::l

a11 =a, +10d

=~+10(~) 30 60

3

a,= 8

a9 =a, +8d

30

7

4

=17496

r=Jl_2

2--

a, = x; a2

=2x + 7; a3 =10x- 7

hg = __!_ = ~ a9 5

r = az = a3

a1

3

1 a,,=5

hg

=~x-i(~x)= : 7 x

Substitute r in equation 1:

d=2

1 a,.,.,-

=s3

a2

2x + 7

1Qx -7

x

2x+7

Solving for the volume left after the 61h stroke: as= als

2

BD

1 h,,=-a 11

h,,=T

5 = 160r

-

r=0.5

5

h,, =5 h, =

X=

Thus,

1

2

5

8

By inspection, d

=3

fJ!I

a 4 =2+3(3)

a 4 = 11 1

'

5

1994 =

al

a6 =a,r

~ Eq. -2

Thus h4 = - = ,

a4

11

y

mt

=7 12 35-49 7 x2 =-1-2- =-6

Divide equation 2 by equation 1:

=0.9122x r =0.9 a, = 325(0.9) = 292.5

a4 = a,r3 a4 =292.5(0.9t a 4 = 213.33

mt ~ Eq.1

216=al

y = x- 0.08779x

x, = ~~-+49

a4 = 216; aa = 1994

a 4 = a 1r3

a 4 =a, +3d

~ 4(6)( -49)

35±49 12

Sum = 80 + 40 + 20 + 10 Sum= 150

1 h3 = 0.125 =-; a3 = 8

2

X=---

=

1 h2 = 0.2 =- ; az = 5

35 ± ~5)

2{6)

a2 = 160(0.5) = 80 a3 = 80(0.5) = 40 a4 40(0.5) = 20 as= 20(0.5) = 10

; a1 = 2

Thus, the total volume removed after the 61h stroke:

Using the quadratic formula:

5

(~x )(~J

a6 = 0.08779x

6x 2 - 35x- 49 = 0 aa = als

1

as=

4x 2 +28x+49 = 10x2 -7x

a, = 160; as = 5

3

3

(2x + 7) = x(10x -7)

Therefore:

ml

Volume left after 2"d stroke

r=3

2 3

d in equation 1:

a,,

al 1994 al = 216

21

1 1 d=---

60

a,

3

h2 = 2; a2 =

d = a 2 -a 1

d=~ Substitute

1

Let: x = total volume of air in the·tank y = total volume removed from the tank after the 61h stroke

Bil

a,= 28; a2 = x + 2; a3

r = a2

=

a1 Volur. ;e left after 1"1 stroke = x

_13 x = -~3 x

X+2

112

28

X+2

a3 a2

= 112

Day 5- Clock, Variation, Miscellaneous Problems & Progression 111

110 100 l Solved Problems in Engineeruig Mathematics (2nd Edition) by Tiong & Rojas 2 (x+2) =112{28)

0.1(2"-1) 104857.5 = _ _,__-----!._ 2 -1 2" -1 = 1048575

Divide equation 2 by equation 1: a r 11 -1792 _1_=-56

2

(x+2) =56

al

X=54

S

1( 2030 -1) s = --'------'-

2" = 1048576

2-1

rs = -32 Solving for r:

r

r = a2 = 54+2 a, 28

56= a,(-2)

1111

9

a, 0 = 14336

Ell

15

2(2 -1) s = --'-2---1--'-

a,= 9; r = _ 3_

-iJ

=9G~)

a7 = a1r

a1 (r" -1)

,r:-q.2

·r-1

By inspection, r =

S=~

S=-3. 1- _!

log1000001 n=--=---log2 n=19.93. Say n = 20 baskets

a, = 0.1; r =2 S = a1(r"-1) r -1

100/3 ft

.

so[3.J3 = 100. ,n=oo 3

S=~

I!IIIIW 1 1·alaiiil a,= 3 ; a2 = 9 , 3 -

Take logarithm on both sides:

EJI

,>··\ .-· · .

v·······v···\..-·-.,_.... -·

a,=

-1)

3

a,2 =a/, -1792=<>r 11

6

I

j ,f .Iv ........ v ,, ,, .. h.~ r-1 100

S=-32 1-3 S= 100

1 27

=--'---'-

1

log2" = log1000001 nlog2 = log1000001

----+Eq.1

-

Total cost= 4{1,111,110) = P 4,444,440

2" = 1000001

6

56=al

~0(10

1-r 1 -

a7 =56; a12 =-1792

n =6

10 -1 S=1111110

2" - 1 = 1000000

as= a,r4

16 as=9

~~

.

1(2"-1) 1000000=-. 2-1

3

= 10;

:-~·.

So~---''---

a,= 1; r=2.

s

a,= 10; r

5o' ft

a,(r"-1) S=--'----'r-1

r-1

a

2(2 -1) s =--'-----.-:.. 2-1. s =2046

as

-1)

s = 65534 ancestors

10

as =9(

Bill

S=-'--~

By inspection, r = 2

S = a1 (r" --1) r-1

a, =2; r = 2; n = 15 a,(r"

a,= 2; a2 = 4; a3 = 8; n = 10

•

8

8

a10 = 28(2)

• v\

log2" = log1048576 nlog2 = log1 048576 log1048576 n=--"--log2 n = 20 wealthy men

7 a1=-

= a,rg

S = 1,073741,823 organisms

Take log on both sides:

Substitute r in equation 1:

r=2

a,o

=-2

a,(r" -1) r-1

S=2

B:l

a1 =1;r=2

60 20

n=-=30

31 Let: D = total distance traveled by ball D =50+ 2S D =50+ 2(100) = 250 feet

tm 0.133133133133 = 0.133 + 0.000133 + 0.000000133 + ...... Note: The numbers being added are in a GP and 0.133133133133 ... is the sum of an infinite GP.

Solving for the common ratio: · 1 0.000133 r= =-0.133 1000

S=~ 1-r

112. IOOl.Solved Problems in Engineering Mathematics (2"d Edition) by Tiong & Rojas_ 5=0.133 1- ___ 1-

1-r 1

S=_9~33

s9?r

2=2(1-r)

1000

4 =---

999

-2

a1

6

The four repeated digits are 1, 7,8 & 5 1987 = 496.7G;

4

2 r=--=-

Layer I

6

Ell=

1-

Let: x

9 S=2

55

a1 = 1, a2 = -

i,

a3 =

2~

the

6th

40

term of the number series

28

19

13

!

X

in

'--y-~~ -12

-15

By inspection, r =- 5 ;

:1 1 1 1!1!~

=10

,r,l/,, 11 I

Let: y = the

6th

H

term of the number series

,,,·_1

y,

S=~

\.'.- v··-

6

1-r 1

~,·

X=13+(-3)

5;

S=-~

X1

i

.~".. ..

} . . ·y. k

y·

x1/2 y=

1/24

1/6

x113

y

xY.

!I

y

I~, 'Iii

·""-- ·y·--)

X 1/5

riJ'ii!

2~(i)

. ~'I

! .>!' Ill ~;l

1

l,ilr,l

y = 120

~~

s =_1__

1-r

I

i

Ell

S=~-(--1~

j

II I ~ _,·.~ I 1'.~:'I'

-3

-6

-9

X

a1 S=----1-r 1

S=2;a1=

5 7 1 8.

Layer IV

1

EDI

=1984

Total spheres= 10 + 6 + 3 + 1 = 20 spheres

S=-(1) -3 B

Layer Ill

Layer II

496(4)

Therefore, 19841h digit= 19861h digit= 19851h digit = 19871h digit=

£o

1 r=-3 a1 S=·---. 1-r

=0.178517851... '"'"'

The above number is a-repeating decimal number.

ED

a1 =6; a2=-2

a

1785 9999

1 1-- r 3 r=4

s =: 133 ~~~

•

2 = _1__

1000

Day 5- Clock, Variation, Miscellaneous Problems & Progression 11~

l

.JI':rl

',! 1

-

116. 1001 ·solved Problems ~n Engineering Mathematics (2nd Edition) by Tiong & Rojas

Topics

[J Mon

IJ IJ Tue

1

~-

....

Theory

Vl/ed

!_j []

Solutions

D D

D

~

Problems

Notes

Thu

Fri

"'" ,-,,.-~.~""'"-"'-'"-""'"'''"

·~u·m"'""""'"""'"""'""

'""""" ""'•'''""'"'"m""

Venn Diagram Combinatorics Fundamental Principle of Counting Permutation Inversion Cyclic Permutation Assortment Combination Probability Theory Mutually Exclusive Events Independent Events Binomial Distribution Odds For and Odds Against Mathematical Expectation

Sat

What is a Venn Diagram? Venn diagram is a rectangle (the universal set) that includes circles depicting the subsets. This diagram, named after the English logician John Venn (1834- 1923) in 1880, is a way of displaying the events or an experiment. lhe Venn diagram below shows two events A and B and their intersection.

A Venn diagram can be employed for any number of subsets, but more tnan three defeat the purpose of gaining increased clarity. Below is an example of a typical problem that is given in the engineering licensure examtnations.

Problem: lilt• following Venn diagram shows two lltttltt.tily exclusive events A and B

A survey was conducted in a graduating ECE students in a certain university on which board subject they like best. The result is tabulated as follows:

118 _I 00 I Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas Mathematics Math & Electronics Electronics Math & Communications Communications Electronics & Comm All three subjects

55 32 50 28 51 25 10

How many were there in the graduating class? Solution: Use Venn diagram:

.,

. /f

Mathematically,

N=m-n

II

What is Permutation?

''l'/1

Permutation is defined as an ordered arrangement of a finite number of elements, either all of the available n elements or of a part of them. The

'.;.);

't

j

npn =(n-1)!

If taken all at once,

What is the Permutation with Identical Elements?

npn = n! N = 5 + 22 + 10 + 18 + 3 + 15 + 8

What is Combinatorics? Combinatorics is the branch of mathematics that concerns with the selection of objects called elements. Combinatorics traces its history back to the ancient times when it was closely · associated with number mysticism. This branch of mathematics led to the creation and development of probability theory.

What is the "Fundamental Principle of Counting? The Fundamental Principle of Counting states that: "If a thing can be done in m different ways and another thing can be done in n different ways, then the two things can be done in m times n different ways."

The total inversion is 2 + 3 + 2 + 1 =8. So permutation (C, E, D, B, A) is an even permutation

Cyclic permutation is the shifting of an entire order of elements one or more steps forward or backward -the first element taking the position of \he last, or vice versa, without changing the order of the elements in the sequence.

n! nPr=(n-r)!

N = 81 students

The number of permutations is reduced when a collection contains identical elements. The number of permutation n objects, p of one type, q of another, s of another, etc., is given by:

What is an Inversion? If two elemen.ts in a permutation of distinct elements are in reverse order relative to their normal or natural order, they constitute an inversion.

Example: In the permutation (C, E, D, B, A) Element C precedes B and A, therefore has 2 inversions.

= 4989600

Assortment refers to a group of objects selected from a larger group in such a way that an object can be used more than once.

Assortments = (No. of choices for 1st position) x (no. of choices for the 2nd position) x (no. of choices for the 3'd position) x · · · x (no. of choices for last position) What is a Combination? Combination is an arrangement of the selection of objects regardless of the order. The number of combinations of n different things taken r at a time is n! ncr= (n-r)!r!

If taken all at once,

ll

n! p n r-(n-r)!_p!q!s!

A permutation is said to be even if it contains an even number of inversions; it is odd if the number of inversions is odd. The number of transpositions that are required to return a sequence of elements to their natural order is even or odd according to the number of inversions in the arrangement.

11p11

What is an Assortment? Element B precedes A, therefore has 1 inversion.

What is a Cyclic Permutation?

The number of permutations of n different things taken r at a time is

Communications

Element D precedes B and A, therefore has 2 inversions

The number of assortments is expressed· mathematically as:

permutations that contains exactly the same elements but not in the same order are regarded as different.

Electronics

Day 6 -V§!!l_llJ:)iagram, Permutation, Combination & Probability 119 Element E precedes D, Band A, therefore 111 11 11 has 3 inversions. P = (11-11)12121211i111i 111i

I'~

Problem: Find the largest number of permutations of the letters in MATHEMATICS? Solution: The word MATHEMATICS has 11 letters. The letter M occurs 2 times, letter A occurs 2 times, letter T occurs 2 times while the rest occurs once.

ncr =1 What is the relation between Permutation and Combination? The relation between permutation of n thing taken r at a time to the combination of n things taken r at a time is expressed mathematically as follows:

nCr=~ r!

120 · 100 l Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas What

is Probability?

Probability is simple defined as the numerical assessment of likelihood. It is . expressed as a number between 0 and 1, where 0 means an event is impossible while 1 means it is surely or certainly to occur or happen. The total probability of favorable and unfavorable outcomes is 1.

I f

One of the earliest mathematical studies on probability was "On Casting the Die" written by the 16th century Italian mathematician and physician Gerolamo Cardano (1501 -1576). Cardano defined probability as "the number of favorable outcomes divided by the number of possible outcomes". Cardano was regarded as "the father of the theory of probability". The following are some of the important terms used in probability: · A. Experiment - is a controlled study whose outcome is uncertain but not entirely unknown. B. Trial -a recorded result of an experiment. C. Outcome - is one of the possible results from an experiment trial. It also refers to the basic unit of possible occurrences. For example, in. tossing a coin, the outcomes are heads and tails. D. Event- is some combination of possible outcomes in one experiment trial. In picking a card from an ordinary deck of cards, the outcomes are the 52 cards themselves. Examples of events are: "Pick a 5" (one of the four cards numbered 5) "Pick a heart" (one of the 13 cards in the suit of heart) "Pick a face card (one of the 12 cards showing a face i.e. king, queen or jack)

E. Frequency of the outcome - refers to the number of times a certain outcome will occur. For example, if you roll a die 50 times, you are conducting 50 trials of the same experiment. The number of times a "6" comes up is known as the frequency of the outcome.

Day 6- Venn Diaqram, Permutation, Combination & Probability 121 The probability for a mutually exclusive event (E or F) is:

Pe or F = Pe + PF

What are independent events? The relative frequency of the outcome is the ratio of the frequency of the outcome to the number of trials. Mathematically,

RF

=· no. of occurrences no. oftrials

What is the Principle of Probabiitv?

If an experiment has a set of distinct outcomes, each of which is equally likely to happen, then the probability of an event E, is the ratio of the number of outcomes to the total number of possible outcomes.

An event is independent if the outcome of one trial has no effect on the outcome of any other trial. A good example is when a coin is tossed, the outcome has nothing to do with what happened on the previous toss and will not affect the next one. If two events are independent, the probability that the two events occur is the product of their individual prob.abilities. The probability of independent events (E and F) is:

may mean or is synonym<;>us to the word "chance". If a coin is tossed, the probability of coming up head is 50% and the probability of coming up tail is also 50%. This is an example of 1-to-1 odds or even odds. For a game with even odds, a one peso bet is paid exactly one peso upon winning. This is known as the "true odds". However, there are instances that 90 centavos is paid instead to a one peso bet, which would give it an advantage on each game. This refers to the "payoff odds". The payoff odds are not always the same as the true odds. The odds of an event occurs is the ratio of the probability that it will occur to the probability that it will not occur. odds for E = prob. that E occurs prob. that E does not occur

P. . 1-Pe

odds for event E =-e-

Pe & F = Pe x PF

P. __ no. of outcomes e - total outcomes

What is a Binomial Distribution? . The probability that it is not E is:

Pnot E = '1.- Pe

What are mutually exclusive events?

If there are two possible outcomes of an event and the-possibilities of the outcome are independent and constant, the distribution of probabilities is called binomial distribution. Binomial or repeated trial probability:

Two or more events are said to be mutually exclusive if no two of them can possibly happen in the same trial. For example, in picking a card from a deck of cards, it is not possible to pick a card that is a diamond and a heart at the same time. If two or more events are mutually exclusive, the probability that either one or the other will occur is the sum of their probabilities.

p =ncr pr qn-r

where:

p = probability of success q = probability of failure = 1 - p n = number of trials r = number of successful trials

Problema What is the odds for throwing a total of 5 and 10 in rolling two dice? Solution: First, solve the probabilitY of throwing a total.of 5: For a total of 5, it must be (1 and 4), (4 and 1). (2 and 3) and (3 and 2).

4 1 p5 =-=36 g. Then, solve the probab' ity ofthrowing a total of 10: For a total of 10, it must be (4 and 6), (6 and 4), and ( ' and 5).

3·

1

What is an Odd?

P1o = 36 = 12

The term "odd" is an ambiguous word that may refer to the probability that an event occurs, or it can be used to indicate the payoff on a winning bet. It sometimes

Psopo =

1

1

7

g + 12--36

Day 6 - Venn Diagram, Permutation, Combination & Probability 123

122. 1001 Solved Problems in Engineering Mathematics (2"d Edition) by Tiong & Rojas The odds for. throwing a total of 5 or 10 is:

7

odd=~ 1- .2_

36

odd=

7 29

If the expectation is positive, the player will win in the long run and lose if the expectation is negative.

Problem: In tossing a single coin, P 100 bet is placed on heads i.e. if heads comes up, the player wins P 100 and if tails comes up, the player loses P 100. What is the expectation?

This is expressed as 7 -to-29 odds. Tails

y,

What is another formula for "odd for"?

-100 If the·"odds for" an event are given as "a" and "b", then the probability of that event would be: odds for a·n event =-aa+b

Expectation=

(~}1 oo) + (~ )c-1 oo)

deck of cards contains 52 cards divided • into 4 suits; spades, hearts, diamonds and clubs. Each suit contains 13 cards labeled Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, jack, queen and king. Each of the four kings in a deck represents a great leader from history; Charlemagne (hearts), Alexander the Great (clubs), Julius Ceasar (diamonds), and King David (spades). The playing cards were used also to describe a calendar year. The 52 cards represent the 52 weeks in a year. The 4 suits represent the four seasons of the year and the 12 face cards (4 kings, 4 queens and 4 jacks) represent the 12 months ot' the year.

Expectation = 0

Problem: A player in a certain card game lays odds of 3 to 5 for him to win. What would be the payoff for a bet of 3 pesos?

In Casino Filipino, a roulette offers winning a bet on a single number pays 35 to 1. The actual probability is 1 in 38. What is the expectation?

Solution:

Solution:

Problem:

3 -~ Odds for = 3 + 5 - 8 Therefore, it is a 3-to-8 odd. A bet of 3 pesos will have a payoff of 8 pesos.

What is "odd against"? The "odds against" an event are the reciprocal of the "odds for" the event. The odds of 5 to 7 for an event is translated into odds of 7to 5 against it.

What is a Mathematical Expectation? A mathematical expectation is the average amount a player can expect to win or lose on one play in any game of chance. ·

The 35 to 1 is a house (payoff) odds of Casino Filipino. Expectation can be obtained by taking the difference betWeen the product of the first number in the house odds and probability of success, and the product of the second number of the house odds and the probability of failure. 37 Expectation= 35(_!__) -1( ) 38 313 .

Expectat1on . =- 2

38 Expectation = -0.0526 Thus, a player expects to lose to Casino Filipino an average of 5.26 centavos for every peso bet. What are "Card Games"?

It can be found by multiplying the probability of each possible outcomes by its payoff, and then adding these results.

Card games are games which are played using a standard deck of cards A standard

Poker Harid with corresponding number of ways: Hand Royal flush Stra_ight flush · · ~our of a kind · Full house Flush Straight . Three of a kind Two pair Pair None of the above Total

No. of ways 4 36 624 3744 .5108 10200 54912 123552 1098240 1302540 2598960

Poker Hand with corresponding probability: Hand Royal flush Straight flush Four of a kind · Full house Flush StraiQh! Three of a kind Two pair Pair None of the above Total

Probability 0.00000154 0.00001385 0.0002401 0.0014406 0.0019654 0.0039246 0.0211285 0,0475390 0.4225690 0.5011774 1.00000000

Poker Hand with corresponding expected frequency: Hand Royal flush Straight flush Four of a kind Full house Flush Straight Three of a kind Two pair Pair None of the above

Expected frequency 1 in 649740 hands 1 in 72192 hands 1 in 4165 hands 1 in 694 hands 1 in 509 hands 1 in 255 hands 1 in 47.33 hands 1 in 21 hands 1 in 2.37 hands 1 in 2 hands

What are Probabilities with Dices? Dice were invented for sole purpose of gambling. As a matter of fact, dice gaines have so little intrinsic interest that in the absence of wagering they would hardly be worth nl::>llinn

''' '',.

-

Dice were first used by th~ Chinese. The sum of the opposite faces of~ die is always equal to 7. And the sum of all the vertical faces of a die, no matter how it rolls is always equal to 14. When two dice

124 IO
G8

~.8 [ZJ8

Gr:J~r:J G rsJ ~ rsJ l:l [i"7l L:.J l!_!J l:l r-;;1 L:.J l!:.!J

Proceed to the next page for your 6th test Detach and use the answer sheet provided at the last part of this book. Use penCil number 2 in shading your answer.

[ZJ[J 1.··1 LSJ

rel

[i"7l l!._j l!_!J

r:;][i"7] l!.:J l!_!J

rel ~ l!._j l!:.!J

r:eJ fe.eJ l!.:J l!:.!J

•• .• G []

~

~8

1:·:1 G 1::1 G

[i"7lr.l l!_!J L!J

r.=--11-l LUI L!J

•

[]• • •• 1.··1 []

GOOD LUCK I

Topics

D D D D

-m:ribia:

tvlon

Did you know that .. the number 1 followed by 100 zeros is called "google" and the term "google" was coined in the 1930s by the aine-year old nephew of the American mathematician Edward Kasner when he was asked to come up with a ·name for a very large number !

Tue

Theory

~

~uote:

1: :11··.1 1:·:11··.1 fe"il [i"7l

l!_!J l!_!J

~- reel

LUl

l!_!j

I!!]

LUJ

lei

u

.

~

The following are the two-dice probabilities:

9 10 11 12

Probability 1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36

D Thu

D D

- Morris Kline

o·

Solutions

Notes

LUJ L!_..!J

~ [] ••• []• •• •• []• •• •• []• •• •• •• • ~-

8

Problems

l!!lr-::1 LUJ ~ I!!] reel

1: :11:·:1 1:·:11:·:1 1: il

Two-dice sum 2 3 4 5 6 7

"Statistics - the mathematical theory of ignorance.

VVed

Fri

(QJ Sat

:15&: EE Board October 199~ In a class of 40 students, 27 like Calculus and 25 like Chemistry. How many like both Calculus and Chemistry?

A. B. C. D.

Venn Diagram Combinatorics Fundamental Principle of Counting Permutation Inversion Cyclic Permutation Assortment Combination Probability Theory Mutually Exclusive Events Independent Events Binomial Distribution Odds For and Odds Against Mathematical Expectation

10 11 12 13

.2$8: GE Board February 1994 ·' A survey of 100 persons revealed that 72 of them had eaten at restaurant P and that 52 of them had eaten at restaurant Q. Which of the following.could not be the number of persons in the surveyed group who ·had eaten at both P and Q?

A.

B. %57: ECE Board November 1998 A c!ub of 40 executives, 33 like to smoke

Marlboro and 20 like to smoke Philip Morris. How many like both? A.

B

c n

10 11 12 13

C. D.

20 22 24 26

2591 ECE Board November 1992 'The probability for the ECE board examinees from a certain school to pass the subject Mathematics is 3n and for the subject Communications is 5/7. If none of the examinees fails both subject and there are 4 examinees who pass both subject~. find the number of examinees from that school who took the examinations.

126 H>O I Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas A. B.

c. D.

20 25 30 28

/ ,260: EE Board March 1998 In a commercial sur.iey involving 1000 persons ori brand preference, 120 were found to prefer brand x only, 200 prefer brandy only, 150 prefer•brand z only, 370 prefer either brand x or y but not z, 450 prefer brand y or z but not xand 370 prefer either brand z or x but' not y. How many persons have no brand preference, satisfied with any of the three brands? A.

B.

c. D.

280 230 180 130

Day 6- Venn Diagram, Permutation, Combination & Probability 127

2&8: EE Board April199&

26~:

Five different mathematics books, 4 different electronics books and 2 different communications books are to be placed in a shelf with the books of the same subject together. Find the number of ways in which the books can be placed.

A. B. C. D.

A.

B. C. D.

292 . 5760 34560 12870

D.

265: ECE Board November 1998 If 15 peop!e won prizes in the state lottery (assuming that there are no ties), how many ways can these 15 people win first, second, third, fourth and fifth prizes?

A.

B. C. D.

B. C. D.

beginning and ending with a vowel without any letter repeated can be formed from the word "personnel"?

B.

c. D.

40 480 20 312

What is the number of permutations of the letters in the word BANANA?

274: EE Board October 1997

B. C. D.

B.

c. D.

c.·

~

There are four balls of four different colors. Two balls are taken at a time and arranged in a definite order. For example, if a white and a red balls are taken, one definite arrangement is white first, red second, and another arrangement is red first, white second. How many such arrangements are possible?

A.

~

B. C. D.

6 12 ~

275: How many different ways can 5 boys and 5 girls form a circle with boys , and girls alternate?

A B. C. D.

28,800 2,880 5,600 14,400

276: EE Board October 1997 A B. C.

120 130 140 150

1440 480 720 360

2,024 12,144 480 360

In how many ways can a PSME Chapter with 15 directors choose a President, a Vice President, a Secretary, a Treasurer and an Auditor, if no member can hold more than one position?

D.

How many permutations are there ifthe letters PNRCSE are taken six at a time?

A. B. C. D.

36 60 52 42

A.

271: ME Board October 1992

267: EE Board June 1990

A.

2204

A

How many 4 digit numbers can be formed without repeating any digit from the following digits: 1, 2, 3, 4 and 6?

A.

262: How many four-letter words

D.

270: ME Board April1994

266: CE Board November 1996 Neither yes nor no Yes No Either yes or no

2&9: EE Board April1997

A PSME unit has 10 ME's, 8 PME's and 6 CPM's. If a committee of 3 members, one from each group is to be formed,. how many such committees can be formed?

4,845 116,260 360,360 "3,003

In how many ways can 4 boys and 4 girls be seated alternately in a row of 8 seats? 1152 2304

B. C. D.

144 258 720 450

A toothpaste firm claims that fn a survey of 54 people, they were using either Colgate, Hapee or Close-up brand. The following statistics were found: 6 people used all three brands, 5 used only Hapee and Close-up, 18 used Hapee or Closeup, 2 used Hapee, 2 used only Hapee and Colgate, 1 used Close-up and Colgate, and 20 used only Colgate. Is the survey worth paying for?

EE Board June 1990 EE Board April199~ CHE Board May 1994

B.

A.

and 4 engineers be seated on a bench with the nurses seated together is

261: EE Board April i:997

A.

720 120 360 180

264: The number of ways can 3 nurses

A. B. C.

27~:

In how many ways can 6 distinct books be arranged in a bookshelf?

t ~

I

!

360,360 32,760 3,003 3,603,600

There are four balls of different colors. Two balls at a time are taken and arranged any way. How many such combinations are possible?


A

36

Four different colored flags can be hung in a row to make coded signal. How many signals can be made if a signal consists of the display of one or more flags?

B. C.

3 6

D.

12

277: EE Board March 1998 A. B.

c. D.

64 66 68 62

How many 6-number combinations can be generated from the numbers from 1 to 42 inclusive, without repetition and with no regards to the order of the numbers?

A

J

850,668

1Z8 1001 Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas B.

c. D.

5,245, 786 188,848,296 31,474,716

Z78: Find the total number of combinations of three letters, J, R, T taken 1, 2, 3 at a time.

c. D.

Z8~: ECE Board April1998 A semiconductor company will hire 7 men and 4 women. In how many ways can the company choose from 9 men and 6 women who qualified for the position?

A.

7

B. C.

8 9

A. B.

D.

10

C.

D. Z79: ME Board October 1997 In how many ways can you invite one or more of your five friends in a party?

A. B.

c. D.

15 31 36 25

zsoa CHE November 1996 In how many ways can a committee of three consisting of two chemical engineers and one mechanical engineer can be formed from four chemical j;!n'gineers and three mechanical engineers?

A. B. C. D.

18

64 32

120 530

c.

720

D.

320

Z84a ECE Board April1994 There are 13 teams in a tournament. · Each team is to play with each other only once. What is the minimum number of days can they all play without any team playing more than one game in any day?

B.

1390 1240

A. B. C. D.

11

B.

12 13 14

C.

D.

Z85a EE Board October 1996 There are five main roads between the cities A and B, and four between B and C. In how many ways can a person di'ive from·A to C and return, going through B on both trips without driving on the same road twice?

A. B.

c. D.

260 240 120 160

Z86: EE Board Aprll1991 There are 50 tickets in a lottery in which there is a first and second prize. What is the probability of a man drawing a prize if he owns 5 ti't:kets? ·

B.

c. D.

50% 25% 20% 40%

1/36 1/9 1/18 1120

Z88: Roll two dice once. What is the probability that the sum is 7?

%93: EE Board Apll'ill996 The probability of getting at !east 2 heads when a coin is tossed four times is, A. B.

c. D.

A. B. C. D.

1/6 1/8 1/4 1/7

A.

is

c.

B.

D. A.

B.

c.

1/6 1/9 1/12 1/18

290: Determine. the probability of drawing eithel" a king or a diamond in a single draw from a pack of 52 playing cards.

A.

c.

B.

2/13 3/13 4/13

D.

1/13

:1911 A card is drawn from a deck of 52

playing cards. Find the probability of drawing a king or a red card.

A B. C. 0.

0.5835 0.5385 0.3585 0.8535

1:91:1 CE Board November 1998

A coin is tossed 3 times. What is the probability of getting 3 tails up? A B. C.

o.

1ffl 1/16 1M

m

11/16 13/16 114 3/8

::t94: A fair coin is tossed three ti.mes. What is the probability of gel!ing either 3 heads or 3 tail?

289: In a throw of two dice, the probability of obtaining a total of 10 or 12

D.

A.

A.

zsza EE Board April 1997 How many committees can be formed by choosing 4 men from an organization of a membership of 15 men?

A.

680 540 480 840

None of these

Z81a EE Board Aprii199S In Mathematics examination, a student may select 7 problems from a set of 10 problems. In how many ways can he make his choice?

A. B.

1435 1365

Dart- VexmDiagram, Permutation, Combination & Probability 1Z9

118 3/8 1/4 112

295: ECE Board Mare»• 19'96 The probability of getting a credit in an examination is 113. If three students are selected at random, what is the · probability that at least one of them got a credit? A. B. C. D.

19/27 8/27

2/3 1/3

:Z9fl: There are 3 questions in a test. For each question 1 point is awarded for a correct answer and none for a wrong answer. If the probability that Janine correctly answers a question in the test is 2/3, determine the probability that she _gets zero in the test. A.

B. C. D.

8/27 4/9 1/30 1/27

:Z97Z EE Board April :199:1 In the ECE Board Examinations, the probability t.hat an examinee will pass each subject is 0.8. What is the probability that an examinee will pass at least two subjects out of the three board subje~ts?

Z87a Roll a pair of dice. What is the probability that the sum of two numbers is

A.

11?

B.

70.9% 80.9%

130 . rOO! Solved Problems in Engineering Mathematics (2nct Edition) by Tiong & Rojas C. D.

85.9% 89.6%

298: In a multiple choice test, each question is to be answered by selecting 1 out of 5 choices, of which only 1 is right If there are ·1 0 questions in a test, what is the probability of getting 6 right of pure guesswork?

302: ME Board April 1.99& An urn contains 4 black balls and 6 white balls. What is the probability of getting 1 black and 1 white ball in two consecutive draws from the urn?

A. B.

c

D.

A. B. C.

d

D.

10% 6% 0.44% 0.55 %.

:&99: ME Board April1994 From a box containing 6 red balls, 8 white balls and 10 blue balls, one ball is drawn at random. Determine the probability that it is red or white.

A. B. C. D.

1/3 7/12 5/12 1/4

A. B. C. D.

A.

A.

C. D.

25/81 16/81 5/18 40/81

301: CE Board May 1996 A bag contains 3 white and 5 black balls. If two balls are drawn in succession without replacement, what is the probability that both balls are black? A.· 5/16 B. 5/28 c. 5/32 D. 5/14

B. C. D.

1/6 1/4 1/2 1/8

305: From 20 tickets marked with the first 20 numerals, one is drawn at random. What is the chance that it will be a multiple of 3 or of 7? A. B.

c. D.

u

Theory

0

16/81 25/81 20/18 40/81

300: EE BoardOctober 1990 From a bag containing 4 black balls and 5 white balls, two balls are drawn one at a time. Find the probability that both balls are white. Assume that the first ball is returned before the second ball is drawn.

1/2 8/15 3/10 2/5

Venn Diagram Combinatorics Fundamental Principle of Counting Permutation , Inversion Cyclic Permutation j Assortment 1 Combination 1 Probability Theory Mutually Exclusive Events Independent Events Binomial Distribution Odds For and Odds Against Mathematical Expectation

tv1on

303: EE Board October 1.990 From a bag containing 4 black balls and 5 white balls, two balls are drawn one at a time. Find the probability that one ball is white and one ball is black. Assume that the first ball is returned before the second ball is drawn.

304: EE Board October 1.997 A group of 3 people enter a theater after the lights had dimmed. They are shown to the correct group of 3 seats by the usher. Each person· holds a number stub. What is the probability that each is in the correCt seat accordi•1g to the numbers on seat and stub?

B.

Topics

0.24 0.27 0.53 0.04

Problems

Solutions

0

0 I I 0 0 I 0 Tue

\JVed

Thu

Fri

[QJ

Notes

Sat

ANSWER KEY 256.C 257. D 258. 259. D 260. A 261. 262.A 263. 264.C 265.C 266.A 267.C 268. A

c c

c

269. B 270. 271. A 272.A 273. 274. 275. B 276.C 277. B 278.A 279. B 280.A 281. A

c c c

282. D 283. B 284. 285. B 286.C 287. 288.A 289. B 290. 291. B 292.A 293. D 294.

c

c c c

RATING

295.A 296. D 297. D 298. D 299. B 300.A 301. D 302. 303. D 304.A 305 D

c

c:J 43-50

.,

c:J c:J 0

Topnotcher

33-42 Passer 25-32 Conditional 0-25 Failed


,jil

132· lQOl Solved Problems in Engineering Mathematics (2"d Edition) by Tiong & Rojas

•

Day 6- Venn Diagram, Permutation, Combination &Probability 133

ml

Let: x = number of students who like both subjects Calculus

Let:

ED

x = number of examinees who took the examination

Given word:

Math Communication

Chemistry

N = (5!}(4!)(2!}(number of patterns)

2

I s

~

Let:

x =Hx-4)*4+(%x-4)

x = number of executives who smoke both brand of cigarettes Marlboro

Philip Morris

I

4

I

1

J

v

8 7

Any of the five consonants can be filled in the second letter.

•

Any of the remaining four consonants can be filled in the third letter.

- Let: x = number of persons who have no brand preference Brand X

Brand Y

Brand Z

Q

+ 150+100 X=280

(72-x)+x+(52-x)=100 72 + 52 - X = 100

n E

n

n

E

n

n

n J

E

Number of ways the 3 nurses can be arranged = 3! Number of ways the 4 engineers can be arranged = 4!

J

FD

Number of ways the books in ELEC can

N = 5(4)(3)(2) N= 120 ways

I '·I

I

E

N

total nwnber of wctys

E

E

Five patterns

l1c arranged= 4!

l11• arranned = 2!

j

n

N = 360,360 ways

Number of ways the books in COMM can

X =24

n

J

Number of ways the books in MATH can be arranged = 5!

The survey is not worth paying for. One error is that according to the said survey, there are 6 people who used 9.!JJ.h_ree brands but only ~~ used !he brands Hapee and Close-up.

n

E E

N;, 15(14)(13)(12)(11)

Six patterns

Note:

;E

•

COMM ELEC COMM MATH ELEC MATH

-~

\_

fll

E

E

N =720 ways

ELEC COMM MATH COMM MATH ELEC

E.·

E. n

N = (3!)(4!)(5)

MATH MATH ELEC ELEC COMM COMM

1000 =X+ 120+ 50+ 200 + 100

p

E

n n

N = (3!)(4!)(number of patterns)

Ell

= 40

X= 13

x = number of persons who have eaten in both restaurants

E E E

n n E E

Let: N = number of words

(33- X)+ X +(20- X)= 40

Let:

E

Let: N = total number of ways

N= 40 ways

Ell

n

The remaining bowel can be filled in the fourth letter.

N = 2(5)(4)(1)

33+20-x

Ill

·x=-x-4 X=28

Bil

I

Four-letter word Note: Any of the two vowels can be filled in the · first letter.

(27 ~ x)+ x +(25- x) = 40 X= 12

N = 34,560 ways

Number of vowels= 2 (E & 0) Number of constants= 5 (P, R, S, N & L)

I

27 +25-x = 40

N = (5!)(4!)(2!)(6)

PERSONNEL

:I


l34 100 l ·Solved Problems in Engineering Mathematics (2"ct Edition) by Tiong & Rojas

!,

N = 2" -1

N = npn = n! = 6!

ml

N=720

N=nPn

Given word: BANANA Number of A's = 3 Number of N's = 2

•

p!q!

N

GoBO? 0 08oG

3!2!

08 GO

e

OG

Number of ways of selecting a mechanical engineer: 31 N = C = . =3 2 3 1 (3-1)!1!

8 '~

\'~~

.

I

This seat is permanently occupied by one of the children!

"'--..__

I. ~;c I Tr7;s I ~f]

J

N = 6(3) = 18 ways

.~

Ell

10! N= 10c1 = ( 10 _7 )!7!

Number of ways the girls can be arranged = 5!

N=120 ways

N = 15(14)(13)(12)(11)

N=(4!)(5!)

N= 360,360

N = 2880 ways

ml

ml N = 4p1 + 4P2 + 4P3 + 4P4

1

Number of ways of selecting a chemical engineer: 41 N - C = =6 1- 4 2 (4- 2)!21

Number of ways the boys can be arranged = (5-1)! =4!

Cit! ~~

-

ml

G

N= 60 ways

N = 10(8)(6) N=480 ways

= 2" - 1 =2

5

N = 31 ways

B0

N=_I1J__=~

1

ml

N = 12 ways

N= 720 ways

=23 -

N = 7 ways

4! N= 4p2 = ( 4 -2)!

n!=6!

ml

ml

N=(4!)(4!) N = 576 ways

N= 15 C 4 =~ (15-4)!4! N = 1,365 committees

4! N= 4c2 = (4 -2)12!

Number of games that can be played per day: 13 N = - = 6.5

2

~

6 games/day

Number of days needed to complete the tournament: 78 N=-=13 days 6

ED Number of ways to travel from A to B = 5 Number of ways to travel from 8 to C = 4 Number of ways to travel from C to 8 without using the same road to travel from 8 to C =3 Number of ways to travel from B to A without using the same road to travel from A to B =4 N = 5(4)(3)(4) = 240 ways

•

P = probability of winning a prize in the lo_ttery

N=--4_!_+ __4_!_+ __4_!_+ __4_!_ (4-1)! (4-2)! (4~3)! (4-4)!

N =6 ways

ml

N = 64 signals

Ell

Number of ways of hiring men: 91 N1 = C7 = . =36 g (9- 7)!7!

- c-~

N-42 6-(42-6)!6! Num_ber of ways the 4 boys can be arranged= 4!

Total number of games 131 N= 13 C 2 = . =78 (13- 2)!2!

Number of ways of hiring women: 6! N.,"' .C 4 =------=15 • (> (6-4)14!

N = 5,245,786 ways

Number of ways the 4 girls can be arranged = 4!

N

j

2 1 P=-=50 25 P = probability for the man to win = number of tickets he bought x. probability of winning a prize

P= 5(_!_) .25 p =0.20

:16( 15) = 540 ways

!

136 1001 Solved Problem~ngineering Mathematics (2nd Edition) by Tiong & Rojas

•

2

Total number of trials

2

r----

2

f-·

3

4

3

=36

2. I 3

·r=a 4

5

6

I

.3

4

5

6

5

•

Let: p = probability of getting a head in a single tl)row of a fair coin

--+-----1

L- L-+--J---1! I

4


q = probability of getting a tail in a single throw of a fair coin

6

P = number of successful trials

5

total number trials

-

6

.

4

~'---

Number of trials with a sum of 11

=2

P = number of successful trials_ total number trials

2

1

36

18

II!.'P.fJII

laiiill

34!]

Total number of trials = :36

2

2

1 9

P=--=36

•

3

--1

·~

6

Number of trials with a sum of 7 = 6

P = number of successful _trials total number trials

6 1 P=-=36

PKorD =PK +Po-PK&D

F.ll Total number of trials = 36 Number of trials with a sum of 10 or 12

=4

0

3 0 3 1) 1) (1)2 p3H=3C3 (2 '(2 1

3H =8

p

3c3(~J(~J =(1{iJ

3

P3T =

=(1)

1

8

1 1 p = p3H +P3T =B+S

p:;;:~ 4

P=n CrPr qn-r

where: p =

1

2

; q = 1 ; n = 4; 2

P2H =

4c2(iJ(iJ

p2H:;;:

4! ( 1) (4-2)!2! 2

r=2

4

:;;:.S3

13

ED Note: Probability of getting a predit = 1/3 Probability of not getting any credit

P3H =

PK = probability of drawing a king PR = probability of drawing a red card PK & R = probability of drawing a king at

the same time

a red card

4c3(~J (~J 1

=4

4

26

52

52 52

PKorD

= 0.5385

3

4)

2

1

3! ( 12 (3-1)!1! 21 = 21

P2 = probability that exactly two students

(21)2(1)2

got a credit

1) (2) = c (3 3 = 2

4 2

1 :;;:16

P

3

1

2

2) = 21.6

3! ( (3-2)!2! 21

P3 = probability that all three students

2

=--+----

P,

2

PKorD =PK +PR -PK&R

R

1) (2) = c (3 3 = 1

4! ( 1) p3H:;;: (4-3)!3! 2

1) p4H:;: ( 2

= probability that only one student got a credit

4

P4H=4c 4

= 2/3

Let:

P1

Note: In a pack of 52 playing cards, there are 4 king cards, 26 red cards and 2 king &.red cards at the same time ·

Koro

2

p3T :;;:_

•

4 13 1 16 R ·=-+-----=Koro 52 52 52 52

6

. 1 1 :n= 3:r= 3 where:p=2";q=

8

PK = probability of drawing a king Po = probability of drawing a diamond ·PK & o = probability of drawing a king at the same time a diamond

•

(1) (1) (1)2 2 2

r=3

1

RKoro =-

5

1 : q= : n=3; 2 2

P=-.

4

4

1

3

P=3C3

Note: In a pack of 52 playing cards, there are 4 king cards, 13 diamond c.ards and 1 king & diamond card at the same time

P=-=--

where: p=

P=n CrPr qn-r

=(1)

P=j, CrPr qn-r I

Ill

p"" Pm + P3H I'~

+ p4H-

3 1 1 8 +4- +16-

got a credit

11 16

r3

=3

J( J

c3 (i j

= (1 {

2~) = 2~


138 IOOl·Solved\Problems in Engineering Mathematics (2nd Edition.) by Tiong & Rojas P = probability that at least one student got a credit

P = probability of passing at least two Subjects

_ __g+~+~ p = p1 + p2 + p3 - 27 27 27

P=0.384+0512

19 p = 27

p = 0.896

p2

4

6

10

9

P =-x24 p =1 90

=~ 9

P = probability that both balls drawn are all wh1te

Assume the first draw is white and the second draw is black:

Let:

Note: The only way she c'an get zero is, if all her 3 answers were wrong.

Let: p = probability of getting a correct answer p 2/3

=

p = probability of getting a correct answer p = 1/5

3

3

3

1 P=3C3 ( 1 ) ( 2 ) =(1) ( 1 ) =27

ED

1

= 10 c61 5

Probability of failing in any of the three subjects is 0.2

Let: P1 = probability of passing exactly two subjects ·

= 3C 2 (0.8) 31 (3 -2)121

2

(o.2f

(0. 128) = 0.384

P2 = prol;>ability of passing all the three subjects 3

0

P2 = 3C 3 (Q8) (02) =(1)(0512)

9

1\

6

4r

5) 5;

P1- -5P2 = probability of drawing a black ball in the second draw Note: The 1st ball was not returned in the bag before the 2"d ball was drawn

P = probability of getting a red or a white ball from tile box

p2

_ number of red or white balls Ptotal number of balls -------~~--·-·-

7 12

P = probability that both balls drawn are all black

1m

5 4 P=PxP=-X-

Let

P= 20

=~

56

14

2

8

1\ssume the first draw is black and the :;econd draw is white:

P2 = probability of drawing a white ball 111 the second draw

p1

= pblack X pwhite 9 9

p = 20 1 81 Assume the first draw is white and the second draw is black:

7

mJ

_5 P1 - 9

Assume the first draw is black and the second draw is white:

1

7

P=-=-

P1 = probability of drawing a white ball in the first draw

•

4 5 p =-x-

=~

1

24 24 p = p1 + p2 = 90 X90 p =0.533

8

p = 0.0055

14 24

Let:

P1 = probability of drawing a black ball in the first draw

Let:

Probability of getting a passing score in each subject is 0.8

4

P = probability that one ball is black and the other is white ·

Let:

l5 !

mJ

Notes:

6

10

p = 24 1 90

•

)v ( 4 ~4

p = nC,p' qn-r 3

p2 = pwhite X pblack 1

81

P = probability of getting 6 correct answers out from 10 questions

101 P= (10-6)!61

0

5

p = 25

q = probability of getting a wrong answer q = 4/5

p

3

5

P=P1 X p2 =g-Xg

P =-x-

q = probability_.of getting a wrong answer q = 1/3

pt-

P1 = Pblack X pwhite 1

ml

ED

P1

Note The 1st ball was returned in the bag before the 2"d ball was drawn

P2 = pwhite X pblack

..

5 4

'")• =-x 9 9 p2 = 20 81

Let: P

=probability that one ball is black and the other is white

p2 = 0.512

j

[

140 100·1 Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas

I

P=P1 +P2-20 20 --x 81 81 P= 40 81

D D D D D D D D

ID

Mon

Probability that A is correct= 1/3, assuming he is to sit down first

Tue

Probability that B is correct= 1/2, assuming he is to sit down after A Probability that C is correct = 1, assuming he is the last to sit down Let: P =probability that A; Band C

~re

correct

1 1 P=-x- x 1

3

I~

Topics

2

1 P=6

•

Theory

Wed

Problems

Thu

Solutions

Fri

[I]

~ Sat

Notes

Numbers from 1 to 20, which is divisible by 3 = 6 numbers (3,6,9, 12, 15, 18) Numbers from 1 to 20, which is divisible by 7 = 2 numbers (7,14)

Total numbers from 1 to 20, which is divisible by 3 or 7 = 8 numbers Let: P = probability that the ticket nu!llber is divisible by 3 or 7 P = successful outcomes total outcomes

8

P=20

2 P=-

5

j

T T

Venn Diagram Combinatorics Fundamental Principle of"Counting Permutation Inversion Cyclic Permutation Assortment Combination Probability Theory ·Mutually Exclusive Events Independent Events Binomial Distribution Odds For and Odds Against Mathematical Expectation '!

142 100 l Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas

Topics ' "

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Thu

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What is Geometry? ,,.

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Branches of Geometry Basic Postulates of Euclid Geometric Elements and Figures Types of Angles Units of Angles Polygons Classifications of Triangles Other Types of Triangles Types of Quadrilaterals Bramaguptha's Theorem Ptolemy's Theorem Terms Related to Circles Theorems Involving Circles Ellipses

Geometry is the branch of mathematics which deals with the properties of shapes and spaces. The term "geqmetry" was derived from the Greek words, "ge" meaning earth and "metria" meaning measurement. Euclid (c. 330- c 275 B.C.) in his best known book in geometry "Elements" give more emphasis on Plane geometry while Archimedes (287- 212 B.C.) contributed so much to Solid geometry. What are the different branches of

2. Solid Geometry - deals with the properties of geometrical shapes of three dimensions, such as cones, pyramids, cylinders, prisms, spheres, etc. 3. Euclidean Geometry- a geometry that is based on the assumptions of Euclid, who, about 300 B. C., collected the mathematical knowledge of that time in the 13-volume "Elements". The problems of this type of geometry can be solvd by logical reasoning from an initial core of postulates (axioms), a method commonly referred to as the classical' axiomatic method of Euclid.

Q~s>metry?

1. Plane Geometry- deals with the pmperties of plane figures or qt~ometncal shapes of two dimensions, :;11d1
4. Non-Euclidan Geometry- a geometry that is not based on the assumption of Euclid .. 5. Projective Geometry- deals with the study of those properties of plane frqures that arf' rmch;mqed when ;1

··- Da_y_I -- Plane Qeq_m~.!Dt:..~

144 1001 oSo1ved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas

~

What are the Basic Geometric Elements and Figures?

given set of points is projected onto a second plane.

1. Point- a.dimensionless geometric

6. Trigonometry - a geometry which

figure having no properties other than location or place.

specializes on the study of triangles. This subject is divided into two branches, namely Plane. Trigonometry and Spherical Trigonometry.

5. Straight angle- equal to 180° or n radians.

2. line -.the shortest distance between

--~

any two points. A line is implicitly a straight line. A line segment is a piece of a line with definite endpoints.

7. Analytic Geometry- a geometry which deals with geometric problems by using the coordinates systems, and transforming them into algebraic · problems. This type of geometry was invented by Rene Descartes. Plane analytic geometry deals primarily with the analysis of equations in two variables while Solid analytic geometry deals with equations hi three variables.

lines or two planes that meet. The point where the two line me~t is known as the vertex. Angle comes from Latin "angulus'' for "sharp bend". ·

r cr

What are the different types of angles according to their measure?

8. Differential Geometry - a geometry

J

1. A unique straight line can be drawn between any two points. 2. Such a line can be extended indefinitely in either direction.

3. A circle can be drawn in a plane using a given point (a center) and a given distance (a radius).

4. All right angles are equal. 5. Given a line and a point not on the line, there .exists exactly one line parallel to the original line passing through the given point Postulate number 5 above is commonly known as the "parallel postulate"

2. Acute angle- greater than 0 but less than goo or

2: 2

..-

/

( ~-o2:__).

\___,..J - - -

5. Vertical angles - angles formed by two intersecting lines. Vertical angles are

7. Full angle or Perigon- equal to 360" or 2rr radians.

radians

.&__

G What are the other types of angles?

3. Right angle- equal to goo or ~

1. Adjacent angles -two angles with a common leg.

radians

k_

l:_ 2.

4. Obtuse angle- greater than 90° or ~ 2 radians but less than 180° or n radians

Complementary angles- two angles whose surn is a right angle (90°).

L, /

LL~--

l

s_/

equal',

-........,·:r-£ ( ~\A A\ / ;r 1-

~

0~ '

Euclid's Elements is based upon five basic postulates or assumptions which are as follows:

~

1. Null or Zero angle -equal to 0.

What are the Basic Postulates of Euclid?

----'~

4. Explementary angles -two angles whose sum is
6. Reflex angle- angle great than ·180° or n radians but less than 360° or 2rr radians. 270"

3. Angle - The opening between two

that applies differential and integral calculus to cuiVes, surfaces and other geometric entities.

3. Supplementary angles -two angles whose sum is a straight angle (180°).

/

7"~

Angles A and B are vertical angles. These angles are called vertical angles because each side of one is an extension through the vertex of a side of the ot~1er Wh~..!!..tl!.~

A bisector is a straight line which divides a geometric figure into two equal figures. An angle bisector is a light which divides an angle into two equal angles.

;· Angle bisector

I

146 . I 00 Solved Problems in Engineering Mat!tematics (2nd Edition) by Tiong & Rojas What are the different units of angles?

A regular polygon is a polygon having all sides equal and a!: angles equal.

1. Degree- The degree is a non-S! unit approved in 1969 by the General

Table showing th•; regular polygons:

Conference on Weights and Measures. This unit was established by the Babylonians more than 4000 years ago. A degree is divided into 60 minutes; a minute is divided into 60 seconds. 2. Radian -The radian is the standard angular measure in the International System.of Units (SI). A radian is the angle between two radii of a circle which cut off on the circumference an arc equal in length to the radius.

3. Gon - The gon or centesimal degree is 1/4001h of the full circle. A right angle is 100 gons A gon is divided into 100 centesimal minutes; one centesimal minutes is divided into 100 centesimal seconds. This unit is also known as grad or grade and is used in surveying works and aircraft navigation. 4. Mil- The mil is a·measure of an angle which is 1/6400 of the full circle. A right angle is 1600 mils. The following is a table showing different measure of an angle with corresponding value in one revolution. Unit Degree Radian Gon Mil

One Revolution

360 2n 400 6400

Name Equilateral triangle Square Regular pentagon Reqular hexagon Regular heptagon ReQular octaQon Regular nonagon ReQular decagon Regular hectagon · ReQular meqagon. ReQular googolgon

Sides

Angle

3 4 5 6 7 8 9 10 100 106 100100

6Qo goo 108° 120° "'128.57° 135° 140° 144° 176.4° 179.99964° "'180°

D~Plane

Reentrant angle is the inward-pointmg angle (anglt= A 1n the f1gure) of the concave polygon wh1le the other angles are called salient angles.

.'

I

The diagonal of a polygon is the line connecting two opposite vertices. The number of d1agonals in a polygon can be calculated using the following formula:

3 4 5 6 7 8 9 10 11

12 100 1000000 10100 n

Name Triangle Quadrilateral or TetraQon Pentagon Hexaqon Heptagon OctaQon Nonagon Decagon UndecaQon Dodecagon Hectaqon MeqaQon Googoigon n-gon

A convex polygon is a polygon having each interior angle less than 180° while a concave polygon is a polygon having an interior angle greater than 180°

What is a Polygon?

Concave polygon

A triangle is a polygon with three sides. All the three sides are .contamed in a plane. What are the classifications of triangles according to their sides?

equal and all angles are equal. An equilateral triangle is also equiangular. Each interior angle is 60° 2. Isosceles triangle- two sides equal and corresponding two angles are also equal

Diagonals= -'2(n2

3. Scalene triangle - no two sides are equal.

3)

What are the classifications of triangles according to their angles?

where n = number of sides The interior angles of a polygon are the angles inside the polygon. The sum of the mterior angles of the polygon can be calculated using the following formula:

1. Acute triangle.- each interior angle is less than a right angle. 2.. Right triangle -one angle is a right angle. 3. Obtuse triangle - one angle is greater than right angle.

Sum of interior angles= ( n- 2)180° where n

I he deflection angle of a polygon is the ,mgle subtended by the prolongation of 'n1e s1de to the next s1de.

~;lllll

of dofll)C[IOII cHID IUS

Note: A triangle which is not a right triangle is known as oblique triangle. -Both acute triangle and obtuse triangle are oblique triangles. What are other types of triangles?

=number of s1des

A polygon is a plane figure with three or more angles. It has as many sides as angles. The sides of the polygon are straight lines. The term "polygon" comes from Greek words "poly" which means "many" and "gonia" which means "angles". Convex polygon

What is a triangle?

1. Equilateral triangle- all sides are

Polygons are named according to the number of sides or vertices. Number of sides

Geometry 147

'·'

360"

1. Egyptian tri;mgle- a triangle with sides 3, 4, 5 units. 2. Pedal triangle- a triangle inscribed in a given triangle whose vertices are the feet of the three perpendiculars,to the sides from some point inside a g1ven triangle. 3. Golden triangle- an isosceles triangle with sides is to its base in the golden ratio; its angles are 72°, 72", 36"

Day 7 - Plane Geometry 149

148 100 l Solved Problems in Engineering_ Mathematics (znct Edition) by Tiong & Rojas Note: The formulas to use in solving problems involving triangles are found in Chapter 9 (Plane Trigonometry) of this book What is a quadrilateral? A quadrilateral is a polygon with four sides. A quadrilateral is also known as quadrangle or tetragon. What are the types of quadrilateral?

1. Square- a regular quadrilateral. All its sides are equal and all angles are equal to right angle. A square is a rectangle with all sides equal.

aD

A= ab 'h

b

-------

i

4. Parallelogram - both pairs ot opposite sides are parallel. Another term for parallelogram is rhomboid.

A= bh

A Given base and altitude

0

B. Given two diagonals and their

where: d1 and d 2 are lengths of diagonals B. Given four sides and opposite angles: (Using Bramaguptha's formula)

r

7

A=

...... d1 ..... ·

{-">
9. Cyclic quadrilateral- a quadrilateral whose vertices lie on a circle

What are the formulas used in solving -

!!!e areq_of a 9!J.?drilateral?

£" 7

where: d1 and ch are lengths of diagonals

a

A= absine

C. Given side and included angle:

I ---:

tJ

a

b

GLih

2

·A=a sin0

8

A= a

A= ~(s-a)(s-b)(s-c)(s-d) -abcdcos 0 2

where:

t)

=A+C 2

2

A.=

A

1

--(B + b)h 7

or

t)

=

a+b+c+d S=------2

7. Cyclic quadrilateral 5. Trapezoid

a

a

2

C Given two sides and included angle:

1 A= 2(d 1 )(d 2 )

adjacent sides are equal in pair.

-~(d 1 )(d 2 )sine

where: d 1 and d2 are lengths of diagonals

dz \ ..

·For rhombus, 0 is a right angle.

8. Deltoid -concave quadrilateral whose

~(d 1 }(d 2 )sin0

included angle:

h

b

7. Kite- convex quadrilateral whose adjacent sides are equal in pair.

A:=

B. Given two diagonals

5. Trapezoid-- only two sides are parallel. 6. Trapezium- if no two sides are parallel.

d~·

b

3. Rhombus

A =bh

3. Rhombus - all sides are equal but no angle equal to right angle.

aD

A. Given two diagonals and included angle:

A. Given base and altitude:

b

2. Rectangle - a right-angled parallelogram.

1. Square

6. Trapezium (General quadrilateral)

4. Parallelogram or Rhomboid

2. Rectangle

~_:+:_Q 2

Day 7- Plane GeolllEillY 151

150 1001 Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas where: n = number of sides

What is Ptolemy's Theorem? A= .J(s-=a)(s- b)(s- c)(s- d) where: s =a+ b + c + d 2 The radius of the circle circumscribing the quadrilateral may be cal<;ulated using the following formula:

This was named after the mathematician astronomer, and geographer, Ptolemy Alexandria.

J( ab + cd)( ac + bd)( ad+ be)

What are the formulas used in solving the areas and perimeters of other regular polygons?

r=~--~~~~--~

4R

where: A = area of quadrilateral

2. Regular polygon inscribed in a circle

"The sum of the products of two pairs of opposite sides of a convex cyclic quadrilateral is equal to the product of the lengths of the diagonals."

perpendicular to the radius of the circle. 3. Secant of a circle - a !ine cutting the

circle in two places. 4. Diameter of a circle -the longest

chord of a circle that passes through the center.

cl

5. Radius -the distance from the center

to the circle. It is one-half of the diameter. A=

1. Regular polygon

a

1 . (360) n 2 nr 2

sm

P =2nrsin(

8. Quadrilateral circumscrib;ng a circle

6. Chord- the segment of a secant bounded by the circle.

1~0) secant

where: n = number of sides

a

Wl:tiit is a perimeter?

A perimeter is the distance around a two-

a A= A = rs "'

v'abed

where: s = a + b + c + d

dimensional shape. The perimeter of the square is the sum of the length of its sides. The formula for perimeter will vary depending on the geometric figure or poly~ on.

1 0

~na 2 cot( ~ ) P=na

What is a circle?

where: n = number of sides

2 What is Bramaguptha's Theorem?

A circle is a plane curve that is the locus of all points in the plane equidistant from a given point, called center. The term "circle" comes from the Latin ·'circus", which refers to a large round or rounded oblong enclosure in which the famous Roman chariot races were held. The circle is a conic section whose eccentricity is zero.

2. Regular polygon circumscribing a circle

"In a cyclic quadrilateral having perpendicular diagonals, the perpendicular to a side from the point of intersection of the diagonals always bisects the opposite side" Bramaguptha (A.D. 598- 665) was a Hindu astronomer and mathematician who became the head of the observatory at Ujjain- the· foremost mathematical center in ancient India.

What are common terms related to a circle? A= nr

2

tanC~O)

1. Arc of a circle- length of circle between to points on the circle or between two radii.

1

P = 2nrtan(. ~0 I '- n J

2. Tangent of a circle- a line touching the ctrcle in one place A tangent is

j

I i

6. Circumference- the perimeter of a circle. This is also known as periphery .. The circumference of a circle is 2n(radius) or n( diameter) . 7. Radian- the measure of an angle whose arc length is equal to the radius of the circle. 8. Sector of a circle- area bounded by two radii and the included arc. 9. Segment of a circle- area bounded by a chord and the arc subtending the chord.

~9b 1 Solved Problemsi!l Engineering Mat!:ematics (2~~ Editio~by Tiong & Rojas

3. Inscribe angles subtended by the

10. Central angle --an angle whose vertex is at the center of a circle and whose sides are the radii. The central angle is an angle subtended by an arc.

360

e is in degrees

4. Area of segment of a circle:

'--

=-ts

e =a(a+b) e = 13 =goo

A.= Area of sector - Area of triangle AOB The angle 0 is the central angle while is the angle subtended by chord "x".

p

What are the for111ulas used in sqjyin.Q_
circle1 1. Area of a circle:

A= ltr~

or

A=

.

6. Secant- Tangent Theorem:

diameter of a circle are right angles.

A= _:
where:

11. Angle subtended by a chord - an angle whose vertex is along the periphery or circumference and its sides are chords.

Day 7 -Plane Geometry 153

What are useful theorem involving a circle?

What is an ellipse?

Therefore, the triangle formed in the figure is a right triangle. 4. Chord Theore;n:

1. If a central angle and a peripheral angle are subtended by the same arc, then the central angie is twice as large as the peripheral angle

ab = cd

Ellipse is a locus of a point which moves so that the surn of its distances to the fixed points (foci, denoted as F in the figure) is constant and is equal to the length of the major axis.

v

v

----r--.. . .·

rt02

4

~

where: r = radius D =diameter

2. Circumference of a circle: C =2nr

or

5. Secant Theorem:

C=:rcD

2, Inscribed angles subtend the same arc are equal.

·-·~

a

a

Ellipse has two axis of symmetry, a major axis (2a), betvveen the vertices (V) of the ellipse, and a minor axis (2b), which intersect at the center of the ellipse. Area of an ellipse, A A= n:ab

3. Area of sector of a circle:

c

0=13

where: a = semi-major axis b = semi-minor axis

1 A= --rc

2

a(a+b)=c(c+d)

i A ""-r2e

2

where:

e is in radians

t,l,

II ,.

154 1001 Solved Problems in Engineering Mathematics (2"d Edition) by Tiong & Rojas Proceed to the next page for your 7th test. Detach and use the answer sheet provided at the last part of this book. Use pencil number 2 in shading your answer.

i

·i

,I.. '··

,t',

GOOD LUCK I

Topics

~ Mon

~ribia:: Did you know that. .. the symbols + and for plus and minus, respectively were introduced by German mathematician and astronomer, Johannes Regiomontanus in 1456!

D D D D D DI D D

'/-:

Tue

,:'',

I

Branches of Geometry Basic Postulates of Euclid Geometric Elements and Figures Types of Angles Units of Angles Polygons Classifications of Triangles Other Types of Triangles. Types of Quadrilaterals Bramaguptha's Theorem Ptolemy's Theorem Terms Related to Circles Theorems Involving Circles 1 Ellipses

!

"Man is like a fraction whose numerator is what he is and whose denominator is what he thinks of himself. The larger the denominator the smaller the fraction." -Tolstoy

Theory

Wed

Problems

Thu

Solutions

Fri

Notes

Sat

30&: ECE Board November 1998 Find the angle in mils subtended by a line 10 yards long at a distance of 5000 yards. A. B. C. D.

1 2 2.5 4

307: ECE Board April1999 Assuming that the earth is a sphere whose radius is 6400 km. Find the distance along a 3 degree arc at the equator of the earth's surface. A. B. C. D

335.10 km 533.10 km 353.10km 353.01 km

:J08: EE Board April199:1 1111'
A. B. C. D.

16.85 em 17.85 em 18.85 em 19.85 em

309: ME Board April1990 A rat fell on a bucket of a water wheel with diameter of 600 em which traveled an angle of 190° before it dropped from the bucket. Calculate for the linear em that the rat was carried by the bucket before it fell.

A B. C. D.

950 965 985 995

310: ECE Board November 199:1 Given a circle whose diameter AB equals 2 m. If two points C and D lie on the circle and angles ABC and BAD are 18° and 36°, respectively, find the length of the major arc CD.

156 100 l Solved Problems in Engineering Mathematics (2"d Edition) by Tiong & Rojas

A. B. C. D.

1.26 m 1.36 m 1.63m 1.45 m

3u:: A certain angle has a supplement 5 times its complement What is the angle?

A. B. C. D.

67.5° 58.5° 30° 27"

31::&: ECE Board November 1998 Each angle of a ·egular dodecagon is equal to

A. B. C.

D.

135° 150° 125° 105°

313: CE Board May 1997 How many sides have a polygon if the sum of the interior angles is 1080°? A. B.

5 6

C.

7

D.

8

314: ECE Board March 199& The sum of the interior angles of a polygon is 540°. Find the number of sides. A. B. C. D.

12 24 20 48

3:t:Z: EE Board April 1991 From a point outside of an equilateral triangle, the distances to the vertices are 10m, 18m and 10m, respectively. What is the length of one side of a triangle?

317: How many diagonals are there in a polygon of 20 sides?

A. B. C.

D.

200 170 100 158

A.

A. B. C. D.

17.75 18.50 19.95 20.50

m m m m

318: ME Board April1999 Find each interior angle of a hexagon.

3::&3: EE Board April1991 The sides of a triangle are 8 em, 10 em and 14 em. Determine the radius of the inscribed circle.

A. B. C. D.

A. B. C. D.

90° 120° 150° 180°

319: EE Board April1994 Given a triangle, C = 100°, a= 15m, b = 20m. Find c. A. B.

C. D.

26m 27m 28m 29m

2.25 em 2.35 em 2.45cm 2.55 em

3::&4: CE Board May 199& What is the radius of the circle circumscribing an isosceles right triangle having an area of 162 sq. em.?

A B. C.

D. 3::&0: CE Board November 1994 In triangle ABC, angle A = 45° and C = 70°: The side opposite angle Cis 40 m long. What is the length of the side opposite angle A?

3::&7: ECE Board March J99& A circle with radius 6 em has half its area removed by cutting off a border of uniform width. Find the width of the border.

12.73 m 13.52 m 14.18m 15.55 m

B. C. D.

1.76cm 1.35 em 1.98cm 2.03 em

3%8: ME Board April :199& The area of a circle is 89.42 sq. inches. What is its circumference?

A. B. C. D.

32.25 33.52 35.33 35.55

in. in. in. in.

I

3::&9: ECE Board April 1991 A square section ABCD has one of its sides equal to x. Point E is inside the square forming an equilateral triangle BEC having one side equal in length to the side of the square. Find the angle AED. A. B. C. D.

130° "140° 150° 160°

3ZS: EE Board April 1991 The sides of a triangle are 8 em, 10 em and 14 em. Determine the radius of the circumscribing circle.

330: CE Board November :1.995 The area of a circle circumscribing about an equilateral triangle is 254.47 sq. m. What is the area of the triangle in sq. m?

A. B.

A.

ii r

3 4

5 6

315: ECE Board April :1991 Fioo the sum of the interior angles of the vertices of a five pointed star inscribed in a circle. A. B. C. D.

A. B. C. D.

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _.:::D~a::..;yc....:..7_-Plane Geometry 157

150° 160° 170° 180°

A. B. C. D.

m m m m

3::&1: CE Board May i995 In triangle ABC, angle C = 70°, A= 45°, AB = 40 m. What is the length of the median drawn from vertex A to side BC?

C. D.

7.14 7.34 7.54 7.74

em em em em

3261: CE Board May 1996

Two sides of a triangle are 50 m and 60 m long. The angle included between these sides is 30° What is the interior angle opposite the longest side?

A.

~.3m

B. C.

~.6m

A.

~.9m ~.2m

n c

D.

31&: ME Board April1999 How many sides are in a polygon if each interior angle is 165 degrees?

26.1 27.1 29.1 30.1

(l

93."14" 9) l4° 90.? 11" llCI J()'

B. C. D.

100.25 102.25 104.25 105.25

3311 CE Board May 1.995 What is the area in sq. em of the circle circumscribed about an equilateral triangle with a side 10 em long? A. B. C.

104.7 105.7 106.7

D.

107.7

'if

1~8 1001 Solved Problems in Engineering Mathematics (2"d Edition) by Tiong & Rojas 33:11 CE Board November 199:1 The area of a triangle inscribed in a circle is 39.19 cm 2 and the radius of the circumscribed circle is 7.14 em. If the two sides of the inscribed triangle are 8 em and 10 em, respectively, find the third side.

A. B. C. D.

11 em 12 em 13 em 14 em

C. D.

64 1t 16n

3371 ECE Board }lllovember 1993 The arc of a sector is 9 units and its radius is 3 units. What is the area of the sector in square units?

A. B. C. D.

12.5 13.5 14.5 15.5

3331 CE Board November 1994 The area of a triangle is 8346 sq. m and two of its interior angles are 37"25' and 56° 17'. What is the length of the longest side? A. B. C. D.

171.5 m 181.5 m 191.5m 200.5 in

A. C.

59.8 89.5 58.9

D.

81:;CJ

B.

Two perpendicular chords both 5 em from the center of a circle divide the circle into four parts. If the radius of the circle is 13 em, find the area of the smallest part. 30 31 32 33

2

cm cm 2 cm 2 2 cm

A circle having an area of 452 sq. m is cut into two segments by a chord which is 6 m from the center of the circle. Compute the area of the bigger segment.

A. B. C. D.

354.89 363.68 378.42 383.64

sq. sq. sq. sq.

m m m m

339: ECE Board April 199:1 A swimming pool is constructed in the shape of two partially overlapping identical circles. Each of the circles has a radius of 9 m and each circle passes through the center of the other. Find the area of the swimming pool. A. B. C. D.

The distance between the centers of the three circles which are mutually tangent to each other externally are 10, 12 and 14 units. The area of the largest circle is

727t 23

1t

Determine the area of the quadrilateral shown. OB = 80 em, AO = 120 em, OD 150 em c;nd q, = 25•.

A.

W.5~

B.

21.5cm

c. D.

n.5~ n.5~

A trapezoid has an area of 36 m 2 and an altitude of 2 m. Its two bases have ratio of . 4:5. What are the lengths of the bases? A.

12, 15

B. C.

7, 11 8, 10

D.

16, 20

Tf::>,o

;V

'\1

<

•

=

A rhombus has diagonals of 32 and 20 inches. Determine its area. A. B. C. D.

380m 2 2 390m 2 400m 2 410m

2

360 in 280 in 2 2 320in 400 in 2

If the sides of a parallelogram and an included angle are 6, 10 and 100°, respectively, find the length of the shorter diagonal.

A.

Find the difference of the area of the square inscribed in a semi-circle having a radius of 15 m. The base of the square lies on the diameter of the semi-circle.

B. C. D.

A. B. C. D ...

2721.66 2271.66 2172.66 2217.66

cm 2 cm 2 cm 2 2 cm

347: CE Board Oc:tober 1997

343: EE Board Marc:h 1998

340: ME Board April 1991

10.63 10.37 10.73 10.23

Find the area of a quadrilateral have sides 12m, 20m, 8 m and 16.97 m. if the sum of the opposite angles is equal to 225°, find the area of the quadrilateral. A. B. C. D.

!:!i,

'I

'r'·j

I 1!,,,

1 1·i

i

'I'

'j:

2

100m 124m2 168m2 158m2

i 1,

348: ME Board Oc:tober 1996 ME Board April :1997 The area of a regular hexagon inscribed in a circle of radius 1 is A. B.

c. D.

,,

1.316 2.945 2.598 3.816

'I

349: EE Board April 1990 2

Find the area (in em ) of a regular octagon inscribed in a circle of radius 10 em?

I

3451 CE Board Nove.mber 1996 .

A. 33fn ECE Board April 1998

A. B.

Problem 346: EE Board October 199:1

A rectangle ABCD which measures 18 em. by 24 em. is folded once, perpendicular to diagonal AC, so that the opposite vertices A and C coincide. Find the length of the fold.

344: ECE Board April 1998

3351 EE Board Ai"ll'il 199:1

A. B. C. D.

341: ECE Board November 1995



334: ECE Board April 1998 The angle of a sector is 30° and the radius is 15 em. What is the area of the sector in cm 2?

Day 7- Plane Geometry 159

B. C. D.

171.5 cm 2 172.5 cm 2 173.5 cm 2 174.5 cm 2

Find the area of a quadrilateral having sides AB = 10 em, BC = 5 em, CD= 14.14 em and DA = 15 em, if the sum of the opposite angles is equal to 225•. A. B. ('

n

A. B.

C. D.

283 289 298 238

96 sq. em

350: GE Board February 199:1

100 sq. em 94 sq. em qg sq. em

A regular hexagon is inscribed in a circle whose 'diameter is 20m. Find the area of the 6 segments of the circle formed by the sides of the hfiXilCJOil

I

llnl"l''l:;;'r'

l

iiJj!

I '•I! I

.:

1

:_ '.l.__ ·. ~-h

I

Ill'~ ·

160 1001 ~olved Proble!ns in Engineering Mathematics (2nd Edition) by Tiong & Rojas A.

B. C.

D.

36.45 63.54 45.63 54.36

sq. sq. sq. sq.

m m m m

351: EE Board Aprii199:J Find the area of a regular pentagon whose side is 25m and apothem is 17.2 m.

A. B. C. D.

1

li'

355: EE Board March 1999 Determine the area of a regular 6-star polygon if the innerregular hexagon has 10 em sides. A. B. C. D.

441.66 467.64 ·519.60 493.62

II

cm 2 cm 2 cm 2 cm 2

Topics

2

1075 rn 2 1085 m 2 1080 m 2 1095 m

Mon

0

TtJe

3S:l: ME Board October 1!.99&

0

The area of a circle is 89.42 sq. inches. What is the length of the side of a regular hexagon inscribed in a circle? A. B.

c. D.

5.533 5.335 6.335 7.335

Theory

[1

in. in. iio. in.

Problems

353: EE Board April1990

Solutions

In a circle of diameter of 10m, a regular five-pointed star touching its circumference is inscribed. What is the area of that part not covered by the star? A. B. C. D.

40.5 45.5 50.5 55.5

sq. sq. sq. sq.

0 306. B 307.A 308.C 309. D 310.A 311. A 312. B 313. D 314.C 315. D 316. B 317. B 318. B

354: EE Board March 1998 A regular pentagon has sides of 20 em. An inner pentagon with sides of 10 em is inside and concentric to the larger pentagon. Determine the area inside and concentric to the larger pentagon but outside of the smaller pentagon. A. B. C. D.

430.70 573.26 473.77 516.14

cm

3

3

cm 3 cm 3 cm

Thu

0 Fri

[]

ANSWER KEY 319. B 332. D 345.8 320. D 333.8 346.A 321. A 334. 347. 322.C 335. B 348. 323.C 336.C 349. A 324.A 337. B •350.0 325.A 338. B 351. A 326. D 339. D 352. B 327.A 340.C 353. 328. B 341. 354. D 329. 342. D 355. 330. D 343. 331. A 344.

c

c

I~

Sat

Notes

m m m m

[WedJ [l

;I

Branches of Geometry Basic Postulates of Euclid Geometric Elements and-Figures Types of Angles Units of Angles ' Polygons Classifications of Triangles Other Types of Triangles Types of Quadrilaterals Bramaguptha's Theorem Ptolemy's Theorem Terms Related to Circles Theorems Involving Circles Ellipses

c c c

c c

c c

RATING

I

'

;

0 ~3-50 c::J c::J 0

Topnotcher·

33-42 Passer

25-32 Conditional 0-25 failed


:l

il. iI•l

II i

i

j

il :'

t

;,j[,,

162 100.1 Solved Problef!lS in Engineering Mathematics (2nd Edition) by Tiong & Rojas

Day 7- Plane Geometry 163

Ell

Ill 5

Note: A dodecagon has 12 sides, n

5

9 = (n- 2)(180') n

5

8 = (12-2)(180") 12

tan8=-5000 8 = tan·

1

5 = 0.0573 5000

28 = 2(0.0573) = 0.1146" = . x 6400 mils 0 1146 360' 20 = 2.037 mils

-

2

n rad) 360°

1080 = (n- 2)(180) n=8sides

Ill S=(n-2)(180") 540=(n-2)(180)

8=36'

n-2 =3

28=72"

n = 5 sides

a

Let:

c = length of. arc CD

C = 18.85cm

C=r8

360°

C = 3oo(190o x 2n: rad) 360°

n

N=-(n-3)

2 20 N =-(20-3) N = 170 diagonals

a Note: A hexagon has 6 sides, n = 6. 8= (n-2)(180")

6

•

Ell 28 = 360

e =the angle go• - e = the complement of angle e 180•- e =the supplement of angle e

Elll

a

e =120·

Let:

C=995cm

n =24 sides

8 = (6-2)(180')

C=1.26m

c =re

15n = 360

n

C=1(n·x~) . 360°

Ell

n

2

18+ e + 36 =go

C=45(24~x 2nrad)

= (n-2)(180")

165n = 180n- 360

n·2=6

C =335.10km

C=re

165

S=(n-2)(180")

Refer to the first figure:

Ill

S= (n-2)(180')

'looo<..-,.:

Ell

For the same intercepted arc (arc CD), the value of the central angle is twice that of the inscribed angle. If one side of the triangle inscribed m a circle is equal to the diameter of the circle, then the said triangle is a right triangle.

C =r9 C = 6400.(3° x

Theory:

a n

a= 1so·

c

( --)

=12.

5 0=36

c

Az=s:B c =?

By cosine law:

Let: S = sum of the interior angles of the five

180 - e = 5(90 -e)

vertice!O

180-8 = 450-58 48= 270

e = 67.5' c

j

2

2

c=Ja +b -2abcosC

s;:; 50-:;; 5(36')

c=~(15) 2

:; - 11l0

c =27

2

+(20) -2(15)(20)cos100'

!'',1

,j:l, I'

I

'. '.·I'

""

Day 7- Plane Geometry 1&5

164 1001 Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas

•

c

x. = fc 2 + (at2/ - 2( c)( a/2)GOS B

A= Js(s-a)(s-b)(s-c)

x=~(40) 2 +(15) 2 -2(40)(15)cos65'

A= J16(16- 8)(16 -1 0)(16 -14) A=39.19cm

x=36.3m

S=

Mf _......_---:c--~40 \ _ ___: 'o B

sinA

a

c

I

4r

By sine law: sin 30

sin (60 + ~)

10

18

B

a+ 4.158 + 90 = 180

By sine law:

sin 10•

sin45"

40

a

a=30 m a/2=15 m

sin 30•

X

10

By cosine law:

S=

2

2r = .J2x 2.

x =19.95m

2

11,1 'i :1

II'!; 1

1!:1' Iii.' 2 -

2(50)(60)co:;30'

lL II''

1:,.

r i

ml

b = 10

2 ac:8

j

A

It

=----

S=:16cm

sin30"

60 30.06 B=86.38

• 8+10+14

c = 30.06m

sillB

2r=J2(1W r=12.73cm

a+b+c

r

I',,II:

By sine law:

d=Jx 2 +x 2

A+B+C=180 45 + B + 70 = 180

B=65°

c = ~ (50) + (60)

Using Pythagorean theorem:

sin 85.842•

•

b =60

c=Ja 2 +b 2 -2abcosC

2 x =18 em

J

30"

By cosine law:

1 162 = -(x)(x)

a =85.842'

a

·~

A= -(r<~se)(height)

2

\l.,ll' . :[ .

B

C

1

a + ~ +60 + 30 = 180

c

II II

39.19 = 8(10)(14)

Note: Since the isosceles right triangle, is inscribed in the circle, then its hypotenuse must be equal to the diameter of the circle

~=4.158'

sinA

II

A= abc

•

c

sinC

I

II

4r r=7.14cm

sin45 J a=40 - ( sin70 a-= 30.1m

By sine law:

~·,I

A= J16(16 -8)(16 -1 0)(16 -14)

sinAJ a=c ( sinC

ml

2

A=39.19cm 2

•

sinG

2

A= Js(s-a)(s-b)(s-c)

A=rs 39.19 = r(16) r=2.45 em

By sine law:

8+10+14

=----

S = 16 em

2

b

ml

a+b+c

Day 7 - Plane Geometry 167

166. 100 I Solved Problems in Engineering Mathematics (2"d Edition) by Tiong & Rojas Note: Since half of the area was removed, then the area (A} left is also one-half of the total area.

30' +a+a=180o

A

111

a=75

c

Ill

60' +2a+ 8=360°

A=.:!_[ n(R/ J=:_[n(6) J 2

2

60 + 2(75) + El = 360.

2

A= 18n But "A" is also equal to the area of the small circle. (Refer to figur~ A= nr 2 18n = nr

Bd"

•

Note: Since an equilateral triangle, then angles A, B and C are equal to 60°.

:f

.A +B+C = 180

c =86.301'

2

A =-(x) sin A 2 1 2

By sine law:

A= -(10) sin 60°

2

x=6-4.24 A= nr

r=1.76cm

El = circumference of the circle

C = 2nr C = 2n(5.335)

By cosine law:

Solving for area of circle:

x=) r 2 + r 2 - 2(r)(r)(cos 9)

A= nr 2

2

J:JA

X

::;:-acsinB

2 1

8346 =-(0.609c)(c)sin 56.283) 16,692 =0.5065 c 2 c=181.5

'I

.

1

b = 10

2 1

A= abc

2

.,.- tJs

1

Atriangle

II'

A= ~(15.59) 2 (sin 60°)

--

Jc

sin37.416 a =0.609c

I!

A= ~(x)(x)(sln A) X

sin86:301 (

Ill

Ill

Note: Since the triangle is an equilateral triangle, then angles A, B 'and C are each equal to 60 degrees.

1

a=

A =104.7cm 2

x=15.59m

X

i

l

2

A= 7t(5.774) 2

2

x.=A9) +(9) -2(9)(9)(cos120°)

C=33.52in

n.--

4r r =5.774cm

e= 120°

r =5.335in

a

4r

3

89.42 = nr 2

c

43.3 = (10)(10)(10)

360° 9=-

sin37.416•

'!l'r

A= abc

r=9m

A= nr 2

sin86.301•

A=43.3cm 2

2

254.7 = nr 2

c~

'DA

37.416 +56.283 +C = 180

.1

2

x=R-r

•

'

c =?

A= 3r25' = 37.416° B = 56°17' = 56.283°

A

r =4.24cm

Let: C

'

9 =150'

A =105.24 m

2

4r 39.19 = -~O)c 4(7.14)

X

Note: Triangle CDE is an isosceles triangle.

l j

cco14crn

1 . A =-r 2 e 2 1 A =--.(15) 2

, I'I::if I r:

2(30. x--J 7t

180'

A =58.9cm 2

I

lflilll' ,l ,11'1 ,

1

I j

I

LLJ ttl

II:

168 ._LOOl Solved Problems_in Eng_!~eering Mathematics (2nd Edition) by Tiong & Rojas

•

•

12 = y+5

B


y=7

·A =As -2A 1

A 2 = Ac - A1 = 452-84.44 A 2 = 363.56 m 2

5 A= 66-2[;(base)(height)]

A= 66-(7)(5) = 31 cm 2

1 A=-rC 2

1

.

A=-(3)(9)=13.5sq. umts 2

.

Ell

5

stna =--

~

4.5 cose=-=60° 9

13 o;

6

=22.62"

Let: 4.5 A = area of the pool A, = area of triangle ABC Ac = area of the sector As= area of the segment (shaded portion.)

2a+0=90 2(22.62) + e = 90

.9=44.76°

Let:

=

2

1 9=-(13) 1 As =-r 2

r2 +r3 =10-tEq.2 r1 +r3 =14-tEq.3

2[ 44.76•x-1t ]

2

A1 = area of the smaller segment A2 = area of the bigger segment Ac = area of the circle As = area of the sector A, = area of the triangle ABC

r1 +r2 =12-tEq.1

Let: A area of the shaded part As == area of sector ABC A, = area of triangle COB

180"

2

2

=_!.(9) 2 (120"

Subtract Eq.3 by Eq.2:

(r1 +r3 )T(r2 +r3 )=14-10

A 5 =66cm 2

A. =Ac -A 1 1 1 =-r 2 (29)--r 2 sin29

r1 -r2 =4-+Eq.4

A. = 1tr

2

2

452 = 7tr

2

1

.

x-n:-} 180"

--
r=12

A, =49.75 m 2

Add Eq.4 and Eq.1: (r1

·~r2

cos8=~=~ r

)+(r1 -r2 )=10+4

'

5

=21t
r1 =8 r2 =12-8=4 r3 =10-4=6

~c

By Pythagorean theorem:

Let: A = area of the largest circle

{13) 2 =(5) 2 +(y+5) 2

A=

1tr/ =1t(8)

2

=2(1tr 2 )-2A.

9=600

2r1 = 16

'

A'=2A- -2A.

12

1 2 1 2 • A 1 =A. -A, =-r (29)--r stn29

2

2

=_!.(12) 2 (120/' x-1t-}

2 1 --(12) 2 sin 120" 2 .A 1 ::::84.44

= 647t

144=(y+5) 2

.'~

180'

A =409.44 m 2

Ill • x/2

x/2

:11!

170 1001 Solved Problems in Engineering Mathematics (2"d Edition) by Tiong & Rojas

r2 = x2

a

+(;r

Day_J-:()l_ane Geometry 171

II

a

i',l

A=

15 =1.25x = 13.416

-abcdcos 2 9

x

Let: A, = area of the semi-circle Az = area of the square A = difference of A1 and Az 1 A=A 1 -A 2 =-1tr 2 -x 2

2

1 A =-1t(15) 2 -(13.416) 2

(22.07- 5)(22.07 -14.14)

b

1

2

Ill 24

2

o~c X

24-x

a

4

b

5 4

29 + 100 + 100 = 360

5

9=8o·

Substitute Eq.2 in Eq.1:

By cosine law:

36=;(~b+b}2)

d2 =6 2 +10 2 -2(6)(10)cos80. d = 10.73

2

x =324+576-48x+x

4 5 36 =-b+-b 5 5 9 36=-b 5 b=20m 4 a=-(20)==16m

Ell

Length of the fold= 2y = 2(11.25) Length.of the fold= 22.5cm

'!1:'1

:-=oo

i ~I ,:1 r'

l

By secant law:

a =5

.AO • BO =CO • DO 120• 80 =C0•150 C0=64 em

(s- a)(s- b)(s- c)(s- d)

Let: A = area of the quadr lateral A, = area of triangle 1100 Az = area of triangle FOG

abcdcos 2 9

A =A 1 -A 2

A

X

a+b+c+d S=----

(18.75) 2 =(15) 2 +y 2 y = 11.25

V(

li

llillllil!ili

L 'Ill, 1',

l1

~~~

I

A=

2

+(y)2

t'J

Ill

X= 18.75

=(;r

• ~

c =15

x2 =18 2 +(24-x) 2

A=100cm 2

Total interior angle= 360'

5

d=-J18 2 +24 2 =30

-(5)(14.14)(15}(10)cos 2 112.5"

Total interior angle= (n- 2)180 =(4-2)180

A~B

18

A= 1(22.07-15)(22.07 -10)

10

A=-(a+b)h~Eq.l

a=-a~Eq.2

A =173.44

x2

ii'I'

(s- a)(s- b)(s -c )(s- d)

2

2

Ill

Substitute:

2

1

A=-d 1 d2 2

s=

5+14.14+15+10

2

1

A= -(32)(20) 2 A=320in 2

A +C 225" 9=--=--

2 0=112.5"

2

=22.07

1 1 A = -(AO)(OO)(sin E - -(BO)(CO)(sin 9) 2 2 ·1 A= -(120)(150)(sin '5") 2 1 --(80)(64 )(sin 25") 2 A= 2721.66 cm 2

,,'i ,,

ir I'

':

;I,!

,,

172 1001 Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas .

II

A1

=

(21)

Let: . r 2 Sln9

Ah = area of the hexagon At = area of triangle Ac = area of the circle A = area of the 6 segments of the circle

2

A 1 =.2,(1) sin60°


•

2

A0~'~--7b-=~2~0---

A 1 =0.433

360° 9=-=60°

oo

6

Atotal =6At = 6(0.433) (s- a)(s- b}(s- c)(s- d) A= abcdcos 2 9 a+b+c+d S=----

2 S=

2

A total = 2.598

A 1 =(;} sin9

•

360° 9=--=60°

6

1 2 At =-(10) sin6o·

= 7tr2 89.42:;; 1tr 2 Ac

2 At =43.3m 2

r = 5.335in Ah = 6A 1 = 6(43.3)

12 +20+8 + 16. 97 = 28.485

By cosine law:

Ah = 259.8 m 2

2 5 9- A+C - . -_ 22 • =112.5• - 2 2 Substitute: (28.485 -12)(28.485- 20) A=

-(12)(20)(6)(16.97)cos 2 112'.5" A=168m

2

• Let: Atota~

Let:

=area of the hexagon

A:: 1tr 2 - Ah

Atotat :;; area of the octagon At = area of triangle

A= 54.36

360° 9=--=45° 8

I (28.485- 8){26.485 -16.97) A1

x=Jr 2 +r 2 -2(r)(r)cos9

A =Ac -Ah

= 1t(10)2 - 259.8

6

2

-

2

2(5.335) cos60

x =5.335in

m2

•

Ill

. =(1)r 2 stn9

2

1

2

A 1 =-(10) sin45°

2

. A 1 = 35.355 cm 2 b=25 A total =SA, = 8(35.355)

Let:

A 10181 = 262.84 cm 2

Ah = area of the hexagon At = area of triangle

•

1

1

2

2

A 1 =-(base}(height)=-(25)(17.2) A 1 =215m 2

At = area of triangle 360° 9=-=60°

x = b(5.335)

I

A= 5A 1 = 5(215)

I

A=1075 m 2

!

j

.Let: A = area not covered by the star As = area of the star Ac = area of the circle

9

360" 29=-

X

5

a

8=36c

=

8/2 18 9/2

"'-~ ________ [}liy_7,_ !'lal\f?

174 . 100 1 Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas

Qeometry 175

e

9+-+a =180

A= n(;-(base)(height)J = n(;xh J

2 36 +18 +a= 180 a= 126~

A =;xCt:neJ nx 2 A=-4tan9

By sine law: sin18"

sin126"

X

5

X=

180

180

9=-=~=30"

n 6 6(10) 2 A1=--4tan30" A 1 = 259.8 cm 2

x = 1.91 m 1

1

2

2

A 1 = -rx sine= -(5)(1.91 )sin 36 A 1 =2.806m 2 X2=

x= 10

20

\ll

A=Ac -10A 1 180" 180° 9=--=--=36" n 5

=7tr 2 -10A 1 = 1t(5)2 -1 0(2.806) A=50.5m 2

Note: By inspection, the triangle must be equilateraL

Let: A1 = area of the inner pentagon A2 = area of the outer pentagon A = area of the shaded portion

Ell

1

A 2 = -x 2 sin 60

A =A 2 -A 1

2 1

2

2 nx ___ nx =--2 . _1_

= x/2 x/2

X

h

2h

10

x/2

tan9=-=~

il'

= -(10) 2 sin 60 2

4tane 4tane 5(20) 2 5(10) 2

ll·il!l

II, I' I

A 2 =43.3

---

4 tan 36 4tan36 A= 516.14 cm 2

A=A 1 +6A 2 = 259.8 + 6(43.3)

A =519.6 cm 2

Ell

X

h=-2tane 360" 180' 9=-=2n n Let: A = area of an •n• sided polygon

Let: A1 = area of the hexagon A2 = area of the triangle A = total area

IIIII: , ... 1

j

I d

178 l 00 l Solved Problems in Engineering Mathematics (2nd Edition) by 'riong & Rojas

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Polyhedrons Platonic Solids Properties of Platonic Solids Prisms Cubes & Parallelepiped Cylinders Pyramids Cones Frustum of Pyramid or Cone Prismatoid Sphere Zone Spherical Segment & Sector Spherical Pyramid & Wedge Torus Ellipsoids

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Convex polyhedron is one that lies r~ntirely on one side of a plane that contain ;111y of its faces while a concave polyhedron has at leart one face so located that there are parts of the l'olyhedron on both sides of a plane Lorrt.Hninn that face. If a line that connects .'"Y lwrrrts on the surface of a 1" >lylir ·dr '>ll 1s cornpletely rnside or on the i'' 1lylu ·rlr orr, lilr, polylwdron IS convc:x, "ll11·1wr:;r rl r:; <.<•llC.lVI'

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Every polyhedron in three-dimensional space consists of (two-dimensional) faces, (one-dimensional) edges and (zero-dimensional) vertices. The faces of the polyhedron intersect each other along the edges, which meet at the corners known as vertic.es What are the 5 regular polyhedrons? A regular polyhedron is a solid with all its faces identical regular polygons. There are only five regular convex polyhedra, namely the tetrahedron, the hexahedron (cube), the octahedron, the dodecahedron and the icosahedron. These solids are also known as Platonic solids in honor of Plato (427- 348 B.C) The following are the Platonic solids with its properties:

Day 8- Solid Geometry 181

180 1001 S~lved Problems in Engineering Mathematics (2"d Edition) by Tiong & Rojas 1.

3.

Tetrahedron

5.

Octahedron

Icosahedron

Surface area, A: A=2(ab+bc+ca) Right prism is one which has its lateral faces perpendicular to the base.

e

Faces No. of faces No. of edges No. of vertices Polygon angle Sum of angles Radius of circumscribed sphere

Triangle 4 6 4 60° 180°

Faces No. of faces No.ofedqes No. of vertices Polyqon angle Sum of anqles Radius of circumscribed sphere

eJ6 4

--

12

Total area

6

2e

Volume

4.

2

Trianqle 20 ~----1 12 60° 300°

~~2(5+J5)

~t+3.J5

./3

5e 2.f3

Volume

Dodecahedron

6

2

Total area

!e3.J2 3

~e3.J2 12

2.

e.J6 -

Total area

e2.f3

Volume

e../2 ___?__

Radius of inscribed sphere

e.J6

Radius of ir.scribed sphere

Faces No. of faces No.ofedqes No. of vertices Polygon angle Sum of anqles Radius of circumscribed sphere Radius of inscribed sphere

Trianqle 8 12 6 60° 240°

~~3 (3 + .J5)

What is a prism?

Hexahedron

A prism is a polyhedron with two faces (bases) parallel and congruent and whose remaining faces (lateral faces) are parallelograms.

e Faces No. of faces No. of edges ~_of vertices ,.l'_l)lygon angle

Square 6 12 8

~ofanqies

270°

e~

.!-Radius of inscribed sphere

J

e

2

i

6e 2 eJ

Total area ------

~..Qt:l_angle

goo

Radius of circumscribed sphere

~olume

Faces No. of faces No. of edges -~o. of vertices

~-

l

~-~---

Sum of angles -Radius of circumscribed

Cube is a prism with all six faces a square. It is a regular hexahedron.and one of the five Platonic solids.

e

Pentagon 12 30 20 108° 324°

Rectangular parallelepiped is a prism with all six faces a rectangle.

~--··

Total area

-·-·

/

Volume

71 c

~J3(1 + JS) a

I

A= h Pb where: B = area of the base h =altitude of prism Pb =perimeter of base Oblique prism is one in which the lateral faces are not perpendicular to the base.

~,Ih Volume of prism, V

V =Bh = Ke

A= e Pk

Volume, V:

Gt/ 'I ( 1 1)

Lateral area of prism, A

Lateral area of prism, A

---·--·------·-·-··· e:~

V=Bh

'---------"b

~ J.~9.~~~-~I5

!--.------·----··---·

Volume of prism, V

--

~_her~---~ ----·------------Radius of inscribed sphere

Ih

V =abc

7 J~,) I

j

where: B = area of the base h = altitude of prism

182 ·100 l Solved Problems in Engineering Mathematics (2"d Edition) by Tiong & Rojas

=area of right section =lateral edge PK =perimeter of right section

K e

183

Oblique cylinder is one which has its cylindrical surface not perpendicular to the base.

r

1 3

V=-Bh

Volume of cylinder, V

=altitude of cylinder

K = area of right section e = lateral edge What is a pyramid?

What is a c_ylinder? Cyiir;ader is a solid bounded by a closed cylindrical surface and two parallel planes.

Right cylinder is one which has its cylindrical surface perpendicular to the base.

+rz + Rr)

where:

What is a frustum of a pyramid? r~

8 = area of the base h

V = 1t3h (Rz

R = radius of the lower base r = radius of the upper base h = altitude of the frustum of a cone Frustum uf a pyramid is a portion oft~- . what is a prismatoid? pyramid included between the base and a section parallel to the base. Prismatoid is a polyhedron having for bases two polygons in parallel planes and for lateral faces triangles or trapezoids with one side lying in one base, and the opposite vertex or side lying in the other base of the polyhedron.

where:

s(h1 + hz: h3 + h·-)

or

where: B = area of the base h = altitude of cone

V = 8h = Ke

v "' ~( 81 + 82 + ~8182) where: 81 and 82 = area of the bases h = altitude of the frustum of cone

Volume of cone, V

)

v"'

_,}

,_

Truncated prism is a portion of a prism contained between the base and a plane that is not parallel to the base.

Volume of the frustum of cone, V

--lh

Pyramid is a polyhedron of which one face, called the base, is a polygon of any number of sides and the other faces are triangles which have a common vertex.

~l·.L

Volume of the frustum of pyramid, V

Ih

)"

v =~(81 +82 + ~8182)

A

where: 81 and 82 = area of the bases h = altitude of the frustum of pyramid

Volume of pyramid, V

Volume of cylinder, V V=_!Bh

3

V=8h Lateral area of cylinder, A

What is a frustum of a cone?

where: B = area of the base h = altitude of pyramid

Frustum of a cone is a portion of the cone included between the base and a s.ection parallel to the base.

A = (circumference of base )(h) What is a cone? where: 8 = area of the base h = altitude of cylinder

l

Cone is a solid bounded by a conical surface (lateral surface) whose directrix is a closed curve, and a plane (base) which cuts all the elements.

d

\

J

Volume of the prismatoid, V

L

V=s{A 1 +4Am+A2 ) where: At and A2 = end areas Am = area at the midsection (at half oft) L = distance between end areas This formula is known as the Prismoidlll Formula.

184 · lOQ 1 Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas

Dav 8- _SoU!iGeometrv 185

What is a sphere?

B

lh

Sphere is a solid bounded by a closed surface every point of which is equidistant from a fixed point called center.

Cross-section of torus: :+-- generating axis

€J,~~=t)

Volume of spherical segment, V

Volume of torus, V Volume of spherical pyramid:

rch2

-

1

V =-{3R-h) 3

Great circle

V=

540 Volume of sphere, V

What is a spherical sector?

4 3

Spherical sector is a solid generated by rotating a sector of a circle about an axis which passes through the center of the circle but which contains no point inside the sector.

V=-nR 3 Surface area of sphere, A

Jh

A=4'ltR 2 What is a zone?

What is a spherical wedge? Spherical wedge is a portion of a sphere bounded by two half great circles and an included arc.

where : A

What is a spherical segment? Spherical segment is a solid bounded by a zone and the planes of the zone's base.

where: R = distance from axis to center of generating circle R = radius of generating circle

What is an ellipsoid?

Volume of spherical sector:

1 3

A= 2nRh

A =4tt2Rr

Ellipsoid (Spheroid) is a solid formed by revolving an ellipse about its axis.

V=-AR

Area of zone, A

Lateral area of torus:

where: E = spherical excess of polygon ABCD in degrees

Zone is that portion of the surface of a sphere included between two parallel planes.

.!h

=

V 2'lt 2Rr2

3

itR E

= area of zone

Volume of spherical wedge: V= 1tR39

.

.

maJOr axiS

270 Volume of general ellipsoid, V

What is a spherical pyramid?

What is a torus?

Spherical pyramid is a pyramid formed by a portion of a surface of a sphere as base and whose elements are the edges from the vertices of the base to the center of the sphere.

Torus is a solid formed by revolving a circle about a line not intersecting it.

4 3

V =-ltabc Prolate spheroid is a solid formed by revolving an ellipse about its major axis.

4 2 V ::,...nab

3

·"

..

186 1001- Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas Oblate spheroid is a solid formed by revolving an ellipse about its minor axis.

4 2 V=-na b 3

Topics

0

Proceed to the next page for your 8th test. Detach and use the answer sheet provided at the last part of this book. Use pencil number 2 in shading your answer.

Polyhedrons Platonic Solids Properties of Platonic Solids Prisms Cubes and Parallelepiped Cylinders Pyramids Cohes Frustum of Pyramid or Cone Prismatoid Sphere Zone Spherical Segment & Sector Spherical Pyramid & Wedge Torus Ellipsoids

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Theory

Did you know that ... the Pascal's triangle which is used to determine the coefficient of a binomial expansion was named after the French mathematician, philosopher and physicist E31aise Pascal but die:! not claim recognition for the discovery because such triangle was first introduced by a Chinese mathematician, Chu Shihchieh in 1303 !

Problems

0 0

Solutions

Notes

~uote: "Mathematics consists of proving th~ most obvious things in the least obvious way."

0 0 0 0

Wed Thu Fri

Sat

31!0&: ME Board October 1.99:1

A circular piece of cardboard with a diameter of 1 m will be made into a conical hat 40 em high by cutting a sector off and JOining the edges to form a cone. Determine the angle subtended by the sector removed.

- George Polya

A. B. C. D.

358: CE Board May 1995 A sphere having a diameter of 30 em is cut into 2 segments. The altitude of the first segment is 6 ern. What is the ratio of the area of the second segment to that of the first?

. A. B. C. D.

144° 148° 152° 154°

4:1 3:1 2:1 3:2

359: CE Board November :199& :li>S7: CE Board November :1994

What is the area in sq. rn of the zone of a spherical segment having a volume of 1470.265 cu. rn if the diameter of the sphere is 30 m?

A

j

2

,.ll

465.5 m 5G5.5 m 2 f!G~>-~i m 2

IJ

!l~lfifJ

rn

If the edge of a cube is increased by 30%, wj how much is the surface area increased?

A. B. C. D.

W% ~% ~%

W%

2

LL

188· 1001 Solved-Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas 3601 ECE' Board April1995

3&51 CE Board November 1997

Each side of a cube is increased by 1%. By what percent is the volume of the cube increased?

Find the volume of a cone to be constructed from a sector having a diameter of 72 em and a central angle of 210".

A. B.

c. D.

1.21% 2.8% 3.03% 3.5%

3&11 ECE Board November 199Z Given a sphere of diameter, d. What is the percentage increase in its diameter when the surface area increases by 21 %?

A. B.

c. D.

5% 10% 21% 33%

3&ZI ECE Board November 199Z Given a sphere of diameter, d. What is the percentage increase in. its volume when the surface area increases by 21 %?

~

5% 10% 21%

~

~%

A.

a

3&31 EE Board October 1991 How many times do the volume of a sphere increases if the radius is doubled?

A. B. C. D.

D.

3

cm 3 cm 3 cm 3 cm

A conical vessel has a height of 24 em and a base diameter of 12 em. It holds water to a depth of 18 em above its vertex. Find the 3 volume (in cm ) of its content. 188.40 298.40 381.70 412.60

368: CE Board May 1995 A. B. C. D.

4 times 2 times 6times 8 times

3&41 CE Board May 1997 A circular cone having an altitude of 9 m is divided into 2 segments having the same vertex. If the smaller altitude is 6 m, find the ratio of the volume of the small cone to the big cone. A. B. C. D.

0.186 0.296 0.386 0.486

What is the height of a right circular cone having a slant height of diameter of 2x?

A. B.

2x 3x

C.

3.317x

D.

3.162x

J10x

and a base

3&9: CE Board November 1995 The ratio of the volume to the lateral area of a right circular cone is 2:1. If the altitude is 15 em, what is the ratio of the slant height to the radius?

A. B.

5:6

C. D.

5:3 5:2

5:4

I}:

14 15

375: CE BOard November 1995

2

B.

3

C. D.

4 5

3

cm 3 cm cm 3 3 cm

3731 EE Board April1993 :•. In a portion of an electrical railway .cutting, the areas of cross section taken every 50 mare 2556, 2619, 2700, 2610 and 2484 sq. m. Find its volume.

D.

A. B. C.

3.50 3.75 4.00 4.25

3761 CE Board May 199&

What is the volume of a frustum of a cone whose upper base is 15 em in diameter and lower base 10 em. in diameter with an altitude of 25 em? 3018.87 3180.87 3108.87 3081.87

A circular cylinder with a volume of 6.54 cu. m is circumscribed about a right prism whose base is an equilateral triangle of side 1.25 m. What is the altitude of the cylinder in meters?

D.

37Z.: EE Board April 199Z

c.

te

413 3 311 ft 313 ft3 391 ft3

c.

A.

A. B.

A. B.

1~

The volume of the frustum of a regular triangular pyramid is 135 cu. m. The lower base is an equilateral triangle with an edge of 9 m. The upper base is 8 m above the lower base. What is the upper base edge in meters?

A. B. C. D.

the triangular base and have the height of 8.6 ft., 7.1 ft. and 5.5 ft. respectively.

13

371: CE Board November 1995

3&71 CE Board November 199&

A. B. C. D.

A regular triangular pyramid has an altitude of 9 m and a volume of 187.06 cu. m. What is the base edge in meters?

C.

Find the volume of a cone to be constructed from a sector having a diameter of 72 em and a central angle of 150" 5533.32 6622.44 7710.82 8866.44

3701 CE Board November 1994

A. B.

3

12367.2 cm 3 13232.6 cm 3 13503.4 cm 3 14682.5 cm

3&61 CE Board May 1998

A. B. C. D.

Day 8- Solid Geometry· 189

522,600 m 3 520,500 m 3 3 540,600 r'n 534,200 m 3

374r ME Board April199& Determine the volume of a right truncated triangular prism with the following definitions: Let the corners of the triangular base be defined by A, B and C. The length of AB = 10ft., BC =9ft. and CA =12ft. The sides A, 8 and C are perpendicular to

A circular cylinder is circumscribed about a right prism having a square base one meter on an edge. 'T:he volume of the cylinder is 6.283 cu.m. Find its altitude in meters.

A. B. C. D.

4.00 3.75 3.50 3.25

377: CE Board November 1997 The bases of a right prism are hexagons with one of each side equal to 6 em. The bases are 12 em apart. What is the volume of the right prism? A. B.

c. D.

1211.6cm3 2211.7 cm 3 1212.5 cm 3 11.22.4 cm 3

378: EE Board April199& Two vertical conical tanks are joined at the vertices by a pipe. Initially the bigg~r tank is full of water. The pipe valve is open to allow the water to flow to the smaller tank until it is full. At this moment, how deep is the water in the bigger tank? The bigger tank has, a diameter of 6 ft and a height of 10 ft, the smaller tank has a diameter of 6

190 1001 Solved Problems in Engineering Mathematics (2"d Edition) by Tiong & Rojas ft and a height of 8 feet. Neglect the volume of water in the pipel.ine. A.

'1200

B.

~

C.~ D.~ ~79:

The central angle of a spherical wedge is 1 radian. Find its volume if its radius is 1 unit.

A.

2/3 1/2 3/4. 2/5

B. C. D. ~80:

A regular octahedron has an edge 2m. Find its volume (in m 3 ).

A.

3.77 1.88 3.22 2.44

B.

c. D. ~81:

CE Board May 1996

38~: An ice cream cone is filled with ice cream and a surmounted ice cream in the form of a hemisphere on top of the cone. If the hemispherical surface is equal to the lateral area of the cone, find the total volume (in cubic inches) of ice cream if the radius of the hemisphere is 1 inch and assuming the diameter of hemisphere is equal to the diameter of the cone.

A. B. C. D.

Topics

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3.45 3.91 4.12 4.25

·u

D D D ~ D D Di

384: ME Board April1997 A cubical container that measures 2 inches on a side is tightly packed with 8 marbles and is filled with water. All 8 marbles are in contact with the walls of the container and the adjacent marbles. All of the marbles are of the same size. What is the volume of water in the container? 0.38 in 3 2.5 in 3 3.8 in 3 4.2 in 3

A. B. C.

D.

Tue

Polyhedrons Platonic Solids Properties of Platonic Solids Prisms Cubes and Parallelepiped Cylinders Pyramids Cones Frustum of Pyramid or Cone Prismatoid Sphere I Zone I Spherical Segment & Sector j Spherical Pyramid & Wedge Torus . Ellipsoids

Theory

VVed

Problems

Thu

Solutions

Fri

Notes

Sat

A mixture compound of equal parts of two liquids, one white and the other black, was placed in a hemispherical bowl. The total depth of the two liquids is 6 inches. After standing for a short time, the mixture separated, the white liquid settiing below the black. If the thickness of the segment of the black liquid is 2 inches, find the radius of the bowl in inches.

~8S:

A. B.

A.

D.

7.33 7.53 7.73 7.93

~82:

CE Board November 1996

B.

C.

CE Board May 1997

The corners of a cubical block touched the closed spherical shell that encloses it. The volurrte of the box is 2744 cubic em. What volume in cubic centimeter inside the shell is not occupied by the block?

C. D.

ANSWER KEY 356. A 357. B 358.A 359. D 360. 361. B 362. D 363. D 364. B 365.

2714.56 3714.65 4713.56 4613.74

c

c ,,

The volume of water in a spherical tank having a diameter of 4 m is 5.236 rna Determine the depth of the water in the tank.

A. B. C. D.

1.0 1.2 1.4 1.8

I

4

366. 367. 368. 369. 370. 371. 372. 373. 374. 375. ~

c c B D A

B

c A

B

c

'

376.A 377. D 378.A 379.A 380.A 381. A 382. A 383. B 384.C 385.

c

,,

RATING

c:J c:J c:J 0

26-30 Topnotcher 20-25 Passer 15-19 Conditional D-14 Failed


11111;:'11,11:.·,11111


•

A= 21trh = 21t(15)(6) A= 565.49 m 2

:II

.193

•• @ AI

I I

r

2\ =8 =[~) =( v1 r, \) 3

v2

V 2 =SV1

A2= 1.21A1

Ill

A2 -r·~;l2 -

x=~ =f
A1· . d1

2

x=30 c1 = circumference of the circle c2 = circumference of the base of the cone C = length of arc

C=C 1 -C 2 = 21t(50)- 21t(30) c =401t

:: =

A1

d2 =1.1d1 .

2--m:.h 1

•~ .

401t =(50)0 401t 180 0=--x50 1t

•

X2

X1

A

1

[::)' t:J'

1

x

'

~ ..

.

X1

3

X2

3

3

1 01 V2 =[2) =( · ' ] =1.0303 1

X2

V1

=(1.21~1

A1

.

312

~

_ -1.331

'

Thus, V2 is increased by 3.03%

= circumference of the base of the

cone C = length of arc of the sector C =C 1 -C 2

re =2n:r-2:n:x

Thus, V2 is increased by 33.1%

36(210 x-1t-)· =21t(36)-27t(X) 180" x=15cm

ml!f\

C9 r,

..

=circumference of the circle

v2 =1.331V1

·II

h=~

1

x vV2 = 1.0303 v,

= 0.296

4

Let: c1 c2

X2=·1.01 X1

1404 = 45h 2 -h 3 h=6 m

-[A

2 ) v2 - -

X1

1th2 1470.265 = -[3(15)-h]

312

3

•

\:E;}

~=[2)2 =(1.3']2 ·=1.69

II

3

~ = [-~) = (~) v2 112 9

•

Thus, A 2 is increased by 69% .

1th2 V=-(3r-h)

3

~·

X2=1.3x1

A 2 =1.69A 1

0=144'

=11

Thus, d2 is increased by 10%

X1

X1

C=re

6

A 2 :A 1 =4:1

/

= 27tr- 21tX

J:: =r~~

A2 2--m:.h 2 24 -=--=-=4

II'

h = .[362 -15.

2

r2"' 2r,

h=32.726cm

I illl

~!

i

194 I 00 i Solved Problems in Engineering Mathematics (2"d Edition) by Tiong & Rojas 1

1

3

3

V = -nx 2 h = -1t(15) 2 (32.726)

6

V =7710.88cm 3

•

24 18 r =4.5

~

Let: c1 = circumference of the circle = circumference of the base of the cone C = length of arc of the sector

.

50.625 = 35.074 + 0.433x

C=C 1 -C 2 re=2nr-2nx 36(150" x-n-)=2n(36)-2n(x) 180° x=21cm

=-/r 2 -x 2

h = -/36 2 -21 2 h=29.24cm

1 V =-nx 2 h

V =-n(21) 2 (29.24)

3 V = 13,503.44 cm 3

0 =(x-3)(x +12) x=3 em

mJ

3

Ox

V

=;-1[1;x SinSJh

187.06 =

2 .

~(; }

Let:

---

h=3x

AL = nrl 1

A1 =area of the lower base A2 = area of the upper base A 1 = nr,Z = n(5) 2 = 25 n

'

.

A2

=nr/ =n(7.5) 2

= 56.25n

'

h V=-(A 1 +A 2 + -/A 1 A 2

)

3

V =-nr 2 h

Note: 8 = 60°, since equilateral triangle. Substitute:

3

v

-=2~asgiven

AL V=2AL

Let: A1 = area of the lower base A2 = area of the upper base

1

..!_ 'n/\h=2'7tt:_L 3 rh =6L L

h

15

r L

6

6

2

2 1 2 2 A 2 = -(x) sin60o = 0.433x

2 V-=

h :l

(A 1 +A 2 +-}A 1 A 2

25 V = - ( 25 1t + 56.25 n +-} (25n )(56.25n:))

3 V

A 1 =-(9) sin60°=35.074

----2

T25

sin 60°(9)

ED

h 2 = 9x 2

5

2

x=12m

+x2

10x 2 =h 2 +x 2

3

1

e "'60°, since equilateral triangle.

1


•&

)

2

0 = x 2 + 9x -36

X

=h2

l

+ 2

V=-Bh

..

(...fi0x)2

2

+ 3.897x

X

Note:

@

ml

c2

h1= 24

8[35074+0.433x 135 =-

3 )(35.074)(0.433x

V =381.7cm 3

'

•

Substitute:

1 2 1 2 V = -nr h 2 = -n(4.5) (18) 3 3

r

h

195

•

By ratio & proportion:

)

=3108.87 cm 3

•

Note: Since the areas being cut is at the same distance, then the given solid is a prismatoid. And since there are five different areas being cut then, this solid is equivalent to two rrisrnatoids

Day 8 - Solid Geometry 197

196 1001 Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas h V =-(A 1 +4Am +Az)

6 where: A1 = area of the first base (base 1) A2 = area of the second base (base 2) Am = area of the middle section h = distance between base 1 and base 2

•

v =(~d

1t nd,Z 2 v1 =-1 ( - ) h1=-(6) (10)

)h

3

2

h

6.283=.::.(1.4142) h 4 h=4m

•

Let V1 = total volume of the two prismatoid

v,

2

6

x2 =r 2 +r 2 -2r 2 cos9

Ill

(1.25)2 = 2r 2 - 2r 2 cos 120 1.5625 = 3r 2 r=0.72m

h21

A a+b+c

9+12+10

2

2

S=---

··1h3

c

s =15.5

nx 2 A=-4tan9

n

6 2

6 6 A= ( )( ) = 93.53 cm 2 4 tan 30• V = Ah = 93.53(12)

•

V:-:1122.4cm 3

•

A =)s(s-a)(s-b)(s-c) A= j15.5(15.5- 9)(15.5 -12)(15.5 -10)

A = 44.039 ft

.

2

•

r

= (10)

18.849

6.54 = 1t(0.72)h h=4m

~ =(h~ 94.247

18o· 180° 9=-=--=30°

V=1tr 2 h

····t::

b

= 75.398 ft 3

V = V1 - V2 = 94.247-75.398

x=6 Let: A = area of one base x = length of each side of the base

By cosine law:

...••-·····...~

1(1td/) 1t 2 (8) V2 =- - h 2 =-(6) 3 4 12

By similar solids:

360° 9=-=120° 3

V1 =522,600m 3

..."···

3

v =18.849 ft 3

100 +-[2700 +4(2610)+ 2484] 6

.li... a . c .......··· B . . •••••··•••. ··

12

V1 = 94.247 tt

V2

100 =-[2556+4(2619)+2700]

h1

4

3

h

h =~200ft

1tR 3 9

V=-2700

V=

180 1t(1 )3 1 rad x -) ( 1t ·~ 180 270'

270

2

V =- cubic units 3

Top view

V =A(h 1 +h; +h3

J

d=~(1)

2

+(1)

Ell

2

d=1.4142

v = 44.039( 8 ·6 + 7~1+ 5.5 J V=311ft 3

Let:

v1 =total volume of the bigger tank

V2 = total volume of the smaller tank V := volume left in the bigger tnnk

. '.!·.~

J i

•

198 1001 Solved Problems in Engineering Mathematics

J2

(2nd

Edition) by Tiong & Rojas

Day 8- Solid Geometry 199

Jth2

V=-x 3

Ell

V=-(3r-h)

3

3

x=2

7th2

J2(2)3 V=-__:___:_-

5.236 = --(3(2)- h)

3

3

5 = 6h

V=3.77m 3

-

h

.J x 2 + x 2 + x 2

d=

.[3(14)2

d = 24.24 em

'

2

d=

3

r=12.12cm

h=1m x=2

ED

ml

V=Vs -Vc

Let: r = radius of each marble Vw = volume of water inside the cube Vc = volume of the cube Vm = volufl1e of each marble

,y

3

v=(;nr .)-x3 3

v=(;n(12.12) )-14

3

V =4,713.55em 3

x =4r 2 =4r

Let:

r = 1t2

vb = volume of the black mixture Vw = volume of the white mixture

!Vw =Vc -8Vm Let: Ac = surface area of the cone Ah = surface area of the hemisphere V1 = total volume Vc =volume of the cone Vh = volume of the hemisphere

Vb =Vw V 1 =Vb+Vw V 1 =2Vw

Yw

V.,

Ac =Ah V 1 =2Vw 2

[ 1r(h ) ) -1r(h1-(3r-h -(3r-h 2 ) 1) = 2 3 3 2

1t(6)2

-

[ 1t( 4}2

-(3r-6)=2 -

3

l

-(3r-4)

3 36(3r-6) = 32(3r-4)

108r-216=96r-128 12r = 88 r=7.33in

J

3

Vw =(2)

Substitute:

2

=x

~L =2(4\rl)

-s(;nr -s(;n(0.5) 3

)

3

3

)

=3.8in 3

Ill

2 L = 2r = 2(1) r=1

L =2in

h=~ = )(2)2 -(1)2 h = 1.732in

Let: V = volume inside the sphere but outside the box V. = volume of the sphere Vb = volume of the box

V1 =Vc+Vh

2

1(4 nr V 1 = 1 nr h+;

3

3

2

3J

v, =(;}1) {1.732)+ :1t(1)

vw = x3

3

2744 = x 3

V1 = 3.91in 3

x

14 em

,,

I,

;,.,,lr

J -~"Et.


Topics

0 0

Mon

Tue

Theory

0 0 0

Problems

Solutions

Notes

~ VVed

0

Thu

0 0 Fri

Sat

What is Trigonometry? Trigonometry is the study of triangles by applying the relations between ·the sides and the angles. The term "trigonometry" comes from the Greek words "trigonon" which means "triangle" and "metria" meaning "measurements.

I

What are the two branches of trigon(lmetry? Trigonometry is divided into two branches, namely:

'i

!i Plane Trigonometry deals with triangles in the two diiT!::nsions of the plane. Spherical Trigonometry concerns with triangles extracted from the surface of a sphere

I j

Plane Trigonometry Solutions to Right Triangles The Pythagorean Theorem Special Triangles Solutions to Oblique Triangles Law of Sines & Cosines Law of Tangents Trigonometric Identities Exponential Form of Identities Other Parts of Plane Triangle Radius of Inscribed Circle Radius of Circumscribing Circle Properties of Triangles Points in a Triangle Conditions for Congruency Conditions for Similarity What are the two general classifications of plane triangles? There are two general types of triangles, namely: Right triangle - a triangle that has a right angle. Oblique triangle- a triangle that does not have a right angle. Acute triangle and obtuse triangle :;~re oblique triangles. Isosceles triangle (triangle with two sides er;ual and two angles equal) could be a right triangle or an oblique triangle.

' 204 1061 Sohred Problems in Engineering Mathematics (2nd Edition) by Tiong & Ko]as ' What are solutions to a plane right triangle?

Day 9- Plane Trigonometry 205

3. Special Triangles .

a A. Egyptian triangle:

sinA

LJ.

1. Fundamental Trigonometric Functions side opposite

= sinB = sinC

1603).

1

te co

a2

f

= b2 + c 2 - 2bccosA

2

b =a2 +c2 -2accosB

e2 = a2 + b2 - 2abcosc

side adjacent side opposite

_ hypotenuse sec e - . . s1de adJacent hypotenuSe esc e = --:-''-'------,... side opposite ·

Pythagorean theorem states that "In a right triangle, the sum of the squares of . the sides is equal to the square of the

/"1

hypotenuse• 0

L_ja

'2~ ~.1

This solution to an oblique triangle is used when the given are two sides and the included angle. This was demonstrated by the French mathematician Francois Viete (1540-

.J3

1603). What are solutions to a plane oblique triangle?

This was first described by a Danish mathematician and Physician Thomas Fincke (1561-1656) in 1583.

1. Laws of Sines This solution to an oblique triangle is used when the given are:

1

_a_-b_ a+ b

A. two angles and one opposite side

or

\

= tan 2(A-B)

secA= cosA

1

1

1

2.

cscA= sinA

Even-odd identities sin(-e) "".~sine. cos'(-9) =cose tan(-e) =~tan a cot(-e) =-cote sec(-e)

=sece

B

~ c

sine= cos(90~e) cose = sin(90- 9) tane

=cot{90- 9)

cote= tan(90 -e) seQe ::;;cs9(90- e)

esc a... s~~(9o -.e) 4. Pythagorean relations sin 2 A+ cos2 A =1

tan

1+ cot2 A cse2 A

(A+B)

2

Trigonometric identities are equations that express relations among ttigonometric functions which are true for all values of the variables involved.

c

a +'b*•c

3. Cofunction identities

il

II f

=

· The following are the trigonometric identities:

1+tan2 A "'sec2 A

i

5. Sum of angles formulas sin(A+B)= sinAcosB+c:oSAsinB cos(A +B)= cosAcosB-sinAsinB tan( A+ B)= tan A+ tanB · 1-tanAtanB

A

1

I t,l

What are trigonometric identities?

This was demonstrated by Ptolemy of Alexandria in 150AD.

2

a

-~1.____

B. two sides and an opposite angle

.b 2

1 .cosA=secA

esc( -9) =-esc

2. The Pythagorean Theorem Pythagorean theorem is the most renowned of all mathematical · theorems. It is considered as the most proved theorem in mathematics. This was formulated by Pythagoras .(c.580 - c 500 B.C.) about 500 B.C.

cotA

3. Laws of Tangents

C. 30° - 60° Right Triangle

1 =tanA

sin'A= cscA

tan A"' cotA

A. two sides and the included angle or B. three sides This was demonstrated by the French mathematician Francois Viete (1540-

tane = side opposite side adjacent

1

This solution to an oblique triangle is used when the given are:

B. 45°- 45° Right Triangle

. side opposite . s1n 9 = -:---'-'--hypotenuse side adjace. nt cos e = ---=--,-hypotenuse ·

1. Reciprocal relations

2. Laws of Cosines

3

side adjacent

c

b

- 11

Day 9 - Plane Trigonometry 207

20& 1001 Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas triangles. This phenomenon is known as the ambiguous case.

11. Difference of two functions

6. Difference of angles formula

2cos~(A + B)sin~(A- B)

sin(A- B)= sinAcosB -cosAsinB cos( :A ·B)"' cos A, cos 8 + sin A sin 8

sin A- sinS=

tan(A .. B)= tl;inA.~tan B · · · 1+tanA tans

cos A- cosB = 2sin1(A+ B)sin.!(A- B)

2

2

c

tan A-· tanS= _s_in_,(_A_-_B..:.) cosAcosB Note: sinp = sin(180- f3)

7. Double angle formulas 12. Product of two functions

sin2A = 2sinA.cosA cos2A = cos2 A - sin2 A tan2A = 2tanA 1·tan2 A

What are the other parts of general triangle?

2sinAsinB =cos( A· B)- cos( A +B) 2sinAcosB =sin( A+ B)+ sin( A -B) 2cosAcosB =cos( A+ B)+ cos( A -B)

1. Height of a general triangle

~

8. Powers of functions What are the exponential forms of the fundamental trigonometric functions?

sin 2 A"" .2(1•-cos2A)

2

cos 2 A 2

tan, · ·

A

The following are the trigonometric functions expressed in terms of exponential functions and the imaginary unit.

,t- C;QS2A .,, . 1+cos2A

9. Functions of half angles

2

sin A 1+cos A

10. Sum of two angles

r

2

co~AcosB

=

ArRIANGLE

a+b+c

cotx = i( eix + e-ix} e•x -eix

R=

2. Equilateral triangle

c 2

where: e = base of the natural logarithm i = imaginary unit

' .1 1 cosA+cosB"" 2cos-(A +B)cos-(A -B) 2 . 2 sin(A+B) . . . t 8= tan A +an

1. General triangle

2. Median of a general triangle

sin A+ sinB = 2sin1(A +B)cos.!(A -B)

2

let R be the radius of the circumscribing circle and r be the radius of the inscribed circle.

a+b+c 2 Note that he is perpendicular to side c.

tanx = -i(eix -e-ix} e•x + e-ix

. 2

=

What are values of the radius of inscribed circle and the radius of circumscribing circle in a triangle?

where: s =

2

A ·l~cos.A cos-=±

tan~= 1- cosA 2 sin A

=./~b[1-(a:bJ]

h -. 2Jf>(s- a)(s'-b}(s- c:) . cc

sinx = eix- e·ix 2i COSX = eix + e-•x

sin~ =±l~cosA 2

tc

c

=~·(1+cos2A) 2 ,

2

~

What is an ambiguous case?

me

If you are given a triangle with two sides and an opposite angle known, you can solve one of the two unknown angles using the law of sines. Since sin e = sin (180- 8), a unique solution is obtained only in right triangles. When rlmlinCJ with oblique triangles, you will f111d lw"

c 2 2 2 =1.J2a + 2b~2 · . c.

3. Angel bisector of a general triangle

j

abc 4ATRIANGLE

208 · 100 I SOlved Problems in Engineering Mathematics (284 Edition) py Tiong & Rojas

,,.aJi

.... a,/3 R=3

6 3. Right triangle

Day 9 •· Plan£.!!!9.!?nOI!!_et!)!'~~J.

--------------

.~

a+b -+ c an d a 1s · th e s1·de where: s = -

fi.--·------·

2 physically tangent to the circle

A= y\s- a)(s- b)(s- c)(s -d) Note: A+ C = 1so• B + 0 = 180"

c

A= .Js
1. Given two diagonals and an included angle

where: s= a-t:b+c 2

Ptolemy's Theorem states that • The sum of the two pairs of opposite sides of a convex quadrilateral inscribed in a circle is equal to the product of th1:1 lengths of the diagonals"

4. Triangle inscribed In a circle

ac + bd "' did2

A=

Whi!t~l.f!m.Q..rtant ~arQJ.!2~

5•. Triangle circumscribing a circle

AL A=

a

___.JD

1.

The sum of two sides of a triangle is greater than the third side and their difference is less than the third side.

J(s :a)(s -b)(s- c)(s- d) .. abcdcos2 9

.2.

The perpendicular bisectors of the sides and the blsectors of the angles of a triangle, meet in points which are the center of the circumscribing circle the inscribed circle, respectively.

3.

The altitudes of a. triangle meet in a point.

4.

The median of a triangle are concurrent at a point which is 2/3 of the distance from any vertex to the midpoint of the opposite sides.

where:

a+b+c+d

A=rs

S=----·-

2

2. Given 2 sides and Included angle

O='~~= B+D 2 2

where: s= a+b+c 2

3. Cyclic quadrilateral (All vertices lie on a circle)

6. Triangle with escribed circle

b

···················

~ r

Using Heron's Formula: Named after Heron of Alexandria (1'1 Century AD)

_ d

t ...._

3. Given 3 sides

c

This theorem w<)s named after the geographer, mathematician and astronomer. Ptolemy or Claudh.1s Pto:~maeus (<:.100- c.16U A.D.) of Alexandria.

triangle?

•. ......_WIT A .. .... l

A=tabstne

sine

b

b

~

1d2

abC A=. 4r

1. Given bllse and altitude

b

Id

2. Given 4 sides and 2 opposite angles

What are the formulas for the area of a triangle?

LJ·£

where : d1 and d2 are diagonals of a quadrilateral

c

.

...•.

A= r(s-a) /./

j

Circumcenter- the point of concurrency of the perpendicular bisectors of the sides of a triangle.

!I

![' 210 .1 00 1 Solved Problems in Engineering Mathematics ~2nd Editioll) by Tiong & Rojas lncenter- the point of concurrency of the angle bisector of a triangle. Orthocenter - the point of concurrency of the altitudes of a triangle.

Proceed to the next page for your 9th test. Detach and use the answer sheet provided .at the last part of this book.· Use pencil number in shading your ansWer.

?

GOOD LUCK!

Topics

What are the conditions for two triangles to be congruent(:::~)? Two triangles are congruent if the elements of one triangle are equal to the corresponding parts of the other triangle. The following are the conditions for congruency:

1.

Two sides and the included angle are equal.

2.

Three sides are equal.

3.

On~

side and adjoining angles are equal.

D D

~ribia: Did you know that... Pythagoras was the first to connect mathematicS and music. In fact, he is credited with discovering octave and the fifth interval of a note !

Tue

D

Theory

The following are the conditions fbr similar triangles: Two corresponding sides are proportional; included angles are equal.

2.

Three sides are proportional.

3.

Two angles are equal.

[g Wed

D D D D D

- Albert Einstein

What are the conditions for two triangles to be similar?

1.

Plane Trigonometry Solutions to Right Triangles The Pythagorean Theorem Special Triangles · Solutions to Oblique Triangles La"'\S of Sines & Cosines Law of Tangents Trigonometric Identities Exponential Form of Identities Other Parts of Plane Triangle Radius of Inscribed Circle Radius of Circumscribing Circle Properties of Triangles Points in a Triangle Conditions for Congruency Conditions for Similarity ·

Mon

Problems

Thu

Solutions

Fri

Notes

Sat

386: ECE Board April :1999 Sin (B -A) is equal to _ _ , when B = 270 degrees and A is an acute angle. A. B. C.

D.

-cosA cos A -sinA sinA

387: ECE Board "pril :1999 2 If sec A is 5/2, the quantity 1 -sin 2 A is equivalent to

A. B. C. D.

2.5 1.5 0.4 0.6

:188: ECE Board April :1999 (cos A) 4 - (sin A) 4 is equal to _ _ .

j

1\

r:o~;

ll

C()~; :)/\

c

~~Ill ~)/\

D.

sin 4A

389: ECE Board April :1999 Of what quadrant is A, if sec A is positive and esc A is negative?

A.

IV

B.

II Ill I

C. D.

390: ME Board October :1996 Angles are measured from the positive horizontal axis, and the positive direction is counterclockwise. What are the values of sin B and cos B in the 4th quadrant?

A. B. C. D.

sin B > 0 and sin B <0 and sin B > 0 and sin' B < 0 and

cos cos cos cos

B<0 B<0 B>0 B>0

'1/\

I

. . t!

212. 100 t Solved Problems in Engineering Mathematics (2"d Edition) by Tiong & Rojas

Da~ Plan_eTrigonometry

:!91: ECE Board November 1998 Csc 520° is equal to

397: EE Board April1992 Find the value of A between 270° and 360° if 2 sin 2 A- sin A =1.

B. C. D.

A. B. C.

A B.

403: CE Board November 199:! Find the value of y in the given: y = (1 + cos 28) tan 8.

D.

cos 20° esc 20° tan 45° sin 20°

:!921 ECE Board Aprii199:J Solve for 8 in the following equation: Sin 28 =cos 8 A. B. C. D.

30° 45° 60° 15°

:J93: CE Board November 1993 If sin 3A = cos 68, then A.

B. C. D.

A+ B = 90° A+ 2B =30° A+ B =180° None of these

394: EE Board October 1996 So lye for x, if tan 3x = 5 tan x.

A B. C. D.

20.705° 30.705° 35.705° 15.705°

39S: EE Board October 1997 If sin x cos x + sin 2x = 1, what are the values ofx? A. B. C. D.

32.2°, 69.3° -20.67", 69.3° 20.90°, 69.1 o -32.2°, 69.3°

396: EE Board April1991 Solve for G is esc (11G -16 degrees)= sec (5G + 26 degrees). A. B. C. D.

7 5 6 4

degrees degrees degrees degrees

C. D.

300° 320° 310° 330°

A. B. C. D.

:J98: CE Board November 199:! If cos 65° + cos 55° = cos 8, find 8 in radians. A. B. C. D.

0.765 0.087 1.213 1.421

A.

fj ).

B. C. D.

8/11 8/19 8/15 8/17

c. D.

6° 3°

410: EE Board March 1998 Solve for x in the equation: arc tan (x + 1) + arc tan (x- 1) = arc tan ( 12).

2

B.

sin 2 8

C.

sin

D.

sec e

2

e sec 28

2

406: ME Board October 199S Simplify the expression

2

A. B.

cos 8 cos e

C. D.

sin 8 sin 8

411: ECE.Board November 1998 Solve for A for the given equation cos 2 A = 1 - cos2 A. A. B. C. D.

45, 45, 45, 45,

125, 125, 135, 150,

225, 225, 225, 220,

335 315 315 315

degrees degrees degrees degrees

If tan x = _.!_ tan y = _.!_ what is the value

A.

3'

R.

c

of tan (x + y)?

IJ

'1/7

412: ECE Board April1991 Evaluate the following: sin oo +sin 1 o +sin 2° + ... +sin 89° + sin90° cosoo +cos 1° +cos 2° + ... + co.s89° + cos90°

2

402: CE Board November 199:l't

yo

2'

1.5 1.34 1.20 1.25

sec 8- (sec e) sin 28

407: ME Board April1998 112 Arc tan [2 cos (arc sin [(3 ) I 2]) is equal to

A

A. B. C.

D.

If sec 2A = _ __.:!___ , determine the angle sin 13A · A in degrees.

so

D.

sin 8 cos 8 tan 8 cot 8

0.149 0.281 0.421 0.316

A.

40_1: EE Board March 1998

B. C. D.

A. B. C.

Simplify the equation sin 8 ( 1 + cot 8)

4/3 5/4 4/5 3/4

A.

7t

4

2


B.

30" 45° 60° 90°

(2x) + arc tan (x) = -

40S: ME Board April1996

The sine of a certain angle is 0.6, calculate the cotangent of the angle.

A.

2 2 2 2

A. B. C. D.

409: ECE Board March 199& Solve for x in the given equation: Arc tan

sin 8 cos 8 sin 28 cos 28

Find the value of sine+ cos e tan e cose

Find the value of sin (arc cos

B. C. D.

408: EE Board October 1992 Evaluate arc cot [2cos (arc sin 0.5



A.

1/6 2 1

213

n/3 n/4 )[/16 11/?

A. B. C.

D.

413: ECE Board April1991 Simplify the following: cos A +cosB sin A+ sinB - - - - , - - - + ---,----=sin A -sinB cosA-cosB A.

J

1 0 45.5 10

0

Day 9 - Plane Trigonomet!Y 215

ZI 4 1001 Solved Problems in Engineering Mathematics (2"ct Edition) by Tiong & Rojas B.

sinA

C.

1

D.

cosA

A. B. C. D.

A.

39.49 35.50 30.74 42.55

B. C.

D.

19.7", 20.1•, 21.1•. 22 3",

307.4 309.4 321.8 319.2

mph mph mph mph

4141 ECE Board April1991 2sin8cos8-cos8 Eva Iuate: -------;:,....-----,;o--2 2 1- sine+ sin 8 - cos 8 A.

sin 8

B.

cos 8

C.

tan 8

D.

cot 8

D.

If sin A= 2.5J 1x, cos A= 3.06x and sin 2A = 3.939x, find the value of x? 0.265 0.256 0.562 0.625

B. C.

D.

A.

A. B.

6 and 12 3 and 9 5 and 11 4and10

C. D.

C. D.

A.

B.

so·

3.68 4.03 5.12 4.83

422: ME Board April

54.23 48.23 42.44 46.21

m

D.

427: CE Board November 1997 Points A and B are 100 m apart and are of the same elevation as the foot of a building. The angles of elevation of the top of the building from points A and Bare 21" and 32• respectively. How far is A from the building in meters? A.

B. C. D.

An aerolift airplane can fly at an airspeed of 300 mph. If there is a wind blowing towards the cast at at 50 mph, what should be the plane's compass heading in order for its course to be 30•? What will be the plane's ground speed if it flies in this course?

The captain of a ship views the top of a lighthouse at an angle of so· with the horizontal at an elevation of 6 meters above sea level. Five minutes later, the same captain of the ship views the top of the same lighthouse at an angle of 30• with the horizontal. Determine the speed of the ship if the lighthouse is known to be 50 meters above sea level.

m m

24 ft, 53.13° 24 ft, 36.87° 25 ft, 53.13° 25ft, 36.87"

A. B. C.

D.

The angle of elevation of the top of tower B from the top of tower A is 28• and the angle of elevation of the top of tower A from the base of tower B is 45•. The two towers lie in the same horizontal plane. If the height of tower B is 120 m, fine! the height of tower A.

C.

C

A

m/sec m/sec m/sec m/sec


B.

66.3 m 79.3 m 872 m

0.265 0.155 0.169 0.210

An observer wishes to determine the height of a tower. He takes sights at the top of the tower from A and B, which are 50 feet apart, at the same elevation on a direct line with the tower. The vertical · angle at point A is 30• and at point B is 40•. What is the height of the tower?

4261 CE Board November 1997

A. B.

259.28 265.42 271.64 277.29

4281 ECE Board November 1991

A.

:i99~

90.7 m

m

A wire supporting a pole is fastened to it 20 feet from the ground and to the ground 15 feet from the pole. Determine the length of the wire.and the angle it makes with the pole.

c.

A man standing on a 48.5 meter building high, has an eyesight height of 1.5 m from the top of the building, took a depression reading from the top of another nearby building and nearest wall, which are and respectively. Find the height of the nearby building in meters. The man is standing at the edge of the building and both buildings lie on the same horizontal plane.

m m

4251 ME Board November 1994

A ship started sailing S 42.35' W at the rate of 5 kph. After 2 hours, ship B started at the same port going N 46.20' W at the rate of 7 kph. After how many hours will the second ship be exactly north bf ship A? A. B.


m m

angle of elevation of the sun is 61 •. If the pole is leaned 15" from the vertical directly towards the sun, determine the length of the pole.

421: GE Board August 1994

c. o. go·

76.31 73.31 73.16 73.61

ECE Board April 1994 A pole cast a shadow 15 m long when the

Two triangles have equal bases. The altitude of one triangle is 3 units more than its base and the altitude of the other triangle is.3 units less than its base. Find the altitudes, if the areas of the triangles differ by 21 square units.

D.

30" 45° 60°

ECE Board April 1998

~;

C.

If conversed sin 8 = 0.134, find the value of8. ·

so•

D.

B.


A. B.

C.

364m 374m 384m 394m


32.47• 33.68• 34.12• 35.21•


A. B. C. D.

A man finds the angle of elevation of the top of a tower to be 30°. He walks 85m nearer the tower and finds its angle of elevation to be so•. What is the height of the tower? A.

Solve for the value of "A" when sin A = 3.5 x and cos A= 5.5x.

C.

42~:

Points A and B 1000 m apart are plotted on a straight highway running East and West. From A, the bearing of a tower C is 32• W of N and from B the bearing of C is 26• N of E. Approximate the shortest distance of tower C to the highway. A. B.

4151 ECE Board April1994

A. B.


D.

D.

85.60 feet 92.54 feet 110.29 feet 143.97 feet

4~0:

ME Board April199~

A PLDT tower and a monument stand on a level plane. The angles of depression of the top and bottom of the monument viewed from the top of the PLOT tower at 13° and 35" respectively. The height of the tower is 50 m. Find the height of the monument

Zll 1001 oSolved Pr~blems in Engineering Mathematics (2"d Edition) by Tiong & Rojas

A

29.13 m

B.

30.11 m

A. B. C. D.

c. ·32:t2m' D. ~ 33.51 m

240 420 · 320 200

431: ECE .Boar«t November 1:998 If an equilateral triangle is circumscribed about a circle of radius 10 em, determine the side of the triangle. A.

Topics

D D Mon

34.64 em 64.12 em 36.44 em 32 ..10 em

B. C. D.

Tue

oz: EE .Board Oetober 1997

D

The two legs of a triangle are 300 and 150. m each, respectively. The angle opposite the 150m side is 26". What is the third side?

A. B.

197.49 218.61 34{78 282.15

C. D. 43~

Theory

~ Wed

EE Board October 1997

Problems

[=] ra,,

Thu

Solutions

Fri

r-Jotes

Sat

C. D.

120m 130m 125m 128m

386. A 387. 388. B 389. A 390. D 391. B 392.A 393. B 394. A 395. 396. B 397. D 398. B

c

434: EE Board October 1:997 The sides of a triangle are 195, 157 and 210, respectively. What is the area of the tria~~gle?

A. B. C.

D.

73,250 10,250 14,586 11 ,260

sq. sq. sq. sq.

I

-----:--··--~

ANSWER KEY B.

Plane Trigonometry Solutions to Right Triangles The Pythagorean Theorem Special Triangles Solutions to Oblique Triangles Laws of Sines & Cosines Law of Tangents Trigonometric Identities ' Exponential Form of Identities Other Parts of Plane Triangle Radius of Inscribed Circle Radius of Circumscribing Circle Properties ofTriang les Points in a Triangle 1 Conditions for Congruency Conditions for Similarity

D DI D DI

m m m m

The sides of a triangular lot are 130 rn., 180m and 190m. The lot is to bl'l divided by a line bisecting the longest side and drawn from the opposite vertex. Find the length of the line.

A

~-~-----·---·-·

c

units units units units

435: ECE Board April1998 The sides of a triangle are 8, 15 and 17

unies. If each side is doubled, how many square units will the area of the new trioogle be?

j

399. D 400.A 40'1. B 402. D 103. 404.C 405.A 406. B 407. B 408.A 409. 8 410. B 411.

c

c

412. A

413. A 414. D 415. A 416. B 417. 418.A 419. B 420. D 421. B 422. 423. D 424.A

c

c

·· 425. D 426. B 427.A 428.C 429. B 430. D 431. A 432.C 433. 434.C 435.A

c '

~---·----·--·---. RATING

0 0

1

13-50 Topnotchel'

33-42 Passer·

I [ ] 25-32 Conditional

0

0-25 Failed

IfFAILED, repeat the test.

rull

!l,i( 218 lOOi Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas· ·

II

Ill

•

a

sin (270. -A)= sin 270' cos A In the 4th quadrant:

3A =90' -68

2

+cos A= 1 1-sin 2 A=cos 2 A

- 12 --------

A+2B=30'

ml

5/2

ml

Complementary angles have the same values of their sine functions. Thus, the other angle is equal to: 90" - 20.9" = §JU.:

esc(11G-16" )=sec(5G+26')

tan 3x = 5 tan x tan (2x + x) = 5 tan x tan2x+tan x - - - - - = 5 tan x 1-tan 2x tan x tan 2x +tan x = 5 tan x- 5 tan 2x tan 2 x

Thus, sin B < 0 and cos 8 >0

1

1

sin(11G-16')

·--

cos(5G+26')

cos(5G+26' )=sin(11G-16'

)~Eq.1

Note: sine = cos (90° - a ) Let: 9 = 11G-16"

tan 2x = 4 tan x -5 tan 2x tan 2 x

lliiiiill cos 4 A-sin 4 A=cos 2 Acos 2 A - sin 2 A sin 2 A = cos 2 A(1- sin 2 A) 2

esc 520' =esc (520' - 360' )

2

- sin A(1- cos A)

=cos'2A-~

-sin 2 A+~ = cos 2 A -sin 2 A

•

=esc160'

sin(11G-16' )=cos(106' -11G)~Eq.2

4~=( 2 ~ J(1+5tan 2 x)


5G+26' =106' -11G 16G=80' G=5'

Substitute: 2 sin 8cos8=cosa 2sin8 ;,1

!llllllillll

•

IDI

sin x cos x + sin2x = 1 ~ Eq.1

sin 2 A- 0.5 sin A= 0.5

Note: 2 sin x cos x =sin 2x sin x cos x = 0.5 sin 2x ~ Eq.2

By completing square:

,,

2 sin 2 A- sin A= 1

In the 4th quadrant: hypotenuse c , =adjacent side b hypotenuse c c escA= =-=-opposite side -a a

Cbs..,(5G+26' )=~(106' -11G)

. X= 20.7048°

Note: sin 29 = 2 sin 9 cos 9

-a

4 tan x =tan 2x (1 + 5 tan 2 x)

2=14tan 2 x tan x = 0.377964473

esc 160' = C$C 20'

sin 29 =cos 9 b

sin (11G -16' ) =cos (90 -(11G -16)

4-4tan 2 x =2+10tan 2 x

esc160' =esc(180' -160')

Ell

Note: cos 2A = cos2 A- sin2 A

4 tan x =tan 2x + 5 tan 2x tan 2 x

1-tan 2 x

Thus esc 520° = esc 20°

cos 4 A- sin 4 A= cos 2A

secA=

=20.9'

•

A =30' -28

-b b . Opposite side s1nB= =-=-hypotenuse c c adjacent side a cos B . hypotenuse c

a

-

X

sin 3A =sin (90'' - 68)

-b

sin(270' -A)=-cosA

sec A 1-sin 2 A=0.4

2x=41.8'

sin3A =cos6B

-$in A cos 270' = (-1 )cos A- sin A (0)

sin 2 A

I

Day 9- Plane 'I'l"igonometry 219


sin9=0.5

0.5 sin 2x +sin 2x = 1

8=30'

1.5 sin 2x = 1 sin 2x = 0.6667

(sin A- 0.25) 2 = 0.5 + (0.25) 2 (sin" A- 0.25) 2 = 0.5625 sin A -0.25 = ±0.75 sin A =-0.75 + 0.25

1' 1111''1' 1'1 ,,:

,I

1

A =sin·1 (-0.5) A =-30' or

A= 360-30 = 330'

ill j

'


220. l 001 Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas

•

Note:

a=~ =J5 2 -3 2 a=4

=(1-sin 2 8)-sin 2 8

cos 65 + cos 55 = cos e

e = 5" 1t

e=5" x - 1800 e = 0.087 rad

•

b~?~ a =15

e=cos-

15)

1

(

~ j~Asgiven

15 adjacent side cose=-= · 17 hypotenuse

•

y= (1 +cos 28) tan 8

3

2

Note: sin

e = cos (90• - 9}

Substitute:

b=~ =J17

-15

2

90-13A =2A

b=8

90=15A

. opposite side b stne = =hypotenuse c 8 sine=-

•

•

1-tan xtan Y

1

-+-

5

=~

1-(;)(;J

hypotenuse

b~3bh

tan (x + y) = 1

•

y = ( 1 +cos 2e} tan e

a=?

8=tal')-1 [2cos(60' )]

=2cos8sin8

8=tan- 1 (1) 1t

•

a= cot-1 [ 2 cos (sin-1 0.5)]

cos8 sine Cos-e_tan8 =--+ Cos-a_ cos = tan8+tan 8

8 = cot- 1 [ 2 cos (30

a

8=cot- (1.732) cot 8 = 1.732 1 --=1.732 tan8 1 tane=(-1.732

= sin 2 8 + sin 2 8 cot 2 8

~[coseJ 2

=sin 2 8+

~

=sin 2 8+cm~ 2 8 X= 1

J

8=30"

•

tan-1 2x+tan-1 x =2:or45' 4 Let: A= tan-1 2x tanA=2x

SeC

tl sin 2 8 7

-csec0(1-sin 0)

l]

1

x = sin 2 8(1 + cot 2 8)

X= sec; 8-

1t

•

sin 8+cos8 tan 8

•

J]

8=45' x - - = - rad 180' 4

D 1

sine= 0.6 or~= opposite side

sinS J = 2(cos ) 8) ( Cbse_

x =2tan8

A=6"

tan (x + y) = tan x +tan y

17

e = tan-1 [ 2 cos (60" l]

x=

sin 13A = cos 2A

•

=2(1-sin 8)tan8

y =sin28

cos 2A sin 13A sin 13A = cos2A

Cbs-a_

e = tan-1 [ 2 cos( sin- 1 ;

= (2- 2sin 2 8) tan 8

I .: i

1 = - -- ( cos 'a e)

\

=(1+1-2sin 2 8)tan8

1 sec2A=--sin 13A 1

e)

X= COS8

Substitute:

4

cote=~

Cbs._ (90 -13A} =do&_ 2A 2

cos 2e = 1- 2sin 2 8

. adjacent side a cote= =opposite side b

0.99619=cose

x=sece(cos 2

cos 29 = cos 2 8- sin 2 e

B=>=tan-1 x tan B = x


Z22. 100-1 Solved Problems in Engineering Mathematics (2"a Edition) by 'l'iong & Rojas Substitute: A+B=45 tan (A+ B)= tan 45 tan A +tanB -----1 1-tanA tanB 2x+x --=1 1- 2x(x)

= --'------'---'---'2(12)

-2±34 x=---

x=~----------~

24

(sin A- sin B)(cos A- cos B)

-2+34 x=--=1.33 24

(cos 2 A- cos 2 B) +(sin 2 A- sin 2 8) (sin A- sin B)(cos A· cos B) 2

2

-3±J(3) -4(2)(-1) = -3±4.123

4

-3+4.123 X

4 =0.28

tan·1 (x + 1) + tan·1 (x -1) = tan-1 12 Let:

2

cos A= 1-cos A 2cos 2 A =1 cos 2 A = 0.5 cosA=±0.707 cosA =0.707

A =ta'ril-1 (x+fj'

tan A =(li! + 1'

•

.

2

eos O+cos 1+ ... cos 2 89+cos 2 90

tan(A+B)=tan(tan- 1 12)

Thus, the above equation can be written in the following format:

--~-=12

tan A +tanB

(sin 2 0 + sin 2 90) + (sin 2 1+ sin 2 89)

1-tanA tanB (x + 1) + (x -1 )

+... +(sin 2 44+sin 2 46)(sin 2 45)

=

12

1- (x + 1)(x -1) 2x = 12-12(x 2 -1) 2x=12-12x 2 +12 12x 2 +2x-24=0

a

x=

2 sine cos e -cos

a

1-sin e + sin 29 -cos 29 cos 9(2 sine -1) (1- cos 2 9) + sin 2 9-

sinS

X=---~----~-

sin29 + sin 2 e- sine cos 9(2 sin e -1)

•II ! I

provided, A+ B = 90

A+B=tan-1 12

sin 2A = 2 sin A cos A 3.939x = 2(2.511 x)(3.06x)

a

X=

Note: sin 2 A +cos 2 B =1andcos 2 A +cos2 B = 1

Substitute:

Substitute:

coversed sin

e = 0.134

Note: coversed sin 9 = 1 - sin 6 Substitute:

cos 9(2 sin e -1 )

1

tanB = x-1

Note: sin 2A = 2 sin A cos A

3.939\ = 15.367x 1. 3.939 x=-15.367 X =0.256

~= 2

B = tan - (x -1)

sin A= 2.511x cos A =3:06x sin 2A = 3.939x

(sin A- sin B)(cos A- cos B) 1-1 x=---------------(sin A -sin B)(cosA -cos B) x=O

A =135" or 225°

sin 2 0 + sin 2 1+ ... sin 2 89 + sin 2 90

X=

2

(cos A+ sin A)- (sin B + cos B)

A =45" or335 cos A =-0.707

a

2

2

x=----------~-------~

2

a

x=--------------~

Ill


X

sin A +sin B x= +~·------sin A - sin B cos A - cos B (cos A+ cos B)(cos A· cos B)J ( +(sin A+ sin B)(sin A- sin B)

-2± J(2) -4(12)(-24)

X

2x 2 +3x-1::r0

2(2)

cos A +cosB

2

3x = 1-2x 2

X:

1111


x=--~------~--~-

(cos2 0 + cos 2 90) + (cos 2 1+ cos 2 89) +... + (cos 2 44 + cos 4 46)(cos 2 45) X= 1

"

2 sin 2 e- sin e cos9(~)

coversed sine= 0.314 1-sin9=0.314 sin9=1-0.314 sin9=0.866

X=--------sinS(~) x =cote

a

9=60'

ml

'I

I'

II

I '

•I

f;!:'I ,I

,1,.!

sin A= 3.5x; cos A= 5.5x sin A

3.5x

cosA 5.5x tan A= 0.636363 A= 32.4T

!'iillil 1:

1~1!

j

224

J_Q_OJ oSolv~d Pro~em~ in Engineering Mathef!lati~~nd Edition) by Tiong & R~~ X

X

~Jso-h

\l\j50

1\ I :h, \ L ih\

X.

b

x=8.816m 50-·h

h1 = b + 3

X

h2 =b-3

!an 50 - -----··

50-h tan 50 ---·--·--

b

2

c i

d.

32°

'

l>A

•

85+x sin 13

Substitute x in Eq.2:

N

B ?t

-~-

v

50

sin111.7

:>in8.3

BC Note:

7t = total distance traveled by ship B 10 + 5t = total distance traveled by ship A d

----------sin 111.7 V=--(50) sin.3 V = 321.8 mph

(j o: :·\7·~ I'll

sin 42.35' 71

sin 46"70' I()

I

15

•

e + 61 + 90 + 15 = 180 6=14

By sine law: h

By sine law:

852.719

d =: 852.719(sin26

*

..~···..··

ml

By sine law:

BC = 852.'119m

sin26

h = 42.5 (tan 60) h = "73.61m

a=111.7

sin 58

sc

x(~~ 60 )

x =42.5m

a+8.3+60=180

46°20'

=

a +p +60 = 180

10

sin 26

sin 60

13 =8.3

h 2 =7-3=4units

By sine law:

,d

,,I

\ tan30 85+x =3x

300

50

= 42

0=96

1000

I

to Eq.2:

(85 + x) tan 30 = x tan 60

b=7

9 + 26 + 58 = 180

sin 96

h

By sine law:

h 1 =7+3:::1Qunits

,.-<_t\•: ei

1000

"'

tan60=-

2

_;;;.

I--

{~

Equate Eq.1

b(b + 3)::: b(b- 3) + 42

6b

B cf

i

B5+x h = (85 + x) tan 30-+ Eq.1

x

tyz_ + 3b = ~•. - 3b + 42

/

I"

h tan30=--

1 1 --bh1 =--bh2 +21

10.506 '0 50- h h = 39.49m

./

X

A 1 =A 2 +21

tan50° (8.816):c::50-h

N

•

85 +X

= 4.03 hrs

h = x tan60 -+Eq.2

8.816

a

2.3231 = 9.354 _t

~ ill

~h~h

sin42.35') {1 0 + 5t) -_-.----- :: 7t ( stn46.20' 9.354 + 4.6771 = 7t

A

L ____~ -

50

tan 80 = --·

Iii,

- - - - - - - - - - - - - - - - - - - - = D a y 9-~ane Trigonome~!!_

r~r

sin14"

sin61°

15

X

II

x = 54.23m

I

II·

'ol

II j:

i

lb

ZZ6 . 100 1 Solved Problems tn Engineering Mathematics (2nd Edition) by Tiong & Rojas

•

I_

•

Equate Eq.1 to Eq.2: 120-h tan 28'

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _D_a..._y_9_-_Pl_a_ne Trigonometry 227

h

/ l ~h~h 11

---tan 46

5(@+

J·

tan28' 120-h= - h ( · tan46' 120-h =0.513h h=79.31m

tan 40'

x=J(15) 2 +(20) 2 X= 25ft

s

~44

S

~h

11h

X

100 +X

120 h

tan21° = - 100+x h=(100+x)tan21' ~Eq.1 X

......---1,20-h

./"1

&_j

X

X

120-h tan28 = - X

120-h

x=--~Eq.1

tan28" h tan46' =-

x h

x=--~Eq.2

tan46"

tan32' h

1144

h

h

=x

=x tan 32'

+X

41

=a-h =m--tan40 h

~

Eq.1

·I

tan30' = - 50+x h

nil

X =----50~

Eq.2

tan 30 ·

X

Equate Eq.1 to Eq.2:

44 tan60=x x =25.4m

~&J

~

X

~~

15 tane=20 9=36.87

•

X

6X

•


X

I

h

h

- - =----50 tan400 tan30" 119175h=1.73205h-50 h = 92.54 ft.

44 tan30=-S+25.4 s + 25.4 = 76.21 S=50.81m

•

s

V=t 50.81 V=·-5(60) V =0.169m/s

35°

~ Eq.2

ml

Equate Eq.1 to Eq.2: ·

X

I

X

3°

(100 + x) tan 21' = x tan 32'

J

tan 32' 100+x=x - ( tan 21' 100 + x = 1.6278x x=159.286m Thus:

¥: + 100 = 259.286 m

h

~~ 50

50 tan35' = x X'"71.407m

50

228 I 00 ["Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas By sine law:

50-h tan13. = - -

x

150

-~

c

--------

50-h tan13'' = - - · 71.407 tan13. (71.407) =50- h

b=30

sin 26° sin 92.75° c =341.78m

c= 34

m1

h =33.514m

c

A

1111

Day 9 -Plane Trigonometry 229

S=

a +b +c

2

16 + 30 +34 -----

2

8=40

Be{ \

~·-

w

b

vu..

"oA

c=190 "'"¢ <

Bd

_'_?QC

By cosine law: 2

2

180 2 =130 2 +190 2 -2(130)(190)cosB B =65.35'"

r tan30" = - 0.5x 10 tan30' = - 0.5x x =34.64cm

•

r~ 0.5x.

By cosine law: x 2 = a 2 + (c/2) 2 - 2a(c/2)cos B

x2 = 130 2 + 95 2 - 2(130)(95) cos 65.35' x =125m

•

c

Bcf ----=c==-:-?-~r::.J_,:' 'o A

c=210

By sine law: S= 150

a+b+c

2

300

sin26'

sinB B = 61.2!5 A+B+C=f80 26 +61.25 +C = 180

c =92.75

A= .J 40(40 -16)(40- 30)(40 -34) A= 240 square units

2

b =a +c -2ac cosB Note: Since equilateral triangle, A = B ::: C = 60°

A= .J s(s- a)(s- b)(s- c)

195+157+210

=----2

s =281 A= .Js(s-a)(s-b)(s-c) A= .J281 (281-195)(281-157)(281- 210) A= 14,586.2 square units

"'l

!IIi

il

232 100 !'Solved Problems in Engineering Mathematics (2nd Edition) by Tionsz & Rojas

I

[] Mon ,, < •

' , •. .,. ..~

~-

·< -'·

:v "~

~ • \

0

'

>

·,. . r

Tue

~.

-

·. ' ~

101 0

I

.-

---'

Theory

Wed

[j

[QJ

Problem<;

Thu

0

Solutions

0

Notes

0 0 Fri

Sat

Topies Spherical Trigonometry Great Circle & Small Circle Poles and Polar Distance Spherical Wedge & Lunes Propositions of Spherical Triangle Solutions to Right Spherical Triangle Napier's Rules Quadrantal Spherical Triangle Solutions to Oblique Spherical Triangle Law of Sines & Cosines Area of Spherical Triangle Terrestrial Sphere GMT & UTC Latitudes and Longitudes

I [ii

I

'r:

:!11 ',

I 111,1

I

:11 'I

What is SQherical Trigonometry?

·,,._,..,,,.,

:I

I ·'

i

Spherical Trigonometry is the part of trigonometry which deals with triangles on the sphere. The object of this subject is to study the relations connecting the sides and angles of a spherical triangle'. This is of great importance for calculations in astronomy and navigation.

Circle A

What is a Great Circle'?

1{ c'

II

Great circle is a circle whose center coincides with the center of the sphere On the surface of the sphere, the closest analogue to straight lines is great circles.

A great circle is a circle on the surface of a sphere that has the same diameter as the sphere, dividing tl1e sphere into two equal hemispheres. A great circle is the intersection of a sphere with a plane going tluouqh its center. J\ great circle is the l
Circles A and circle B are example of great circles because their centers coincide with the center of the sphere. The m~idians and the equator of the earth are also examples of the great circles.

I

.I.

, I

~ !I

.,

~~.~~

I!

Great circle paths are used by ships and aircmfts where currents and winds are not a siqnificant factor. For aircraft traveling w<>;l\;rly between r:on11nents in the

:lillii

I I,

234 1001 Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas northern hemisphere these paths will extend northward near or into the arctic region, while easterly flights will often fly a more southerly tract to take advar>+age of the jet stream. What is a Small Circle?

Day I 0 -Spherical Trigonometry 235

What is a Polar Distance?

What is right spherical triangle?

The polar distance of a circle is the least distance on a sphere from a point on the circle to its pole. The polar distance of a great circle is goo.

A right spherical triangle is one with a right angle.

I

ill I I

A spherical triangle with two right angles is called birectangular spherical triangle while the one with three right angles is known as trirectangular spherical triangle.

;!,

r:

A small circle of the sphere is the circle constructed by a plane crossing the • sphere not in its center. Circle C belowis an example of a small circle. '

'111111111'11

j

:I

A

Center of Sphere What are the important propositions of a spherical triangle?

I

B

The following are the important proposition of a spherical triangle:

p1 is the polar distance of the small circle and P2 is the polar distance of the great circle. P2 is an arc that measures goo.

1.

If two angles of a spherical triangle are equal, the sides opposite are equal; and conversely.

2.

If two angles of a spherical triangle are unequal, the sides opposite are unequal, and the greater side lies opposite the greater angle; and conversely.

What is a Spherical Wedge?

What is a Pole?

A line through the center 0 of a sphere perpendicular to the plane of a circle on the sphere cuts the sphere into two points called poles.

Two diametral planes that form an angle divided the sphere into four spherical wedges, opposite wedges being congruent. The surfaces of the we_dges are called sphericallunes or digons.

,II' I

3.

The sum of two sides of a spherical triangle is greater than the third .side.

C

a

What are the solutions to right spherical triangle?

A right triangle can be solved by using the Napier's Rules. Napier circle (sometimes called Neper's circle or Neper's pentagon) is a mnemonic aid to easily find all relations between the angles and sides in a right spherical triangle.

a+b >C

tune 4.

Pole

The sum of the sides of a spherical triangle is less than 360 degrees.

0° < a + b + c < 360° 5.

The sum of the angles of a spherical triangle is greater than 180 degrees and less than 540 degrees. 180° < A + B + C < 540"

6.

What is a Spherical Triangle? Pole

Spherical triangle consists of three arcs of great circles that form the boundaries of a portion of a SP.herical surface.

The sum of any two angles of a spherical triangle is less than 180 degrees plus the third angle. A +B< 180° +C

j

Napier's Rules: Hule no. 1 (Tan-ad Rule): The sine of any middle part is equal to the product of the cosines of the opposite parts. Rule no. 2 (Co-op Rule): The sine of any middle part is equal to the product the tangent of the adjacent parts.

of

il I

Day 10 -Spherical Trigonometry 237

Z36 1001 Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas How to apply the Napier's Rules? Consider the right spherical triangle below:

A

Important Rules: 1.

B

C

2.

a

Convert it to its equivalent Napier's circle:

A. Law of Sines:

sinb "'sinBsinc

3.

In a right spherical triangle an oblique angle and the side opposite are of the same quadrant. When the hypotenuse of a right spherical triangle is less than 90°, the two legs are of the same quadrant and conversely. When the hypotenuse of a right spherical triangle is greater than 90°, one leg is of the first quadrant and the other of the second and conversely.

What is a quadrantal spherical triangle? A quadrantal spherical triangle is a spherical triangle having a side equal to

goo.

B. Law of Cosines for sides:

=cosbcosc + sinbsinccosA

cosa

cosb == cosacosc + sinasinccosB

=cosacosb + sinasinbcosC

cosc

C. Law of Cosines for angles:

A meridian is half of a great circle on the earth terminated by the north pole and the south pole. In October 1884, it was agreed upon that the reference meridian is the one that passes through the Royal Greenwich Observatory at Greenwich, England. This meridian is called .the Prime Meridian or sometimes known as the Greenwich Meridian.

cos A= -cos8cosC + sirt8sinG~cosa. cosB = -cosAcosC + sinAsinCcosb cosC = -cosAcos8 + sinAsin8cosc

What are the formulas for area of spherical triangle?

A

If "b" is considered as the middle part, then the opposite parts are

sina sinb sine sinA = sinB =sinG

I

The equator is the great circle whose plane is perpendicular to the axis. The equator has a latitude of zero degree.

A

c and B while the

adjacent parts are A and a. Applying rule no. 1: sinb =tan A tan a

~

Eq. 1

But TanA=tan(90-A)

B

C

~8

a

c

Substitute Tan A in equation 1 sinb = tan(90- A)tana

An oblique spherical triangle is one with no angle equal to a right angle.

. 1 But tan(90-A)=cotA=-tanA . 1 smb =--tana tan A sinbtanA = tana

What are the solutions to an oblique spherical triangle?

A

sinb = cosBcosc

But cos(90- B)= sinS cos{90- c)= sine

A=-.-·-·

taoo

where: R = ~adius of the sphere E spherical excess in degrees E =A+ 8 + C -180°

=

What Is a Terrestrial Sphere? Terrestrial sphere refers to the earth (though slightly ellipsoid) as sphere with a radius·of 3959 statute miles.

Applying rule no. 2:

sinb = cos(90- B)cm:(90- c)

1tR2E

What Is an oblique spherical triangle?

B

c

a

The terrestrial sphere rotates about a cjiameter called its axis which pierces the .sphere in the north pole and the south pole.

The meridian on the side of the earth the lie opposite to the prime meridian is known as the international date line. This imaginary line on the surface of the earth offsets the hours that are added or subtracted as one travels east or west through successive time zones. If one crosses the international dateline precisely midnight, going westward one skips and entire day while going eastward, the day repeats.

l ~

238 1001 Solved Problems in Engineering Mathematics (2"ct Edition) by Tiong & Rojas What is a Greenwich Mean Time (GMT)? Greenwich Mean Time I GMT) is the mean solar time at the Royal Greenwich Observatory, in Greenwich, England. Theoretically, noon Greenwich Mean Time is the moment when the sun crosses the Greenwich meridian (and reaches its highest point in the sky in Greenwich). Because of the earth's uneven speed in its elliptic orbit, this even may be up to 16 minutes off apparent solar time. GMT became a world time and date standard because it was used by the Britain's Royal Navy and merchant fleet during the nineteenth century. In January 1, 1972, GMT was replaced as the international time reference by Coordinated Universal Time (UTC}. The UTC is maintained by an ensemble of atomic clocks (a much more stable timebase) around the world. · UTC uses a 24-hour system of time notation, "1:00am" in UTC is expressed as 0100, pronounced as "zero one hundred". Fifteen minutes after 0100 is expressed as 0115; thirty-eight minutes after 0100 is 0138 (usually pronounced as "zero one thirty-eight"). The time one minute after 0159 is 0200. The time one minute after 1259 is 1300 (pronounced "thirteen hundred"). This continues until 2359. One minute later is 0000 ("zero hundred"}, and the start of a new UTC day.

A parallel of lati~de or briefly parallel, is the small circle cut from the earth by a plane parallel to the equatorial plane. latitude

Colatitude is the complement of latitude.

I

DayJO- SpliE!rical Trigonometry 239 What are Terrestrial Sphere constants?

.!

-:~

: 1minute: ; angle

N,,rth

'\

(+)

31)

·~~-·L

····:·:r>~~:··,_.;.,:.~.~·· · :.-:.·.~· \· · · · · ·. . . . \

Equator If 1:.

':1 .•

J ..·

T'~':'''l

I

1

Radius of earth = 3959 statute miles 1 minute on the great circle arc = 1 NM 1 NM (nautical mile)= 6080 feet = 1852 meters 1 statute mile = 5280 feet = 1760 yards 1 statute mile = 8 furlongs = 80 chains ·

S·>Ut.L

Tl laid out on a circle of constant latitude. l.Dngitude 180

The!. statute mile (sometimes known as the international mile) is typically meant when the w9rd "mile" is used without qualification. The nautical mile is used universally in aviation, naval and maritime purposes.

Proceed to the next page for your 1Oth test. Detach and use the answer sheet provided at the last part of this book. Use pencil number 2 in shading your answer. GOOD LUCK I

90

·; •. !

'. ·/~~- .. '

East (+)

0 Prime meridian

Did you know that..: in the ancient times, the number 40 was used to indicate "many or too many" and the number 1001 ·signifies a kind of "finite infinite" as in Alibaba and the 40 thieves, Moses leave his people for 40'days and 40 nights and the famous Arabian "A thousand and one nights"

®uotr: "To be a scholar of mathematics, you must be born with talent, insight, concentration, taste, luck, drive and the ability to visualize and guess"

Lf',t o

')(!

'Ql:ribia:

''(

\1 Nautical mile

The UTC is also kncwn as "Zulu Time" or "World Time". The latitude of a point is the angular distance of the point from the equator, and will be considered positive if the point is north of the equator and negative if the point is south of the equator. This angular measurement ranges from 0° at the equator to 90° at the poles. One minute of the of the arc of latitllde is approximately one nautical mile or 1852 meters. A degree of latitude always corresponds to about 111 km or 69 miles.

1l

- Paul Halmos

·-----'=D-"'aY.JQ.- SEherical Trig_9~!.Y 24_1 441: Solve for side b of a right spherical triangle ABC whose parts are a ::: 46°, c "' 75° and C =90°

0

tv! on

0 lJ 0

Tue

Theory

Problems

0

Solutions

0

Notes

V'./ed

[Q] Thu

0 Fri

0

Sat

436: If Greenwich Mean Time (GMT) is 6 A.M., what is the time at a place located 30° East longitude? A.

7A.M.

B.

BA.M.

C.

9A.M.

Spherical Trigonometry Great Circle & Small Circle Poles and Polar Distance Spherical Wedge & Lunes Propositions of Spherical Triangle Solutions to Right Spherical Triangle Napier's Rules Quadrantal Spherical Triangle Solutions to Oblique Spherical Triangle Laws of Sines & Cosmes Area of Spherical Triangle Terrestrial Sphere GMT & UTC Latitudes and Longitudes

C.

30 minutes

D.

1 hour

B.

68°

48° 74°

A.

45"

B. C. D

90° 60° 30°

437: If the longitude of Tokyo is 139°E and that of Manila is 121°E, what is the time difference between Tokvo and Manila?

73.22

D.

1 hour and 1 hour and 1 hour and 1 hour and

12 minutes 5 minutes · 8 minutes 10 minutes

spherical triangle whose given parts are A

= B = 80° and a

A.

B. 438: One degree on the equator of the

C.

earth is equivalent to _ _ in time.

D.

A. B.

1 minute 4 minutes

158°12' 162°21' 168°31' 172°12'

What is the value of the side opposite the right angle?

A. B.

83°30' 84°45'

C.

86°15'

D.

85°15'

448: ECE Boar4 April 11.997 The area of sphencal tnangle ABC whose parts are A= 9::1°40'. B = 64°12', C = i 16°51' and the radius of the spl1ere i~ 100 rn is:

443: Determine the value of the angle B of an isosceles spherical triangle ABC whose given parts are b = c 54°28' and a =

=

A. B. C.

D.

15613 sq. m. 16531 sq. m. !8645 sq. m. 25612 sq. rn

92°30'. A. B.

C. D.

449: A spherical triangle has an· area of 327.25 sq. km. What is the radius of the sphere if its spherical excess is 30°?

89°45' 55°45' 84°25' 41°45'

A. B. C. D.

45°54' 80°42' 97"09' 72°43'

445: Solve for angle C of the oblique spherical triangle ABC given, a = 80°, c 115° and A = 72°

74.33 75.44 76.55

440: Solve the remaining side of the

A. B. C.

44:.t: Given a right spherical triangle whose parts are a = 82°, b = 62° pnd C = 90°.

A.

20 !\m

B. C. D.

22 krn 25 krn 28 krn

4$0: EE Board. April :1:99•1

=

A. B. C. D.

74°

444: Solve the angle A in the spherical triangle ABC, given a = 106°25', c = 42°16' and B = 114°53'.

439: A spherical triangle ABC has an angle C 90° and sides a = 50° and c = 80° Find the value of "b" in degrees.

D. 4A.M.

··'

90°57' 98°45'

C. D.

.~.·

D.

447: What is the spherical excess of <> spherical triangle whose angles are all right angles?

A.

Topics

c.

= b =89°

A.

B. C. D.

=

A B. C.

61° 85° 95° 119°

D.

441&1 Determine the spherical excess of the spherical triangle ABC given a = 56°, b 65" and c 78°

=

A

:n"33'

ll

(ill":l7'

A ship on a certain day is at latitude 20°N and longitude i 40°E. After sai!ing for 150 hours at a uniform speed along a great circle route, it reaches a point at latitude 10°S and longitude 170°E. If the radius of the earth is 3959 miies. find the speed in miles per hour.

17.4 15.4 16.4 19.4

Day 10- Spherical Trigonometry 243

•

sin co-c = cos a cos b cos c = cos a cos b

diff. in til11e

Topics

D D D D D ~ D D D Mon

Tue

Wed

Theory

Problems

Thu

Solutions

Fri

Notes

Sat

436. B 437. A 438. B 439. B 440.C

441. B

442.

c

443. D 444.C 445. D

c

c:J 13-15

c:J c:J 0

Topnotcher

9-12 Passer b-8

Conditional

0-b Failed


30°-

oo

-------

24 360" diff. in time = 2 hours Note: The time in the place is 4 hours ahead of GMT because the place is at the East. Thus, the time is 8 AM.

••

diff. in time , diff. in longitude 360° 139°-121 o

----=~---

24 360° diff. in time= 1.2 hours = 1 hour and 0.2(60) min = 1hour and 12min

•

•

c il:·

sin co-B =tan co-a tan c/2

f3

=cot a tan c/2

1 cos B =--(tan c/2) tan a 1 cos 80 = --(tan c/2) tan 89 tan c/2 = cos 80 tan 89

c/2 =84.26° c = 168.52" or 168° 31'

Note: 360 degrees = 24 hours

'

b = 74.33''

cos

24 diff. in time

RATING

446.A 447.8 448. B 449. 450.C

360°

24 diff. in time

Spherical Trigonometry GreatCircle & Small Circle Poles and Polar Distance Spherical Wedge & Lunes Propositions of Spherical Triangle Solutions to Right Spherical . Triangle Napier's Rules Quadrantal Spherical Triangle Solutions "to Oblique Spherical Triangle Laws of Sines & Cosines Area of Spherical Triangle Terrestrial Sphere GMT & UTC Latitudes and· Longitudes

ANSWER KEY

=diff. in longitude

cos 80 = cos 50 cos b

time=1 \

Ill

A

241m;_) x (- x [60 - minJ -. 360\

. 1 ~

time =4 minutes

•

c

a=46°

A sin co-c = cos a cos b cos c = cos a cos b cos 75 =cos 46 cos b b=68.12' or68'07' a=so·

.,,1'1'11111 l

II

I

•I

It

'

244. 1001 Solved Problems in Engineering Mathematics (2"d Edition) by Tiong & Rojas;

' cos b = cos a cos c + sin a sin c cos B

A

cosb = (cos 106 25')( cos 42 16') +

(Sir 106

25' )(sin 42 16' )

Day 10 - Spherical Trigonometry 245

•

•

B=72°

E=(A+B+C)-180 E = (93'40' +54· 12' + 116° 51') -180

(cos114 53')

c

E =94°43'

b =11843'

A

a=82° Using law of sines:

sin co-c = cos a cos b sinA

cos c = cos 82 cos 62

• c

sin 106°25' sin A

sin b sin 114°53'

sin 106°25'

sin 118°43'

----

A

A=82'51'or A= 97"09' (its supplement)

•

a=92°30'

8=72°

cos a= cosb cosc +sin b sin ccosfl,. cos 56 =(cos65 cos 78) ·· + (sin 65 sin 78 cos A) A =5T53' Using law of sines: sine sine sin C

1

A

tan 54'28'

sinC

2

sine sin C

B =41"75'or41"45'

sin 115°

II

8=114°53'

A

b

Using law of sines:

[ t a92'30') n--

sinA

b

Using law of cosines for sides:

sinS

sin 80°

sinA

sinb sin B

sina sin 57" 53'

sin65"

sin 56"

sina sin 72°

8=67"48'

1tR 2 E ~t(100) 2 (94.43') A=-------180" 180° A.= 16,531.17m 2

•

I

,I,,

E =(5T53'+67 48' +8T52')-180

C = 118 ° 56' (its supplement)

E=33"33'

•

I

~tR 2 E A= 180°

327.25 =

~tR

2

1.,

I

!',,I

(30°)

180° R=25km

•

North

:: \h;~~~·~:~}"N) u \···... _ C

I

I,

,II " i:li·llll.i'

B(170J'E, 10°8)

1'.1

'i

~>b-l

E = (A + B +G) -180

C=61"04'or

Note: Since side c is greater than side a then angle C is greater than angle A. Thus use = 118°56'

c

sin a sin 57°53'

sin 56° C =>8T52'

cos B = cot c tan a/2

cosB=

sinA

sin 78°

sin co-B = tan co-c tan a/2 1 cos B =--(tan a/2) tan c

c

Using law of cosines for sides:

sinS

-------

c=86.25° or86'15'

b=65°

South

A

E =(A+ B + C)-180

E=(90 +90' +90 )-180 E=90

B C a=3o•

cos b = cos a cos c + sin a sin c cos B ;11:

,,

246 100 1 Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas sin co-c = cos a cos b cos c = cos 30 cos 30

c =41.4096°

1\

C=41.4096\ X ( 60NMJ

Topics

c = 2,484.58 NM

v

D

distance

Mon

time

2484.58

D D D D ~ D D D·

.

= 16.56NM/hror 150 V = 16.56 knots

V

Spherical Trigonometry Great Circle & Small Circle Poles and Polar Distance Spherical Wedge & Lunes Propositions of Spherical Triang~e Solutions to Right Spherical Triangle Napier's Rules Quadrantal Spherical Triangle Solutions to Oblique Spherical Triangle Laws of Sines & Cosines Area of Spherical Triangle Terrestrial Sphere GMT & UTC Latitudes and Longitudes

Tue

Theory

Wed

Problems

Thu

Solutions

Fri

Notes

Sat

"'

,,.•,,-o·,..,

?

,,

~

·' '

, , ,,. '

,,

;;

'

·,.

····'00 '*'"'"•"<

•. • • .

~ <

< '~

,A,,,;,

,.,.

~

<

248 100 l Solv:ed Problems in Eng!!'!~~g Mathematics (2nd Edition) by Tiong; & Rojas

E'ii'IW I Topics

0 0 0 ~J l] 0 0 ~ 0 0 Mon

'I

~

..: ;,,

~

< '

~· 0'

Tue

~il

'

Theory

V\fed

Problems

Thu

Solutions

Fri

Notes

Sat

Analytic Geometry Rectangular Coordinates System Distance Between Two Points Slope of Line Angie Between Two Lines Distance BetWeen a Point & a Line Distance Bel.\-IJeen 2 Parallel Lines Division of Line Segment Area of Polygons by Coordinates Equations of Lines Conic Sections or Conics General Equation of Conics Geometric Properties of Conics Equations of Circles

Wha!Js Analytic Geometry? Analytic geometry deals with geometric problems using coordinates system thereby converting it into algebraic problems. Rene Descartes (1596- 1650, Cartesius in Latin language) is regarded as the founder of analytic geometry by introducing coordinates system in 1637.

First Quadran!

Second Quadrant

P(5,3)

abscissa

2

I

I

ordinate

0

I t--4>~---t

-4 -3 -2 -1 Rectangular Coordinates System (Also known as Cartesian Coordinates System)

Th;ro ""''""'

-1

~

•

2

I

I

I

3

4

Fomth Qoed"ot

l•x 5

Day 11 -Analytic Geometry (Points, Line11 & Circles) 251

250 1001 Solved Problems in Engineering Mathemati<;:s (2"d Edition) by Tiong & Rojas

A line parallel to the x-axis has a slope of

Z·

Point 0 is the origin and has coordinates (0,0). The x-coordinate or ~bscissa is always measured from the y-axis while the y-coordinate or ordinate is always mei;lsured from the x-ax1s. The point P has 5 and 3 as abscissa and ordinate, respectively.

(Xt, Yt)

zero while a line parallel to the y-axis has a slope of infinity {co).

•...

"······-~-~

For parallel lines with slopes of m1 and m2, respectively, the slopes are the same.

(X2, Y2, Z2)

Ax+By+C=O

~--~--~--~--~x

m~=m2

What is distance between two points in

i elane? Consider two points whose coordinates , are (Xt, y,) and (x2, y2), respectively. A right triangle is formed with the distance between two points being the hypotenuse of the right triangle.

For perpendicular lines with slopes of m, and m2, respectively, the slope of one is the negative reciprocal of the other.

y d = J(x2- x1)

2

2

2

+ (Y2- Y1) + (z2- z1)

y

(x2, Y2)

.~//./'!'··········· (XI, Y1

) I...

X2- XI

Using Pythagorean theorem, the distance between two points can be calculated using:

Line 2

d ..·

...···········' (X1, Y1)

X2- X1

Line 1 ~~

The angle, e between these lines (line 1 and line 2) may be calculated using the . following formula:

What is distance between two I?Oints in space?

1:

~

,j

2

This formula is known as the distance · formula.

rise t:..y slope=m = - =-run t:..x where: !!.. denotes an increment

x2

-x~

tane = m

m = Y2 -y2 x2 -x~ I· I

-m~

Ax;t-By+C1 =0

tane "' 1 + m m 2 1

t~

The (perpendicular) distance, d, between the two lines is:

c,":'c2

I

,I '.',

But

m2

J

tane = Y2 - Y2 Consider three axes namely x,. y and z and two points with coor<;linates (x,, Yt. Zt) and (x2. y2, z2), respectively.

What is distance between two.Qarallel lines? Consider two parallel lines with equations as shown in the figure.

----~-----------------------X

+(Ya -Y,)

if B is positive and the point is above. the line or to the right of the line if B is negative and the point is below the line or to the left of the line If otherwise

Consider two lines with slopes of m, and m2.

,l

d;:J{; -xt)

+ What is angle between two lines?

y

•I

~---+-----------------------X

2

+

I

Consider two points whose coordinates are (x,, y1) and {x2. y2}, respectively. A line is formed by connecting the two points. The slope of the line is defined as the rise {vertical) per run (horizontal)

4

use

1

m2 =--. m, What is slope of a line (m)?

d=Ax +By1+C ± A~ +82

\

d= What is distance between a point and a

±.JA2 +B~

!!.!!!1 Consider a point with coordinates (x1, y,) and a line with equation Ax + By + C = 0.

Use the sign (either+ or-) that would make the distance positive.

Day 11 -Analytic Geometry (Points, Lines & Circles) 253

252 1001 Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas What are the coordinates of a point that divides a line segment? Consider two points with coordinates (x,, y,) and (x2. y2). The line segment formed by these two points is divided by a point P whose coordinates are (x, y). Let r, and rz be· the corresponding ratio of its length to the total distance between two points.

Y./\ . . ,\

(h Y2)

E. Intercept Form:

(XJ, Yt)

~+1'.=1 a .. b

~~y~

The arrow shown in tire figure moving in a counterclockwise direction ir1dicates that the vertices must be written in the equation below in counterclockwise direction. .~

//

A=

(x,, y,)

..~

.~

.!((XWz +i<2Y3 +x3y,) 1 2 -(y,x2 +y2x3 +yax,)j

Conic section (or simply Conic) is the locus of a point which moves so that its distance from a fixed point (focus) is in constant ratio, e (eccentricity) to its distance from a fixed straight line (directrix) . The term "conic" was first introduced by a renowned mathematician and astronomer of antiquity, Apollonius (c.255- 170 B.C.) Also, the term ~conic section" was due to the fact that the section is formed by a plane made to intersect a cone. Circle

The abs.cissa of the point, P is:

$

What is line?

K= (Xl2)+(XifQ

r1 + r2

A line is defined as the shortest distance between two points. The followi'ng are the equa~ons of the lines:

The ordinate of the point, P is:

$

A. General Equation:

Y"' (y,r2)+(Y:l1) r1 +r2

Ellipse is produced when the cutting plane is not parallel (o~ inclined) to the base of the cone.

What is a Conic Section? ~z.Y0

A-~lx, . ~2·.··/~3 )'1r·· - 2 Y1/ Y2"1_-a ~Y1 . . ~

~P(x,y}

Circle is produced when the cuttiryg plane is parallel to the base of the cone.

What is the General Equation of a Conic Section:

Ax2 + Bxy + Cy 2 +Ox+ Ey + F =0 When B is not eq·ual to zero, then the principal axes of the conic are inclined (not parallel to the coordinates axes). The curve can be identified from the equation given by determining the value of the determinant; 8 2 - 4AC.

8 2 -4AC <0 =0 >0

Ellipse

y:::::Y2+Y1

2

C. Slope-Intercept Form:

y=mx+b What is the formula for the area of a polygon using the coordinates of its vertices? Consider a polygon whose vertices have coordinates of (x,, y,), (Xz. Yz) and (x3, y3).

D. Two-Point Form:

Parabola

1E Hyperbola

_ Yc y, (x ... x,)

y-y,-·xz-x,

Conic Section Ell~

Parabola Hyperbola

Eccentricity

< 1.0

= 1.0 > 1.0

When B is equal to zero, then the principal axes of the conic are parallel to the coordinates axes (x and y axes). To ' identify the curve, compare the coefficients of A and C.

B. Point-Slope Form: y-y,=m{x-x 1)

and

Hyperbola is produced when the cutting plane is parallel to the axis of the cone.

Ax+By+C=O

If the point, P is at the mid-point of the line segment, then the abscissa and ordinate of the point are the following:

x, +X2 X= 2

Parabola is produced when the cutting plane is parallel to the element (or generatrix) of the cone.

~~

If A = C, the conic is a circle. If A C but the same signs, the conic is an ellipse. If A and C have different signs, the conic is a hyperbola. 'If either A or C is zero, the conic is a parabola.

*

The conic sections have geometric properties that can be used for some engineering application such as beams of sound and reflection of rays of light.


254. 100 1 Solved Problems in EngineeljC,Q Mathematics (2nd Edition) by Tiong & Rojas Circle reflects rays issued from the focu;s: back to the center of the circle. · ·

If D & E

= 0,

Center (h,k)

center is at the origin (0,0)

If either D or E, or both D & E "" 0, the center is at (h,k).

2.

Standard Equations:

Radius ( r)

I

~I/

;• I

X


x2 *·Y2 =rt· C(h,k) Ellipse reflects rays issued frorn the focus Into the pther focus.

GOOD LUCK I

y '(ll;ribia:

'•.,

k

0

What Is a Circle? A circle is a locus of a point that which moves so that it is equidistant from a fixed point called center. 1.

General Equation:

J!- +Yt +DX+Ey+F = 0

2A

D2 +E 2 -4AF r = .1---.-4A 2

Parabola reflects rays issued from the focus as a parallel (with respect to I~ axis) outgoing beam.

Hyperbola reflects rays issued from the focus as if coming from the o~er focus.

k= -E

2A

y

C(O,O)

8···/······~

h"' -D

i

X

h 2

{-J(-.h) +(y-kf

Did you know that. .. nature seems to know its math and never wastes its · resources! The reason why bubbles are spherical because with this shape, bubbles can enclose the most volume with the least material. This was first shown by Archimedes.

=

r2

When the equation given is general equation rather than standard equation, the center (h,k) of the circle and its radius (r) can be determine by converting the. general equation to standard using the process known as completing the square. Or using the following formulas: General equation:

·Afi!.· +Qy2 +'Ox +Ey + F =0

"Can we actually know the universe? My God, it's hard enough finding your way around Chinatown!" -Woody Allen

Day 11 -Analytic Geometry (Points, Lines & Circles) 257 45•1 If (-2,-4) is the midpoint of (6.-7) and (x,y), then the values of x and yare

Topics

D D D D

Analytic Geometry Rectangular Coordinates System Distance Between Two Points Slope of Line Angle Betw~en Two Lines Distance Between a Point & a Line Distance Between 2 Parallel Lines Division of Line Segment Area of Polygons by Coordinates. Equations of Lines Conic Sections or Conics General Equation of Conics Geometric Properties of Conics Equations of Circles

Mon

Tue

Theory

Problems

D

Solutions

Wed

D Thu

~ Fri

[] D Notes

A

x= 2, y

=1

B. C.

X= -10,

y

0.

X:::;

10, y = -1 ·8, y -1

=

45'71 ECE Board November 1998 Determine the coordinates of the point which is three-fifths of the way from the point (2,-5) to the point (-3,5).

A.

(-1,1)

B.

(-2,-1) (-1,-2) (1,-1)

c. D.

4S8a ECE Board April 1998 The segment from (-1 ,4) to (2,-2) is extended three times its own length. The terminal point is

A. B.

c. D.

Sat

X::

=-1

(11,-24) (-11,-20) (11,-18) (11,-20)

4S9t The points (a, 1), (b,2) and (c;3) are collinear. Which of the following is true? 451: ECE Board A.pril :t999 The linear distance between -4 and 17 on the number line is

C. D.

19 -5or19

B.

c. D.

13 21 --17 -13

A. B.

4601 If the slope of the line connecting the origin and point P is 3/4, find the abscissa

c.

(2, (3, (3, (2,

-2) -2) -3) -3)

D.

(-2,5).

455: lEE Board April 1995 The line segment connecting (x,6) and (9,y) is bisected by the point (7,3). Find the values of x and y.

11 9 10 8

A. B.

5 or-5 ·5 or 19

of P if its ordinate is 6. A.

2

B.

6 7 8

c. D.

4••a ECE Board April 1999

453: If the distance between points (3,y) and (8,7) is 13, then

=

equidistant from (1, -6), (5, -6) and (6, -1 ).

452: EE Board Apri!l 1.994 Find the distance between A (4,-3) and B A. B. C. D.

c-b=c-a c-b b-a c-a =·a-b c-a = b- a

B. C. D.

454: Find the coordinates of a point

A.

A.

y is equal to

A.

14,6

B. C.

33, 12 5, 0

D.

14,6

Find the inclination of the line passing through (-5,3) and (10,7).

B.

14.73 14.93

C.

14.83

D.

14.63

. A.

4•2: Find the angie formed by the lines 2x + y -8 = 0 and x + 3y + 4 0.

=

A. B.

c. D.

30"

35" 45°

so·

4•:J• Find the angle between the lines 3x + 6 and x + y = 6.

2y

A.

=

B.

12•20' 11•19'

D.

14°25' 13.06'

c.

4•4a What is the acute angle between the lines y·= 3x + 2 andy= 4x + 9? A. B.

C.

4.4" 28.3" 5.2c

D.

18.6"

465: EE Board October 1997 Find the distance of the line 3x.+ 4y from the origin.

B,

4 3

D.

2 1

A.

c.

=5

4 ... CE Board November :1'!'92

=

The two points on the lines 2x = 3y + 4 0 which are at a distance 2 from the line 3x + 4y-6 = 0 are

A. B. C.

D.

(-5, 1) and (-5,2) (64,-44) and (4,-4) (8,8) and (12, 12) (44,-64) and (-4,4)

4•71 CE Board November :1992 The distance from the point (2, 1) to the line 4x- 3y + 5 0 is ·

=

A. B.

1 2

D.

4

c. 3

258 . 100 1 Solved Problems in Engineering Mathematics. (2nd Edition) by Tiong & Rojas 4681 CE Board November 1996 Determine the distance from (5, 10) to the linex- y = 0.

A. B.

c.

3.33 3.54 4.23

D.

5.45

4 5 6 7

ME Board April 1998 Find the slope of the line defined by y - x


1 1/4 -1/2 5+x

0?

What is the x-intercept of the line passing through (1,4) and (4,1)?

.

3/2 2/3 -3/2 -2/3

A. 3 B.

c. D.

4

In a cartesian coordinates, the vertices of a triangle are defined by the following points: (-2,0), (4,0) and (3,3). What is the area?

4711 EE Board April 1995 Find the distance between the lines, 3x + y - 12 = 0 and 3x + y - 4 = 0.

A. B. C. D.

8 sq. units 9 sq. units 10 sq. units 11 sq. units

4761 EE Board April 1994

A. B.

c. D.

16

J10 4

J10 8

J10

47:&1 ME Board October 1996 What is the length of the line with a slope of 4/3 from a point (6,4) to they-axis? A. B. C. D.

Given three vertices of a triangle whose coordinates are A (1, 1), B(3,-3) and (5,-3). Find the area of the triangle.

12

J10

10 25 50 75

A. B. C. D.

5

D.

6

A. B. C. D.

3x + y3x- y + X+ 3y + X- 3y-

1 =0 1=0 1 =0 1 =0

481: ECE Board April1999 If the points (-2,3), (x,y) and (-3,5) lie on a straight line, then the equation of the line is

A. B. C. D.

2y -1 = 0 2x + y -1 = 0 X+ 2y -1 = 0 2x + y + 1 =0 X-

The equation of a line that intercepts the xaxis at x = 4 and the y - axis at y =- 6 is,

4771 ECE Board November x99o In a cartesian coordinates, the vertices of a square are: ( 1,1 ), (0,8), (4,5) and (-3,4)~ What is the area? blnits units units units

478: EE Board April1997 A line passes thru (1 ,-3) and (-4,2). Write the equation of the line in slope-intercept form.

A. B. C. D.

y- 6x =0 y = -6 l X+ y = -6 6x + y =0

intercept and y-intercept are -2 and 4, respectively.

48ZI ME Board April :1998

3 4 5 6

A. 20 sq. B. · 30 sq. C. 25 sq. D. 35 sq.

B.

Find the equation of a straight line with a slope of 3 and a y-intercept of 1.

5 6.

4.5



484: Find the equation of the line passing through the origin and with a slope of 6?

485: Find the equation of the line if the xA.

c. 4

4701 CE Board May 199:1

=

A . ., -4 =X B. y =-X - 2 C. y =X- 4 -D. y- 2 =X


4741 CE Board November 1995 What is the slope of the line 3x + 2y + 1 = A. B. C. D.

Find the distance between the given lines 4x- 3y = 12 and 4x- 3y -8.

=

5. A. B. C. D.

4691 The distance from a point (1 ,3) to the line 4x + 3"y + 12 = 0 is A. B. C. D.

47~1


A. B. C. D.

3x + 2y = 12 2x- 3y = 12 3x- 2y = 12 2x- 3y = 12

A. B. C. D.

y- 2x-4 y + 2x-4 y- 2x + 4 y + 2x + 4

=0 =0 =0 =0

486: ECE Board April 1998 Determine B such that 3x+ 2y ·- 7 = 0 is perpendicular to 2x- By + 2 = 0. A. B. C.

5 4 3

D.

2

487: The line 2x - 3y + 2 = 0 is perpendicular to another line L 1 of unknown equation. Find the slope of l.,. A. B. C.

312 -3/2 2/3

D.

-2/3

488: A line through (-5,2) and (1 ,-4) is perpendicular to the line through (x,-7) and (8,7). Find X.

A.

B. C: D.

-4 -5 -6 -19/3

483: A line with an inclination of 45• passes through (-5/2,-9/2). What is the xcoordinate of a point on the line if its corresponding y-coordinate is 6? A. B. C.

6 7 8

f)

<)

489: CE Board May 1996 What is the equation of the line that passes thru (4,0) and is parallel to the line X- y- 2 = 0? A. B. C. D

X- y + 4 = 0 x+'y+4=0 X .... y --4 =.- 0 X y 0

260 100 l Solved Problems in Engineering Mathematics (2"d Edition) by Tiong & Rojas 49e: Find the equation of the line through point (3, 1) and is perpendicular to the line X+ 5y + 5 = 0.

49&: CE Board May 1.997 Find the slope of a line having a parametric equation of x =2 + t and y = 5

-3t. A. B. C. D.

5x- 2y = 14 5x- y = 14 2x- 5y = 14 2x + 5y = 14

491.: Find the equation of the perpendicular bisector of the line joining (5,0) and (-7,3) A.

B. C. D.

8x + 2y + 11 = 0 8x - 2y + 11 = 0 8x - y + 11 = 0 8x + y + 11 = 0

492: Which of the following lines is parallel to the line 3x- 2y + 6 = 0? A. B. C. D.

3x + 2y - 12 = 0 4x- 9y = 6 12x + 18y = 15 15x-10y-9=0

49:J: The equation of the line through (-3,5) parallel to 7x + 2y - 4 = 0 is

3

C.

-2

D.

-3

C. D.

7x 7x 7x 2x

+ 2y + 31 = 0 - 2y + 30 = 0 + 2y - 4 = 0 + 7y + 30 = 0

A. B.

C.

3

D.

4

A. B.

C. D.

2x + 3y = 0 4x- 5y = 22 4x + 5y = 2 5x + 4Y = 7

C.

A. B.

3 4

C. D.

5 2

A. B.

A. B.

C. D.

Theory

Problems

C. D.

2x+y-2=0 2x - Y,- 2 = ·o 2x- y + 2 = 0 2x + y + 2 = 0

Wed Thu

[IJ Solutions

0

Notes

499: ECE Board April 1.998 Find the area of the triangle which the line 2x - 3y + 6 = 0 forms with the coordinate axis.

passing through (-2,6) with the x-intercept half the y-intercept? x-y=6 2x + 2y + 2 = 0 3x- y + 2 = 0 2x + y- 2 = 0

0 0

4x- y = 14 4x + 4y = 14 X+ 4y = 12 x-4y=-14

soo: ECE Board November 1998 A line passes through point (2,2). Find the equation of the line if the length of the !ine· segment intercepted by the coordinates axes is the square root of 5.

49S: What is the equation of the line

Tue

498: ECE Board Aprill:999 Two vertices of a triangle are (2,4) and(2,3) and the area is 2 square units, the locus of the third vertex is A. B.

0 0 0 0

Mon

1 2

494: What is the equation of the line joining the points (3,-2) and (-7,6)?

Topics

497: CE Board May 1.998 Find the slope of the line having a parametric equation y = 4t + 6 and x = t + 1.

D.

A. B.

2

A. B.

Fri

0

Analytic Geometry Rectangular Coordinates System Distance Between Two Points Slope of Line Angle Between Two Lines Distance Between a Point & a Line Distance Between 2 Parallel Lines Division of Line Segment Area of Polygons by Coordinates Equations of Lines Conic Sections or Conics General Equation of Conics Geometric Properties of Conics Equations of Circles

Sat

ANSWER KEY 451.B 452.C 453.D 454.C 455.C 456. B 457.A 458. D 459. B 460. D 461. 8 462.C 463.8

464.A 465. D 466. B 467. 8 468. 8 469. 8 470. 8 471. D 472. A 473.A 474.C 475. 8 476. 8

477.C 490. B 478. B . 491. B 479. 8 492. D 480. 8 493.A 481. D 494.C 482.C 495. D 483. C . 496. D 484. A 497. D 485. 8 498. D 486.C 499.A 487. B 500. 8 488.C 489.C

RATING

c:J c:J c:J 0

43-50

T~pnotcher

33-42 Passer 25-32 Conditional 0-25 Failed

IfFAILED, repeat the test.

262 l 00 l Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas

ml

ml

4

0


Let Xm and Ym. the coordinates of the midpoint

17 Xm-

'----y---__)

x,

Iii

P2(-3,5)

+X.2 __

2

2.1

7= x+9

The linear distance from -4 to 17 is equal to 21.

2 x=5

P2(-2,5)

(1 ,~6)

(5,-6)

Ym = Y1 + Y2

2 d, =d2 2 +-(-y-+6) 2 = Jr-(x---5)7" 2 -+(-y-+-6)7' 2 J'"(x---1)::-(x-1) + ~ =(x-5) 2

2

+~

'JR.- 2x + 1 ='JR. -1 Ox+ 25 8x=24 x=3 P,(4,-3)

3= 6+y P,(2,-5)

2 y=O

•

X

2

4

6 8

y= Y1r2 +y 2 r1

+(5+3) 2

•

2

2

2

2

(3-5) +.(y+6)

Let: Xm and Ym. the coordinates of the midpoint Xm

(3/5) + (2/5)

•

P,(-1,4)

= x, +x2 2

10y::: -30

P2 (x 2 ,y 2 )=(8,7)

• ::: 6+ X 2

y =-3

2

Using distance formula:

Thus the pointis at (3, -3)

-x, )2 +(y2 -y, )2

mJ

13 2 =(8-3) 2 +(7·y) 2

X =-10

Ym = y, + Y2

P,(x,6) 6

169:::25 + 49 -14y + y 2 2

2 2 =(x-6) +(y+1) 2 2 =(3-6) +(y+1)

4+ ) \ +12y+36=9+ ) \ +2y+1

P1 (x 1 , y 1 ) = (3, y)

O=y

P,(6,-7)

(x-5) +(y+6)

d=~(x2

r1 + r2

Substitute x = 3:

d=10

= -5(2/5)+5(3/5)

y =1

2 2 2 2 (x-5) +(y+6) =(x-6) +(y+1)

d=~(x2 -x, )2 +(y2 -Y, )2

(3/5)+(2/5)

-1

2 = J,-(x---6--=)2:-+-(-y-+-1)""" 2 J,-(x---5--=)2:-+_(_y-+-6)-=-

Using distance formula:

2(2/5) + ( -3)(3/5)

-~~~~~

r1 +r2 X=

d2 ::: d3

d=J(-2~4} 2

x 1 r2 + x2r1

4

-4=

-14y··95

O=(y-19)(y+5)

I

y=19or

y =·-5

I

o

2 -7 + y -

2 I

I

I

2

4

6

10.-1

1---..

P2(9,y)

y::: -1

x-

r1 +r2 2

::: -'-(--'-1)("'--3-'-)_+ d _:X2:..:(d-'-) d+3d

-3'li.+'Cl_x 2 2=--4'Cl_ X1

~

x1 r2 + x2 r1

=11

.

'

.

264 . 100 I Solved Problems in Engirieering Math:emaacs (a" Edition) by Ttong~ Rojas

•

Substitute:

y= Y1r2. + Y2r1 r1 +r2

3 6-0 ----

4(3d)+y2(d) -2= . d+3d

4

3x+2y=6

x-0

2

x=8

4)(

m1 =-3/2 x+y=6 y=-x+6 m 2 =-1

a

Y2 =-20

•

Let 9 = angle of inclination _ ry
'Y

Let: 9 = angle ~tween the two lines

Note: P1 (-5,3); P2 (10,7)

9 =tan·1

Since collinear,

m1 = m2.

m 1 =m 2

2-1

3-2

b-a

c-b

1

1

--=---=-b-a c-b c-b=b-a

•

X2

-1- (-3/2) ) 1 + (-3/2)(-1)

•

9=11.3099" or 11" 18' 35"

Ill

2x+y-8=0 y = -2x +8 m1 =-2

Y=~x+2

m1 =3

1 4 y=--x-.3 3 m 2 = ·113

5

d (-)1

Note: The negative sign only denotes that the point is below',the line.

•

Let: 9 = angle between the two lines

-m, )

9=tan·1[ m2 1+m1m 2

Let: 9 = angle between the two line£!

-m~

e =tan·

1

whenx~O.y = -J.JJ = • 2.0

4 3 ( -----) 1+(4)(3)

'9=4.398'" 9=tan·1 [ -113-(-2) )

-x 1

=

t-13 2 +4 2

y=4x+9 m2 =4

x+3y+4 =0

9=tan·1[ m2 ) 1+m,m2

m=Y2·Y1

[ 1+m m 1 2

.

9 = 14.93'

x2 -x1

m2 -m 1 )

9=tan -1 (

9 = tan· 1 ( 4/15)

m=Y2·Y1

d= A(xt)+B(y 1 )+C __:..±....:..J.:.,:A=:2::::+::8=;:2:-d= 3(0)+4(0)-5 = -5

Y2 -y, 7-3 m=--=--=4/15 x 2 -x 1 10+5 tan9=m

Given two points, the slope of the line is,

3

Y=~-x+3

3x=24

-2= 12M_+ M,y2

Day ll -Analytic Geometry (Points, Lines & Circles) 265

when y=O, x

+d,

=A(x 1)+B(yd+C. .JA2 +82

Ill

1+ (-~)(-1/3) 9=45

P1 (0,0)-+x 1 =Oandy 1 =0 P2 (x,6)-+X 2 =xandy 2 =6 m=3/4

Givenline:3x +4y-5 =0 A =3;B=4;C::: -5 Given point: (0,0) XI "'0;y1 tt:Q

J

2=

3(x 1)+4(y 1 )+(-6)

. ./32 +42 10=3x1 +4y 1 -6 16=3x1 +4y 1· y1

:t

4-0. 75x 1 -+ Eq.1

266 . .I 001 Solved Problems in Engineering Mathematics (2"d Edition) by Tiong & Rojas 2x 1 +3y 1 +4 =0 --)Eq.2

!Q

[4x- 3y + s =

~


d

2x 1 +3y 1 +4=0 2x 1 +3(4-0.75x 1 )+4;::0

•

Given line: 4x + 3y + 12 = 0

whenx~O.y ~1.67

· when y-o. x "" -/.25

Day II -:-Analytic Geometry (Points, Lines & Circles) 267

P(2,1)

A= 4; 8 =3; C = 12 Given point: (1,3)

Given line 1: 3x + y -12 = 0

A=3;8=1;C 1 =-12 Given line 2 : 3x + y - 4 =0

x 1 =1;y 1 =3

-0.'25x 1 =-16

•

A= 3; B = 1; C 2 = -4

x 1 =64 y 1 =4-0.75(64) Y1 = -44 Thus, the first point is at (64,- 44) A(x 1)+8(y 1)+C -d1 = -.J-,.A=::=2=+=8::.=2=-_2 = 3(xd+4(yd+(-6) -)32 +42 -10=3x 1 +4y 1 -6 -4=3x 1 +4y 1 Y1 =-1-0.75x 1 -+Eq.3 Substitute Eq.3 in Eq.2: 2x 1 +3y 1 +4=0

_j A2

../16

-~42 +(-3)2

a

10 d=-

-5

•

2

A(x 1)+8(y 1)+C d =--'--';::===='==-±-JA2 +82

Given line : x - y = 0

d=

Given point: (5,1 0)·

+J42 +32 25 d=5 d = 5 units

x;=5;y 1 =10

10 8

y 1 =-1-0.75(4)

6

Y1 = -4

when x "(), y ·0 whenx~J.y of

Thus, the second point is (4, -4).

Given line : 4x- 3y + 5 = 0 A =4; 8 =3; C = 5 Given point: (2,1) x1

= 2; Y1 = 1

Using point slope form:

y- y 1 = m(x- x 1 ) 4

y- 4 = -(x- 6)--+ Equation of line 3

•

Thus, at x = 0; y = - 4 Using the distance formula:

Given line 1 : 4x- 3y -12 = 0 A=4;8=-3;C 1 =-12

A(x 1)+8(y 1)+C

d = ---;::==:::=:~ ±-JA2 +82 d = 1(5)+(-1)(10)+0

-~f

+(-1)

-5

d=-

-.fi

2

(0,-y)

4(1) + 3(3) + 12

__:_'--========-

A=1;8=-1;C=O

2x 1 +3(-1-0.75x 1 )+4=0

ml

4

d = (-) 2 units

x1 =4

J32;"12

d=--units

--'--7::=====-

-0.25x 1 = -1

=--4_-(_-1_2_)

+82

8

4(2)+(-3)(1)+5

d-

c 2 -C 1

d=

A(x 1)+8(y 1)+C d = _..:__~===-- ±-JA2 +82

d=~(x2

2 -x1 ) +(y2 -y1 )2

Given line 2 : 4x - 3y + 8 = 0

d=~(6-0) 2

A =4; 8 =-3; C 2 = 8

d = 10 units

d=

C 2 -C 1 -JA2+B2

20 d=-. 5 d = 4 units

8-(-12)

= --;=:====

~42+(-3)2

+(4+4) 2

•

y-x = 5 y=x+5

From the point-slope form:. y = mx + b

d = 3.54 units lly inspection, the slope (m) is equal to 1


•

l

1 [(1(-3)+3(-3)+5(1))

A=;- -(1(3)+(-3)(5)+(-3)(1))

3x:t-2y+1=0 2y = -3x-1

3

1

From the point-slope form: y

=mx + b

•

A=~l

x,

2 Yt

x2 Y2

P2(4,0) x3 Y3

x, Yt

I

P2(4,5)

2 0 0 3 0

A =~[·(-2(0)+4(3J+3(0))

l

2 -(0(4)+0(3)+3(-2)) A

=9 square units

•

11 x 1 x 2 A=2 Yt. Y2

x3 Y3

•

Y2- Yt = -( x-x 1 )

x2 -x 1

P2(3,-3) P3(5,-3) x3 x1 Y3 Yt

A-111 3 5 .1, 21-3-31

,.

2+3 y.+3=-(x-1) -4-1 y+3=-1(x-1) y+3 =-x+1 y =-x-2

x2 - x1

5-3 y-3=--(x+2) -3+2 2 y-3 =-(X +2) -1

-y+3=2x+4 2x+y+1=0

~

....

(4, 1)

" .

A=; -(1(4)+5(0)+8(-3)+4(1))

Using two point form:

x2 Y2

x1 Yt

A = 25 square units

Y2- Yt y-y 1 =--(x-x 1 )

x2 - x1

_1,140-311 A-215841

y-y 1

_ 11 x 1 A-2 Yt

x4 Y4

P1 (1,-3)-+x 1 =1andy 1 =-3 P2 (-4,2)-+x 2 =-4andy 2 =2

Pt(1, 1)

••

Using two point form:

1-4 y-4 =-(x-1) 4-1 y- 4 = -1 (X -1) y-4 = -x+1 x+y=5

1[(1{5)+4(8)+0(4)+(-3)(1) )]

. A-_1,-243-21

Geometry (Points, Lines & Circles} 269

Y2 - Yt y-y 1 =--(x-xd

Pa(3,3)

P1(-2,0)

~ytic

P1 (-2,3)-+x 1 =-2andy 1 =3 · P2 (-3,5)-+x 2 =-3andy 2 =5

Using two-point form:

Pa(0,8)

By inspection, the slope is equal to -3/2.

•

•

P1 (1,4)-+x 1 =1andy 1 =4 P2 (4,1)-+x 2 =4andy 2 =1

A = 4 square units

y=--x-2 2

Day 11 -

Substitute y intercept:

•

x-interc~pt ~~

a

b

=Oto solve for the x-

x+y=5 x+0=5 x=5

•

r

[Y-~

By inspection, a

=4 and b =6

Using the intercept form:

As given: m

=3; b =1


X y -+-=1 a b X

y

-+-=1 y=mx+b y=3x+1 ~x-y+1=0

4 -6 6x-4y=24 3x-2y=12



•

7 3 y=--x+-

P1 (-5/2,-9/2) - H 1 = -5/2 and y 1

P2 (x,6)~x 2

= -9/2

=xandy 2 =6

9=45°

2

2

1 Since perpendicular, m 2 = - -

m,

By inspection, m, = -3/2

2

2

1 1 m 2 =--=--=1 m1 (-1)

B

B

P3 (x,-7) ~ x 3 = x and y 3 = -7

2x-By+2=0 tan9= Y2 -y 1 X2

X

-(-5/2)

1 = 21/2 x+5/2 x=8

•

P(O,Q)~x 1

y=-x+-

-x 1

tan45° = 6-(-9/2)

=Oandy 1 =0

m=6

Using point-slope fonn: y-y 1 =m(x-x 1 ) y -0 = 6(x- 0) · y=6x

P4 (8,7)~x 4 =8andy 4 =7


m2

Using point-slope form:

2

1= 7+1

B -3/2 B=3

8-x x=-6

•

Using slope-intercept form: y = mx + b 2x-3y+2 = 0 3y =2x+2

2

2

y =-x+3 3

)

y-1=5(x-3) y -1 = 5x -15

x4 -x3

m,

By inspection, m,

5x-y = 14

•

•

P1 (5,0)~x 1 =5andy 1 =0

Using slope-intercept form: y = mx + b

P2 (-7,3) ~ x 2 = -7 and y 2 = 3

x-y-2=0 y =x-2

m,

=Y2 -y 1 = 3-0 x2 - x1

By inspection, m = 1

mz = m,

=-~

-7-5

m,

= 2/3

b

X y -+-=1 -2 4 -2x + y = 4

y -2x-4 = 0

Using slope-intercept form: y = mx + b


Using point-slope fonn:

m,

- -1- - -1m2 -

m2

m1 =-3/2

•

2/3

)

y-0=1(x-4)

y =x-4

•

Using slope-intercept form: y = nix + b

P1 (-5,2)~x,-=-5andy 1 =2 P2 (1,-4)~x 2

y-y 1 =m(x-x 1

=1andy 2 =-4

X +5y

+5 = 0

1 5 y =----X--· 5 5

4

1 Since perpendicular: m 2 = - -

P(4,0), thus x1 = 4 and y, = 0

X y -+-=1

•

m,

1 1 m2 =--=---=5 m1 (-1/5)

y-y 1 =m(x-x 1

= Y4 -y3

Since parallel:

Using the intercept fonn:

a

1 Since perpendicular: m 2 = - -

P(3, 1), thus x; = 3
By inspectiQn, niz = 2/B

y -6x =0

•

By inspection, m 1 = -1/5

Y2 -y, -4-2 m1 = - - - = - - = - 1 x2 - x1 1-(-5)

3x +2y -7 = 0

1 1 m 2 =--=---=4 m1 (-1/4)

Dayll-.AnlllYlic GeometrY
272 100 1--Solved Problema in Engineering Mathematics (2"ll Edition) by 'l'iong &: Rojas

x, =- 3 andy,= -5

Solving for the midpoint of P, and P2:

P(-3,-5), thus

x 1 -x 2 5-7 x=---.=--=-1


2

2

y-y 1 =m(x-x 1 ) 7 y.+5=--(x+3) 2 2y +10 = -7x-21 7x+2y+31=0

_y,-y2 0+3 3 y---=-=-

2

2

2


•

y-y 1 =m(x-x 1 ) y-312 4(x +1) 2y-3=8x+8 8x-2y+11=0

=

•

P1 (3,-2)-.x 1 =3andy 1 =-2

P2 (-7,6)--+x 2 =-7andy 2 =6

Using two poi.ntform:

Using slope-intercept form: y

=mx + b

Y2 -y, y-y 1 =--(x-x1 ) x 2 -x 1 6+2 y+i=-·-(x-3) -7-3

3x-2y+6 =0

3

y=-x+3 2

8

=312 Since parallel: m2 =m, By inspection. m,

From the choices, the answer is choice D.

•

15x-10y-9=0

3

9

2

10

y=-x--

1

a =_;b

By inspection, m2 =.312

•

y+2=-(x-3) . -10 -10y-20=8x-24 8x+10y-4=0 4x+5y-2=0

2 b=2a

Using slope-intercept form: y 7x+2y-4 =0

7

4

2

2

y=--x+-

=

By inspection, m, -712 Since parallel:

mz • m,

=mx + b

Using the intercept form: X

-2

I

·I

•

6

-+-=1 a 2a -4+6 --=1 2a a=1 b=2

Substitute:

- 11 x1 A--

~.!=1· a b X y

2

-+-=1

1 2 4

2x+y=2 2x+r-2=o

Substitute: x =- 2, y = 6 and b = 2a:

-2 6 -+-=1 a 2a

-2

2 4 3

x1 Y1

I

x 21 y 4

(2(3)-2y.f.4x) ]

=[ -(4(-2)+3x+2y) 4 =6 -2y +4x+8-3x-2y

•

O=x-4y+10

x=2+t t •x-2 --+Eq.1 y =5-3t--+Eq.2

Substitute Eq.1 in Eq.2: y=5-3(x-2) y=5-3x+6 y=-3x+11

A

=

By inspection, m -3

•

y =4t+6 --+Eq.1 x=t+1 t • x -1--+ Eq.2


x2 x3 x1

2 Y1 Y2

_ 1,-2 2--

Y3

2 -2,

Y1

I

X

(-2(4)+2y+3x)] 4= [ · -(3(2)+4x-2y) 4 =-8+2y+3x -6-4x+2y O=-x+4y-16 O=x-4y-18

Suggested answer Is choice D.

y •4(x-1)+6 y •4x-4+6 y•4x+2 By lnapectlon.

=.!1 x1

2 '3 .4 y 3

y

-+-=1 a b

=_:12

x3 Y3

x2 Y2

2 Y1

m• 4

27 4 100 1 Solved Problems in Engineering Mathematics (2nd· Edition) by Tiong & Rojas

•

X y -+-=1

a

Given line--+ 2x-3y +6 =0

Atx

=0:

2x-3y+6=0 2(0)-3y +6·= 0 y=2

b

y -+-=1 -2 1 -x +2y = 2 x-2y+2 = 0 X

Topics

0

Mon

0

0 0 0

0

~

[I]

0

At y =0:

Tue

2x-3y+6=0 2x-3(0)+6 = 0 X =-3 Thus the x-intercept (a) is 3 units and the y-intercept (b) is 2 units.

0

By inspection, the x and y intercepts are: a= 1; b = -2

y -+-=1

b

Notes

X y -+-=1

2

2

A = 3 sq. units

•

By inspection, the x and y intercepts are: a= -2; b = 1 Using the intercept form:

1

Fri

Solutions

X

1 1 A =-ab =-(3)(2)

Thu

Problems

Using the intercept form:

a

VVed

Analytic Geometry Rectangular Coordinates System Distance Between Two Points Slope of Line Angle Between Two Lines Distance Between a Point & a Line Distance Between 2 Parallel Lines Division of Line Segment Area of Polygons by Coordinates Equations of Lines Conic Sections or Conics General Equation of Conics Geometric Properties of Conics Equations of Circles ·

Sat

-2

2x-y=2 2x-y-2=0 ... ., '<

* .,., . . ·, .. "''

0

,,

'

,.

~

• • • ,.

~

... "' "'

,, '

"

i ·' < ~ "' .,

'\'"'. " .. ·•' .., ' ,, ~ '

·}.·

T T

' .., ., <. '

" ·•' " "· '

<. "'' "'

~

~ , • • • ·""·'

"·,',

276· 100 1 Sol'ved Problems in Engineering Mathematics {2nd Edition) by Tiong & Rojas

-·~~

.<"'

,,.

~,,~<

'"~"·~·~·"·.>"<·

Topics

D D [I] D D D D D D ~ Mon

Tue

,,.,·_.

'

.,. «

~.

{ '• o. '

y

«

,, ""

4

'

,.,,

•

Theory

Wed

Problems

Thu

Solutions

Fri

Notes

Sat

1, General equations:

What is a Parabola? A parabola is a locus of a point which moves so that it is always equidistant to a fix,ed point called focus and it a fixed straight line called directrix. y l

~

Parabolas - General & Standard Equations - Eccentricity & Latus Rectum Ellipses - General & Standard Equations - Major & Minor Axis - Eccentricity & Latus Rectum Hyperbolas - General & Standard Equations - Transverse & Conjugate Axis - Eccentricity & Latus Rectum Polar Coordinates -Radius Vector - Polar Angle Relation Between Rectangular and Polar Coordinates

A. Axis parallel to the y-axis:

y

----t----t------~x

•

directrix

-

X

Ax2 + Ox + Ey + F

=0

B. Axis parallel to the x-axis:

<"!"'""'"'

y

"<····

··~

I

whQre

a

= distance from vertex V to focus F

ci I

di~;(;IIICi' fiOIII fHlillllo diWc!riX fo• • If d1· .I dill "

. ----x


=

Opens down

C/ + Ox + Ey + F 0

y

Day 12 -An(llytic Geometry (Parabola, Ellipse & Hyperbola) 279 When the equation given is a general equation rather than standard equation, the vertex V(h,k) of the parabola and its focal h;mgth or focal radius "a" can be calculated by converting the general equation to standard using the process known as completing the square.

B. Axis parallel to the y-axis: Opens upward

2. Standard equations:

--------~~~~--~

X

y

Ve1iex (V) at origin (0,0)

A. Axis along x-axis:

The following formulas can be obtained

Opens to the rrght

. x2 =-4ay

y

~~------~r----.

X

Vertex (V) at (h,k) 2

2

h = _·E_-_4_C_F

A. Axis parallel to the x-axis:

y =4ax, ---+------·~

For axis horizontal: C/ + Dx +

k = -E

2C

4CD

(x.., h)2

Opens to the right

X

·

Ey + F = 0 -D a= 4G

=4a(y - k) A~ + Dx + Ey + F = 0

For axis vertical:

y Opens downward

h= -0 2A

y

Opens to the left

k::

..

0

2

""4AF

4AE

-E

a== 4A

y X

What is an Ellipse? ---4---------+----~

y 2 "'~4ax ------~~---·~

X

X

(y-k) 2 =4a(x-h)

Opens to the ~eft

(x- h)2

An ellipse is a locus of a point which moves so that the sum of its distance to the fixed points (foci) is constant and is equal to the length of the major axis (2a).

=-4a(y- k)

P(x~

y

I

._

B. Axis along y-axis: The eccentricity of the parabol;. is the ratio of the distance to the focus to the distance to the directrix.

Opens up

y X

v

v

e=.!.

d

a

Since f = d, then:

(y- k)2 ""-4a(x- h)

x~ ':'
D

e=1 The latus rectum of a parabola is a line that passes through the focus and perpendicular to the axis of the conic.

LR=4a

directrices

1. General Equation:

Ax2 +Cy2 +Ox +Ey +F =0

281 1001 Solved Problems in Engineering Mathematics (2nci Edition) by Tiong & Rojas

=

Nole: d1 + d2 2a. The major axis the distance from V1 to V2.

=2a, is

Major axis is vertical: y

Day 12 -Analytic Geometzy(P~ill:>oJ~Ellipse &_!ln>_er:l:>ola) 281

(x-ht +(y-kf = 1 b2 a2

Since a < 0 and c < a, then the eccentricity of an ellipse is always less than 1.

When the point is located along the minor axis as shown in the following figure:.

e <1' The eccentricity of .an ellipse is the ratio of the distance to the focus to the distance to the directrix.

P(x,y) X

The latus rectum of an ellipse is a line that passes through the focus and perpendicular to the axis of the conic.

~[

LJ·

x2·

y2

z+z=1 b a

2w2

2

b +C~ =a

2

Center, C at (h,k)

.Et-_J:.d

Major axis is horizontal:

y

Whe'l the point P(x, y) is the minor axis:

will •a• is always greater than "b".

=

If D & E 0, center is at the origin (0,0). If either 0 or E, of both 0 & E "' 0, the r..enter is at (h,k).

2. Standard Equations:

~ri=~

a

directrices

c

The relationship between a, b and c is

.

----~r---~--~-----------x

where: a = semi-major axis b = semi-minor axis When the equatiOn given is a general equation rather·tl'lan standard equation, the center (h,k) of an ellipse and its focal length "c" can be calculated by converting the· general equation to standard using the process known as completing the square. · · The following formulas can be obtained:

Ce!lter, C at (0,0) Mljor axis is horizontal:

General equation:

(x~bf + (Y-:f =1 a2

y

c

b

Ax'- + Cy2 + Ox + Ey + F =0

a D

Major axis is vertical: --~----------.----------r---x

directrices

h::~

2A

k=

-a: 2G

c=.Ja2 -..tr

y

a

e=Q Iff= c, then eccentricity of an ellipse, e is 2 x2 -+y a2 ·b2 = 1

. c·

X

e=-a

What is a Hyperbola? A hyperbola is a locus of a point which moves so that the difference of the distanGes to the fixed points (foci) is constant and is equal to the length of the transverse axis (2a) .

Day 12 -Analytic Geometry (Parabola, Ellipse & Hyperbola) 283

282 100 1 .Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas directrices

asymptot'

Transverse axis vertical:

Transverse axis

2 X - 2 =1 -

i_

Transverse axis

a2 b

y

Transverse axis vertical: I

•

• .,

)I('

I Y6 •

• ...

X

Transverse axis

y

--· ··-~-·- _... _ :

)I{

a• i

Conjugate axis

~-7'

Transverse axis

Ax

2

-

=

Gy2 +OX,+ Ey + F 0'

Transverse axis is the axis that passes through the foci, vertices and the center of the hyperbola while the conjugate axis is the one that is perpendicular to the transverse axis.

Length of the transverse axis

= 2a

Length of the conjugate axis

= 2b = 2.../A

Length of the conjugate axis = 2b = 2.JC

2

Y

. 2 X

a2-b2='1

where: A and C are the numerical coefficients (absolute value) of x2 and y2 , respectively.

The eccentricity of a hyperbola is the ratio of the distance to the focus to the distance to the directrix. y

Center, C is at (h,k) Also, the relationship between a, b and c is Transverse axis horizontal:

Transverse

b I

•

•

VI

"*'

I Y6 •

~

2. Standard equations:

.--

Center, C at (0,0) .· Transverse axis horizontal:

a c

y

where: A and C are the numerical coefficients (absolute value) of x2 and y2, respectively.

Transverse axis

Also, the relationship between a, band cis B. Transverse axis -vertical:

---t-----------+---~x

= 2.../A

y

= 2a = 2.JC

X

(y-k)2- (x-h)2 =1 a2 b2

• E!-2 + b,~: =42 Length of the transverse axis

~

··f<· ··~···.-r'\1\ · A.:~ + ·o"~+ • eY:+·· F·=v · ". • .~Y

1. General equations:

A Transverse axis - horizontal

i

--+-------~~----~x

b I

•

•

VI

)I("

I 116 •

'

.. X (x-h)2 _(y-k)2 =1 a2 b2

. Conjugate axis

I a . ·.., ~

c e=a

or

e=oa

Since a > c and D > a, then the eccentricity of a hyperbola is always greater than 1.

e >1

X

Day 12 - Analyti(: QE!Qrnetry (Parabola, Ellipse & Hyperbola) 28J;

Z84 . 100 1 Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas The latus rectum of an ellipse is a line that passes through the focus and perpendicular to the axis of the conic.

Proceed to the next page for your 12th test. Detach and use the answer sheet provided at the last part of this book. Use pencil number 2 in shading your answer. GOODLUCKJ

(r.e) (x,y)

y X

QI:ribia: Did you know that... the oldest known example of numeration using place value is not the Roman nor the Arabic numeration but the Babylonian or Mesopotamians' sexagesimal system of numeration which dates back to the 2nd millennium B. C. ! This system of numeration still survives today (i.e. 1 hour = 60 minutes and 1 minute = 60 seconds, and 1 degree 60 minutes)

b Polar angle is sometimes called the vectoral angle, the argument, the amplitude, or the azimuth of a point.

~

=

,;t.R=2~2-

I ....po e

. 'a'

where: a = semi-major axis b = semi-minor axis When .the equation given is a general equation rather than standard equation, the center (h,k) of a hyperbola can be calculated by converting the general equation to standard using the process known as completing the square.

r . polar ax1s

~uott:

•

"Every new body of discovery is mathematical in form, because there is no other guidance we can have. •

What is the relation between polar coordinates and rectangular coordinates?

- Charles DarWin

Relationship between polar coordinates and rectangular coordinates: ·

x:rC0$6

y =rt~ne

The following formulas can be obtained:

h= -Q

2A

k=

-e

2t

What are Polar Coordinates? Polar coordinates (r,e) refers to the coordinates of a point in a system of coordinates where the position of a point is determined by the length of ray segment (the radius vector) from a fixed origin (the pole) and the angle (the polar angle) the ray (the. vector) makes with a fixed line (the polar axis).

r·=Jx*~y2

I

~

Day 12- Analytic Geometry (Parabola, Ellipse & Hyperbola) 287 506: CE Board May 1:996 How far from the y-axis is the center of the 2 curve 2~ + 2y + 10 x - 6y - 55 0?

=

Topics

D

Parabolas - General & Standard Equations - Eccentricity & Latus Rectum Ellipses - General & Standard Equations -Major & Minor Axes - Eccentricity & Latus Rectum Hyperbola - General & Standard Equations -Transverse & Conjugate Axes - Eccentricity & Latus Rectum Polar Coordinates - Radius Vector - Polar Angle Relation Between Rectangular and Polar Coordinates

Man

D Tue

D D [I] D D D Theory

Problems

Wed

Thu

Solutions

Fri

D

~

Notes

Sat

SOJ:: CE Board May 1:995 What is the radius of the circle x 2 + y2 - 6y = 0?

A. B.

2 3

5

A. B. C. D.

502: CE Board November 1:995 What are the coordinates of the center of the curve~+/- 2x- 4y- 31 = 0?

A.

(-1 ,-1)

B.

(-2.-2) (1,2) (2,1)

C. D.

so~: A circle whose equation is ~ +

4x + 6y -'-23 = 0 has its center at:

A. B.

(2,3) (3,2)

D.

(-3,2) (-2,-3)

504: ME Board April 1:998 What is the radius of a circle with the ff. equation: ~- 6x + y2 - 4y- 12 = 0

c. 4 D.

C.

3.46

7 5 6

SOS: ECE Board April 1:998 The diameter of a circle described by 9~ + 9y2 = 16 is

A. y2

+

C:

4/3 16/9 8/3

D.

4

B.

A.

-2.5

B.

-3.0 -2.75 -3.25

C. D.

l-

B. C.

D.

l

C. D.

A.

(X - 2) + (y - 2)2 ::: 5

B.

(x - 2) 2 + (y + 2) 2 = 25 2 2 (X - 2) + (y + 2) = 5 2 2 (x- 2) + (y- 2) = 25

C. D.

A. B. C. D.

A.

A. B. C. D.

2

x + / + 8x - 1Oy - 12 = 0 ~+ +ax- 10y + 12 = 0 ~ + / + 8x + 10y- 12::: 0 ~+l-ax+ 10y + 12 o

l

=

51:4: ECE Board April 1:998 Find the value of k for which the equation ~ + y2 + 4x - 2y - k = 0 represents a point circle.

2.1 2.3 2.5 2.7

509: ME Board Oetober J:996 2 The equation x + y2 - 4x + 2y - 20 describes:

2

SJ:~r Find the equation of the circle with the center at (-4,-5} and tangent to the line 2x + 7y - 10 = 0.

7.07 7.77 8.07 7.87

508: CE Board November 1:99~ The shortest distance from A (3,8) to the circle~+/+ 4x- 6y = 12 is equal to

A. B.

25 1t 27 rr

SJ:2r Determine the equation of the circle whose radius is 5, center on the line x = 2 and tangent to the line 3x - 4y + 11 = 0.

507: What is the distance between the + 2x + 4y - 3 centers of the circles x2 + = o and x 2 + ax - 6y + 7 = 0?

A.

C. D.

=0

A circle of radius 5 centered at the origin. An ellipse centered at (2, -1 ). A sphere centered at the origin. A circle of radius 5 centered.at (2,-1 ).

5

B.

6 -6 D.. -5

C.

515: ECE Board April 1:999 3x2 + 2x- 5y + 7 = 0. Determine the curve. A. B.

Parabola Ellipse Circle Hyperbola

uo: EE Board April 1:997.

C. D.

The center of a circle is at (1, 1) and one point on its circumference is (-1 ,.-3). Find the other end of the diameter through (-1 ,3).

51:6: CE Board May 1:99~, CE Board November 1:99~, ECE Board April

A. B. C. D.

(2,4) (3,5) (3,6) (1,3)

sur Find the area (in square units) of the circle whose equation is~+ y2 = 6x- By. A

R

20 rr 7/rr

1:994 The focus of the parabola y2 = 4x is at

A. B. C. D.

(4,0) (0,4) (3,0) (0,3)

51:7: CE Board November 1:994 Where is the vertex of the parabola ~ 4(y- 2)?

=

__________ pay .~.:~nalylic Geornetry__(~ar~~~~Eilips~!_f:!.'[f~f::~E-~!

A

a

~

n

~~ ~~ ~~

SZ:JI ECE Board Aprll1998

8,

Find the equation of the axis of symmetry of the function y =2x2 -7x + 5.

C.

D.

A.

Find the e~uation of the directrix of the parabola y = 16x.

B. C. D.

7x + 4 = 0 4x + 7 0 4x-7 =0 x-2 = 0

~

n

What is the area enclosed by the curve 9x2 + 25y 2 - 225 ::: 0?

directrix y =2. Find its equation.

x=2 x=~

x=4

A.

x=4

B. C. D.

5191 Given the equation of a parabola 3x + 2y2 - 4y + 7 0. Locate its vertex.

=

A. B. C. D.

(5/3, 1) (5/3, -1) (-5/3, -1) (-5/3, 1)

In the equation y curve facing?

x2 + 12y -14x + 61 = 0 x2- 14y + 1·2x + 61 0 x2- 12x + 14y + 61 0 None of the above

=

= -x2 + x + 1, where is the

A. B. C. D.

5Zftl ECE Board Noveaaber 1997 Compute the focal length and the length of the latus rectum of the parabola y2 + 8x . 6y + 25 0. 2,8

4, 16

D.

1, 4

=

5

=

D.

c.

A.

Find the location of the focus of the parabola y2 + 4x- 4y- 8 0.

c.

I

16,64

B.

x=-3 x= 3

D.

y=3

A. B.

C. D.

A B.

=-8(x-

C. D.

1). What is the equation of its directrix?

.J5

(2.5, -2) (3,1) (2,2) (-2.5,-2)

n.3

2

2x - 4y 2 = 5 4x 2 + 3y 2 = 12 2 2x + =3 2 x + 2y· = 4

st

3 4 5 6

(3,-4) (3,4) (4,-3) (3,5)

c. y =-3

533: EE Board October 1.997 Find the major axis of the ellipse x2 + 2x- By+ 1 = 0.

SZ8& ME Board October 199'7

A.

B.

The general equation of a eonic section is given by the following equation: A>t! + Bxy + Cy2 + Ox + Ey + F 0. A curve maybe identified as an ellipse by which ofthe following cOndition&?

=

A.

B2 -4AC <0

I 11

2 10

c

4

D.

6

2 3

D.

4 5

lt99~

x2

y2

64

16

.

53$1 An ellipse with an eccentrir..:ity of 0.6!:i and has one of its foci 2 units from the center The length of the latus rectum is nearest to

A.

3.5 units 3.8 units 4.2 units 3.2 units

B. C. D.

S361 An earth satellite has an apogee of 40,000 km and a perigee of 6,600 km. Assuming the radius of the earth ns 6,4CO km, what will be the eccentricity of the elliptical path described by the satellite with the center of the earth at one of the foci?

A B.

0.49

0.46

D.

0.52 0.56

c

S3jt: The equation 25x2 + 16y2 - 150 x + 128 y + 81 =0 has its center at

sa?• Given a parabola ( y- 2 )2

5ZZI EE Board October 1997

A. B.

I

'I

=

B.

20

~.8

D.

B.

A

c.

S3l.: The lengths of the major and minor axes of an ellipse are 10m and 8 m, respectively. Find the distance between the foci.

=

A

B.

c.

A. B. C.

y2 - 11y + 11x- 60 = 0 y2 -11y + 14x- 60 = 0 y2 -14y + 11x + 60 0 None of the above

5z:ll CE Board ala,- 1995

C. D.

50.2

D.

What is the length of the length of the latus rectum of the curve x2 20y?

.J20

~1

B.

5301 ECE Board Aprill.998 Point P(x,y) moves with a distance from point (0, 1) one-half of its distance from line y = 4. The equation of its locus is

=

x-axis, vertex at (-1, 7) and one end ofthe latus rectum at (-15/4, 3/2). Find its equation.

Upward Facing left Facing right Downward

A.

A

5Z5r A parabola has its axis parallel to the

5ZOI ME Board Aprll1997

A. B. C. D.

.

S:l9: CE Board November 1.994

=

5Z4& A parabola has its focus at (7. 4) and

a

S34: CE Board May

The length of the latus rectum for \lh': ellipse ··- •·--- , 1 IS equal to.

~~

518: ECE Board Aprll1994, ECE Board Aprll1999 ·

A

2 8 - 4AC = 0 8 2 - 4AC > 0 8 2 - 4AC = 1

S~7:

A

4/ ~

ECE Board Aprill998

The major axis of the elliptical path in which the earth moves around the sun is approximately 186,000,000 miles and the eccentricity of the ellipse is 1/60. Determine the apogee of the earth.

B.

C. D.

93,000,000 miles 91,450,000 miles 94,335, iOO miles 94,550,000 miles

S38: CE Board November :199% The earth's orb1t is an ellipse with the sun at one of the foci. If the farthest distance of the sun from the earth is 105 . 50 million km and the nearest distance of the sun from the earth is 78.25 million km, find the eccentricity of the ellipse.

290· A. B.

c. D.

roo l Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas 0.15 0.25 0.35 0.45

539: An ellipse with center at the origin has a length of major axis 20 units. If the distance from center of ellipse to its focus is 5, what is the equation of its directrix? A.

B·:

X= 18 X= 20

C. D.

X= X=

15 16

540: What is the length of the latus rectum of 4J2! + 9l + Bx - 32 = 0? A. B.

c. D.

2.5 2.7 2.3 2.9

541: EE Board Odober 1993 4J2! - / ='16 is the equation of a/an A. B. C. D.

parabola hyperbola circle ellipse

542: EE Board October .1993 Find the eccentricity of the curve 9J2! - 4l -36x +By= 4 A.

a

1.00 1.~

~

1.~

D.

1.ro

A.

2

B. C. D.

3 4 5

sso: ME Board April1997

545: CE Board May 1996 What is the equation of the asymptote of

xz 9

C. D.

B.

C. D.

D.

4J2!+/+16=0 4J2! + 16 = 0 ,(! - 4/- 16 = 0 2 4x = 16

l-l

D.

l

,(! + J2! + / ,(! + y2 2 x +

l

2

+z + z2 + z2 2 +z

+ 6x - 4y - 8z + 6x - 4y - 8z + 6x- 4y + 8z + 6x- 4y + Bz

= 36 =7 =6 = 36

r - 8 cos e = o r - 6 cos e = r- 12 cos e = o r- 4 cos e = o

o

5531 ME Board October 1996 What are the x and y coordinates of the focus of the conic section described by the following equation? (Angle e corresponds to a right triangle with adjacent side x, opposite side y and the hypotenuse r.) rsin 2 cos

a=

A. B. C. D.

and P2 {5,1,-4). A.

11

J11

f12

The semi-transverse axis of the hyperbola x2 Yz .

c.

o

D.

5441 EE Board. October 1994

B.

o

How far from the x-axis is the focus F of the hyperbola J2! - 2l + 4x + 4y + 4 = 0?

D.

A.

5J2! - 4y 2 + 20x + 16y - 16 = 0 5J2!- 4y2 + 20x- 16y- 16 = 5J2!-4y2 -20x+16y+16=0 5J2! + 4y2 - 20x + 16y- 16 =

12

B.

10.5

a

548: Find the distance between P1(6,-2,-3)

B.

c.

D.

B. C. D.

Find the polar equation of the circle, if its center is at (4,0) and the radius 4.

with vertices at (-4,2) and (0,2) and foci at (~5.2) and (1 ,2).

B. C.

JjQ1

10

5521 EE Board April1997

5471 Find the equation of the hyperbola

A.

A

C.

A. B. C. D.

Find the equation of the hyperbola whose asymptotes are y = ± 2x and which passes through (5/2, 3). A.

9

at (-3,2,4) and of radius 6 units is

546: EE Board A.pril1994

x + 5y - 2z = 9; 3x - 2y + z

= 3 and x + y +

A.

z = 2 is at A.

B. C.

D.

(1/4, 0) (0, rr12) (0,0) (-1/2,0)

sS4: Find the polar equation of the circle of radius 3 units and center a.t (3,0).

549: The point of intersection of the planes

B. C.

(2, 1-1) (2,0,-1) (-1,1,-1) (-1,2,1)

D.

---=11S 9 4

l

Given the polar equation r = 5 sin a. Determine the rectangular coordinates (x,y) of a point in the curve when a is 30•.

A. B.

551: The equation of a sphere with center

2x- 3y = 0 3x- 2y = 0 2x- y = 0 2x + y = 0

c.

4.5 3.4 2.7 2.1

yz 4

the hyperbola - - - = 1 ? A. B.


What is the radius of the sphere center at the origin that passes the point 8,1 ,6?

543t CE Beard November 1995

A.

Day 12 -Analytic Geometry{Parabola, Ellipse & Hyperbola) 291

r=3cose r = 3 sine r=6cose r =9 sine

(2.17, 1.25) (3.08, 1.5) (2.51, 4.12) (6,3)

Day 12 -Analytic Geometry (Parabola, ElliiJse &Hyperbola) 293

•

ED

x 2 +y 2 -6y=O

0 0 0 0 0

tvlon Tue

0 0

Theory

Problems

Solutions

0

Notes

Wed

Thu

Fri

~ Sat

Topics

·--

Parabolas - General & Standard Equations - Eccentricity & Latus Rectum Ellipses -General & Standard Equations - Major & Minor Axes - Eccentricity & Latus Rectum Hyperbola -General & Standard Equations - Transverse & Conjugate Axes - Eccentricity & Latus Rectum Polar Coordinates - Radius Vector - Polar Angle Relation Between Rectangular ahd Polar Coordinates

x2 +y 2 -6x-4y-12=0

By completing square: By completing square:

x 2 +(y-3) 2 =0+(3) 2 (x-3) 2 +(y-2) 2 =12+(3) 2 +(2) 2

x2 +(y-3)2 =(3)2

I

(x-3) 2 +(y-2) 2 =25 Standard equation with center at (h,k):

(X- h) 2 + (y- k) 2 = ,.Z

By inspection, r

(x-3) 2 +(y-2) 2 =(5) 2 Standard equation: (x- h) 2 + (y- k) 2 = r"

=3

By inspection, r = 5.

mJ

a

x 2 +y 2 -2x-4y-31=0

9x 2 +9y 2 =16


16

x2 +y2 = -

(x-1) 2 +(y-2) 2 =31+(1) 2 +(2) 2

9

(X -1 ) 2 + (y- 2) 2 = 36

2

x +y2=(;r 2

2

Stanaaro equation: (x- h) + (y- k) = r"

ANSWER KEY

501. B 516. l\ 502.C 517.8 503. D ·5"18. 0 504.C 519. D 505.C 520. 0 506.A 521. B 507.A 522.C 508.A 523.C 509. D 524.A 510. B 525. c 511 c 526.A 512. B 527. 8 5i3. c 528.A 514. D 529.A

531. D

546. D

532.A

547. A

515.A

545./\

530. B

533. c 534. c 535. A 536. D 537. D 538.A 539. B 540. B 541. B 542. A

By inspection, h = 1 and k center is at (1 ,2)

RATING

548. B 549.A 550. c 551. 8 552.A 553.A 554.C 555.A

c:J 46-55

c::J 0 0

= 2,

thus the

•

Topnotcher

33-45 Passer

0-26 Failed



(x + 2) 2 + (y + 3) 2 = 23 + (2) 2 + ( 3 )2 (x + 2) 2 + (y + 3) 2 = 36 Standard equation: (x- h) 2 + (y- k) 2 = r"

c

543. 544. B

By inspection, h = - 2 and k center it at (-2, -3).

2

4

8

3

3

By inspection, r =- and d = 2r =-

x 2 +y 2 +4x+6y-23=0

27-32 Conditional

Standard equation with center at (0,0):

x +~l=r"

= - 3, thus the

•

2x 2 +2y 2 +10x-6y-55=0 2 x +5x-3y-27.5=0

+l


(x+2.5) 2 +(y-1.5) 2 =27.5+(2.5) 2 +(1.5 )2 (x+2.5) 2 +(y-1.5) 2 =36 Standard equation: (x - h) 2 + (Y - k) 2 = r" /

'I


= - 2.5 and

By inspection, h

Day 12- Analytic Geometry (Parabola, Ellipse & Hyperbola) 295

x =d-r

k = 1.5 C2(4,3)

y-axis

X

=7.071-5

X=

2.071

x 2 +y 2 -6x+8y=O

Ill

By completing square: (x-3) 2 +(y+4) 2 =(3) 2 +(4) 2 (x-3) 2 +(y+4) 2 =25or(5) 2

x 2 +y 2 -4x+2y-20=0 x2 + y2

C,(-1,-2)

-

4x + 2y = 20

By inspection, the radius is 5.

By completing square: Note: The distance of the center of the circle from they-axis is equal to h. Thus, the answer is 2.5 unit length.

•

x 2 +y 2 +2x+4y-3=0 x 2 +y 2 +2x+4y=3

By completing square: (x+1) 2 +(y+2) 2 =3+(1) 2 +(2) 2 (X + 1 ) 2 + (y + 2)

2

=8

By inspection, the center of the first circle is at C,(-1, -2).

x

2

d = ~(x2 -x,)2 +(y2- y,)2

(x-2) 2 +(y+1) 2 =20+(2) 2 +(1)

d = ~(4 + 1)2 + (3 + 2)2

(x-2) 2 +(y+1) 2 =25or(5) 2

d=7.071

•

By inspection, the center is at C(2, -1) and the radius, r = 5 .

•

x 2 +y 2 +4x-6y=12


P,(-1 ,-3)

+y -8x-6y+7 =0 x 2 +y 2 -8x-6y=-7

Using midpoint formula:

x = x1 +x2 2 1 = -1 + x2 2


(x-4) 2 +(y-3) 2 =-7+(4) 2 +(3) 2

= 18

By inspection, the center of the second circle is at C2(4, 3). Using distance formula to solve .for the distance between c1 & c2.:

= n(5)2

. A = 25n square units

•

By inspecJion, the center is at (-2, 3) and the radius, r = 5.

(3,8)

2

A= nr 2

(x + 2) 2 + (y- 3) 2 = 12 + (2) 2 + (3) 2 (x + 2) 2 + (y- 3) 2 = 25 or (5) 2

2

(X- 4 ) 2 + (y- 3)

2

X2

Solving for distance between (-2,3) and (3,8):

=3

2

.

Try circle 1:

-d= A(x1)+B(y1)+C

±~A2 +B2 Note: d is negative since the point is below the line.

y = y, +y2 2

f

Note: There are two possible circles

1= - 3 +Y2

2

d=v(x2 -x1) +(y2-Y1)

2

d = ~(3 + 2)2 + (8- 3)2

Y2 =5

d = 7.071 Thus, the point is at (3,5) Let: x = shortest distance

't

Given line: 3x -4y + 11 = 0 A=3;B=-4;C=11 Center of circle 1: (2, y) X1

= 2; Y1 = Y

!~!!'"J.99J_.§_'?}v~d I:r.£!?~~_in Engi~!l_l9 M~pematics (2nd Edi!ionl_ey Tiong ~ Rojas Substitute:

Solving for the equation of the circle with center at (-4, -5):

~ ::l(2) + H>(y) + 11

-5·-

-~~;)2-~(-4?~-

(x-hf +(y-k) 2 =r 2 (x+4) 2 +(y+5) 2 =7.28 2

6-4y+11 -5 ·;: ,,,_, ____ _

x2 +8x+16+y 2 +10y+25=53 x2 +y 2 +8x+10y-12=0

··5 y =-2

SCJiving for the equation of the circle with center at (2, -2):

1111

{x- h) 2 + (y- k) 2 ""r 2

x2 +y 2 +4x-2y-k=O

(X- 2f + (y + 2) 2 (X- 2)

2

2

::5 2

t

whenrO.x~·5

Given line : 2x + 7y -1 0 = 0 A=2;B=7;C=-10 Center of circle: (-4,-5) x1 =4;y 1 :::-5 Solving for the distance from the center of the circle to the line: r = j\(x1) + B(y1) + C

JA2 +fil ~-~2

=(-)7.28

& Hyperbola) 297

y2 = 4ax

By inspection 4a = 16, then a= 4.

•

3x + 2y2 - 4y + 7 = 0 2

3 2

7 2

y -2y+-x+-=O By completing square:

~-

2

3

2

By inspection, r = .Jk + 5 .

•

Note: For a point circle, r = 0.

Standard equation: (x- h)2

+ ( 1) 2

2

(....rk+5f

(x+2) +(y-1) 2 =k+5 or

::-~X .. .?. +(1 )2 2

0 =.Jk +5 k:: -5

•

Note: .. If A = C, the conic is a circle. If A 'I C but the same sign, the conic • is an ellipse. • If A and C have opposite signs, the conic is a hyperbola. If either A or C is zero, the conic is a " parabola. Given equation: 3x 2 + 2x- 5y + 7 = 0 By inspection C = 0, thus the curve is a paratJola.

(y -1)

= 4a(y- k) where: h and k are the coordinates of the vertex By inspection h = 0 and k = 2, thus the vertex is at (0,2)

•

2

.

3 5 :::::--x-2

Given equation: x2 = 4(y - 2)

7

Y -2y=--x-2 2

Thus, the focus is at (4,0)

2

(x + 2) + (y -1 ) 2 = k + (2) 2

General equation of a conic section: Ax 2 + cy2 + Dx + Ey + F = 0

+\/\<') +(7)

Standard equation:

(y -1 )2

--...:

r= 2(-4)+7(-5)-10

Given equation:/= 16x


[2x+ 2'-t~ 1o= EJ ~-t whenx=O,y"/.43

~.

•

Ellips~

By inspection, the equation of the directrix of the given parabola is x = -4.

x +y +4x-2y=::k

+ (y + ;?) 2 = 25

Note: Since this equation is in the chOices, there is no need to get the equation of the second circle.

Ill

Day 12- Analytic Geometry (Parabola,

2

2:: -%(X+~)

Standard equation: (y -kf = -4a(x -h) By inspection, h = - 5/3 and k = 1, thus the vertex is at (-5/3, 1)

•

y=-x 2 +x+1

2

x - x = -y + 1

Given equation: y2 = 16x Standard equation:

y2 = 4ax

By inspection 4a = 16, thus a= 4.

a·

~

By completing square: (x -1/2l = -y + 1+ (1/2)2 =-y+5/4 (x-1/2)2 =-1(y-5/4)

~-+u =

{X..f1)2 t4~(ywk~

(x.-h)-z

=: -4~{y4<+

By comparison to the standard equations, the given parabola is facing downward!

Day 12- Analytic Gee>rn~try (f'ara):)o@L~se & Hyperbol~ 299


•

Standard equatian: x = 4ay Length of latus rectum = 4a

•

= 20

2

Ill

x 2 -14x + 49 = -12y -12

=r+~

2

By inspection, 4a

2 (X- 7) = -4(3)(y + 1)

(x-~r =t-%+(~r

Given equation: x 2 = 20y

Focal length = a = 2 Length of latus rectum = 4a = 8

Substitute:


x 2 -14x+12y+61=0

16

(x--~r =i(Y+~) Standard equation: (x- h)2 = 4a(y- k)

•

(y- 2)2 = -8(x -1) Standard equation: (y- k)2 = -4a(x- h)

V(-1,7)

By inspection, the vertex is at (1, 2). Also 4a = 8, thus a = 2 .

By inspection the vertex is at (7/4, -9/8)

~

·-/ +4x -_4y -8 = 0 y2 - 4y = -4x + 8 Solving for a:


15 11 a=--1=-

(y- 2)2 = -4x + 8 + (2)2

4

= -4x +12 (y- 2)2 = -4(x- 3)

V(7/4,-9/8)

714

Standard equation: (y- k)2

Refer to the figure, the axis of symmetry is,

x=7 4

V(3,2)

Standard equation: (y- k)2

2

(y- 7)

4x-7 =0

y 2 -14y+11x+60=0

•

•

~

y 2 + 8x - 6y + 25 = 0 y

(y- 3)2

By inspection, 2a = 6 and a

2x 2 -7x 2

7

=y-5 y 5

x --x=--2 2 2

=3

2

•

For an ellipse, the discriminant must be less than zero (B2 - 4AC < 0).

•

-

9x 2 + 25y2 = 225

x2

6y = -8x - 25

y2

-+-=1 25 9 x2 y2 --+-=1 (5)2 (3)2

= -8x- 25 + (3)2 = -8x -16

x2

(y-3)2 =-8(x+2)

By inspection, a Standard equation: (y- k)2

= 5 and b = 3

= -4a(x- h) A= 1tab = 1t(5)(3)

IJy inspection,

y2

Standard equation: 2 + 2 = 1 a b

The coordinates of the vertex is at (7,-1). Standard equation: (x - h)2 = -4a(y- k)

Refer to the figure, the equation of the directrix is, x = 3.

Note: This is an equation of an ellipse


Thus, the focus is at (2,2).

y =2x 2 -7x+5

=-4( ~1 }x + 1)

y 2 -14y + 49 = -11 X -11

a

•

= -4a(x- h)

Substitute:

= -4a(x- h)

By inspection the vertex is at (3, 2). Also 4a = 4, thus a = 1.

4

4a = 8 thus, a = 2

A= 47.12 square units

' Day 12 -Analytic Geometry (Parabola, Ellipse & Hyperbola) 301 300 1001 Solved Problems in Engineering Mathematics (2"d Edition) by Tiong & Rojas

•

~

mt

(x-3)2 (y+4)2 _ --+----1 16 25

P(x,y)

cb "'? (0,1)

By inspection, h = 3 and k center is at (3, -4 ).

2

c

= - 4,

a

c =ea 2=0.65a

thus the

a= 3.0769 a-c

b = Fa~ = )3.0769 2(2.338) 2 2b 2 LR=--=---a 3.0769 LR =3.55

x + 4y2 - 2x- By + 1 = 0 x 2 - 2x + 4(y 2 - 2y) = -1

4


x + y2 - 2y + 1 = ..!. (16 - 8y +

4

l)

4x 2 + 4y2- By+ 4 = 16- By+ y2 4x 2 +3y2 = 12

•

1)2

---+(y-1)2

4

a

2

-

2

2

:::

2.33B

Assuming the radius of earth and sun to be very small compared to a:

"'

=1

a

Satellite

a

a=5 b=4

• 40,000

c=1a -b =15 2 -4 2 =3 2

2

Distance between foci= 2c = 2(3) = 6

Ill

12,800

2a = 40,000 + 12,800 + 6,600 a=29.700

(x- 3)2 + (y + 4 )2 = 1 64 16

c =a- (6.600 +6,400) c ::29,700- (6,600 + 6,400)

ml

2

25x + 16y2 -150x + ~ 2By + 81 = 0 2

2

Standard equation: (x- h) + (y- k) _ 1 a2 ~-

16,700

a+c=105.5~Eq.1

a

29,700

a-c =78.25 ~ Eq.2

e =0.56

2 a = 64 ~a= 8;

25(x- 3f + 16(y + 4)2 = -81 + 25(3)2 +16(4)2

Length of latus rectum:

=400

c

e=-=--

By inspection:


25(x- 3)2 + 16(y + 4)2

b2 = 16 ~ b = 4

ml 2a = 186,000,000

2

a" 93,000,000

LR = LR = 2b _ 2(16)

a -

a+c

c=16,700

2

2

25(x -6x) + 16(y + 8y) = -81

a

a Earth

Length of major axis = 2a = 2(2) = 4

Solving for c:

= 93,000,000 + 1,550,000 Apogee= 94,550,000 miles

By inspection, a2 = 4 thus, a = 2

2b=B

Apogee =a+c

6,600

Standard equation: (x-h)2 + (y-k)2 a2 ~-1

2a =10

c = ea 1

c = -(93,000,000) 60 c = 1,550,000

mt

2 2 (X -1? + 4(y -1 ) = -1 + ( 1) + 4(1)2 2 2 (X -1 ) + 4(y -1 ) = 4 (X

e=-

2

(xf + (y -1? = ..!.(4- y)2 2

c

Solving for b:

2

2 J(x-O/+(y-1) =-i(4-y)

a

e=-

•

1

d2 =-d1

a

Earth

2 2 k)- 1 _ Standard equation: (x- 2h) + (y-b a2

d1

..,

--·a-·- = 4 j

Add Eq.1 and Eq.2: 2a = ~05.5 + 78.25

a =91.875

a-c

Day 12- AnalyticQeometry (Parabola, Ellipse &

302 100 1 Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas Substitute a in Eq.2:

2

By inspection, a2

2

"Standard equation: (x- h) (y- k) · a2 +~=1

c=105.5-91.875 c=13.625

2

By inspection, a = 9 thus, a

=4

thus, a = 2 .

Also,b 2 =9 thus, b = 3.

2

2b - 2(4) LR=a- 3

ml

=.,ff+2 =1.73

=3 .

Also, b = 4 thus, b = 2 .

c - 13.625 = 0'.148

zjb

= 2.667

a -~--,,-s<.,~

I

I

)>

x-axis

Ill Solving for c:

General equation of a conic section:

Refer to the figure:

2

Ax +Cy 2 +Dx +Ey +F=O

-d

+d

2a=20 a=10

c

a

e=-=a d (10)2

= 5d

=../4+9 =3.605

Solving for e:

Let: D1 = distance of focus 1 to the x-axis D2 = distance of focus 2 to the x-axis

3.605 e- -c--a 2

D2 = c -1

D1 =1+c=1+1.73=2.73

e=1.8

x=±20

•

2

X i_=1 -g-4

x 2 + 4x- 2(y 2 - 2y) = -4

2

(x + 2)2 - 2(y -1 )2 = -4 + (2)2 - 2( 1)2 (X + 2)2 - 2(y -1 )2 = -2 2

9x -4y -36x+8y = 4 2

(x+2) 2 ---+(y-1) 2 =1

2

9(x - 4x)- 4(y - 2y) = 4 2

4(x 2 + 2x) + 9y 2 = 32

>

By completing square: 2

4(x + 1) + 9y 2 = 32 + 4( 1)2 2

4(x + 1) + 9y 2

'(x +1)2

=36

y2 --+-=1

(y -1 )2 - (x + 2)2 = 1

2

2

9(x- 2) - 4(y -1 )2 = 4 + 9(2)2 - 4( 1 )2 9(x- 2) - 4(y -1 )2 = 36

4

_ (y

Standard equation· (y- k)

·

-1l = 1 9

2

Standard equation: (x- h) _ (y- k)

a2

a2

2

b = 4 thus, b = 2

•

=1

b

4

I ,j

2

'

2 thus, b =

J2 .

2

L:1 -g· 4 X

(x- h)2

-~=1

By inspection, the center of the hyperbola isat(-2,1); a2 =1thus,a=1;

2

---b-2-

=3 ;

2

2

2

2

By inspection, a = 9 thus, a

Length of semi-transverse axis: a = 3

2


(X- 2)

x2 y2 Standard equation: 2 - 2 = 1 a b


2

4x2 + 9y + 8x - 32 = 0

= 1.73-1 = 0.73

rm

Ill

By inspection A = 4 and C = -1 , and since A and C have opposite signs, the curve is a hyperbola.·

By inspection the equation of the directrix is,

9

Note: • If A = C, the conic is a circle. If A '1- C but the same sign, the conic • is an ellipse. • If A and C have opposite signs, the conic is a hyperbola. If either A or C is zero, the conic is a • parabola.

x2 - 2y2 + 4x + 4y + 4 = 0

d=20

•

c=~a 2 +b 2

Given equation: 4x 2 - y 2 = 16

a 2 =cd

303

Solving for c:

c = ~a 2 + b 2

2

e =;- 91.875

Hyperbo~

x2 y2 Standard equation: 2 - 2 = 1 a b By inspection, a2 = 9 thus, a = 3 ; b

2

=4 thus, b = 2

fI

q

304 1001 Solved Problems in Engineering Mathematics (2nil Edition) by Tiong & Rojas

Day 12 -Analytic Geometry (Parabola, F.1l!Ese & H.:meEola) 305.

Thus, the equation is, x2

a

y2

---=1 22 42 x2

y2

P2 (5,1,-4)--~x 2

2

2

•

at (0,0): r 2

=

r2=x2+y2+z2 r2 "'(8)2 + (1 )2 + (6)2

x + y + z = 2---+ Eq.3

•

= x 2 + y2 + z 2

Substitute the coordinates of the given point to the standard equation:

3x- 2y + z"' 3---+ Eq.2

2x +3y =0

Thus the point is {2, 1, -1)

2

Standard equation of a sphere with center

x + 5y- 2z 9 ...-+ Eq.1

2x-3y = 0 or

x=2

2

d=Ji1

a

r 2 =101 Subtract Eq.3 from Eq.1:

Standard equation of the ~symptotes of a hyperbola with center at (0,0): y

=±E.a x

By inspection, the coordinates of the center is at (-2,2); a = 2 and c = 3.

By inspection:

E.a = 2 ---+ b = 2a

=

4y -3z = 7 -l- Eq.4

b = ~c 2 - a 2 = ~3 2 -2 2 b = -J5

Multiply Eq.3 by 3: 3x + 3y + 3z

Standard equation: (x- 2h)2 -(y- -k)2 =1 a b2

=1

(x+2)

2

2 2

2

(y-2) _ (

2

(x+2f

(y-2)

2

4y-3(%-%v)=7

_

------1 4 5 2 5(x + 2) - 4(y- 2)2 20

25 9 - -2 - =2 1 4a

2

a=2

b =2a =2(2)

=-

Substitute to the standard equation:

(x + 3}2 +(y -2)2 + (z

9 15 4y--+-y=7

= 2 2 5(x + 4x + 4)- 4(y - 4y + 4) =20

2 2 8y - 9 + 15y = 14 y=1

2

5x + 20x +20-4y 2 +16y -16 = 20

3 5

2

5x -4y2 + 20x +16y -16 = 0

z=---(1)=-1 2 2

b=4

.

-4Y =62

x2 +6x+9+y2 -4y +4 + z2 -8z +16:::: 36

Substitute in Eq.4:

-!5)2

(5/2)2 _K=1

As given the center is at (-3, 2, 4), thus 3, k = 2 and I = 4 and the radius is 6. h

3 5 z=----y

------1 22

16 = 4a

5y+2z=3

b

=r2

where: (h. k, I)= coordinates ofthe center

(3x + 3y + 3z)- (3x- 2y + z) = 6- 3

a

Substitute b = 2a and the coordinates of point (5/2, 3) to the standard equation:

(2a)

=6 ---+ Eq.S

2 2 ( y-k) _ (X • h) - - 2-.. - - - 2 - - 1

•

(x- h)2 + (y- k}2 + (z -1)2

Subtract Eq.2 from Eq.5:

Substitute: xz y2 Standard equation: 2 - 2 a b

p:..fi'01

(x + 5y -2z)-(x + y + z) 9-2

Given equation of asymptotes: y = ±2x

4a

.

d = J(5-6) + (1 + 2) + ( -4 +3)

2 y=±-x 3

2

2

d=v\~z-x1) +(y2-y,) +(z2-z1)

b y=±-x

a

=5;y 2 =1;z 2 =-4

~--2

Ill

y in Eq.3:

x+(1)+(-1)=2

P,(6,-2,-3)..-+x 1 =6;y 1 =-2;z 1 =-3

----=1 4 16 4x2 -y 2 =16 Equation of the asymptotes of a hyperbola with center at (0,0):

Substitute z and

•

=

x2 +y 2 +z2 +6x-4y-8z 7

Standard equation: (x - h )2 + (y - k)2

=r2

Substitute coordinates of the center and radius as given:

-4)2 +(y -0)2 = 4 2 2 x - Bx + 16 + y2 16 (X

=

II

' I

:~

306 lOO(Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas x 2 + y 2 = 8x ~ Rectangular equation

ml . Standard equation: (x- h) 2 + (y- k) 2 = r 2 Substitute coordinates of the center and ra~ius as given:

Topics

D D D D l_j D D D

(x- 3)2 + (y- 0) 2 = 32 Refer to the triangle:

y~

x2+y2=r2 x =rcos9 Substitute:

Refer to the triangle:

x2 + y2

y~ x.

x + y2 =6x

r =8cos9

ra = 6()1_ cos 9)

r-8cos9=0

r sin 2 e = cos e

Tue

2

r'a = 8()1_ cos 9)

Refer to the triangle:

=r2

x=rcos9

x2 +y 2 =8x

•

Mon

x2 - 6x + 9 + y2 =9 x 2 + y2 =·6x

X

r = 6cos9

y~ X

r(~J =(7)

•

Refer to the triangle: x=rcos9

y =r sine

Parabolas - General & Standard Equations - Eccentricity & Latus fjectum Ellipses -General & Standard Equations - Major & Minor Axes - Eccentricity & Latus Rectum Hyperbola ... General & Standard Equations -Transverse & Conjugate Axes - Eccentricity & Latus Rectum Polar Coordinates - Radius Vector - Polar Angle Relation Between Rectangular and Polar Coordinates

y~

Theory

Wed

Problems

Thu

Solutions

Fri

Notes

~ Sat

.x

y2 =x

Given: r = 5 sin 9 and 9 = 30°

Note: This equation is a parabola with vertex at (0,0) and opening to the right.

Substitute: x=rcose

Standard equation:

y2 = 4ax

By inspection, 4a = 1 thus, a = 1/4

x = (5 sin e)cos e = 5 sin 30° cos 30; X=

2.17

y =r sine y = ( 5 sine) sine = 5 sin 2 30° y =1.25 Thus, the point is at (2.17, 1.25).

Thus, the focus is at (1/4, 0).

+

'

A

~

308 . 100 1 Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas

Topics

[QJ Mon

0 0

Tue

Theory

<";.<·, ,,,

0 0 0

Problems

Solutions

Notes

~

~

Wed

0

Thu

0 Fri

0

Sat

What is Calculus?

What,is Differential Calculus?

The term "Calculus" was derived from a Latin word "calx" which means "stone" and from a Greek word "chalis" which means "limestone".

Differential Calculus is a branch of mathematics which deals with derivatives and limits.

In 1684, a German mathematician and philosopher Gottfried Wilhelm von Leibniz published his early work on calculus, while an English astronomer, physicist and mathematician Isaac Newton made an early study on the subject in 1665 but did not published his work until 1704. These two rn
<

Differential Calculus Limits Theorems of Limits One-Sid~d Limits Continuity Special Limits Derivatives -Algebraic Function -Exponential Functions - Logarithmic Functions -Trigonometric Functions - Inverse Trigonometric Functions - Hyperbolic Functions - Inverse Hyperbolic Functions

1111 · ,;ulljed Calculus is divided into four ,,,,.,~.;

lldllldy, differential calculus,

inft·qra! calculus, differential equations .l!rtl c.llculus of v
What is a Limit? Let f be a function that is defined on an open interval containing a, but possibly not defined at a itself. Let L be a real number. The statement limf(x) = L

x->a

defines the limit of the function f(x) at a point a. L is the value that the function has as the point a is approached. What are the Theorems of Limits?

lhu followin(J are important theorems and pr<>Jll'tlr•·'; 11! lrr;~rt'; con:oirl.-:rinq thr: fiiJJ(lWIII
j

·

310 100 ~ Solved Problems in Engineering Mathematics (2na Edition) by T1ong & Rojas lim f(x) = L

and

lim g(x) = K

What are One-Sided Limits?

1. If lim f(x) exists then it is unique. X->8

lim[f(x)+g(x')] = limf(x)+ limg(x) X-Joa

, X-J-3

X-+3

Suppose f is a function such that it is not defined for all values of x. Rather, it is defined in such a way that it "jumps" from ·one y value to the next instead of smoothly going from one y value to the next.

3.

x-Joa

lim f(x) = f(a) x--+a

The following are the three conditions being satisfied:

1. f(a) exists, that is, f is defined at a, lim log. x = +oo

X-Joa

x->0

2.

lim f(x) = ~i~ f(x) =!:: x->ag(x) limg(x) k X->8 1

6. lim--=~ x->a

7.

What is a Derivative?

3. the two numbers are equal.

X-+8

lim[f(x)·g(x)] = L·K X->8

5.

lim[f(x)j"

x ...... a

The figure above shows that y defined for all values of x.

=f(x) is not

Continuity test at a point x

X-+3+

9.

What are Special Limits? 1.

lim sinx = 1 x-+0

= lim[f(x)J" fern> x~~a

lim ex"

2. The fi9ure above shows that y = f(x) "jumps" from a positive value to a negative value.

= c lim x" = ca"

11.

lim X-----+a

X

'1fW = \Jnllim f(x) X-+a

If f(x)::; h(x)::; g(x) for all x in an open interval containing a, except possibly at a, and if lim f(x) = L = lim g(x) then X-Joa

limh(x) = L

(1 + ~)" n =e

3.

n-+oo

4.

lim{h n)n = e

lim

X-+3

n-+0

3.

Likewise, the statement lim f(x) = L X->8 means that as x af?proaches "a" from the left hand side or from negative infinity, the function f has the. limit L.

5.

de =O dx d du dv -(U+V)=-+dx dx dx d dv. du -(uv)=U-+Vdx . dx dx du dv

~(~) = vdX -udx

d n n-1 du - u =nu dx dx du

6.

lim log)'= +oo

8.

X H·OO

lim log. x =

x->a

x

if and only if lim, f(x) ·~ lim f(x) X >il

J

~o

2vu

7.

X--+-oo

limf(x) = L

~JU = d~ dx

lim ax= 0

v2

dx v

lim ax= +oo X-HOC

Iff is defined in <:~n open interval containing a, except possible at a, then

X· >
4.

•

5. For a > 1, then:

X--+8

•

2.

1

x approaches "a" from the right or from positive infinity, the function f has the limit

R.

= f(a)

when n is positive integer

12

Algebraic Functions:

lim 1- cosx = 0

x-Joa+

10. Iff is a polynomial function then X-->8

A.

X

The statement lim f(x) = R means that as

X---4-8

lim f(x)

The following are the derivatives of the different functions:

1.

o

~i~[cf(x)] = c[~~f(x)J X--78

X-+3-

The derivative of a function expresses its rate of change with respect to an independent variable. The first derivative of a function is the slope of the tangent line to that curve defined by the function.

(K ot 0)

X--+0

8.

= a:

lim F(x) = lim F(x) = F(a)

y

(K ot 0)

K

g(x)

lim f(x) exists, x--+a

X

lim[f(x)•g(x)] = limf(x)-limg(x) X-Jo3

lim IQg. x = -CJJ

X->+oo

----------4r-----------X-43

lim ax= +oo X->-oo

lim[f(x)- g(x)] = L- K X->8

4.

lim ax= 0 X--++«·

~

lim[f(x)-g(x)]=limf(x)-limg(x)

x-+a

A function f is continuous at a point a if

y

lim[f(x) + g(x)] = L + K X->8

6. For 0 < a < 2·, then:

What is a Continuity?

X-+a

X-Joa

2.

Day_l3 -Differential Calculus (Limits anci Derivatives) 311

-CJJ

9.

d (u)

dx

c =c1 dxdu

du d c c dx dxu=lT

3~2

B.

1001 Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas Exponential Functions:

dx

dx

F.

10. _E_(e") =e" du dx _ dx C.

Hyperbolic Functions: . 26.

~(sinhu) = coshu du

logarithmic Functions:

27.

~(cosnu) = sinhu~~

11

d log du . ·-·(log ) a edx .u =-~ u

28.

12

d log du . -(log ) 10 e dx 1ou = .u dx

30.

du d -. dx(lnu)=·dx

31.

13

Proceed to the next page for your 13th test Detach and use the answer sheet provided at the last part of this book. Use pencil number 2 in shading your answer.

d -1 -1 du 25. -(esc u ) = - - = dx uN -1 dx

-~(a")= a" Ina~-~

9.

Day 13- Differential Calculus (Limits and Derivatives} 313

29.

dx

GOOD LUCK I

dx

m:ribia:

dx dx d 2 du --,(tanhu)=sech u-dx . dx d 2 du -(cothu) = -csch u dx dx d du -(sechu) = -sechutanhudx .dx d du -(cschu) =-cschucothudx dx

Did you know that. .. the most proved theorem in Mathematics is the Pythagorean Theorem which has more than 370 different proofs! All the 370 different proofs are found in the book entitled "The Pythagorean Proposition" which was published in 1940

c!euote: "The mathematician does not study pure mathematics because it is useful; he studies it because he delights in it and he delights in it because it is beautiful."

u

D.

G.

32.

d . du 14, -(s1nu) =cosudx dx 15.

16.

H. 18. 19. E.

Inverse Hyperbolic Functions:

Trigonometric Functions:

d (COSU ) =- -SinU-· . du dx dx d du 2 -(tanu) =sec u-dx dx d · 2 du --(cotu) =-esc u dx dx d du -(secu) = secutanudx dx d du --(cscu) = -cscucotu-dx dx -

Inverse Trigonometric Functions: 20.

~(sin-1 u) = _1__ du

21.

du -~(cos- 1 u) =- c--1 . -u2 dx

dx

~(arcsinhu) = ~ dx

- Henri Poincare

u2 + 1

d 1 33. -{arccoshu) = ~; Jxl > 1 2 dx u -1 d 1 34. -(arctanhu) =--· lxl < 1 dx · 1- u2 ' d 1 35. -(arccothu) =- ; lxl > 1 dx · 1- u2 d 1 36. -(arcsechu) = - - - ; O
*0

~1-u2 dx

dx

22.

· "1/l1 d _1 _ _ _1___ du dx (tan u)- 1+ u2 dx

23.

d~ (cor-1 u) = t+ u2 dx

24.

:x (sec-1 u) = uf:l1

--1

'""'

du

1

du

a; _1

:111

Day 13- Differential Calculus (Limits and Derivatives) 315 5&o: EE Board April :1995 1- cosx Evaluate:

A. B.

Topics Differential Calculus limits Theorems of Limits One-Sided Limits Continuity . Special Limits Derivatives - Algebraic Functions - Exponential Functions - Logarithmic Functions -Trigonometric Functions - Inverse Trigonometric Functions - Hyperbolic Functions - Inverse Hyperbolic Functions

Mon

D Tue

D D D D D D D Theory

Problems

Soi~Jons

Notes

Wed Thu

Fri

Evaluate: Lim

X--+1

A

1/5

B. C. D.

2/5 3/5 4/5

2

x +3x-4

2 -1/2

A. B. C.

0

D.

1/7

Undefined Infinity

1 0

2

D.

Infinite

(x-4)

------o:'-----'2

Undefined

3/5

B.

2

4 6

cos (x!- + 2)

B. C. D.

B.

ECE Board November :199:1

3x + 6 3x- 3

6x- 3 6x + 3

C. D.

4x{x2 + 1) 4xlog10 e x2 + 1 log e(x)(x!- + 1} 2x(x!- + 1)

24

Differentiate (x2 + 2)

B.

26 28 30

A.

Evaluate:

Lim (2- x) X--~

x-2

A.

0

C. D.

D.

5&9: EE Board October :1997

559: ECE Board April :199~ x2 -4 Evaluate: M = Lim - -

A

C.

A.

C. D.

X--+2

2x cos (x!- + 2)

- cos (x!- + 2) cot (x2 + 2) 2x sec (x2 + 2 ) tan (x!- + 2)

A.

Infinity Zero

x-4

0 1 8 16

(x - x -12}

A B.

,I

5&8: EE Board October :1997 Differentiate y = log 1o (x!- + 1)2

Evaluate: Lim (x!- + 3x - 4) X--+1

x -16

II.,

5&&: EE Board October :1997 Differentiate y = sec (x!- + 2).

A.

56:&: EE Board Octuber :1994 . 3x 4 - 2x 2 + 7 Evaluate: Lrm ---,3, - - - 5x +x-3 X-+oc

5&~:

1r

5&7: CE Board November :1994

t.

A. B. C.

-ex sin x!ex (cos x!- - 2x sin x!-) ex cos x!- - 2x sin x!-2xex sin x

What is the derivative with respect to x of (x + 1)3 - x3 ? ·

5&4: ECE Board November :1994

557: ECE Board April :1998 Evaluate: Lim x--+ 4

C. D.

1/2

D.

A. B.

2

A. B. C. D.

B.

5&:1: ME Board October :1997


X--+4

A

Compute the following limit: X+4 Limit: X-+oc x-4

Sat

Evaluate the Lim

_x_2_

0

C.

D.

55&: CE Board November :1997 x2 -1

Lim X--+0

5&5: EE Board October :1997 Differentiate y = ex' cos -?-.

I

:

I

!I

1

B. C. D.

ez. e'l/Jr

0

1

B.

tan~

2

C. D.

112

(x 2 +2)1/2 2 X

(x 2 +2)1'2 2x (x 2 + 2)1'2 (x!- + 2)3/2

570: EE Board October :1997 If y = (f + 2)2 and t = x 112 • determine dy . dx

l

Day 13- Differential Calculus (Limits anciJJerivatives) 317

316· 1001 Solved Problems in Engineering Mathematics (2nd Edition) byTiong & Rojas

A. B.

c.. D.

3 2

-

B.

2x 2 +2x 3 2(x + 2) x5'2 +x1'2

57:1: ME Board April :1.997 What is the first derivative of the expression (xy)x = e?

4 (1-16x2 )o5

c. D.

B.

--4

C.

(1- 4x2)o.s

D.

4 (1- 4x2 )o5

B.

0 X

D.

(1+1nxy). X

-y (1-lnxy) x2

A. B.

B.

(x + 1)2 _ (x + 1)3 X X 4(x + 1)2 _ 2(x + 1)3 X

c.

D.

X

2(x + 1)3

(x + 1)3

2

3

--x---r 3(X+ 1) 1) --x----xz (X+

Find the derivative with respect to x the function .J2- 3x 2

576: ECE Board November :199:1

-2x 2 .J2-3x2

B. C.

-3x

Differentiate the equation: y =_x_ X+1 x2+2x

A.

(x+1t

.J2-3x2 -3x 2

B.

.J2-3X2

c.

3x

D.

D.

.J2-3x2 573: EE Board April :1.995 Find y' if y = arc sin cos x

A. B. C. D.

-1 -2 1 2

A.

D.

--4 (1-16x 2 ) 0 5

X+1 2x

A.

1

10x- 5 6x-10 3x + 10 3~-5x.

A.

6

B. C. D.

7

A B

3x2 -5 8

c; ll

A. B. C. D.

2 -1 -1/2

-2

587: CE Board May :1.998 Find the slope of the curve ~ + 10y + 5 = 0 at point (1 ,0).

A B.

-0.1463 -0.1538 -0.1654 -0.1768

2.21 -4.94

1~

58fn ECE Board November :1998 Find the slope of x2 y = 8 at the point (2,2).

C. D.

1/5 2/5 1/4

2

Find the slope of the tangent to the curve, y = 2x- ~ + x3 at (0,2).

C. D.

1 2 3 4

589: ECE Board April :1999 .Find the coordinates of the vertex of the parabola y = ~ - 4x + 1 by making use of the fact that at the vertex, the slope of the tangent is zero.

3.25

/21

l- 6x +

588: CE Board May :1996

A. B.

=

I

~~

~1~

S8:Jz EE Board October :1997 If y = 4 cos x + sin 2x, what is the slope of the curve when x 2 radians?

J

1~

B. ~

B. C. D.

64~

A C.

A.

578: CE Board November :1993 Find the second derivative of y by implicit differentiation from the equation 4~ + 8y2 = 36.

585: ECE Board November :199:1 Give the slope of the curve at the point

4

2

2 sin (x2 + 2) c~s (~ + 2) -2 sin (~ + 2) cos (~ + 2) 8x sin (~ + 2) cos (~ + 2) -8x sin (x2 + 2) cos (~ + 2)

D.

2x + 1 at x = 1.

x3

582: CE Board May :1996 Find the slope of the ellipse ~ + 4y2- 1Ox - 16y + 5 = 0 at the point where y = 2 + 8° 5 and X = 7.

2x 2 X+1

577: CE Board November :1995 The derivative with respect to x of2 cos (~ + 2) is

A. B. C.

574: CE Board May :1997 Find the derivative of arc cos 4x.

-

X

1/2 1/3 1/4.

-

(1,1): y=--2x+1

58:1: ME Board April :1998 Given the function f(x) = x to the 3rd power - 6x + 2. Find the first derivative at X= 2.

2

1

B. C.

0.

C. D.

D.

y2-5 y2 xy- 5y 2xy

A.

580: ME Board October :1997 Find the second derivative of x 3 - 5~ + x =

A. B.

572: ME Board April <:f98

A.

9

3 Find the derivative of (x + 1)

A.

c. -y

curve y = x 3

32 xy 16 3 -y

575: CE Board November :1.996

C.

-y

584: ECE Board November :199:1 Find the slope of the line tangent to the

3

579: ME Board April :1998 Find the partial derivatives with respect to x of the function xy2 - 5y + 6.

X

A.

9

4y

A.

(2,-3)

:Jls

1001 Solved Problems in Engineering Mathematfcs (2nd Edition) by Tiong & Rojas

B.

(3,-2) (-1 ,-3) D. ' (-2,-3)

595: ECE Board November 1.996

c.

Find the radius of curvature at any point in the curve y + In cos x = 0.

590: ECE Board April 1:999 Find the equation-of the normal to . 5 at the point (2, 1). A. B. C. D.

.

x? + y2 =

A. cos X 8. 1.5707 C. secx D._ 1

......,.,._,_____

~ lv1on

y = 2x 2y 2x + 3y = 3 X+ y = 1 X=

i

What is the equation of the normal to the curve x? + y2 = 25 at (4,3)?

Tue

A.

B. C. D.

·,·,I

..

5x + 3y = 0 3x-4y = 0 3x + 4y = 0 !}x- 3y = 0

59%: EE Board April :1997 Locate the points X: e•

y = f(x) A. B. C. D.

=

of inflection of the curve

J3 2 ± J2 -2 ± J2 2 ± J3

-2±

i.;

l

Theory

Wed

Problems

Thu

Solutions

Fri

Notes

Sat

ANSWER KEY 556. B 557. D 558. 559. 560. B 561. A 562. 563.A 564. B 565. B

points.

B. C. D.

c c

(2,18) & (-2,-14) (2,18)&(2,-14) (-2,18) & (2,-14) (-2,18) & (-2,14)

c

5941 CE Board November :1997 Find the radius of curvature of a parabola

566.C 567. D 568. B 569. B 570.C 571. 572. B 573.A 574.A 575. D

c

576.A 577. D 578. B 579. B 580. B 581. A 582. D 583. B 584.A 585. D

RATING :,"

586. D 587. B 588. 8 589.A 590. B 591. B 592. 593.A 594.A 595.C

p

0

c

";'

22.36 25.78 20.33 15.42


I

~~

c:J 34-40 c:J 24-33

0

Topnotcher Passer

20-23 Conditional 0-19 Failed


y2 - 4x = 0 at point (4,4). A. B. C. D.

Topics ______

.\

593: ECE Board November :199:1 In the curve 2 + 12x- x 3 , find the critical

A.

.

Differential Calculus Limits Theorems of Limits One-Sided Limits Continuity Special Limits Derivatives - Algebraic Functions -Exponential Functions - Logarithmic Functions - Trigonometric Functions - Inverse Trigonometric Functions -Hyperbolic Functions - Inverse Hyperbolic Functions

D D D D D D D D

59:11 CE Board May :1995

~,

'

-~

320 1001 Solved Problems in Engineering Mathematics (2nd Edition} by Tiong & Rojas

•

x2 -1 . L1m -..--::-2 x-.1

•

Apply L'Hospital's rule & substitute x = 4: 2x . L 1m=-= 2 x

x-->«>

lim= 2(4) =8 Substitute x = 1: Lim=

2

(1) ~1

(1)1 +3( ). 1 4

0

=0 .indetenninate

Note: Using L'Hospital's rule, differentiate separately the numerator and denominator and substitute the value of limit to the variable Lim=

x-2

-5

l

-

x = 2:

.I

• x-+o

0

Lim=(2~1)1an.'!~ =1"',

indete1minate

5x +X -3

= oo: .

Lim= 3(oo)4- 2(oo)2 + 7. 5(oo)3 +oo--3 '1\:

-~

-x-2

o·

-

00

Apply L'Hospital's rule & substitute x

_1tx (In

2 ..: ln(2- x)

lim == .~. indeterminate

~r_,

. 1-cosO 0 . d t . L1m = - -2- =- 1n e ermmate (0)

In lim= tan

tanJI!I 2

1

tan n:x

= oo:

12x 3 - 4x Lim = - 15x2 +

(2- x))

2 . ln(2--x) in Lim ::: ------

cot2:.~ 2

1

3

Lim= 2x-1

:..! 7

sinx . Llm=-2x Lim = sin 2(0) = ~ indeterminate 2(0)

. 12(oo ) - 4( oo) = 00 • indeterminate Ltm == 15( 00 )2 + 1 oo

Note:

Apply L'Hospital's rule again:

d(lnu) = - and d(tan u) =-esc u du u

o·

36x 2 -4 Lim=

Apply L'Hospital's rule again: 2

Lim=

Lim x -16 X-+4

Substitute x ::: 1:

In Ltm =In (2-x)

1

•

2

Take In on both sides:

Apply L'Hospital's rule & substitute x = 0:

Lim= 2 (4 )-1

X-+1

Substitute x

Apply L'Hospital's rule & substitute x = 4:

1

Ill Lim (2 _ X)tan 1t~

• x->oo

Lim 1-cosx

4-4

x = oo:

. 3x 4 - 2x 2 ·+ 7 ltm ·--~- 3

lim =2(2}=4

(4~ -4-12 = 0· indeterminate

oo-4 oc.

Lim=2=1 1

1

1

Substitute x = 4:

Lim =(4)2 + 3(4)- 4 Lim= 24

Substitute x = 2:

. 2x L1m=-= 2x

x-4

Substitute x = 4:

x = co:

Apply L'Hospital's rule & substitute

Apply L'Hospital's rule & substitute

Ll m2- - x-+4 x -x-12

Lim=

x-+2

and Deriv!!_tives) 321

X·->4

oo+4 co 'd . =---=-, m etermmate,

. X2 -4 Llm.--

o·

2x 2x+3 2(1)+3

.

. L tm

~Limits

lim x2 + 3x- 4

x-4

Substitute

. (2)2 - 4 0 . d t . • L1m = - - · =- m e ermma,e 2-2

Lim=~-2

•

•

•

. x+4 Ltm - -

1

x +3X --4

Day 13- Differential Calculus

Lim =

COSX

Lim= cosO

---2-=21

Substitute x = 4;

l-

36( oo 4 _ '!!._ indeterminate 30( oo)- - '

Apply again L'Hospital's rule: ltm '-

4 2 16 Lim= ( ) " =.Q. indeterminate 4-4

o·

j

2

l2x

30

72(co) .:: .. --'~ oo

30

·

Apply L'Hospital's rule:

-1

30x

2

X -4

du

-1

lnlim=---.£:._x_ =

csc2 n;(~) -1

2

:n:

1t

2 In Lim= 2/n

· 2-1

-csc2~~)(~)

322. 1OOJ Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas Take exponential on both sides: eln Lim

= e2/n

•

!I

•

•

, y=

where: u = (x 2

Note: d(uv) = udv + vdu

+ 1)2

du = 2(x 2 + 1)(2x) = 4x(x 2 -t·1) 2

where: u =ex; du = ex; v = cos x 2 ; dv = -2x sinx2 • y' = ex(-2x sin x 2)+ cos x 2 ex y'= ex(cosx 2 -2xsin x 2)

y' =log1o e

4x(x + [ ( x2 +

In (xy)x = In e x In (xy) = 1

ml

Note: d (uv) = udv + vdu

y =cos-14x

IB

xy' = - y - y In xy xy'

=- y(1 +In xy) X

•

Y =~2-3x

where: n=112;u=x 2 +2;du=2x y' = ~(>.
Note:

y' = x(x2 + 2f112

= nu"-1 du 2

2

y' = 3(x + 1) (1)- 3{x) (1) y' = 3(x + 1 )2 - 3(x)2

-4

•

y = (x + 1)

du

dJU = 2JU

Note: d(

=---X ··-

y' = 3(x 2 + 2x + 1)- 3x 2 y'= ~ +6x+3-}~

y = [(X112 )2 + 2

y'=6x+3

Y = (x +2)2

v::: x;

•

J

_J

vdu-udv

v2

v 2 = x 2 ; dv = 1 2

2

,_3\(x+1)

y-

sin 1cosx

wtww u

~)

y' = x(3)(x+1) -(x+1) x2

N••terl(sin-

y' = 2(x +2)

3

2

y'=-~2-3x 2

y

5

u = (x + 1)3 ; du = 3(x+1 )

-3x


r

where:

y' = 2~2 -3x 2

t=x 112 ~Eq.1 y = (tz + 2)2 ~ Eq.2

u):::

1

du = 4

X

where: u = 2-3x 2 ; du =- 6x

•

;

~1-16x 2

v· = (1-16x2

2

2

-4

y'=

y' = -y (1 +In xy)

-{)x

y = (x + 1)3 - x3 Note: d(u)"

where: u = 4x; u2 =16x

xy'+y+ylnxy=O

Note: d(u)" = nu"·1 du

•

d{ cos-1u) = ~1- u2

\ [ x\:y ]+In xy(1) = 0

y' = 4x log10 e x 2 +1

y

-du

2

1}

y = sec(x + 2)

y' = 2x sec (x 2 + 2) tan (x 2 + 2)

~sin 2 x

x

Sin X

y' = -1

Note:

Y= (xz + 2)1/2

= sec (x 2 + 2) tan (x 2 + 2)(2x)

-sin x

=--

2

Differentiating both sides:

1)]

2

where: u = x 2 + 2; du = 2x

~1- cos

Take in on both sides:

Ill

Note: d sec u = sec u tan u du

-sin x

, -sin x y=-.-

Note: d(log10 u) = log10 e ( duu)

y =ex cos x 2

y'

Day 13- Qifferential Calculus (Limits and Derivatives) 323.

(xy)x =e.

Y = log1o (x2 + 1)2

Lim= e2 /n,

•

I

cosx; u2

=cos 2 x;

2 (

x + 1)

3

y------2-

x

v1- u2

du =-sin x

3

- -2 x

x2,

, _ 3(x + 1)

~

(x+1)

3

x

(1)


•

J 2 + 2)sin{x + 2) J

2

2

•! ( • •

y' = -4x[ 2cos( x + 2)sin( x + 2) y' =-ax[ cos( x

x2 y= x+1

•

Note: d(~) = vdu -.udv

v

2

v2

xy

!(

xy

4x 2 +8y 2 = 36

x2 + 2y 2 = 9 ~ Eq.1

where: u.,x2 ; du = 2x

Differentiate both sides:

v = x + 1; v 2 = (x + 1)2 ; dv = 1 2

(x+1)2

2

2x 2 + 2x- x 2

2y

, x2 +2x

Take second derivative:

Y = -(x_+_1-) 2

y" =

2

-

5y + 6) = y (1) + 0

-

5y + 6) = y

y' = 4 (-sin x) +cos 2x (2) y' = 2 cos 2x - 4 sin x At x = 2 radians

-M y(1~;

1 00

=3x2 - 1Ox + 1

y' = 2 cos 2[ 2( :

xy']

~ Eq.3

y = 2cos (x + 2)

y' =-4.94

rm.

Differentiate and substitute x

e = 1+cos 26

Note: cos2 . S1mplifying: 2

y-x( -x

y"=-~

2

2

2y

J]

.,

y2

[

= 2:

y'=6

Differentiate and substitute x = 1:

1!1

y' = 3x 2 -2. y' = 3(1)2 -2

y+ X

Y = ~[ 1+ cos i x +2)]

y"=-i[ /y2]

2

=

Y 1 +cos 2(x 2 + 2)

y = 1 + cos (2x 2 + 4)

2l

Note: d(cosu) = -s.inu du where: u =2x 2 + 4; du

= 4x

2

y' = -4x sin 2(x 2 + 2) Note: sin 29 = 2 cos. e sin

2y

e

.9 y"=--34y

As given: y = 2 + 8°· 5 = 4.828, x = 7 Differentiate & substitute x = 7 & y = 4.28

Thus, the slope of the line is equal to 1.

2x + 8yy' -1 0 -16y' = 0

y"= 2y2 +x2 4y3 Substitute Eq.1 in

y' =-sin (2x + 4)(4x)

J

y'(8y -16) = 10- 2x

y":

, 10-2x y=-8y-16

I I

'-

10-2(7) y- 8(4.828)-16 y' = -0.1768

I

j

y' =1 Note: Since the line is tangent to the curve at x = 1, then the slope of the line is the same as the slope of the curve at the given point.

Note: slope = y'

y" = __1_. [2y2 + x2

Y=x3 -2x +1 Note: slope = y'

y' =3(2)2 -6

x2 +4y2-10x -16y +5 = 0

2

Y = 2cos (x + 2)

) ]-4 sin [ 2( 1:0")]

y' = 2 cos 229.183°-4 sin 114.591"

y = x3 -6x +x


2

•

Differentiate:

2

y' = 3x 2 -6 2

Calculus (Limits and Derivatives) 325

y = 4 cos x +sin 2x

2

y" = 6x-10

-X y'=-~Eq.2

y' = .=_--=~(X+ 1)2

•

y'

-2x y'=4y

(x+1) 2

·

2

13.-_Q_if!~enti
y = x 3 -5x 2 + x

2x,+4yy' = 0 2

y' = (x + 1)(2x)- x ( 1) =2x + 2x- x

Day

•

x3

y=--2x+1

4

Note: slope = y' Differentiate and substitute x = 1:

iir'

,I !. {

i

Day 13- Differential Calculus {l.ixnits a.11d Derivatives) 327

326 1001 Solved Problems in Engineering Mathematics (2"d Edition) by Tiong & Rojas Note: slope

= _!(3x 2 )- 2

y'

4

Differentiate and subsfih.ite x c,.O.

y'::: ~(3)(1 )2 - 2 4

y' = 2 - 2x + 3x 2

~

y'=-1

y' = 2- 2(0) + 3(0)

4

•

y'

2

Note: Since the line is normal to the curve at the given point, the slope (m2) of the line is equal the negative reciprocal of the slope of the given curve.

Note: d(uv) = udv + vdu

1 1 - 1 mz = - m1 = - -2 - 2

y' = x 2 (ex)+ex(2x)

x y =8 8

y=x2

y-y1 =m(x-x 1)

y" = x 2 ex + 4xex + 2ex

Thus, the slope of the line is equal to 2.

2y-2 = x-2

Differentiate and substitute x = 2:

1 2

x3

2

- -4±)(4) -4(1)(2) = -4±.J8 x2(1) 2 x=

Differentiate and y' = 0:

Let: m1 = slope of the given curve m2 = slope of the normal line

y' = 2x-4

Differentiate and substitute x = 4 & y = 3:

0=2x-4

Ill

2x +2yy' = 0

x=2

,

x +f-6x+10y+5=0

Differentiate and substitute

x = 1 & y = 0.

y'(2y + 10) = 6- 2x 6-2x '- 6-2(1) y- 2(0)+10

y'=~ 5

4~~

y-y1 =m(x-x1 )

4

-10

3

y-3=-(x-4)

4

Let: m1 = slope of the given curve

4y -12 = 3x -12 Differentiate and substitute x = 2 & y = 1:

3x-4y=O

2x +2yy' =0 ,

X

1m

2

y =--=-y 1

~ ~ 3

Y=x 1 ex

y' =-2 ml c·o •. ;:o

j

X

-4±2.J2

2

=-2±.J2

Substitute the values of x to the given equation to solve for the values of y: y =(-2+J2)2 e(·Z+J2) =0.19or y=(-2-J2) 2 e(·Z-J2) =0.38

1


x2 + Yz = 1

y' = 2y +10

4

3

mz=--=--=m1 -4/3 4

Thus, the vertex is at (2, -3)

•

2x + 2yy' - 6 + 1Oy' + 0 = 0

X

=-- =-y 3

4 m1 =-.3 1

y = (2)2 - 4(2) + 1 =- 3

= y'

Y =2x- x2 + x

y

Substitute x = 2 to the given equation:

2

Note: slope

By quadratic formula:

x2 + y2 = 25

16

y' =-2

O=x 2 +4x+2

•

Note: slope = y'

y=-(2)3

O=x 2 ~ +4x~ +2~

x=2y

Y = x 2 - 4x + 1

, _ S(-2x) _ 16 y ------

At point of inflection, y" = 0:

y-1=-(x-2)

•

Note: slope = y'

x4

Note: Since the line is tangent to the curve at (0,2), then the slope of the line is the same as the slope of the curve at the given point.

d(eu) = eudu

y" = x 2 (ex) + ex(2x) + ex(2) + 2x(e')

~2


2

'

= y'

Note: From the choices, only values of the x-coordinates are given. Thus the suggested answer is choice C.

•

Y= 2+12x-x 3

Note: Critical points are points wherein the slope of the curve is zero.

y' = 12 -3x 2

0 = 12 -3x 2

x2

=4 •

X=

±2

/ 328 . 100 1 Solved Prob1Efms i!l Engineering Mathemati~_:C! Edition) by Tiong & Rojas Substitute the values of~ to the general equation to solve for the\values of y: y = 2 + 12(2)- (2)3

=18 or

y =2+12(-2)-(-2)3 =-14

•

&

12

R:: [ 1+(yiJ

Topia

jy''l -

Thus, the points are (2, 18) and (-2, ··14).

y+ln

Ill

~~

'{

=0 y =-In cos X

COSX

Note: d(ln u) = \]u u

R =I1+(y')2r2

ly''l-·

where: u = cosx; du =-sin x

l-4x =P •

4

2

y=-=2y y .. _ y{O)- 2y' _ 2y' y . ---y2 y2

y"y2 =

Mon

D rue

,:'!'

,,:{I

...

Wed

0

D D D

Problems

Thu

Solutions

Fri

y" =sec2 x

Substitute y' and y" to solve for R:

D

12

r

-2(~)

·0' ... ···

D

~~·~

y, =- (-sinx) - - =tanx cosx

2yy'-4=0

RJ1+(tan4J ·

4 . y"--- y3 Substitute y = 4 to solve for y' andy":

Sat

sec 2 ;

Note: 1+ tan 2 x = sec 2 x 2

Differential Calculus Limits Theorems of Limits One-Sided Limits Continuity Special Limits Derivatives -Algebraic Functions -Exponential Functions - Logarithmic Functions - Trigonometric Functions - Inverse Trigonometric Functions - Hyperbolic Functions - Inverse Hyperbolic Functions

t

3/2

R= ( sec x)

y•::3::3:::.! 4

y

. sec2 x

2

y" = --~ :: ___i_ = _...!_ y

3

(4)

~

3

3

16

R::: sec x

., < ~

sec2 x R=secx

'.,

u

->'

~

t

h

j<- 0< •X

,

·, ,< "·:

X

"{ >

;,,.

0.

'>< o

'i-

~ ~·

> .' "l '(

<

H

~

:

~-..;,:" ~ ~·

>

<

<, ;<

~ .0 ~

~

• '" • "'

« ,• <

>

x.••

~

,< 0

A

"0 :

~

~

0 {, "<,

~ 1-: ,,. '

0

''

_., •.,.

~

"-<: .<• _." --;. " ' "

·~

'< ..... •,-<

< -<•·> "-

t•v.;

~

V X 0

< '', ~

{

0

~·

f

«

o < '

o'

' .... 0 '

'> "< " ' :

:!

'<"

~ !"

'

""'

:·,,..,x

~

~

> )

> 1o .; ·k ~"' ~

0

0.0 'A

[<

_.,.

V

~

Substitute y' and y" to solve for R:

_Hi}T

0

R----

> "(

.0 X

~ "< )' '~'._, ~· ., ~

t:. • • .

1 16 R =22.36

T T '

J

~ « o" <'

, ...

,

,:·,·,

< »

X

<•

•

,->:

t ' ·>

»I

I

t.

:~ Topics

0

Mon

[QJ Tue

0

0 0

0

0

0

0

\1\fed

Theory

Thu

Problems

~

: < ,< <:<

Fri

Solut;ons

~ ~ «:" (;- 9

Notes

"~· ::. , ;_

f « ~ .; "

>. """'"

:<<,<';.< .~,.;

Sat

What are Maximum and Minimum Values?

y

If a function f is defined on an interval!, then:

1. f is increasing on I if f(x,) < f(x2) whenever x,, x2 are in I and x1 <. x2.

y

~

/(!~1

Xi

~ ·

!f(x2)

X

j X·;

X2

Decreasing function 3. f is constant if f(x,)

=f(x2) for every

x,, x2 in I.

• f(x,)

.--t-- •

Maximum and Minimum Values Local Maximum and Local Minimum Steps in Solving Maxima/Minima Problems Steps in Solving Time Rates Problems Relation Between Variables Under the Condition of Maxima/Minima

X

X2

Suppose f is defined on an open interval! and cis a number in I, then:

Increasing function

;>

f is decreasing on I iff(x,) > f(x2) wlwnever x,, X:> in I.

1. f( c) is a local maximum value if f(x) :s: ~(c) for all x in I. ·.,

2. f(c) is a local minimum value ifl(x) > f(c) for all x in I.

Day 14.:.. Differential Calculus (Maxima-Minima & Time Rates) 333 332 l 00 1 Solved Problems in En

4 How to Solve Maxima and Minima Problems?

1. 2. 3. 4.

The following are the steps in solving problems involving maxima and minima:

Draw a figure when necessary. Formulate equation. Differentiate with respect to time. Substitute the condition I instant to the equation.

1. Draw a figure when necessary. 2. Determine which variabl~ (the·

Important: Substitute the given values only after differentiating.

dependent variable) is to be maximized or minimized Formulate equation Reduce to one variable. Differentiate Equate to zero

What are some of the Relationships between the Variables and the Maxima/Minima Values?

3.

4. 5. 6.

In step no. 5, the formulas in finding derivatives will become very useful. It is recommended to memorize all the formulas as mentioned in the previous chapter.

Largest rectangle that can be inscribed in a right triangle with the sides of the rectangle parallel to the legs of the triangle.

:I, . ~ !_

5.

The following are the relationships between the variables under the condition of maxima/minima: Largest rectangle inscribed in circle.

b

Largest rectangle that can be inscribed in an ellipse.

a

X=

2.

Minimum point···········' slope= 0

0 0

---~~~

no ·fence needed

y

i

X X=2y

6.

Minimum Point Inflection Point

Rectangle with given area and minimum perimeter to be fenced along 3 sides only.

y

The maximum rectangle is a square.

1

X

9.

-'----~

.fi

slope= 0 ·············Maximum point

First Derivative

Y=2

a

When the first derivative (slope) fs equated to zero, it results to either maximum point of minimum point.

I Maximum Point

Rectangle with given area but with minimum perimeter.

Dy X=y

b 2 h.

X=-

b

1.

8.

Largest rectangle that can be inscribed in semicircle.

0

How to Solve Time Rates Problems? Another types of problems involving the use of derivatives are the Time Rates problems.

3.

b

.fi

7.

Largest rectangle that can be inscribed in a triangle with one side lying on the base of the trjangle.

D

perimeter or maximum area.

y

•~b=C=i

~

In solving a problem under time rates, the following steps are to b~considered:

j

=Y

e

11. Maximum area with perimeter (P) given.

Sector with given area but minimum perimeter.

equilateral triangle

p

X=--

4.268

y X

2 h Y=-2

=45°

X

r=JA e = 2 rad

b X=-

10. Right triangle with maximum

Largest area of a triangle with given perimeter.

c

·······"l~/2 Second Derivative Negative Positive

y=

Day 14 - DifferentialgJlJq_t11~_(Maxima-Minima & Time Rates)

334 100 l.So1v'

12. Maximum light a~ittance for a rectangular win do surmounted with an isosceles tri ngle.

16. Stiffest beam that can be cut from a circuJar section of radius r.

19. Most efficient trapezoidal section.

~-

clock.

It is

\X

t<=

X

~

~----

y = x.../3 -•

:

I....,

,..l'

X= 2y

h·'

X

j

e is maximized,

X= ~Y1Y2

23. Parallelepiped with maximum width at top = sum of sides width at top = 2x

It

r

1\ :

y

20. Length of rigid beam that can pass a perpendicular hallways.

X

18. largest rectangle that can be inscribed in a given ellipse.

·······

' ~

L=a+b

X

.

X=Y=Z

Z

:. a cube

y

24. Open squar~ container with maximum volume.

X=2bH

y=2a~

LJJ .

...

· 14. Maximum length of line segment tangent to an ellipse.

volume.

e = 12oo

"Strength is proportional to the product of breadth (x) and square of the depth (y)" b

a,

.... ··"

......... .

~ .·

L = ~I{ a2/3 + b2/3 )2

X=2y 's_u_rf=-a_c_e_a_r_e_a X= 3

y

.

X

X

21. Minimum length of ladder/rod to be extended from ground to a wall with an intervening fence.

2~.

Location of single stake at ground level to minimize length of wire.

15. Rectangle of maximum perimeter inscribed in a circle of radius r.

stake

h2

X

h1

~ r}

x=y

r

base (not side) from an elliptical section.

........

X

Best view means

Ym]Y ...........

17. Strongest beam that can be cut

········a··········· ················rx/2

~q'T~.w&s:
--= :

j

13. Maximum light admittance-for a Norman wind

h

dl:J

of a regular hexagon.

width at top

~"V"~

y

22. Best possible view of a picture or

"Maximum capacity with minimum perimeter"

"Stiffness is proportional to the product of breadth (x) and cube of width Cv)." X I

335

Aellipse

iC?'"'

_ 1t

,(\rectangle -

2

I i-i

L=

~(a2/3 +b2/3t

d

dh1 x- -- h1 +h2

336. 10.0 l Solved Problems in Engineering Mathematics (2ild Edition) b~g & Rojas

26. least amount of material to be used for a square base rectangular

30. least amount of material for a given volume.

parallelepiped.

y

r=

Day 14 - Differential Calculus (Maxima-Minima & Time Rates) 337 ,;•,

'.-\.

34. Maximum volume of right circular cylinder inscribed in a sphere of radius r.

h

0

J2

X

X

=2y

x

=t'2{Volume) 31. Maximum v<,>lume of cone with a given slant height.

V

27. Least amount of material to be used for an open top cylindrical

. -

-~1tf3 J27

c-

Proceed to the next page for your 14th

tank.

h

test. Detach and use the .answer sheet

h=~ .j3

provided at the last part of this book. Use pencil number 2 in shading your answer.

e =tan-1 J2

GOOD LUCK! ~

I

r

m:ribia:

32. Volume of largest cone, Vc that can be inscribed in a hemisphere.

r=h

28. Minimum cost for a given volume,

l

v.

1 vc =-V.h 2

clauote:

33. Largest cylinder that can be inscribed in a cone.

r=f{:

Did you know that. .. the word "Algebra" comes from an Arabic word "al-jabr'' meaning to transpose terms from one side of an equation to the other! This was introduced by a Persian mathematician, aiKhowarizmi in around 825 A.D.

"The most beautiful thing we can experience is the mysterious. It is the source of all true art and scienCe. • - Albert Einstein

-:\····--·------·--··r

29. Ratio of the weight of heaviest cylinder, We to the weight of the

y

I

r··-·--··

r.

circumscribing sphere, W,..

we-

1

w:··· ./3

h

Y=3

Day 14- Differential Calculus (Maxima-Minima & Ti1;11e Rates)

&oo: CE Boart~J~ay

1.'9~

.

Find ttie minimum amount of~sheet that can be made into a closed cylindB_r having a volume of 108 cu. inches in square Inches.

Topics

0

Maximum and Minimum Values Local Maximum and Local Minimum Steps in Solving Maxima/Minima Problems Steps in Solving Time Rates Problems Relation Between Variables ·Under the Condition of Maxima/Minima

Mon

[QJ Tue

0

Theory

Problems

0 0

Solutions

Notes

0

Wed

0 0 0

Thu

Sat

59&: ECE Board April 1999

598: EE Board March 1.998

Find the minimL,Jm distance from the point (4,2) to the parabola y 2 = 8x.

A triangle has variable sides x,y,z subject to the constraint such that the perimeter is fixed to 18 em. What is the maximum possible area for the triangle?

A.

4/3

B.

2J2

A.

f3 D. 2/3

C. D.

c.

597: EE Board April 1.990 The sum of two positive numbers is 50. What are the numbers if their product is to be the largest possible. A.

~&~

B. C. D.

W&~

25&25

W&W

B.

15.59 18.71 17.15 14.03

cm 2 cm 2 2 cm cm 2

599: EE Board October 1997 A farmer has enough money to build only 100 meters of fence. What are the dimensions of the field he can enclose the maximum area? A. B.

c. D.

\.

25m x 25m 15mx35m 20mx30m 22.5 m x 27.5 m

C. D.

1.22 1.64 2.44 2.68

&os: EE Board April 1.997

A. B. C. D.

125.50 127.50 129.50 123.50

&OJ.I ME Board April 1.998 A box is to be constructed from a piece of zinc 20 sq.in by cutting equal squares from each corner and turning up the zinc to form the side. What is the volume of the largest box that can be so constructed?

A. B. C.

599.95 592.59 579.50 622.49

cu cu cu cu

in. in. in. in.

The cost of fuel in running a locomotive is proportional to the square of the speed and is $ 25 per hour for a speed of 25 miles per hour. Other costs amount to $ 1QO·per hour, regardless of the speed. What is the speed which will make the cost per mile a minimum? A.

B. C. D.

40 55 50 45

&o&1 ME Board April199&

&0:&1 EE Board April 1.997

The cost C of a product is a function of the quantity x of the product : C(x) = x2 - 4000 x + 50. Find the quantity for which the cost is minimum.

A poster is to contain 300 (em square) of printed matter with margins of 10 em at the top and bottom and 5 em at each side. Find the overall dimensions if the total area of the poster is minimum.

A. B. C. D.

A.

&071 An open top rectangular tank with

D.

Fri

A. B.

339

B.

C. D.

27.76 20.45 22.24 25.55

em, em, em, em,

47.8 35.6 44.5 46.7

em em em em

&031 CE Board No-vember 1.99& A norman window is in the shape of a rectangle surmounted by a semi-circle. What is the ratio of the width of the rectangle to the total h1eight so that it will yield a window admitting the most light for a given perimeter?

A. B.

C. D.

1 1./2 2 2/3

1000 1500 2000 3000

square bases is to have a volume of 10 cu. m. The materials for its bottom are to cost P 15 per square meter and that for the sides, P6 per square meter. Find the most economical dimensions for the tank. A. B. C. D.

1.5m x 1.5m x 4.4m 2m x 2m x 2.5m 4m x 4m x 0.6m 3m x3m x Um

&o8: ME Board October 199& What is the maximum profit when the profit-versus-production function is as given below? P is profit and x is unit of production. 8

6041 CE Board May 1.998 I Jetermine the diameter of a closed •;ylindrical tank having a volume of 11.3 cu. m to obtain minimum surface area.

p A. B. C.

= 200,00- X- ( -1·1 -) \X+ 1

285,000 200,000 250,000

Day 14- Differential Calculus (Maxima-Mi~ma & Time Rates) 341

340 1001 Solved Problems in Engineering Mathematics (2" 0 Edition) by Tiong & Rojas D.


305,000

609: EE Board October 1993 A boatman is at A which is 4.5 km from the nearest point B on a straight shore ~M. He · wishes to reach in minimum time a point C situated on the shore 9 km from B. How far from C should he land if he can row at the rate of 6 kph and can walk at the rate of 7.5 kph? A. B. C. D.

4.15 km 3.0 km 3.25 km 4.0 km

At any distance x from the source of light, the j.r{tensity of illumination varies directly a~he intensity of the source and inversely the square of x. Suppose that there is a 1ght at A, and another at B, the one at B having an intensity 8 times that of A The distance AB is 4 m. At what point from A -.Qil1 the line AB will the intensity of ··· mination be least?

A. B.

C.

~-~

610: EE Board March 1998 A fencing is limited to 20 ft length. What is the maximum rectangular area that can be fenced in using two perpendicular corner sides of an existing wall?

A wall "h" meters high is 2 m away from the building. The shortest ladder that can reach the building with one end resting on the ground outside the wall is 6 m. How high is the wall in meters?

A B. C. D.

il

B. C. D.

2.34 2.24 2.44 2.14

A. B.

C. D.

A. B. C. D.

A M% B.

j 'I

A B.

~% ~% ~%

D.

2 and 0.5 2 and 1

6:zo: CE Board November 1998

61.6: ECE Board April 1998

A. B.

C. D.

\.

0.64 m/min 0.56 m/min 0.75 m/min 0.45 m/min

3 and 2 3 and 1

A statue 3 m high is standing on a base of 4 m high. If an observer's eye is 1.5 m above the ground, how far should he stand from the base in order that the angle subtended by the statue is a maximum.

h. What percent is the volume of the largest cylinder which can be inscribed in the cone to the volume of the cone?

m/s m/s m/s m/s

Water is flowin~ into a conical cistern at the rate of 8 m /min. If the height of the inverted cone is 12 m and the radius of its circular opening is 6 m. How fast is the water level rising when the water is 4 m deep?

C.

10 kph 13 kph 11 kph 12 kph

61%: ECE Board November 1998 Given a cone of diameter x and altitude of

C. D.

The coordinates (x,y) in feet of a moving particle P are given by x = cost- 1 and y = 2 sin t + 1, where t is the time in seconds. At what extreme rates in fps is P moving along the curve?

3.64 3.94 4.24 4.46

619: 'E Board October 1993

611: EE Board October 199% 6J.S: EE Board April 1997

crosses a street at 9 m/s at the instant that a car approaching at a speed of 4 m/s is 12 m up the street. Find the rate of the LRT train and the car separating one second later. A.

3.41 m 3.51.m 3.71 m 4.41 m

Water is pouring into a conical vessel 15 em deep and having a radius of 3:75 em across the top. If the rate at which the water rises is 2 em/sec, how fast is the water flowing into the conical vessel when the water is 4 em deep? A. B. C. D.

2.37 5.73 6.28 4.57

After t hours, there are t +

3

m /sec m 3/sec 3 m /sec 3 m /sec

Jt

gallons in

the pool. At what rate is the water pouring into the pool when t = 9 hours? A. B. C. D.

8 fps 8.25 fps 8.33 fps 8.67 fps

6181 An LRT train 6 m above the ground


The cost per hour of running a motor boat is proportional to the cube of the speed. At what speed will the boat run against a current of 8 km/hr in order to go a given distance most economically?

Water is pouring into a swimming pool.

~

~1.15 m 1.33 m

C. 1.50 m D. . 2~

120 100 C.· 140 D. 190

6:u: ME Board October 1996

D.

A B.

A. B.

·,i

Problem6171 A man walks across a bridge at the rate of 5 fps as a boat passes directly beneath him at 10 fps. If the bridge is 10 feet above the boat, how fast are the man and the boat separating 1 second later?

7/6 gph 8/7 gph 6/5 gph 5/4 gph

6:Z:ZI A helicopter is rising vertically from the ground at a constant rate of 4.5 meters per second. When it is 75 m off the ground, a jeep passed beneath the helicopter traveling in a straight line at a constant rate of 80 kph. Determine how fast the distance between them changing after 1 second.

A.

B. C. D.

12.34 m/s 11.10 m/s 10.32 m/s 9.85 m/s

6%31 ECE Board November 1991. A balloon is released from the ground 100 meters from an observer. The balloon rises directly upward at the rate of 4 meters per second. How fast is the balloon receding from the observer 10 seconds later? A. B. C. D.

1.68 m/sec 1.36 m/sec 1.55 m/sec 1.49 m/sec

'&24: ECE Board April 1998 A balloon is rising vertically over a point A on the ground at the rate of 15 ft./sec. A point B on the ground level with and 30 ft 'from A. When the balloon is 40 ft. from A, at what rate is its distance from B changing?

A. B. C. D.

13ft /s ,15ft/s 12ft /s 10ft/s

342 1001 Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas &25: CE Board May 1997 Car A moves due East at 30 kph at the same instant car B. is moving S 30• E, with a speed of 60 kph. The distance from A to B is 30 km. Find how fast is the di~tance between them separating after one hour.

&29: ECE Board November 1998 What is the allowable error in measuring the edge of the cube that is intended to hold 8 cu. m., if the error of the computed volume is not to exceed 0.03 cu. m? A.

A. B. C.

D.

36 38 40 45

B.

kph kph kph kph

C. D.

&2&: CE Board November 199& A car starting at 12:00 noon travels west at a speed of 30 kph. Another car starting from the same point at 2:00 P:M. travels .. north at 45 kph. Find how (in kph) fast the two are separating at 4:00P.M.?

&30: EE Board October 1993 A standard cell has an emf "E" of 1.2 volts. lfthe resistance "R" of the circuit is hm/sec, at increasing at the rate of 0.0 what rate is the current "I" hanging at the instant when the resist ce is 6 ohms? Assume Ohm's law E IR.

A.

A.

B.

c. D.

49 51 53 55

B. C. D.

Topics

0.002 0.003 0.0025 0.001

-0.002 amp/s 0.004 amp/se -0.001 ampdsec 0.003 amp/~c

D Mon

&Q] Tue

D D D D D D D

I: ,J !~·

~~

&27: CE Board May 199& Two railroad tracks are perpendicular to each other. At 12:00 P.M. there is a train at each track approaching the crossing at 50 kph, one being 100 km and the other 150 km away from the crossing. How fast in kph is the distance between the two trains changing at 4:00P.M.?

A. B. C. D.

&28: CE Board May 1995 Water is running into a hemispherical bowl · having a radius of 10 em at a constant rate 3 of 3 cm /min. When the water is x em. deep, the water level is rising at the rate of 0.0149 em/min. What is the value of x? 3

B. C. D.

2 4 5

Wed

Problems

Thu

Solutions

Fri

Notes

Sat

ANSWER KEY

67.08 68.08 69.08 70.08

A.

Theory

11 I

'

'i!! t

"1.

., 1 ' 11

'>

596. B 597.C 598.A 599.A 600.A 601. B 602.C 603.A 604.C 605.C

606.C 607. B 608. B 609. B 610. B 611. D 612. A 613.C 614. B 615. D

c c

616. 617. 618. A 619.A 620.C 621. A 622. 623. D 624.C 625. D

c

----·-.~---


RATING

626. B 627.A 628.C 629. 630.C

c

c:J c:J c:J c:J

30-35 Topnotcher 21-29 Passer 18-20 Conditional 0-17 Failed

If FAILED, repeat the test

.


•

'I

3

2=L 32 y3=64 (4,2)

'

1 l

y=4

Day 14- Differential Calculus (Maxima-M~-~'rilne Rates) 345

•

Thus, the size of the field is 25 m x 25 m.

Note: For maximum area, the triangle must be an equilateral triangle.

Substitute y = 4 in Eq.2:

•

h

' )2 (4 =2 y=-a

Substitute x

=2 and y =4 in Eq.1:

d=J<'X~4) +(y-2r ~Eq.1

2

d=J8 =~4(2)

y2 =8x

y2

A=.!x 2sin9 A= i<6rsineo

8

Ill


··K~ -·r Note:

x and y = two positive numbers P = product of x and y

+(y-2f

:t

x+y=50 y =50-x ~Eq.1 P=xy ~Eq.2

l

f-4)(¥)+~(1) 2./( ~ -4J +(y-2r

2(f-4 )(¥ o- •. o=(

Substitute Eq.1 in Eq.2: P = x(50-x) P=50x-x 2 Differentiate: dP -=50-2x dx 0=50-2x x=25

)+2(y -2)(1)

f: -·r

+(y-2f

Substitute x = 25. in Eq.1:

~ -4)( 2i)+
y3 0=--y+y-2 32 '

,,

•

X

y

!

A

y

'

':'J

Differentiate:


A=15.59cm2

Let:

du d(-!U) =2Tu

d'= 2(

A= 2xrh + 2(nr2} ~ Eq.2

2

d=2../2

x=--~Eq.2

108 = nr2h _ 108. E h --~ q. 1 xr2

3x=18 x=6

d = ~(2- 4) +(4--.:-2)2

2

V=nr2h

X

X

2x+2y=100 y=50-x ~Eq.1 A=xy~Eq.2


=2\~( ~~~ )+2(nr

216 A=-+2xr2 ~Eq.3

r

Differentiate: dA _ -216 +4xr dr r2

--

-216 +4xr 0=-2r 216 =4xr

r2 _r3 =216

A= x(50-x) A=50x-x2 Differentiate:

4x r=2.58 in Substitute r in Eq.3:

dA =50-2x dx 0=50-2x x=25

y=50-25· y=25

Substitute x


y = 50-25 y=25

=25 in Eq.1:

216 >2 A = - + 2n(2.58, 2.58 A= 125.544 in2

2 }

. Day 14- Pifferential Calc.ulus (Maxima-Minima & Time Rates) 347

346. 100 i Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas

Ill

o· J:

Ill

10

X

20

...' f

X

h

20 +y

5.'

h

:.···1''1

'.

). --::

•· }

10

'•

y

.

:5

X

d

II

A =.::_d2h

4

2x

10 +X A=(10+x)(20+y)

X

A= 200+10y + 20x + xy--+ Eq.1 ,,.,

2

'!.

20-2x V = (20- 2x)(20- 2x)(x) V

X

=(20- 2x)2 (x)

~j

I


V =(400-80x +4x 2 )(x) V = 400x- 80x 2 + 4x 3 --+ Eq.1

'

Differentiate:

=200+1o(_3 ~ 0 )+20x+ \( 3~0 )

I

~

,'~i

A= 500 +

dV 2 - = 400-160x +12x dx 0 = 400 -160x +12x 2

3000

dA dx

=0-~+20 x2

1

2

nx2 --+ Eq.2


i II

Differentiate:

0 =- 3000 +20 x2

x = 10 in (absurd)

A=7tdh+2( A= 1td(

x 2 = 150

x =3.33in

x =12.24cm

2 A = 45.2 + 1td

d

v = 400(3.33)- 80(3.33)2 +4(3.33)3 = 592.59 in3

Solving for h:

Thus, the dimension of the poster is, -~

= total area

'·

(1 0 + 12.24 )(20 + 24.5) or 22.24 em by 44.5 em

Ratio= 2X h Ratio= 1

Differentiate: dA = -45.2 +.::_(2d) dd d2 2

1t d= 2.432 in

y=0.14P

300 =24.5cm y = 12.24

2

d

y = 0.5P- 2.57(0.14P)

Substitute x = 12.24 in Eq.2:

:~;)+2(~ )d2

3 45.2 d=~

x=0.14P

Substitute x = 3.33 in Eq.1:

~d2 )

-45.2 + d 0=-2- 1t.

dA -=P-7.14x dx O=P-7.14x

3000 =20 x2

3x =10

=area with the picture

A = y(2x) +

2

4x=40

Let: A Ap


A= Px -5.14x 2 +.::_x 2 2 A = Px- 3.57x 2

,li~

~d2 )--+ Eq.2

y = 0.5P- 2.57x--+ Eq.1

A= 2x(0.5P- 2.57x)+ 2x2

Differentiate:

(4x -40)(3x -1 0) = 0

•

P-2X-1tX y= •.., 2

}~

+ 20x--? Eq.3

X

By factoring:

V

A= 1tdh + 2(

P = 2x + 2y + 1tX

l'

·~

A

4 45.2 Eq. 1 d -----? 1td2

1 P = 2x +2y +-(27tx)

AP =xy 300 = xy 300 y=-. -?Eq.2

11.3 =.::.d2h

Let: P = perimeter

•

Let: C = total cost per hour N speed in miles per hour

=~= x+y

2(0.14'R_) 0.14'R_+0.14"R:

=

C = fuel cost + others C = kN2 100--+ Eq.1

+

Day 14 -

348 100 i Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas

C = 15x2 +'24xh--+ Eq.1

2

25 =k(25)

V=x 2h

1

k=25

h=

•/1

10

\'~

~Eq.2 2 X


Let: x = total cost per mile

C:; 15x2 + 24x(:~)

N

_ 6 -7.5

75 J<4.5)2 +(9-.x)2 =-·- (9-x)

6

~(4.5)

p

= 199,999.457

p

=200,000

11 ) · 0.371 + 1

(9-x)2

8

x=3km

a

I y

L-'

4000x + 50

dC -=2x-4000 dx

x+y=20

30

A= xy

= 2 in Eq.2: Let: T = total time needed T = time to row + time to walk

10 =2.5m =(2)2

d

•

h

Differenti.ate: -1.1 1)7( +1) )

dP _ 1.1 1 8 dx - - - ( X +

X

Let: C = total cost

\

(X

2

Differentiate:

F-5) +(9-4 2

!

8

P=200,000-x- ( - 1.1 ) -+Eq.1 x+1

A= x(20-x) A= 20x-x 2

X

T=-+6 7.5 T=

Ill

x=2000

~Eq.2


~~

2x =4000

y=20-x~Eq.1

B

Thus, the dimension of the tank is, 2 mx2 m x2.5 m

0=2x~4000

X

4.5

x=2m

h

= 36

9-x =6

•

3 __ _ 240

Ill

2

0.5625(9- x)2 = 20.25

240 =30x

X -

2

+ (9- x) = 1.25(9- x) (1.25)2(9- x)2 = (4.5)2 + (9- x)2

=0.371

Substitute x in Eq.1:

7

Substitute x

-

·~ it 'll

dC = 30x- 240 x2 dx 240 0=30x--· x2

N=50mph

C=x

'i

9-x

X+ 1 = ~8(1.1 )

P=200,000-0.371-(

Differentiate:

0 = _!_+ (-100) 25 N2 2 N =2500

2

~;,!lll1

75 ·

~(4,5f +(9-x)2

8

X

- - - + -2-

6~(45) 2 +(9-x)2

+ 1)9 = 8( 1.1 )8

X

C=15x2 + 240

_!_N2 +100 x= 25 N N 100 x=-+25 N dx _ 1 (-100) 25

(X

!

C=_!_N2 +100 25

x = total cost per hour . miles speed I n - hour

-1-8(~) (~) x +1 (x + 1)2

0 = -1 + 8(1.1 )8 (X+ 1)9

10= x 2h

Substitute k in Eq.1:

dN

0=

(Maxima-Minima & Time Rates) 349

~-(~9=-x:fo)== + _1 = 0

7

C = x2 (15)+ 4(xh)(6)

Fuel cost =kN2

Differ~ntial Calculus

6

X

dA =20-2x dx 0 = 20-2x

+ 7.5

du

l

Note:

f

Differentiate:

X

d( .JU) =2.JU

=10

Substitute x = 10 in Eq.1:

dT -(

11

<.ix-- 6')

y = 20 -10 =10ft 2(9-x){-1)

2J(4s)2 +(9-x)2

1

+7.5

Substitute x and y in Eq.2: A= (10)(10) = 100 ft2

' 350 100 1 SOlved Problems in Engineering Mathematics (2"d Edition) by Tiong & Rojas

Ill Let: C V

V

=C(OSt per hour =speed of the motorboat

c, =total cost

=nr2h-+ Eq.1

Ratio=:!___= n(-;..,

~=-r

Ratio= 0.44

h _ RH-Hr where: k = proportionality constant _ distanee = ~ -+ Eq.2 t- speed V-8

--R~-+

•

E q. 2

Substitute Eq.1 and Eq.2 in Eq.3:

H V ._ -nr 2H -nr- R

x

o

B ,,~ - l':'l. ~

4-x

•. •

E4

)o

I

X

E = intensity of illumination

3

A & B = luminous intensities of the two light sources respectively. k = proportionality constant

3

dC1 _ (V- 8)(kS3V 2)- kSV 3 (1- 0) dV (V -8)2 = (V- 8)(kS3V 2)- kSV 3 0 (V -8)2

kSV3_ = (V- 8)(3R&_~) V =3V-24 2V=24

2 dV -=2 3nr H dr nrH ---

R

x2

=2~)(

x2

Differentiate:

2 r=-R 3

dE= -kA(2x) + (-8kA)(2)(4-x)(·1) dx

•

0

R_

3

Substitute r and h in Eq.1:

r(

h

2 V =n(_i.)R H

27

R Let: R radius of the cone H = height of the cone r = radius of the inscribed cylinder h = height of the inscribed cylinder V = volume of cylinder

x4

(4-x)4

= -kA(2x) + (8kA)(2)( 4- x} 4 4 x (4-x)

k1$'2_$- (8k1\}('2_)~ -~--

V=n(~R iH)

H

x3

(4·x)"-

8 --3 (4-x)

=

2 Vc .!.nR H 3

dL

-hcose

-2(-sine) sin 2 e cos 2e = -hcose + -2(-sine) 0 sin 2 e cos2 e hcose 2(sine) sin 2 e = cos2 e

-=~+-~~.:..

de

3

h = 2(sin e) cos 3 e h =2tan3 e-+ Eq.2

(4- x)3 = 8

--;(3"

Substitute L

J(4~xr =~

.) ..

3

X

Let: Vc = volume of the cone

_2_ cose

Differentiate:

R

h = 'R_H- H(2/3 'R_) = _!H

y

L=x+y h 2 L=--+-·--+Eq.1 sine cose

(4- x)2

Substitute r in Eq.2:

v =12 kph

COSO=~

(4-x)2

E = kA + k(8A)

R-

x= _h sine

Y=

E=kA+~

0 = 2nrH- 3nr2H 3\r"»-H_

sine=~

Let:

Differentiate:

kV S c,=-V-8

=

~

•

,.. - 2 (RH-Hr) V -nr --R

c,=kv (~) V-8

A ·~~ - ""' ~


c1 = Ct -+ Eq.3 3

a

)~

~7t~

vc


H H-h Hr =RH-Rh

C = kV 3 -+ Eq.1

Day 14 ~Differential Calculus (Maxima-Minima & Time Rates) 351

·,

4-x =

2

X

4 _ x =2x x. l.JJ m

=6 and h =2 tan 3 e in Eq.1: 3

6 = 2tan e +-2-. sine 3

cose

= 2tan ecos8+2sine 6 sinecose 6sinecose = 2tan 3 ecos0+ 2sine

(Maxima-Minima~ Ti.~-R.~tes)

Day 14- Differential Calculus 3

6sinecos9 = 2( sine ) ccis9 + 2sin9 cose

a

du

(x2 + 13.75 )3- 3x(2x)

=-----·-------

(x2 +13.75t

s2 = 125(1)2 + 1oo

Substitute:

SubstituteS::: 15 in Eq.2:

2

. 9 ) +1 3cos9= ( ~.

de

du Cix"'1;u 2

cos a

sin2 9 + cos2 9

Substitute 9 In Eq.2:

X

55

tan(a+9)= ·

tan a+tan 9 5.5 =1-tanatane x 2.5 t 9 x+ an 2.5t 9 1--an

3.-108

5.5

/ £·----· _/I

dt

dy =2cost dt

dy =2(1)= 2 dt Thus, the extreme rates are 2 and 1.

L--~" __;1=1ot /~

-X

~-osition of the boat

V1 = 10

I

l_· v~fter I seconds

'~

1 9 = tan- (x2 +;31x3.75)

Jl

2

:11

S '= S/ + S/ + 10

2

:/ oc(10!)2 +{5tl +10 2

du Note: d(tan-1 u) = 1+ u2

s7

.·

.

2 +-1-:-3-.7-5' u =-x"""

if=

9x

=(12-4ti" +(9t)2 +6 2

2

..

(x 2 + 13.75)'

Differentiate both .sides with respect to

2s( ~~}=194t(*)-s6(*) +O dS 194t-96 = - - - - - - _, Eq.2 dt 2S

-

Substitute 1 = 1 in Eq.1:

·125t2 + 100---+ Eq.1

I )ltferentiate both sides with respect tot:

where:

3x

-----

S2

S 2 = 97f - 96t + 180 __,. Eq. 1

.···

13.75

3x tan9= x2+13.75

y =2sint+1

V1 = 4 [POl·.sitton of the c~r . after t seconds

2

10

X+

.--1 . \.. _ _ _

s = 144 - em + 1st2 + 81t 2 + 36

13.75) tane ( x+-.-x- =3

3

~~ is maximum ifcos t = 1

I

X

tane =

-

;;;::..-----------'---

S2=?L .

,;/sz= 9t

s,=~t N2-4t

of!he ~~a:J r.PoSiiic)n aftertseconds

v2 =5

=-x

16

.:.::·:.o·

.~1

X

dx=-(-1)=1

Note:

a

=13.75

..J

~:---~

.

13.75 2.5 + xtane = 5.5 ---tane

~~ is maximum if sin t = - 1

2

X=

X

h=2.24 m

Note:

if;:/ lb&

)3_ -·

2x 2 = x 2 + 13.75

h = 2tan46.1.

dt

osition ~t th$ ii-ai~ .

r:P_.r

l_ aftert seconds

(x + 13.75 ~x(2x) · 0 =--------(x2+13."15f 2

9=46.1.

dx . -=-s1nt

m1

O=du

cos3 9 = 0.333

x = cost-1

dt-=-15

0'~-:;-:;-_-;:l

3cos3 9=1

12~(!L "" 8.33 fps

dS

du

2 cos9

•

Substitute t = 1 in Eq.1:

s = 15

sin9r 6cos9=2 ( - +2 cos a

3cos9=

353

·:f(\~ ) 250t {1) + 0 d:;

I )~it

dt

"

,.

s 2 = 97(1)2 -- 96(1) + 180 s = 13_45 Substitute S = 15 in Eq.2:

dS '194(1)- 96 = 3.64 m/s

>I q 7

d! = --2(13A's)

t:


a

V=

R=6

1

3

Day 14- Differential Calculus (Maxima-Minima & Time Rates) 355

rm

2

nr h ~ Eq. 1


V = 80

15

1

3

1000 m 1 'km.

4

s2: 4.5!

I

I

t

'

V=in(%r h

r -=12 h h

V = __!_h 3 ~ Eq.2 48

.

r=-~t::q.2

Differentiate both sides with respect to

2

2

-

~

t:

dV = 3n h2 dh dt 48 dt


s 2 = 1oo2 +(S1 )2

v1 = 22.22

75

6

s1 = 22.22t

1 '( h ) h V=3n2

= 2 and h = 4:

s 2 = 493. 728t2 + 5625 + 675t + 20.25t 2 s 2 =513.978t2 + 675t + 5625 ~ Eq.1

dV- 3n (4)2(2)

V=~h 3

dt- 48


dV = 6.28 cm3/s dt

12

dV = 3n h2 dh dt 12 dt Substitute dV/dt = 8 and h = 4:

t:


dt

•

2s(

t:

~~) =2(513.978t)oo +67500 dS dt

Let: Q = rate of discharge in gph

= 513.978t+337.5 ~E

Substitute t

Differentiate both Sides with respect to

dQ

dt=

1

1

+

dS dt

= 15t ~Eq. 2

s Substitute t =10 in Eq.1: s 2 =1o,oooo+16(10)2

t:

s

q

_ 2

=1 in Eq.1:

s2 =513.978(1)2 + 675(1) +5625

Substitute t = 10 and S = 107.7 in Eq.2: dS

dt =

16(1 0) = 1.485 m/s 107.7

1m

2Ji.

Substitute t = 1 and S

s1 = 15t = 82.547

in Eq.2:

R=3.75

H-1~ J-~-,

1

t "~

Substitute t

=9 hours:

1

dQ

1

-=1+--=1+dt 2.[9 6

dQ

7

- = - gph dt 6

t:

)=0+32too

s =82.547 m

dh = 0.64 m/min dt

1m

= 1o,ooo + 16t2 ~ Eq.1

8=107.7 m

O=t+Jt

3n 2 dh 8=12(4)

s

2

2s( ~~

s 2 = (S 1 ) 2 +(75+S 2 )2 2

Substitute dh/dt

S =100 2 +(4t)2


S = (22.22t)2 + (75 + 4.5t)2

2

s1 = 4t

~ V2=4.5

~Eq.2

s


s

V1 = 22.22 m/sec

Substitute Eq.2 n Eq.1:

nr 2h ~ Eq.1

X

h

h

r=

V=

1 hf:.. 3600 sec

'km_ X

hi:..

1 .

3.75 r --=-

1m

dS

513.978(1) + 337.5

-~;

dt dS dt

B

-·

82.547 2

10.315 m/sec

S = 30 2 + (S, )2 S

2

s2

= 30~ + (15t)2

=goo+ 22st2 ~ Eq.1

A

356 lO
2s(

~~)=0+4501~


Day 14- Differential Calculus (N.[a_xirna-Minim(l & Time Rates) 357

t:

Substitute t = 2 hours (from 2 PM to 4 PM) in Eq.1:

2S(~T-)=5400t~ -2700~ ' ,~1

dS = 225t ~ Eq. 2 s dt

dS dt

When S1 = 40 ft,

2700t -1350

s

s

40 = 15t

2

,;~.,l ,i

q. 2

s

= 2925(2) + 3600(2) + 3600

Substitute S = 111.8 and t = 4 in Eq.2: dS dt

= 1 in Eq 1:

rm

dS =51 k h p dt

t = 2.667 sec SubstituteS = 30 and t = 1 ir. Eq.2: Substitute t

=2.667 in Eq.1:

s =50ft

dt

;I

dS =45 k h dt p

Substitute t dS

m.l

dS·= 2700(1)-1350 dt 30

s 2 = 9oo + 225(2.667)2

:c

2.667 and S = 50 in Eq.2:

225(2.66 7 ) = 12 ft/s 50

VA= 50

~~

N

·~

rm

A

100

1tX2

V=-(3r-x)

3

.r

f,,

l

50t-150

Q

6. U"1

ml

0

iO

N

4ltt"'-vs=60 B

$ A~

SA = 30t

60

-~

~ SA=30t 9 A'a.:~,--~-

1tX2

>::>s

·--

V=-(3(10)-x)

3

-:

1tX3

Vs= 50

V=10nx 2 - -

0 0

t

m

(/)

-vA=3o

::

Vs=45

"
5000( 4) -12500 111.8

dS = 67.084 k h dt p

dS = 2925(2)+1800 dt 150

= 2700(1f - 2700(1) +goo

25, 000( 4) + 32, 500

S = 150 km

I

S = 30 km

-

S = 111.8 km

2

2

Substitute S = 150 and t = 2 in Eq.2:

Substitute t

s 1 = 15t

~

E

8 2 = 5, 000( 4 )2

3

Differentiate both sides with respect to t Ss =50!

B

Position of carA'i

at2~~-j

s

2

2

=(50t-100) +(50t-150)

S

2

=2500t2 -1 OOOOt + 10000 + 2500t2

dV = ( 2onx _ 3nx dt 3

2

2

Jdxdt

-dV = ( 20nx -nx 2 )dx - ~ Eq.1

dt

dt

-1 5000t + 22, 500 s2 = (60 + SA)2 +Ss2 2

s = (60 + 30t) + (45t)

By cosine law: 2

2

s 2 = 5000t2 -

2

I lifferentiate both sides with respect to

s2 = 36oo + 360at + 9Dat2 + 202st2

2

s = (30t) + (60t- 30)

2

2

S

"'

il

2925t + 3600t + 3600 -~ Eq.1 2

-2(30t)(60t- 30)cos60 s 2 = 9ooe + (60t- 30)2 - 30t(6ot- 30) s 2 =9ooe + 360at 2 - 360at + 9Do -1800f + 900t s 2 =21om 2

-

21oot + 9oo ~ Eq.1


2s(

~~) = 2(2925t)~ + 3600~1J, .!:!_~ "' _2925t + 1800 ~ Eq.2 dt

s

t:

I

/S( ~~)

"

dS dt

~;

Substitute dV/dt = 3 and dx/dt

25000t + 32,500 ~ Eq.1

t:

3 = (20nx -nx 2 )(0.0149)

2(5000t)~ -25000~ 5000t -12500 ~ Eq.2

--s

:;utdihr\(; I= 4 hours (from 12 PM to 4 Ill lq t·

I'M)

= 0.0149 in

Eq.1:

201.342 = 20nx -nx 2 64 = 20x- x2 x

2

-

20x + 64 = 0

(x-4)(x-16)=0

x = 4 em or x = 16 em

358 1001 Solved Pcoblems In Englnee
•

M~

0CL_.don) byTiong & Rojos

V = x 3 ~Eq.1

Topics

Differentiate both sides with respect to its individual variable:

D

dV = 3x dx ~ Eq.2 2

Mon

Substitute V = 8 in Eq.1:

Tue

8= x3

D D D D D D D

x=2 Substitute dV = 0.03 and x = 2 in Eq..2: 0.03 = 3(2)2 dx dx =0.0025

•

.·~

m

il ~~

'"'

l

E=IR 1.2 = IR ~ Eq.1

Differentiate both sides with respect to its individual variable: f\lote: d(uv) = udv +'vdu

o~{:)+R(~!)~Eq.2 Substitute R = 6 in Eq.1: 1.2=1(6) 1=0.2 Substitute I = 0.2, dR/dt = 0.03 and R = 6 in Eq.2: 0 = (0.2)(0.03) +

JJ

·.~

I

(6{ ~~)

dl - = -0.001 amp/sec dt

il

li

T T T T

Theory

Wed

Problems

Thu

Solutions

Fri

Notes

Sat


,.;..r,.,

,,

1'opics

0 0

.

.

.

,.~.

/'

. T. ,T .....

;~ ',;. > S

.,.·~+.!•"\

·~ .« <

···<~=~"~<'<"''~

4

'

'

> , < '"-" I

\> '· ( ·>

.

Mon

.

Tue

.

l~J

.

. !•.

Theory

VVecl

Problems

Thu

Solutions

Fri

Notes

Sat

[] [] D [] [] 0

t'llj i'

~hat

integral Calculus Definite and Indefinite Integrals Fundamental Theorem of Calculus Basic Integrals Formulas - Exponential, Logarithmic, ·rrigonometric, Hyperbolic, Inverse Trigonometric, By Parts, Trigonometric Substitution, Wallis Formula, etc. Applications - Area, Centroid, Arc Length Surface Area, Volume, Work Moment of Inertia, etc. Propositions of Pappus Hooke's law Multiple Integrals

is Integra! Calwlus?

b

Example: Integral Calculus is the branch of calculus which deals with functions to be integrated. Integration is the reverse process of differentiation. The Junction to be integrated is referred to as the integrand while the result of an integration is called integraL

lhe integra! symbol or sign

J

Example:

II

dongated S denoting sum (Latin: summa), was introduced by Leibniz, who named mtegral calculus as calculus smnmatorius.

i!

What are Definite and Indefinite

P<'luuk' tnlegral is an integral that is "' ~fuwd l•y IIH • ltm'f v; dtt"!~ a and b of the Huf<·p• 'IHil 'IIi \l.llf.liil<'

Indefinite integral is an integral with no . restrictions imposed on its independent variables. It is also called antiderivative or primitive integral. Jt(x)dx

is an

lnte.m~l.l~

.·,..

Jt(x)dx

What is the Fundamental Theorem of Calculus? Suppose f is a continuous function in a closed interval [a, b], then '

F'(x) = f(x) for ali x in [a, b). The fundamental theorem of calculus S!
Day 15- Integral Calculus 363

362 1001 Solved Problems in Engineering Mathematics (2"d Edition) by Tiong & Rojas b

Jt(x)dx = F(b)- F(a)

~u~---

19. Jsecut What are the Integrals of the Different Functions? The following are the integrals arranged according to its functions: A.

Basic Integrals:

1.

fdu=U+C

2.

Jadu =au +C

3.

u"du=--+C f n+1

(n ,, -1)

du

4.

f

B.

Exponential & Logarithmic Functions:

u=lnu+C

20. lcscucotudu = -cscu+C

41. fsinh udu = ~sinh2u -iu+C

21 I Jsin 2 udu = .!u- .!sin2u + C 2 4

42. Jcosh 2 udu =.! sinh2u + !u + C 2 4

!2_) Jcos

2

udu =

iu

Using a vertical differential strip:

2

+ ~sin2u + C

43. Jtanh 2 udu = u- tanh u + C

23. Jtan2 udu =tan u- u + C

44. Jcoth 2 udu = u- cothu + C

24. Jcoeudu=-cotu-u+C

F.

Trigonometric Substitution:

D.

44.

J~a 2 -u 2 du

let : u = a sin e

45.

J~a 2 +u2 du

let : u = a tan e

Inverse Trigonometric Functions:

25. Jsin-'; udu = u sin- 1 u + ~ + C

46.

J~u -a du

27. Jtan- 1 udu=Utan- 1 u-Jn~ +C

G.

Integration By Parts:

1

5.

feudu = eu + c

6.

au audu=-+C f Ina

7.

Jue"du = e"(u -1) + C

= u cot- 1 u +In J1 + u2 + C

Jlnudu = ulnu-u+C

9.

f~=lnllnui+C ulnu

30. Jcsc- udu = ucsc- u + lnlu + -Ju 2 1

1

E.

Hyperbolic Functions:

31. fsinh udu =cosh u + C

12. Jcosudu=sinu+C

34. Jcothudu=lnlsinhui+C

13. Jtanudu=ln!secu!+C

35. Jsec hudu = tan- 1(sinh u) + C

16. Jcscudu = ln!cscu- cotul + C

47. Judv

uv- Jvdu

-11

Wallis Formula: Formulated by John Wallis.

.

X

2 48. fsinm ecos" ede =

A·= Jxdy

0

[(m -1)(m ·- 3)· .. 1or2][(n -1)(n- ~)···1or2)

'-'----'---'----~-'-'-----'---"•a

(m + n)(m + n- 2)(m + n- 4)···1or2

B. Polar coordinates:

32. Jcosh udu =sinh u + C 33. Jtanh udu = In !cosh ul + C

15. Jsecudu = ln!secu + tanul +C

Using a horizontal differential strip:

y

H.

+C

11. fsinudu=-COSU+C

14. Jcotudu=inlsinui+C

let : u = a sec e

1

J

Trigonometric Functions:

2

A= Jydx

29. Jsec- udu = usec- u -lnlu + N~~ 1

+C 8.

2

26. Jcos- 1 udu =U cos- 1 u- ~1- u2 + C

28. Jcot- udu

17. Jsec2 udu=tanu+C

40. fcschucothudu = -cschu+C

A. Rectangular coordinates:

0

un+1

C.

39. fsechutanhudu = -sechu+C

18. Jcsc 2 udu = -cotu+C

36.

Jcschudu=lnltanh~~+C

37. Jsech 2udu=tanhu+C 38. Jcsch 2 udu = -coth u + C

y

where: a = ~ if both m and n are even

a = 1 if otherv:ise How to Find an Area using Integration? One import::>nt use of definite integrals is the determination of area between two curves with given equations. This is done by using. a differential area (horizontal or vertical strips for rectangular coordinates of a differential triangle for polar c:o'oldlll
0

X

A=

r2 h·d9

"\

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _..:.D::.:a::.yL'.:.:lS::___-..::In:::t:.:::eSE~Calculus~

364 1001 Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas ,tlow to Find the Location of the Centroid using Integration?

The Second Proposition of Pappus states as follows:

How to Find the Surface Area of a Solid Figure by Integration?

The centroid is the center of mass of a • given figure. The figure below illustrates on how to fin\l the location of the centroid of an arc .. bounded by curves of given equations.

"If an area is rotated about an axis, it will generate a volume equal to the product of the area and the circumference described its centroid."

The surface area of a solid generated by revolving a curve about a certain axis may be calculated using the first proposition of Pappus.

d

The First Proposition of Pappus states as follows:

y

A

e.T;:tt{!tttl :~::18

centroid "If an arc is rotated about an axis, it will generate a surface area equal to the product of the length of the arc and the circumference described its centroid."

.....,

~

X

0

y

y

V ':" J2nyxdx

B. Using a horizontal differential strip (Ring Method):

y 0

-~s,

:

X

x

.

~ f\ ········;···d

X

_ fdA·X X;-

11

_ fdA·r

A

Y=

v

:·······

or

S= J~1+(~~Jdx x, f 1+(ddy )2 dy

y,.

S=

X

Yt

A= JJ1+(~~J

>......!_.........

~

X

A. Using a vertical differential strip (Shell Method):

The figure formed by rotating the horizontal strip about the vertical axis is a cir.cular ring or washer.

where: S =length of arc d = distance from centroid to the axis of rotation

(xz,Yz)

0

V= fdA ·2nd

If the area is not given or cannot be computed easily, the following will be used rn finding the volume of the solid:

A= S ·2nd

----4-------------------------~~

or

where : d = distance from centroid to axis of rotation

\::I

ds

~dy dx

V =A ·2nd

X

0

-· _2 A

How to Find the Arc Length of a Graph?

y

:

: : ''

:

X

----~x

How to Find the Volume of a Solid Figure by Integration? The volume of a solid generated by revolving 11 curve about a certain axis may be calculated using thesecond proposition of Pappus.

+~ dY~ j' :

:

dx ·2nd

•

•

•,.,,.,,,

' ' .c'

"t"" .

v.

•

:

... -

,

J x/ - x/ ~y

. V = n(

f 1'

I IH~ tr~ 1 ure formed by rotating the vertical ·.trrp ;tlmut the veriical axis is a hollow 'vlindrical shell

366., 100 I Solved Problems in Engineering Mathematics (2"d Edition) by Tiong & Rojas How to Solve for the Work Done by Integration?

What are multiple integrals?

Work is defined as the force multiplied by ·--~ce traversed. If the force i.s constant throughout the distance, then work is simply the product of the applied force and the distance traveled. However if the applied force varies with the distance, then the best solution is to apply · integration.

An integral in which the integrand is integrated twice is a double integral. Example:

fJf(x)dx

2

Topics

0 0

If integrated thrice, the integral is called triple integral.

lvlon

Example:

ffJf(x)dx

3

Tue

Work= fF·dX

=force expressed in terms of x

where: F

Example:

What is Hooke's Law? Hooke's Law, named after the English mathematician and scientist. Robert Hooke (1635- 1703) is stated as follows: 'Within elastic limit, the force required to stre~ch (or compress) a spring is proportional to the change in length."

F = kx where: k

GOOD LUCK I

y

X

y

'{[;ribia:

0 0

Solutions

Notes

0 0 0

Thu

Fri

Sat

b:U: ECE Board April 1:999 What is the integral of (3t- 1)3 dt?

A

Did you know that... the propositions of Pappus was inspired by two fruits, namely apple and lemon! Pappus imagined a circle cut by a line not passing though its center and rotate the area of the major part about the said line and produced a solid which looks like an apple while the remaining minor area when rotated about the same line forms what is known as a lemon.

~ Wed

ll. ('>.

12 1

--(3!-4) +C 12 1 )4 +C ~(3t-1

(>.J:Z:

1

~(3t-1)

4

0

= fdA·y

2

'y = JdA·-(~Y 2,

X

(() 10

"It is truth very certain, that when it is not in our power to determine what is true, we ought to follow what is most probable."

1\

:> \/:'

I\

IJ:>

(.

lrr :\ lrr :•

II

~sin( 2x 2 + 7 )+C

B.

~cos( 2x 2 + 7 )+C

3

C.

( sine ) +C 2 4 ( x +7 )

D.

sin( 2x + 7 )+C

+C

2

ECE Board November 1:998

~uote:

- Rene D~scartes

A. 4

4

ll.

633: ECE Board November :1998, ME Board April :1998 Integrate x cos ( 2x' + 7 ) dx.

1 ( 3t-1) 4 +C --

I valuate the integral of dx I (x +2) from -6

---------L--+-----------~----~

lx

Problems

Functions of more than one variable may be integrated with respect to one variable at a time while the other variables are held constant, reversing the process of partial differentiation.

=spring constant

How to Solve for the Moment of Inertia by Integration?

Theory

f J-· Jf(x)dx"


For spring:

0

An integral in which the integrand is integrated to n times is n-fold iterated integral.

jlnt;gral Calcul~ Definite and Indefinite Integrals Fundamental Theorem of Calculus Basic Integrals Formulas - Exponential, Logarithmic, Trigonometric, Hyperbolic, Inverse Trigonometric, By Parts, Trigonometric Substitution, Wallis Formula, etc. Applications -Area, Centroid, Arc length Surface Area, Volume, Work Moment of Inertia, etc. Propositions of Pappus Hooke's Law Multiple Integrals

&34: ME Board April :1995, ME Board April 1:997 Integrate: (7x 3 + 4x2 ) dx.

A

7x 3 4x 2 ----.+-+C

3

2

~

368 l 00 1 Solved Problem~i.Il.~~gineering Math~matig~Jznct l~c!itio~!~~~ Rojas Day 15-:. Integral Calculus 369

B. C. D.

6o39: CE Board May 11.?9"1

7x 4 4x 2 , -+-+C 4 5 7x 4 4x 3 -·+-·--·+C 4 3 4

6.

C.

12

Evaluate fx(x-·5) dx

4

5·'

P..

4x

7x --+C 2

63SI CE Board No·veJuber 5

1!9'9~ 3

D.

0.456 0.556 0.656 0.756

8. C. D.

A. Eva Iuate

D.

'I

B.

xdx

f

O.IJ11 B. . 0.022

63&: CE Boar·d Noyember 1.'11'94

A. B.

0.333

0.233

C.

0.433

D.

0.533

·i2

5

5

D.

Evaluate the integral of sin

c. D.

-2 .fi. COS X + C

A.

B.

6

x dx from 0 to

D.

7t

2n 17

3n 32 5n

C. D.

0.423 0.293 0.923 0.329

A: B. C. D.

765 12n 81

EE Board March 1998

0..278

B. C. D.

0.336

Evauae I t

·1.-'~

0.252 0.305

. ·

&43: ECE Boaa·d November l:9«JJ: 2

Evaluate·the integral Jcos ydy

'II

t,,

A.

2.0

B.

49.7

C. D.

3.0

21og10 edx

y

sin2y

2

4

A.

·'-+-..--+C

B.

y+::>cosy+C

X

6481 CE Board May :1995 What is the integral of cos 2x e

A

ln(eexpx+1)-x+C ' In ( e exp x - 1) + x + C In ( e exp x + 1} square - x + C

1 1/2 0 1/3

c.

10 20 30

D.

40

A.

B.

5.12

sin2x

32

In ( e exp x - 1) square + x + C

65:.1& CE Board November 1'996 Evaluate the integral of {3x2 + 9y2) dx dy if the interior limits has an upper limit of y and a lower limit of 0, and whose outer limit has an upper limit of 2 and lower limit ofO.

f- - - - -

1

J

A.

C. D.

10

3x+4

A.

sin x + C secx+C -sin x + C cscx+C

Evaluate the double integral of r sin u dr du, the limits of r is 0 and cos u and the limits of u are 0 and pi.

B.

0 ·1 2 3

&471 EE Board October 1997

. -1- w1th respect to x an d

·

+c

+C

6511 EE Board April :1997

Evaluate the integral of In x dx, the limits are 1 and e.

Integrate

e•in2x

e•in2x

Find the integra! of [ (e exp x - 1] divided by [ e exp x + 1] dx A. B.

2

evaluate the result from x = 0 and x = 2. 32

C. D.

2

6461 EE Board ApriJ :1997

_?:~~-

&38: ECE Board Aprilll997

B.

J(cos3J.\) dA

768

6421

A.

D.

4

a5n

B.

D

0.4 0.5

n/2.

2.fi.cos.!X+c

B.

~163

0.2

0.3

A.

8

27n

C. A. B. C.

C.

C.

Find the integral of sin x cos x dx if lower limit = 0 and upper limit = 7[/2.

2 -2 .fi. COS X + C

~to~

0

637: CE Board May l'l99&

-2J2 COS-X +C

Evaluate the integral of cos x dx limits from

n/6

A.

_

D.

6501 EE Board April 1997

&41! ECE Board April 1'998 Evaluate

C.

-2 +C

ME Board October :1997 The integral of cos x with respect to x is

1

6451 ME Board October 1997

0.033 0.044

5

What is the integral of sin x dx if the lower limit is 0 and the upper limit is n/2?

.

·----~---

o (>r+ i)8

A.

C. D.

-

6491 ME Board April 1995

n/2?

C.

y + sin 2y + C

e•in2x

B.

Integrate the square root of ( 1 - cosx) dx.

640: CE Boa!.'d November ll'\\"'9&

0.0203 0.0307 0.041"7 0.0543

4


What is the integral of sin x cos x dx 1f the lower limit is zero and the upper limit is

A. B.

y_+ sin2y +C

2

sin •

dx?

6531 EE Board April :1996 x/21 2

Evaluate

e-·-+C 2

JJJzdz r dr 2

0 0 0

A. B.

2/3 4/3

sinu du .

370 . lO

659: ME Board April1.999

1/3 5/3

Find the area bounded by the parabolas = 4y andy =·4.

B.

c

x?:

D.

9 955 5.955 5.595

670: Find the coordinateP. of the centroid of the plane area bounded by the parabola y = 4 - x? and the x-axis.

654: EE Board April1993 Find the area of the region bounded by / c: Sx and y = 2x. ·

A. B.

21.33

A.

C. D.

31.32 13.23

B.

C. D.

1.22 sq. 1.33 sq. 1.44 sq. 1.55 sq.


Ms: CE Board November %994 What is the area bounded by the curve = -9y and the line y + 1 = 0?

A. B.

4 sq. units

C. D.

5 sq. units 6 sq. units

x?

3 sq. units

D.

C. D.

A. B.

666: ME Board April1998 What is the area between y = 0, y = 3x?, x = 0 and x = 2?

75 50 100 25

What is the area (in square units) bounded by the curve y2 = 4x and .; = 4y? A.

B.

5.33 6.67 7.33 8.67

",'li

square square square square


the x-axis on the first quadrant.

A.

B.

·c. D.

8 24 12

x? =4y from x = -2 to x = 2.

D.

6

A. B. C. D.

6&7: CE Board May 1995 What is the area bounded by the curve y2 = x and the line x -4 =0?

=

4.25 2.45 5.24 5.42

a

B.

2a

C.

2

D.

67::&1 Find the length pf arc of the parabola 4.2 4.6 4.9 5.2


a a3

bounded by y = x2 and y = x.

ll.

Find the area in the first quadrant bounded by the parabola = 4x, x = 1 & x = 3

l>

units)·generated by rotating the parabola arc y = x? about the x-axis from x = 0 to x =

1. A. B.

C. D.

5.33 4.98 5.73 4.73


6&9: Locate the centroid of the plane area

t:

9.555

31/3 10 32/3

A

A

l

11

6&8: CE Board November 1.996 CE Board November 1998 Find the area of the curve ~ = a 2 cos 2e.


A.

D.

~.~ ~.~ ~~~~ (~5.V~

B. C.

C. D.

7.67 sq. units 8.67 sq. units 9.67 sq. units 10.67 sq. units

What is the area bounded by the curve y x3 , the x-axis and the line x = - 2 and x = 1?

A.

B. C.

673: Find the surface area (in square

663: ME Board October :1997

Find the area bounded by the parabolas y = 6x - .; and y = x2 - 2x. Note: The parabolas intersect at points (0,0) and (4,8).

11.7 4.7 9.7 10.7

D.

A.

A. B.

Find the area enclosed by the curve x? + By+ 16 = 0, the x-axis, they-axis and the line x-4 = 0.

C. D.

88/3 64/3 54/3 64/5

44/3 64/3 74/3 54/3

B.

C.

A.

B.

M8: EE Board April1997

A. B. C. D.

6711 Locate the centroid of the plane area bounded by the equation y2 = 4x: x = 1 and

66::&: CE Board May 1997

4.

C. D.

A.

Find the area bounded by the line x - 2y + 10 = 0, the x-axis, they-axis and X::: 10.

D.

Find the area bounded by the curve y = x? + 2, and the lines x = 0 and y = 0 and x =

B.


c.

30/3 31/3 32/3 29/3


A.

(0,1.6) (0,2) (1,0)

=


=

C.

(0, 1)

B.

D.

What is the area (in square units) bounded 2 by the curve y x and the line X- 4 = 0?

B.

A.

c.

Mfn CE Board May 1995

A.

665: ECE Board April1998 Find the area (in sq. units) bounded by the parabolas x? - 2y 0 and .; + 2y- 8 = 0.

33.21

0.4 from the x-axis and 0.5 from theyaxis 0.5 from the x-axis and 0.4 from theyaxis O.~i from the x-axis and 0.5 from they<~xis

0 IJ from the x-axis and 0.4 from theyil>
The area enclosed by the ellipse

x2 y2 - +- =1 is revolved about the line x = 3. 9

4

What is the volume generated?

A. -B. C. D.

355.3 360.1 370.3 365.1

675: CE Board May 1996 The area in the second quadrant of the circle X: + y2 = 36 is revolved about the line y + 10 0. What is the volume generated?

=

372 100 l Solved Problems in Engineering Mathematics (2"d Edition) by Tiong & Rojas A.

B. C. D.

2218.33 2228.83 2233.43 2208.53

680: CE Board November 1995 Find the moment ofinertia, with respect to x-axis of the area bounded by the parabola y2 = 4x and the line x = 1.

676: CE Board November 1995 2

=

The area bounded by the curve y 12x and the line x = 3 is revolved about the line x = 3. What is the volume generated? A.

179

B.

181

C. D.

183 185

A. B.

C.

D.

Topics

2.03 2.13 2.33 2.53

D D Mon

Tue

D ~ D D D D D

677: CE Board November 1994 Given the area in the first quadrant bounded by x2 8y, the line y -2 0 and the y-axis. What is the volume generated when the area is revolved about the line y

=

-2 A. B. C. D.

=

=0?

28.41 27.32 25.83 26.81

678: Find the volume (in cubic units) + 6x generated by rotating a circle x 2 + + 4y + 12 0 about they-axis.

=

C.

39.48 47.23 59.22

D.

62.11

A. B.

53.26 52.26

D.

50.26

c. 51.26

Wed

Problems

Thu

Solutions

Fri

Notes

Sat

· Definite and indefinite Integrals Fundamental Theorem of Calculus Basic Integrals Formulas - Exponential, Logarithmic, Trigonometric, Hyperbolic, Inverse Trigonometric, By Parts, Trigonometric Substitution, Wallis Formula, etc. Applications ··Area, Centroid, Arc Length Surface Area, Volume, Work Moment of Inertia, etc. Propositions of Pappus Hooke's Law Multiple Integrals

l

679¥ CE Board May 1!.99$ Given the area in the first quadrant by i! 8y, the line x = 4and the x-axis. What is the volume generated by revolving this area about the y-axis.

A. B.

Theory

~tegral Calculus

=

631. A 632,D 633.A 634.C 635.C 636. D 637.A 638. D 639.A 640. B 641. B 642. D 643.A

ANSWER KEY 644.A 657.A 645. B 658. B 646. B 659.A 647.A 660.A 648.A 661. A 649.A 662. D 650. D 663.A 651. D 664. D 652. D 665. D 653.A 666.A 654. B 667. D 655. B 668. c 656.C 669.A

RATING

670. B 671. B 672. B 673,A 674.A 675. B 676. B 677. D 678.C 679. D 680. B

0 0 0 0

43-50 Topnotcher 30-42 Pa·sser 25·-29 Conditional 0-24 Failed If FAILED, repeat the test.


374 100.1 Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas

IJI

Using Wallis formula: m = 5; n = 3; and 3

I(3t -1) dt = =

i I<3t

i(

3

-1) (3dt)

~ Jc

(3t 1)4 +

12

•

I

nJ2 5

sin xcos

I

--B

Note:

--B

f

dx

X

(2:_)

sin 6 x dx = 5( 3 )( 1) • = Sn 6(4)(2) 2 32

Ju

du

.

I

= (5-1)(5-3)(3-1) •(1) 8(6)(4}(2}_ =0.0417

=lnu

=

-x 7 7(X+1)

J

sin 5 x

ml

I

12

dv = (x-5)12dx; v =

Ixcos(2x + 7)dx

13

..:...(x_~-~-=-)-

5

=~(x-5f3 _ __!_

5

sin xcos x dx

I

ml

5

13

=12( (5 -1)(5- 3)(5 -1)(5- 3))· (1)

7x 4 4x 3 3 + 4x 2 )dx = (7x -+-+ C } 4 3

10(8)(6)(4)(2) =0.2

• nJ2

3

sin x cos x) .dx

I

0

7

7(0 + 1)

4~[(t+1f6 -(0+1f6]

sin

6

x dx

13

= 0.456

l fi

tl ·ii

Jcos 3A dA 8

0

Let: u = 3A; du = 3dA, thus dA = du/3 Change limits: when A = 0, u = 0; when A = rc/6, u = n/2 Substitute: n/2

• I 1

n/6

15

1[(6-st -(s-st]·· -182

5

sin xcos x dx

[

=~(6-5)13 -~(5-5f3

=5; n =5 and

0

. 14]!:6

(x-5) 13 14

13

nl2

12

n/2

-

BIB

x(x-5f3 _ __!_ r(x-5f3dx 13 13 J

numbers

1m

-

5

Using Wallis formula: m

2

-6

0

7(1 + 1)7

2 Jx(x- 5f dx

5(3)(1)

a = 1 since both m and n are odd

=±sin(2x + 7)+C

7

6

0

2

2

+..![~+1r ]

-1

Let: u= x; du = dx

.rr/2

= ± Jcos(2x + 7)( 4xdx)

7

-X 1 [ 11 -.---42 (x+1)--Bt

Note: Judv = uv- Jvdu

dx=~=0.533

0

•

=In(~) =In 2

r,

=0.022

n/2

J(

-X

5

sin x dx

1

_

6

2 Jx(x-5f dx

5

-7

- 7(x + 1)7 + 7 J(x + 1f dx

6

Using Wallis formula: m = 5; n = 0 and a = 1 since m is an odd number

l-1o

= ln(-10 + 2)-ln(-6 + 2) = ln(-8) -ln(-4)

0

-7

IIJ

0

-dx- = ln(x + 2) --B X+2

5

= x[(X+1( ]- J(x+1( dx

n/2

-10

I

'

6 J
0

•

dx X+2

3

0

c .

rm

1

0 n/2

= __2_,(3t -1)4 +

-10

Using Wallis formula: m = 6; n = 0 and a = rc/2, since m is an even number

a = 1, since both m & n are odd numbers

J

J

cos6 u du

0

J

n/2

.:!_

Let:

dv

3

Using Wallis formula: n = 8 and a since n is an even number

0

x; du

3

0

xdx (X+ 1)6 = (x + 1)--B xdx

u

J

n/2

1 cos6 u -du =-

3

=dx

(x t 1) 11 dx; v

cose u du = ..!( 7(5 )(3 )(1) \ 3 8(6}(4)(2))

0

(X I

1)

f

1057t 2304

357t 768

=--=-

= rc/2,

(!:) 2

371 .1 00 l Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas

•

2

J

0

rm

J

2

1 -dx-dx -=3x+4 3 3x+4

1"'2

. X x/<4 d X::: Sin

J

COSX

2

3

. =Sin

=0.305

Note: TCI2

Ill 2 Y)dy

=i i ±

J(1+COS 2y)dy

Jdy +

=

Jcos 2y(2dy)

•

'Zi

e

J'nx dx

let: u =In x; du = dx ; dv X

=dx; v =x

10

10

J

f

21og e dx _ X

dx e• +1

e• + 1, thus e• = u -1; du = e'dx J du

• "cosu

.

=

Jdx X

Jrsinudrdu = J2"

0

0

=

du (u-1)(u:_1+1)

~)d2x

i)+c ~ ~2..[2 cos(i) =2J2(-cos

+C

= 21og10 e lnx1 1 =21og10 e (In 10 -In 1)

=

•

Jcos2x e5in 2x dx = Je•in2x cos:tx dx

1\

I qu
o

A11J

o let: u = sin2x; du =·cos 2x (2)

II

11 n 1

i "j(

cos u- o )sinu du

2

-i "

2

2

Jcos u(-sinu)du

=-..! cos 2 =

3

ul"

3

0

-~(cos3 7t- cos3 o)

1 =--(-1-1) 6 1

1 quate constant:

A

0

0

1=A(u-1)+Bu 1 =Au-A+Bu

=2

sinu du

0

0

f f

1 A 8 ··-=-+u(u -1) u u -1

1

"rr lcosu

J

10

.J2 Jsin ~

=2fi ~sin

J

= ln(e• +1)2 - x +C

dx e• +1 = e•(e• +1)

10 -= - - - 21og10 e -

1

=21n(ex +1)-X+C

dx

Ill

Substitute:

e• -1 --dx = ln(e• +1) e• +1

f

du = u(u-1)

Thus, v.<:sin-='.11-cosx

2

=-ln(e•+1)+x

Je• +1

=1

r;;---

-ln(ex + 1) + ln(e•)

=

2

= In( e~ + 1) +In( ex + 1)- X + c

= e(lne) -e- [ 1(1n1) -1]

sin(~)=P-~sx

=-In( ex+ 1) +In( ex+ 1-1)

-~es•n2x + C

=In (ex +1)- - -

=xlnx-xj~

Ju-1 du

s-du -u-+

-[ -ln(e• +1)+x ]+c

I et: u =

J-b-cosxdx

=

= -lnu+ln(u-1)-

=~ Jes•n 2x (2cos2x)dx

-1 e•dx e• +1 dx =-;.--+1-

·I(,

~··

"dx

fe•

1

= (lnx)(x)- Jx( dxx)

f.J1-cosxdx =

2

Jcosx dx = sinx + C

Jrn x dx =uv- Jvdu

Ill

X

n

• •

=90" and TCI4 = 45"

1

r;;

du u(u-1)

Thus:

e

=j'_+ sin2y +C 2 4

Note:

51

=

7t ( . 7t) 2 - Sin 4

=0.293

3

§..!2- ~s

Jcos2x e

n/4

=-ln(3x+4) 1 1 3 0 1 1 = rn[(3){2) + 4]- rn[(3)(0) + 4]

Jcos y dy =

f

Note: Jeudu = eu + C

1<12

0

2

Day 15 -Inteqral Calculus 377

• fy

=3

2

0 0

r;

2

2

Jc3x + 9y

2

)

dx dy =

0

3

ly + 9y 2 x dy 0

.... ·

378 1001 Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas 2

2

Jx3 + 9y2xl: dy = Jv3 + 9y3) dy 0

0

rm

ID y2 =8x ~Eq.1

2

J1oy d~

=


3

0

x2

~

=-9y ~ Eq. 1

y + 1 =0 y = -1

y =2X ~Eq.2

~Eq.2

Substitute Eq.2 in Eq. 1:

= 10y412

4

0

10(2)4 =-4 =40


(2x)2 =8x

4

x2 = -9(-1)

4x 2 =ax X=2 y = 2x = 2(2) =4

A=2 Jvdx

X=±3

0 4

A=2 Jv'X dx

111

0 3

2

x/2 1 2

=

JJJzdz r dr sinu du

"-A= 2(x)

2

x/21~2

JJ~

0 0

JX

-3

2

2

0 2

JJ2r drsinu du 2

Jr3 11 sinu du

0

0

J-

xl2[13-03J sinu du 3 0

=~

A=

J(~ -2x)dx

1"'2 0

=~( -cos%+cosoo)

A

Another Solution: Use the formula for the area of a parabolic segment

Q

2

-x

3 A= 2J8(2)a'2

A= 3_bh

A=li4X4) 3 A_ 32

-3

1m 2 12

(2)2 -[ 2J8~o)3'2 -(0)2

a

y 2 =X ~Eq.1

0

X= 4 ·-t Eq.2

J

Substitute Eq.2 in Eq.1: A = 1.33 square units

/

-~~ y·

4 y

12

4

A= Jydx 0

EE] .

square units

A = 4 square units

2 2J8x3/2

(;r -x]'

0

3

3

=-(<~~3 -3]+((~3/-(-3)}

J8x2 2x 2 A=----

sinu du

-1Jdx

1-3

3

J

R~

= -[

0

3

2 ,;-(-cosu)

2 3

~ YL) dx

0 2

x/2

0

3

=-

3

A =i(4)3'2 3 A = 32 square umts .

-3

A= J
x/2 3

=2

2

A= Jvdx

0 0

-3

=-J [-1-( ~J}x

Jr drsinu du

x/2 1

=2

2

0

0 0

=

~

2 r drsinu dl;J

22 02 ;

x/2

3

A= J-ydx = J--
3

0 0 0

4

=ix3'21

· . h .==_4 · -r II .. • · · .·· .o

. ··. · ··

380 J..OO I Solved Problems in Engineering Mathematics (2"d Edition) by Tiong & Rojas 4

A=

J(x

2

4

Ill

2

+ 2) dx

A= J( 8x- 2x )dx

0

Day IS- Integral Calcull.!s 381 Substitute Eq.2 in Eq.1:

0

A=~

3

· 2x

\

= 8xz- 2x314

14

2

3 6

=4(4) 2

-~(4)3

0

43 A=( ) +2(3) 3 88 ' A=- square un1ts

10

3

II

rm

2 c- x2 vx - 4 x4 4X=16 64 = x3

=0

when x = O: y = 5 wheny=O:x=-10

rx=-~~

3 64 ' A = - square un1ts

3

y = 6x -x

~

2y

X=4 y=214 y=4 Thus, the parabolas will intersect at (0,0) and (4,4).

2

Solving for the coordinates of the intersection:

x2 - 6x = -y (x-3f = -Y+9

x 2 = 4y

2

(x-3) =-1(y-9)

[X= 10J

10

A= Jydx

= 4(4)

0

X =±4

=

By inspection, the vertex is at (3,9) and the parabola is facing down.

1n 5+~}x 0

2110 .~ 5x + 2~2) o

y = x 2 -2x x 2 -2x = y (x-1)2 = y+1

(x -1)2 = 1(Y+ 1)

102 =5(10)+-·

(0,0)

4

4

4

A=2 Jxdy

A ·= 75 square units

0

0

By inspection, the vertex is at (1,-1} and the parabola is facing up.

=2

=

4

4

0

0

f
4( ~:~~ J %(Y)3'2[ =

3 A= 21.33 square units

4

4

A= Jydx = J
0

4

=

K(

2

2

J

6x - x ) - { x - 2x) dx

0

3 2 A =-(8)(4)

3

x2JX= 2;;~2 -4(:)[ =

:)dx

2(a+b)h

A

~(5+10)(10)

A

75 square units

ml y'

b=8

'\!17

A= 21.33 square units

1

A

~c:J~ II

.0

Ill

h = 10

4(4)3/2 (4)3 =-3--12 A = 5.33 square units

Another Solution: Use the formula for the area of a parabolic segment

A=~bh

Another Solution: Use the formula for the area of a trapezoid

0

0

=~(4)3/2

·0

u

4

A= Jydx= J
y

y

Ill

4x

2/x ~ Eq.1

x2 =-By -16

4y

x2 =-8(y+2)

X

4

'lq ;>

By inspection, the vertex is at (0,-2) and the parabola is facing down.

382 l 001 Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas By 1nspect1on the vertex is at (0,-4) and the parabola 1s facing down.

1 3

A 2 = Jx dy dx

0

x 2 = 2y ~ Eq.2

~[

=

Equate Eq.1 to Eq.2:

4 A 2 = 0.25

2(y-4)=2y

Atatal = JA1j + JA2j

Note: x 2 + 8y+16 = 0 -By=

2

y=2

Atatal = 4.25 square units

0

dx

8 =4y

= 4 +0.25

A= J-ydx

•

By inspection, the vertex is at (0,0) and the parabola is facing up.

o4'_ (1)4

4


2

x

2

A= Jydx = J3x dx

= 2(2)

0

X =±2

Ill

rhus, the two parabolas intersect at points (2,2) and (-2,2).

x2 +16

2

0

x3 =3-

3

=x31:

x2 -y=-+2

(0,0)

8

=(2)3 A = 8 square units

Substitute:

A=

X~ +2}x

~____!_/

0

14

= 3~8) +2X 0 3

= (4)3 +2(4) 24 A= 10.67 square units

•

3

-2

~10 -2

04 -{-2)4 =----

4

X =4 ~Eq.2

2

A= Jydx

y = 2../x


··2 2

3

3

=

A= Jydx = 2 J..fXdx

= ~[(3)3/2- (1)3/2

y2 =4

f(Yupper- Y1ower)dx

n

y=±2

-2

1

= 2x3/2 = ~ x3'213 3/2 3 1

A1 = Jx dy

A1 =-4

y2 =X ~Eq.1

~

Y2 =4x

1

0

=

Iii

=

-2

J

~-

2

J[ 8 - 2x

•

::I [

x 2 +2y-8=0

;

B(2)

A

I(} tili(

3

2

; [ax

x2 = -2(y- 4) ~ Eq.1

2

J dx

7

A = 5.595 square units

x2 =-2y+8

x: ] dx

8 -2x2 -

·J[

~ (2)

4

3 )

l8(-2)-

•;qu;llf~ tlllll~l

~(-2) 3 )]

A= 2 Jydx

a 4

A= 2

J..fX dx 0

~

384 i 001 Solved }>robJems in Engineering Mathematics (2"d Edition) by Tiong & Rojas

•

3

2 A=

2(x) ~ 2

= ~x.3121

3

4 0

x

2

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __,D:::.a= 15- Integral Calculus. 38_5 1

A•x=

x2 =x

•

e o· 30" 45"

so· go· 120" 150" L1ao·

Thus, the parabola and the line intersect at points (0,0) and (1, 1).

o•

1-

13

14

6

3

4

Jr

2

A= Jydx= J
0

dx

_tJ

= x2 2 31o 12 13

=-

=

6

f( y3/2- y2)dy

=

~:~~- y:[ 5/2

0

y

1

1

A•x= Jydx•x= J
0

2 2 ) dx

x

2

8x 3 x5· =16x-3-+5 o 1

0.4

(332)(y

=a 2 [ sin(2)145.)- sin(2)(0·)]

J4-

0

3

I hus, the coordinates of the centroid is "' (0 ~i. 0.4)

0

= jt16- 8x2 + x4 )dx

45" 0

2

2

(1)5/2 (1)3 yoc•------

(j

Jy dx

0

1.

2

A=a 2

0

1

1

0

a cos 20(2d9)

=a sin2ej

1

A•y=2 Jydx•~=

0 1

2.(;) J 2

0

= f{Jy -y)dy•y

45°

=

. 2 2 32 · A=-bh=-(4)(4)=-3 3 3

1

1

0

1

1\•y= Jxdy•y= J(xp·-xddy•y

1

= J(x -x2 )

1

A=-

de

(0,4)

Solving for the area of the parabolic segment:

=--2 3

45°

By inspection, the vertex is at (0,4) and the parabola is facing down.

Thus the parabola intersects the x - axis at points (-2,0) and.(2,0).

1

0

l

=-y+4

~ .(2,0)

0

de

2

x=0.5

45°

45° 2

X

at y = 0, x = ± 2

··-X=---

r ±a ± 0.707 a 0 i i i ± 0.707a ±a 1

i Jr

f(x 2 -x 3 )dx

=-~3- :4[

Y=X =1

y = 4-· x2

x2 =-1(y-4)

0

X=1

r 2 = a 2 cos2e

=2

f(x-x )dx•x 1

co•

•

2

=


3

[

0

1

=y~Eq.1

()

A =32 - square um'ts

A=4

Jydx•x= J(YL -yp)dx•x 0

y =X ~·Eq.2

A =~(4)312 3

1

)=16(2)--8(~)3 + (~5 y=1.6

Thus, the centroid is at point (0, 1.6)

Day 15 -Integral Calculus 387


•

Change the limits:

1

r. A•y=2. J(4x)dx -

.

0

l=4X atx=1,y=±2

s=

-y

JJ4

2

45° ,.------;:2

dY) = 4x 2 ( dx ·

2

+ 4tan 8 (2sec 8 d8)

0 45°

=

3 4

Y=(0,0)

dy -=2X dx

X= 2; 8 = 45o

j(Y )=<1>2 -

dy = 2xdx

when, X::: 0; 8 = 0

=~( 4;2 J[

Thus; the parabola and the line intersect at points (1,2) and (1 ,-2).

Y = x2

1

J

Substitute:

2 2 J4(1 + tan 8) (2sec 8 d8)

1

0 45°

2 A = 27t JJ1 + 4x dx • x

2

2

= J J4(sec 8) (2sec 8 d8)

Thus, the center is at point (3/5, 3/4).

•

0 1

0 45°

=

2

0

0 45°

2)3n

3 =4 J sec 8 d8) (0,0)

4 45°

I 2 s=2 fv1+(:~) dx

-c

4(~sec8tane +~ln(sec8 + tan8)l

2.

0

X

1

= J<2.JX )dx •

x

dx

s

.( dy)2

x2

ldx =4

0

5/2

j(x)

=

-

S=2

3 5

x2 1+"4 dx=2

2

0

-(1+4(0))

312

]

A = 5.33 square units

ml x2 y2 Given equation: - +- = 1

9

4

=1

By inspection: a = 3 and b =2

x =1, y =1

0

4+x2 -dx 4

2

=fJ4+x2 dx=J~(2)2 u2 dx 0

1

Jvdx"~=i Jldx

Q

=~[(1+4(1)f'

12 2

2F? 21¥ J J0

X=-

A•y=

3/2

0

~< 1 >s'2

1

when

Substitute:

= 27t(1+4x2)3'21

x2 y2 Standard equation: 2 + 2 a b

Y= x2

1

=2()(612)

'4.6

•

dy =~ dx 2

1

312

~ln(sec45°

=4y

2xdx =4dy

0

=2 Jx

2

l

-sec45°tan45° 4 1 = [: + tan45°)

0

A•x= Jydx•x

1

= 7t ( 1+4x

0

1

1

= J 2sec8 (2sec 8 d8)

Solving for the area: 1(2 1 4 A=-bh ) =-(4)(1)=2 3 3 3

2)1/2 21t ~( (8x)dx B J 1 + 4x

0

0

Using the second proposition of Pappus:

1

A

Jds• 21td

V=A•21td

0

Let: x =2tane;

fJ--------2 ~~) I

dx == 2sec2 e de

A

1I

II

(

dx • 21tX

= ( 7t8b). 27td

=(~)(3)(2)(21t)(3) V = 355.3 cubic units

388 . 1001 Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas

•

Day

V=(~nr2 )-2n(;: +10)

~

•

Solving for y ·

,

Using the disk method:

x. b V . .

~~!+:.3~x ~

'" '" V~y I

4

V

=n Jy

2

dx

By inspection, d = 3.

dx

0 4

y

~

Ay= JdA•y/2

C(-3,-2)

(0,0)

V =; 222B.91B cubic units

Using the second proposition of Pappur.: 2

=n f(YL- YP )

dx

V = A• 2nd

0

r

· = nr2 • 2nd

2]2

41(

=Jydx•y/2

=n gl2- ~

0

=n(1)2 (2n)(3)

dx

=~ Jldx

x2 =n

X = n(

c r

=~ J
=~(rx-~;)[ (Y)

X3/2

312

)

dx

XS/2

V=A•2nd

~nr )- 2n(y+ 10J

312

= 4n[ (2.Ji2)(3)

-~.Ji2(3)5 12 J

J

•

10

=

X=

4, y = 2

Thus, the parabola <:~nd the line intersect at point (4,2).

(0,0)

(x + 3) 2 + (y + 2) 2

~

=1

I ~:

ll ·~·

~dy

r2

By inspection, h = -3 and k = -2, thus the

y = 2, x ±4

=By

At

(4)5 ]

Standard equation: (x- h)2 + (y- k)2 =

x 2 =By

x2

(x+3) 2 +(y+2) 2 =-12+(3) 2 +(2) 2

V = 1B0.955 cubic units

At

x- 2(3) + 64(5)

4


.Q

Using the second proposition of Pappus:

xs

x 2 +y 2 +6x+4y+12=0

3

=4n[3J12--J12-3/2 5/2 ]1

By inspection, r = 6

V =(

.J12x

0

2

x3

dx

J

•

3

=4n J(3.J12x 112 -

Given equation: x + y 2 = 36

x4 64

= 59.217 cubic units

v = 26.B08 cubic units

0

3n

+

=n [ 4(4 )- 2(3) + 64(5)

=4n J(3- x)( .fi2X) dx

4r y=-

2

(4)3

3

~x'?\(y )=; (! r'~)

2

4

V·+•P3-x)y dx]

3

3

0

Using the shell method:

r J =21 ( r -3

4-

V

•

r

2

Calculus 389

y-axis

46 =..!.(n)(6)2 (2n)( ( ) +10) 4 . 3n

1 4nr

1~- Integral

Substitute r = 6:

Using the ring method:

conter is at (-3,-2). Also the radius of the ctrclc is equal to 1.

2

V=

nJ( XL2 0

Xp

2

)dy

390 100 1 Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas 2

v

=1t K(4)2 -( JSY)2J dy

2 IX= 2 jy (1-

0 2

~ )dy

=2Xy2- :4}y

= 1t J(16-8y) dy

Topics

D

0

0

=1t(16y-

8

~

2

y3

)[

=

2

= 1t[16(2)-4(2) ]

{-4-

y5 4(5)

2

Man

)0

D

1

= 2((2)3 - (2)5)

4

V = 50.265 cubic units

•

lx

Tue

20

D [Q] D D

=2.13

l=4X

Theory

Wed

Problems

Thu

D D

At X= 1, X=± 2

Solutions

Fri

Notes

D

Thus, the parabola and the line intersect at points (1 ,2) and (1 .~2).

·.

' ~ ~ ., ' " ~

tT '

eJ

..

Sat

Integral Calculus Definite and Indefinite Integrals Fundamental Theorem of Calculus Basic Integrals Formulas - Exponential, Logarithmic, Trigonometric, Hyperbolic, Inverse Trigonometric, By Parts, Trigonometric Substitution, Wallis Formula, etc. Applications -Area, Centroid, Arc Length Surface Area, Volume, Work Moment of Inertia, etc. Propositions of Pappus Hooke's Law Multiple Integrals

'

'

2

lx

=2 JldA 0 2

ll

2

= 2 Jy (xdy) 0 2

<·<·'>

2

lx = 2 Jy (xL- Xp) dy 0

"•··~«<'r~oi,.-.

~:<-#0".~:.'<><

"'~''"'")'."•<,·~

·~.

~·"'

·!?;1'<-~""''~h··~·-'"'"·.Cd'''

~,•f.f<~~1<'<'<<'<'>~~&~·'i" .... »OO<'i""ii<.''i"••"'"""""'<

'· " ,. ><-"'-< "

*

"+ .... .,. " " ~ ~ .- •. "'. "-·" ·<·

392 l GO 1 Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas

.;h.

<;-. «

•¥

Topics

.,.,.,.

D D D

~-

~l1on

Tue

Theory '_.,;

" ~ ..;

,, ·' .......

~

"'

~

'

~ .:!"

,·

~

,. ·' "

,, <-. "

~

-'" ".(>

<

<·-->"'·~~

~*~

~

·~«<·0<~'~"''""-

.. ~ / " '

D lQJ D [] ·D D

'

• '" '

·'

'"'''~,.

,.,~

~"~-:%>>".*"<''<'

...

,<(

~

''

y

N

<,•

.....,.':'""' ......

..<.,..«.-~~-;,""~

·~·.• .• :·t~~~~~

'· v

•

,.

\;<'

~

~

..,

VVed

4>:15:

Problems

Thu

Solutions

Fri

Notes

Sat

Differential Equations Types of Differential Equations Orders of Differential Equations Degrees of Differential Equations Types of Solutions of Differential Equations Solutions to First-Order Differential Equations Applications to First--Order DE (Population Growth, Radioactive Decay, Continuous Compound Interest, Flow, Cooling and Heating, Newton's Second Law of Motion, Geometric problems, Orthogonal Trajectory)

What is a Differential Equation? Example: differential equation is an equation that ' ontains one or more terms involving 1lr !rivatives of one variable (the dependent v.1riable, y) with respect to another v.u iable (the independent variable, x).

1\

I xamples:

(-<:!_~~J2Jd2y2 J3 + x3 ~ 0 3

l dx

dx

c. (x - 1) y" + xy' + y (j

ilx

ily

2

its solution y = J2xdx = x + C , where Cis an arbitrary constant.

Whatl!_re the Types

a. (2xy +x 2 ) dx + 2y dy = 0

b.

The differential equation dy = 2x has for dx

I

2

ax y-= (fz

=0

~f

Differential

Equations?

1.

Ordinary Differential Equation- an equation containing only one independent variable, thus having only ordinary derivatives in the equation.

3

llu· :;olul1ons of differential equations are lttll<.llow; and not just numbers like the dq' ·Ill ;tic: r" 111< 1l1oll' •

Examples a, b and cat the left of this page are examples of ordinary differential equations

394 1001 Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas 2.

Partial Differential Equation - an equation containing two or more independent variables, thus having partial derivatives in the equation.

Example: x3 - 3y =ewhere: C =arbitrary constant

2.

·Exampled in the previous page is an example of partial differential equation.

Ill

II

I

I

Example: 2xy + i

- 2=0

The order of a DE is the order of the highest ordered derivative which appears in the equation.

1.

dy = 2x2 + 5x + 3 dx

Variable separable type

jP(x)dx +

2. d3y = 3x2 + 6x + 2 dx 3

Examples: 3

d2 + (d x-Y ..1...) =15 2 dx

3

= 15

What are the Types of Solutions of Differential Equations? 1.

ef(1-n)P(X)dx dx

What are the Applications of First-Order Differential Equations?

1. Population growth problems.

F(x,y) = JN(x, y)ay + k 2

lQ(y)dy-~:c.

dP =kP dt

F(x,y) = jM(x, y)ax + k1

where:

Test for exactness:

OM

Homogeneous type

oN

ay""' ax

=0

where: dP =rate of change of the population dt P =number of inhabitants at any time t k = constant of proportionality

2. Radioactive decay problems 4.

General solution:

The degree of a differential equation whose terms are polynomials in the derivatives is defined as the highest power of the highest order derivative.

2

f<1-n) Q(X)

Standard form:

What is a Degree of a Differential Equation?

J + (dd~ )

y1-n -- ef(1-n)P(x)dx . 1

The Bernoulli Equation is named after the brothers Jakob (1654 -1705) and Johann Bernoulli (1667,... 1748).

F(x, y) = C

fVl(x,y)dx +N(x,y)dy

d2 x ( dx;

·

General solution:

General solution:

. =5x 2 + 1Ox + 3

dx

Exact typf!

General solution:

M(x,.y)dx + N(x, y}dy ""0

P(x)dx +Q(y)dy =0

d2 Second Order: - { dx

Second Degree:

3.

Standard form:

Examples:

First Degree:

Let: f(x,y) = M(x,y) dx + N(x,y) dy f(rx, ry) = r" f(x,y)

Standard form: What are the Solutions to First Order Differential Equations?

Third Order:

Test for homogeneity of degree n:

Particular solution - the solution that has no arbitrary constant.

What Is an Order of a Differential Equation?

First Order:

Day 16- Differential Equations 395

General solution- the solution has at least one arbitrary constant.

Linear type

dQ""kQ dt

Standard form:

=

Substitute y = vx or x vy and the resulting DE becomes a variable separable type·

dy + y P(x)"" Q(x) dx General solution:

Note: First-order differential equations are not always separable. Homogeneous differential equations, however, may .be transformed into separable equations by the substitution of a variable. An expression is said to be homogeneous if all terms have the same degree. The term "homogeneous" is also used to indicate that the right-hand member of a linear differential equation is 0

Ja
y(i.f.) =

where: .1. f. = .10tegra!"mg factor = eJP(x)dx Note: Integrating factor is also known as Euler's multiplier. 5.

Bernoulli's equation type Standard form:

dy dx

1

y P(x)

y" Q(x)

where: dQ - = rate of change of the substance dt Q =amount of the substance present at any timet k = constant of proportionality 3. Continuous compound interest problems

dP =rP dt where: dP - = rate of change of the account dt P = money ·present in the account at a~timer / r =nominal rate of interest per year

396 1001 Solved Problems in Engineering Mathematics (2"d Edition) by Tiong & Rojas 4. Flow problems dQ dt

. - rate of loss rate of gatn

wnere: Q concentration of the mixture at any timet dQ rate of change of the dt concentration of the mixture

=

~I

!

Topics

where: ( dy) dx 1

- =

~ II

(:~)! =-(:;)g

=

D D D D liJ ~ D D D D

=slope of the orthogonal

Mon

trajectory ( dy) dx 9

=slope of the given family of

Tue

curves

5. Cooling and heating problems dT dt

=k(T -t s )

dT =k(t -T) dt $

l3r

1:'/r

Cooling process

Heating process

where: dT rate of change of the body temp. dt T temperature of the body at any timet t. surrounding temperature

- = = =

6.

Newton's second law of motion dV F=mdt where: F =force m = mass of the body

~~ = rate of change of velocity

( dx) 1 dy 9 = ( dy) dx 9


'~i

il,

GOOD LUCK!

\lrribia: Did you know that. .. the fame and influence of Pythagoras spans for centuries! Pythagoras even appears in Shakespear's Merchant of Venice, when reference is inade to the society's belief in the transmigration of souls. ~uote:

"The most painful thing about mathematics is how far away you are from being able to use it after you have learned it." - James Newman

Geometrical problems (Orthogonal trajectory) A curve which intersects all curves of a given family at the same angles is referred to as a trajectory; if the intersection is at right angle, the curve is called orthogonal trajectory.

Theory

Wed

Problems

Thu

Solutions

Fri

Notes

Sat

t.8u Determine the order and degree of 1111~

differential equation

ld_1_ ) -

c. 2r dx == (J?- + 1> dy D.

y dx + (2x- 3y) dy = 0

3

d4 + 5x 2x _r 4 dy

2

.

xy

dx

=0 .

Fourth order, first degree Third order, first degree First order, fourth degree First order, third degree

1\

II <.

1,

(,,.·zz Which of the following equations is .111

7.

Differential Equations Types of Differential Equations Orders of 'oifferential Equations Degrees of Differential Equations Types of Solutions of Differential Equations Solutions to First-Order Differential Equations Application to First Order DE .(Population Growth, Radioactive Decay, Continuous Compound Interest, Flow, Cooling and Heating, Newton's Second Law of Motion, Geometric Problems, Orthogonal Trajectory)

684: ECE Board April1998 2 The equation y = ex is the general solution of:

y'=

72y

B. -y'=

y2x

A.

oxacl DE?

1\ II II

2

(x + 1) dx - xy dy = 0 x dy + (3x- 2y) dx = 0 Jxy dx + (2 + J?-) dy = 0 x1 y dy - y dx = 0

Which of the following equations is a "· 111. rhiP ~~eparable DE?

1.1-1,

11

(x 1 x"y)dy=(2x+xy2)dx

II

( •. •

y) d X

?y dy - 0

C.

Y·- _1_ 2x

D.

y'

=

X

2y

685: EE Board Mareh 1998 Solve the differential equation: x (y - 1) dx + (x + 1) IJy = 0. If y = 2 when x = 1, / determine y when x = 2.

398 100 I·· Solved Problems in En A. B. C.

D.

1.80 1.48 1.55 1.63

Mathematics

~I

I~

If dy = x 2dx; what is'the equation of yin terms of x if the curve passes through (1 '1 )?

A.

x 2 - 3y + 3 = 0

B.

x3 -

y(k)=4-k

B.

y(k)=20+5k

Ill

C.

3

x + 3y + 2 = 0

-~

D.

2y+X 3 +2=0

2

687: ECE BoaJ'd November 1998 Find the equation of the curve at every point of which the tangent line has 'a slope of2x.

B.

X=-y 2 +C y =-X 2 +C

c.

y = y2 +C

D.

2 X= y +C

A.

cosy = cos y = cosy= cos y =

In (c cos x) In (c sin x) -In (c sin x) - In (c cos x)

M9: EE Board Octeber 1997

B. C.

D.

{y-~x 2 +y 2 )

A.

~x2 +y2 +y =C

B.

~x2 +y2 +y =C

C.

~x+y +Y =C

D.

~x 2 -y +y =C

dx-xdy=O

+ +

y=0 y=0

+ y =0 +

y=0

y = C (sec x +tan x) y = C (sec x- tan x)

C.

y = C sec x tan x y = C (sec2 x tan x)

D.

Solve• xy' (2y- 1) = A.

In In In In

(xy) (xy) (xy) (xy)

y (1 - x)

= 2 (x- y) + C = x - 2y + C = 2y - x + C = x + 2y + C

Solve (x + y) dy = (x- y) dx~

=

x4 x 2y = -+C

D.

x3 y=-+C

+c

4 4

4

A.

88.60

B. C.

95.32 92.16 90.72

D.

A.

~x-~y=O

B.

~y-~x=O

C.

~X+~y=O

D.

~X+~y=O

· 702: The population of a country doubles in 50 years. How many years will it be five times as much? Assume that the rate of increase is proportional to the number of inhabitants. A. B. C. D.

100 years 116 years 120 years 98 years

b98: CE Board May 199& What is the differential equation of the family of parabolas having their vertices at the origin and their foci on the x-axis. 2xdx- ydy = 0

B.

xdy + ydx = 0

c

2ydx- xdy = 0

[)

dy --X=O dx

70:J: Radium decomposes at a rate proportional to the amount present. If half of the original amount disappears after 1000 years, what is the percentc>ge lost in 100 years? A. B. C. D.

6.70% 4.50% 5.36% 4.30%

694& EE Board Aprll1996

695& EE Board April 1996

What is the solution of the first order differential equation y(k+1) y(k) + 5.

C.

A.

A. B.

D.

690& ME Board April1996

xy=

3

(y") - xy" + y' = 0 y" - xyy' = 0 xy" - (y') 3 - y' = 0 (y') 3 + (y") 2 + xy = 0

Radium decomposes at a rate proportional to the amount at any instant. In 100 years, 100 mg of radium decomposes to 96 mg. How many mg will be left after 100years?

x4

B.

A. B. C. D.


4

Find the general solution of y' = y sec x

C.

x 2 -2y-1 = 0 2x 2 +2y-2 =0

x3 xy 2 =-+C

A.

Find the differential equations of the family of lines passing through the origin.

69:J& EE Board October 1995

B.

3x 2 +2y-3 =0 2y + x 2 -1 = 0

(x- 1) y" - xy' (x + 1) y" - xy' (x- 1) y" + xy' (x + 1) y" + xy'

= x2.

b971 CE Board May 199"/

692& ECE Board November 1994

D.

Solve the differential equation dy -xdx = 0, if the curve passes through (1,0)?

A.

y(k) =C...: k , where Cis constant The solution is non-existent for real values ofy

691& EE Board Aprll1995

A. B. C.

Solve (cos x cosy- cot x) dx- sin x sin'y dy = 0 x x x x

dy +'i.. dx x

Find the differential equation whose general solution is y =.C1x + C 2ex.

688& ECE Board Aprll1995

A. sin B. sin C. · sin D. sin

D.

Solve

3y + 2 = 0

696: Solve the linear equation:

5

A.

C. 68&: EE Board October 1997

Day 16- Differential Equatl.ons 399

~+f=C

A. B. C.

~- + 2xy + f = C

D,

X: -

~- 2xy2xy +

i l

=C =C

C.99: CE Board November 1995 Determine the differential equation of the family of lines passing through (h, k).

704: ECE Board November 1998. Find the equation of the family of orthogonal trajectories of the system of parabolas y 2 = 2x + C .

1\.

(y-k)dx-(x-h)dy=O A.

y = ce-•

II.

(y-h)+(y-k)=_'j_ dx

B.

c

(x-h)dx-(y-k)dy=O

C.

I)

(x+h)dx-(y-k)dy=O

D.

y = Ce 2 " y = ce• y = ce-2•

.

d

7oor Determine the differential equation of llu: l;tmily of circles with center on the y

~IXIS

400 l 001 Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas

It

~

II

I

70S: According to Newton's law of cooling, the rate at which a substance cools in air is directly proportional to the difference between the temperature of the substance and that of air. If the temperature of the air is 30• and the substance cools from 1oo· to 70° in 15 minutes, how long will it take to cool 1oo• to so·?

709: ME Board April1998 If the nominal interest rate is 3%, how much is P5,000 worth in 10 years in a continuously compounded account?

A. B. C. D.

A. B. C. D.

33.59 min. 43.50 min 35.39 min 45.30 min

706: An object falls from rest in a medium offering a resistance. The velocity of the object before the object reaches the ground is given by the differential equation dV/dt + V/10 = 32, ft/sec. What is the velocity of the object one second after it falls? A. B. C. D.

40.54 38.65 30.45 34.12

A. B. C. D.

171.24 lbs 124.111bs 143.25 lbs 105.121bs

Topics

D

P5,750 P6,750 P7,500 P6,350

Mon

D Tue

71:0: ME Board Oetober 1997 A nominal interest of 3% compounded continuously is given on the account.What is the accumulated amount of P10,000 after 10 years?

D D

A. B. C. D.

Problems

P13,620.1 0 P13,500.10 P13,650.20 P13,498.60

A. B. C. D.

15.45 kg. 19.53 kg. 12.62 kg. 20.62 kg.

708. A tank initially holds 100 gallons of salt solution in which 50 lbs of salt has been dissolved. A pipe fills the tank with brine at the rate of 3 gpm, containing 2 lbs of dissolved salt per gallon. Assuming that the mixture is kept uniform. by stirring, a drain pipe draws out of the tank the mixture at 2 gpm. Fi.nd the amount of salt in the tank at the end of 30 minutes.

I

'·L. ..

Wed

D

(g Thu

liJ D Fri

Solutions

D D

707: In a tank are 100 liters ofbrine containing 50 kg. total of dissolved salt. Pure water is allowed to run into the tank at the rate of 3 liters a minute. Brine runs out of the tank at the rate of 2 liters a minute. The instantaneous concentration in the tank is kept uniform by stirring. How much salt is in the tank at the end of one hour?

Theory

Notes

Sat

ANSWER KEY 681. A 682.C 683.C 684.C 685.C 686.8 687. 688.8 689.C 690. B

c

II

c

691.A 701. 692. A ·· 702. 8 693.A 703. A 694. D 704.A 695.C 705.A 696.B 706. 697.B 707.8 698.A 7.08.A 699.A 709:8 700. 710. D

c

c

Differential Equations Types of Differential Equations Orders of Differential Equations Degrees of Differential Equations Types of Solutions of Differential Equations Solutions to First-Order Differential Equations Application to First Order DE (Population Growth, Radioactive Decay, Continuous Compound Interest, Flow, Cooling and Heating, Newton's Second Law of Motion, Geometric Problems, Orthogonal Trajectory)

I

RATING

c::J 25-30 Topnotcher

c:J 1&-24

Passer

c:J 15-17 Conditional 0 0-:Lif failed If fAILED, repeat the test.

402 I 00 I" Solved Problems in Engineering Mathematics (2"d Edition) by Tiong & Rojas

a

Ill

x(y - 1) dx + (x + 1) dy = 0 xdx +~:cO X+1 y-1

Fourth order, since the highest derivative in the equation is 4th derivative. First degree, since the exponent or power of the 4th derivative is 1.

I

I I

'

(

1 -) dx+--=0 dy 1--

X+1 Integrate:

ED

f

dx-

A DE whose equation isM( x,y )dx + N( x,y )dy = 0 is exact if aM= aN ay Ox Note: From the choices, choice (c) is an exact DE. 2xydx + ( 2 + x

2

a( 2 + xz) =2X

a(2xy) --=2X

ay

Ox

X - In (X + 1} + In (y - 1} = C atx=1;y=2 -ln(1 +1) +ln(2-1)=C

c = 0.307

y = 1.55

1m

2

2

· 2y = N = _sin x sin Y

a( -sin X Sin Y) = -COSX sin Y

ax

ax

x2

2y -1

• -

J(cosxcosy-cotx)Ox.+k 1 =C

y(k + 1) = 20 + 5(k + 1) y(k + 1) = 20 + 5k + 5

·~1nxcosy-lnsinx+k 1

C=3.

=C

~

Eq. 1

y(k+1)=(20+5k)+5

J(-sinxsiny)8y+k 2 =C Thus the solution is,

x3

2 3 3 3y = x + 2 3y + 2 = 0

sinxcosy+k 2 =C

3

x3 -

2

L

X

Differentiate:

yz

xz y' =-y-= y_ 2xy 2x

~

Eq. 2

I ry comparing equations 1 and 1, k 1 = 0 .u1d k2 = - In sin x 1ill IS, the solution is,

Ell

Since, the resulting equation is the same to the given DE, then the assumed equation is the solution to the DE.

1m ( y-

~x 2 + y2 }

dx- xdy = 0

sinxcosy -lnsinx = C

slope = dy = 2x dx

sinxcosy = lnsinx + C

y2 = 2~yy' 2.

y(k+1) = y(k)+5

-sinx(-cosy)+k 2 = C

Y=-+-

y 2 =ex

=0

Assume the solution has the form: y(k) = 20 + 5k

3

Ill

2

x2 -1

Solve y(k + 1) using the assumed solution:

atx=1;y=1

3

lxz2+ 1)dx = ( )dy

0 = x(2yy ')-

x2 1 y=---

ay

,,xact DE.

c

Thus the solution is,

il(cosxcos Y-cot x) = -cosxsin y

3

1 = (1)3 +

2

(cos x cosy- cot x)dx- sin x sin y dy = 0

ay

x3

x = 1; y = 0

o'=l.!t+c 2

Note: aM = aN , then the given DE is an

= Jx dx

y= -+C

2ydx (x 2 + 1)dy 2 y(x + 1) = y(x 2 + 1)

c=

2

at

1 C=--

dy=~dx fdy

2y dx = (x2 + 1)dy

~

xz

y = x2 +C

•

Jdy = Jxdx

Y=-+C

2x 2 y=-+C

M=cosxcosy-cotx at x = 2; y =? 2- In (2 + 1) + In (y- 1) = 0.307 In ( y - 1) = - 0.594

dy- xdx = 0 dy = xdx

2

mJ Note: From the choices, choice (c) is a variable separable DE

~

=2 Jxdx

y _ 1 = e·0.594

A DE is a variable separable if the coefficient of dx shall be functions of x only and the coefficient of dy shall be functions of y only.

J!!'-111

Integrating:

Jdy

f~+ f~= fo X+1 y-1

N=2+~

M = 2xy

dy . -=2X dx dy = 2xdx Integrating:

y-1

dy = 0

}


Since tangent, the slope of the unknown curve is equal to the slope the line

of

''"''' c; 'Inc ·~IIIXCOSY

Ill('"""')

By inspection, this equation is a homogeneous DE, since the coefficient of the dx and dy are of the same degree (degree 1).

404 1O"o I Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas Let: y = vx; dy = vdx + xdv

Substitute (3) in (2): y' = c1 + y" c1 = y'- y"

Substitute: vx dx-

,J;.z + v

-~dx-xdv=O

I

I ~

'

dx+J~=O

JX

,j1+V 2

2

Let: v = tan 9 ; dv = sec e de Jdx + X

-7 Eq. 4

x dx- x(vdx + xdv) = 0

=0

f sec e de 2

0

J1+ tan 2 e

Substitute (3) in ( 1):

y =C1x + y"

In[ x(sece + tane)] = C x(sece +tan e)= C

(x+y)dy=(x-.y)dx

-7 Eq. 5

Let: y

= vx;

x(vdx + xdv) + vx(vdx + xdv) = xdx - vxdx

"'

xdx (2v + 0

dx+ (1+v)dv =O 2

V +2V

-1

X

In y = In (sec x +tan x) + c

dx + _! J(2 + 2v)dv = 0 J x 2 v 2 + 2v -1

y

=C

1 In x + - In

sec x +tan x = C , where: ec

1..

-eJP(x)dx

ef~

i.f. =

=x 2

y(x) = Jx (x)dx + C

2

=c

xy=

- 1) + ~ dv ( 1 + v) = 0

Jdy = Jsecxdx

y

.f _

vxdx + x2dv + v 2 xdx + vx2dv = xdx - vxdx

dy = ysecx dx

Y

(x) = ~

Substitute:

y' = y sec x

secx + tanx

X

i.f.

(x - 1)y" - xy' + y = 0

In

= .! ; Q

Equations 405

i.f. = elnx

Substitute:

•

where: P(x)

dy = vdx + xd

Substitute (4) in (5): y = (y'- y")x + y" = y'x- y"x. + y"

f~ + fsec e de = o lnx + ln(sece +tan e)= C

1m

By inspection, this equation is a homogeneous DE, since the coefficient of the dx and dy are of the same degree (degree 1).

2 2

vx dx- ,J1 + v 2 xdx- vxdx- x 2 dv

Da~J:.)if!erential

•

x4

4

+c

(0 + 2v - 1) = c m

= 'f._;

where: m =slope (constant)

X

Multiply 2 on both sides: where: ec = C Refer to the triangle, substitute values of sec e & tan e to the solution:

·b>:'

•

y

y

= C1 + C2·e· y" =C2 e• y'

= k,

where: k = 2c

2 Jdy -

r-; -

2

(0 + 2v - 1) =C, where C =ek

2

(0,0)

2

x

+ 2xy - x 2 = C 2

-

2xy - y 2 = C

Note: C is an arbirary constant which can

2y + x + C

=in x + In y

ln(xy) = 2y + x + C

xdy- ydx = 0

X

+ Jdx =c

2y -In y -lnx + x = C

0

IP!!IP.II liiliilll

2 x ('i._ + Y -1J = C x x l

rxx

Differentiating: xdy -ydx =

xz

';ubstitute v = 'j_ :

X

x,J1 + v +v = C

ID

x

· xy'(2y -1) = y(1- x)

= y(1- X) dx 2(2y -1)dy = y(1- x)dx 2Y - 1dx - 1 - x dx =0

~xz +Y2 + Y= C

1)

In[~ (0 + 2v -'1)] = k X dy (2y - 1)

xJ1+(~r +~=C

(0 + 2v -

Jln x + In

2

y = C1x + C2e•

=C (sec x +tan x)

1", placed in either side of the equal sign.

Y2 = 4ax

[D ~~ +(~)y=xz

4a=

2

L

X

Differentiating: 2

-7 Eq. 1 -?>Eq.2

-7 Eq. 3

Note: C is an arbirary constant which c.an be placed in either side of the equal sign.

x(2Y9Y)- y dx = l~olo

v (• r )

This equation is a linear DE: Jo(x)(i.f.) dx + C

-7 Gen. Eq.

xz 2xydy -

I

0

dx = 0

2xdy-ydx

=0

•

•

406 · l 00 1 Solved Problems in Engineering Mathematics (2"d Edition) by Tiong & Rojas

II

ll

If

sdg

dQ =kQ dt

In Q = kt + C

elny ::: e-X+C ::: e-XeC

dQ =kQ dt

Let: Q = amount of substance present

= Jkdt

In Q = kt + C

J~ = fkdt

II


~

Gen. Eq.

•

Let: T = temperature of the body

when t ~

= 0; Q = Oo

Differentiating: (x- h)dy- (y- k)dx =

= k(O) + C C =In Oo

0 (x -hf (x- h)dy- (y- k)dx = 0

when t= 0; Q = 100 mg In 100 = k(O) + C c = 4.60517

In 96 = k(100) + 4.60517 k =- 0.0004082

(y- k)dx- (x- h)dy = 0

•

when t = 1000 years; Q

In Q = -0.000693(100) +In Oo elnQ = e-0,0693+1nClg Q = Oo e·0.0693 Q = 0.933 Oo

when t = 200; Q = ? In Q = -0.0004082(200) + 4.60517 Q = 92.16 mg

C(O,k)

•

Let: P = number of population dP =kP dt

r:

>( + (y- kf

= Jkdt

In P = kt' + C

::: r"

~ Gen. Eq.

%lost= Oo - 0 ·93300 x100% Oo %lost= 6.7%

•

=k(O) + C

2x + 2(y- k)y' = 0

C =In Po

+yy' - ky' ::: 0

when t = 50 years; P = 2Po

X

X

k= -+y y' Differentiating: y'(1)-xy"+ '=O (y')2 y

In 2Po = k(50) + In Po k = 0.0138629 when t

= ? Q = 5Po In 5Po = 0.0138629(!) +In Po t = 116 years

· y'- xy" + (y') 3 = 0 xy" - (yf - y'

=0

ml Let: Q

= amount of radium present

Gen. Eq.

= 1oo•

In( 100° - 30°) = k(O) + C c = 4.24849 when t =15 min; T

=70•

In (70·- 30•) = k(15) + 4.24849 k =- 0.0373 when t = ?; T = so•

t = 33.. 59 min. y 2 = 2x+C

when t = 0; P = Po In Po

when t = 0; T

~

In (so• - 30•) = - o.0373(t) + 4.24849

•

Differentiating:

Differentiating:

_dt_= fkdt J T -30• In (T - 30•) = kt + C

= 0.5 Oo

In 0.5Qo = k(1000) +In Oo k :::- 0.000693 when t = 100 years; Q =?

=100; Q =96 mg

when t

dT = k(T- 30•) dt

Gen. Eq. In Oo

y-k m = - - ; , where: m =slope (constant) x-h

y::: ce-x' where: c =ec

2y dy = 2 dx

(:~l =~

+~=32 10

10dV + V = 320 dt

10f~= 320-V

Note: The slope of the orthogonal trajectories is given by: dy 1 (dx) dx=-(dy) =-dye dx

dV dt

c

- 10 In (320- V)

=t + C

fdt ~

Gen. Eq.

whent = 0; V = 0 (at rest) - 10 In (320- 0) = 0 + C c =- 57.6832

:;ubstituting: dy dx = -y ry =- Jdx

y

In y =- x + c

when t = 1 sec; V = ? - 10 In (320- V) = 1-57.6832 In (320- V) = 5.66832 320 _ v = es.sss32 V = 30.45 ft/sec

•

409 l ~0 1 Solved Problems in Engineering Mathematics (2"d Edition) by Tiong & Rojas

Let: Q

Note: This is a linear DE

=amount of salt in the mixture

100 + (3 -· 2)t

= 100 + t =volume of

_.9__ :::: concentratiQn of salt

I Ill ,!1

when t = 10 years; P =? In P

a(i.f.) = JA(t)(i.f.)dt+C 7 Formula for

i.f. =

• r:

dP = 0.03P dt

efP
100+t

d~ = rate of gain dt

rate of loss

2 · A(t) = 6 where: P(t) = - 100 + t ' .f I ..

~9=0-2(-a-) dt 100+t

_

- e

'f 1. • =

e

fP(I)dt .

when

i.f.

In a= -In (100 + t)2 + C 7 Gen. Eq.

J-2too+t

In 10000 = 0.03(0) + C

c = 9.21034 when t

=(1 00 + t)2

.· .i.· ·~

Substituting: when t

=0; a =50 kg. In 50

I~)~

2

= - In ( 100+ 1)2 + C

c:: 13.12236 when t

= 60 min; a = ?

In Q =~In (100 + 60)2 + 13.12236

c 3 Q(100 + t)2 = 2(100 + t) 3 + c when t = 0; Q

= 2(100)3 + c

C=-1.5x106 when t = 30 min;

Let: a = amount of salt in the mixture 100 + (3- 2)t = 100 + t =volume of mixture at any timet

_a_ =concentration of salt 100+t = rate of gain - rate of loss

da (-a-) -=3,2)-2 dt

dQ

~

= 50 50(100) 2

•

c

i

a(100 + t)2 = 6 (100 + t)3 +

a= 19.53 kg.

dt + Q

2

a(100 + t) = J6(100 + t) dt +

Ina= 2.972 ·

~~

a(100 + 30)

2

Q

a =?

= 2(100 + 30)3 - 1.5 X 1()6

=171.241bs.

• r:

dP =0.03P dt

In P

=0.03 Jdt

=0.03t + C

when t = 0; P = 5,000

100+t

(

t =0; P =10,000

i.f. =eln(100+tl'

In a=·· 2 In (100 + t) + C

2 ) 6 100 + t =

=0.03 Jdt

In P = 0.03t + C

i.f. =e21n(tOo+tl

dt +C fada= - f1oo+t

= 0.03(10) + 8.51719

p::; 6,749.30

the general solution of a linear DE

mixture at any timet ·

It

Day IS-Differential Equations 409

In 5000 = 0.03(0) + C

c = 8.51719

·~

l!i ~

r ~,,,. l

i~

I l'' ?cl

l t

)

£.

= 10 years; P =? In P = 0.03(10) + 9.21034

p = 13,498.60

412 1001 So1:ved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas

. ' "' ~·, ,,,~' '"'~*'•'<-A V

, ... "'"' 'o"«,. ~

ll

*

N.

:,0 _.

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<",-

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Topics

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< ·, '

~.-~·.,~

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Tue

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....,.·.,.~,~..,:;. ~

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If "'" ~ « ~ o· '"" '.• » « " ~ .,. ~ ¥. « ?- • ~ '

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~~····~~·

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* "'., • . ., ~· 1~ .&. '" ~

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~

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~ ..'-<<>"

..

~

/ • .;_ *'

>" < >!• ~ ~ 'r ·'II' ... ~

'.~\<

.o .<~ ·~·" ~.,.. .. ,.. : ; . , , . ··~

/

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'**""

/~

~

;·,

~~

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;,

> .->

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'~««·~~%%-0'"'"-''"'""<>:~-~

+,·~"* ~'·~·· .;..,..,.*~''"'~ ¥'\o)~"~".<'

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., :t

............. ~

::.. ~.;;: .. ~ ~:i.-«·<)·,,.>··~·,s.»-lr-i«

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+. .p

;io.<-'"'><,~'1<*!,-:i«<>·-:;•vP<•<~
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.,: « ,.. »

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~

'""••<:• .... ,.. •• ,..

~, .... ,,._.>r~"'·"·"'~"""'"'~

~ "!"

<

'<1·.'<-'J<'~.y

;-·,~ o.. ~'""Y. ~ ~ ~o "'""'

•<

"<

X

X

'v • > '".). ~ "'

',.~···

':

~

Theory

Wed

Problems

Thu

Solutions

Fri

Notes

Sat

i: ..~

Su~cessive integral powers of i or j

What is a Complex Number? By definition, a complex number is any number expressible in the standard form a + bi or a + jb.

>

The value "i" or "j" is the imaginary unit or number. It is equal to the

J=-1 .

Example: Find the value of x of the equation 0. ·' "'" ,,

" 1> .• •

"'" , '

" ·~

·:'«·i<"'ff•.x-

~ .; ,.. '

' ., .... '" .. '<- ~ '

..... -~· " .....

v

i< ;. ~ '

••

'·(.""'''·' •.

x? + .1 =

Simplifying we get x ::: ± Thus the solutions are i and ..,. i.

''~'~-~ ;-;~ ~ ~~-1

=1

j2 =- 1 = jl j2 ) =- j j4 j2( 1 j5 = j3( j2 ) = j j6 j4( j2 ) 1 f i5 ( j j8 j4 ( t) = 1

f

= l) = = == f)== ... etc

J=-1

'

Advanced Engineering Math Complex Numbers Forms of Complex Number Operation of Complex Numbers Matrices Operation of Matrices (Addition, Subtraction, and Multiplication) Transpose Matrix Cofactor of Entry of a Matrix Cofactor Matrix Inverse Matrix Determinants Properties of Determinants Laplace Transforms Laplace Transforms of Functions

What are the Different Forms of Complex Numbers? a. Rectangular form z=a+jb where: a= real part b = imagina;y part

Day 17 -AdvancedEngineering Math 415

414 1001 Solved Problems in Engineering Mathematics (2"d Edition) by Tiong & Rojas b. Multiplication of complex numbers

b. Trigonometric form

z = r cos e + j r sin e

Arc sinh z =In (z± .[zz + 1) sino= ei" -e-iB j2

Rule: Use the polar form

z=rcise

c. Polar form

Arc cosh z = In ( z ± ~)

coso= eio + e-iu

(r1 L91)(r2 L9 2 ) = r1r2 L(6 1+ 62 )

1 (1+Z) Arc tanh z =-In -2 1-z

2

.

c. Division of complex numbers

[

z= rLO

f I l

where: r = modulus or absolute value 6 = argument or amplitude in degrees

Note:


What is a Matrix?

cote= case sine

r1Le1 = 5_ L(91 -62) r2

d. Complex number raised to exponent

Matrix is a rectangular array of real numbers arranged in m rows and n columns. The term "matrix" was introduced by the English mathematician James Joseph Sylvester (1814- 1897) in 1850. The size of a matrix is determined by the number of rows and columns. The expression "m x n" is the dimension or order of the matrix. If the matrix has only one column, it is called a column matrix and if it has only one row, it is called a row matrix. The following is a 3 x 3 matrix or square matrix (i.e. 3 rows and 3 columns).

sec e = _1 case

r2L6 2

Imaginary axis

tan 0 = sine case

esc e = _1 sine

II

"n" bl

'/

h. Hyperbolic functions

Rule: Use the polar form Real axis

a

(rL6)"

0

=r" L (n6)

sin h0 = .ee -2--·e· cos h6

e. n1h root of a complex number

.r""' -/a 2 +b 2

6=

tan· 1 ~

(rLe )*

d. Exponential form 9

where: e = argument in radians O~erations

a. Addition and Subtraction of complex numbers (a1 + jb1) + (a2 + jb2) = (a1 + a2) + j (b, + b2) (a1 + jb1) - (a2 + jbz)

"" rt L

e + k(36oo)

n

where: k = 0, first root or principal root k = 1, second root k = 2, third root k = n -1, n1h root

What are the Mathematical of Complex Numbers?

= (a1 - a2)

2

6


a

z = r ei

0

= e +e-o

f;

Logarithm of a complex number

I

Note

tan he=

~nh0

'

i.

sec he=

_1_ coshe

esc he=

. _1_ sinhe

Inverse trigonometric functions of complex numbers

Rule: Use the trigonometric form Arc sin z = - j In (iz ± ,)1- z2 )

In z = r efl = In r + In efl In z =In r+ j e

Arccos z =- j In Arctanz=

g. Exponential & trigonometric functions of a complex number

+j(b1-bz) ei 0 = cos

e + j sin e

e-ie =coso- j sino

j.

41 4 -2

coshO

cot he = cosh~ sin he

i

5

A= 2 1 [ 3 -2

(z±~)

The first non-zero entry in a row of a matrix is known as the leading entry or the leading element. In the matrix above, 6 is the leading entry. The diagonal from the upper left to the lower right is called the principal diagonal or main diagonal and all entries in the said diagonal are called as diagonal entries. If all entries in a matrix above the main diagonal are zero, then it is said to be a lower triangular matrix, and if all the entries below the main diagonal are zero, the matrix is referred to as the upper diagonal matrix.

-i 1n(1+jz) 2

6

0

A= 2

1

1- jz

Inverse hyperbolic functi6ns of complex numbers

[

~21

3 -2

Lower Triangular Matrix

Day 17 -·Advanced Engineering Malh 417

416 100 i Solved Problems in Engineering Mathematics (2"d Edition) by Tiong & Rojas

6 5 4][0 01 -24

Example: Find A -

4 -3]~1 A=~ [

B= 0

'

Upper Triangular Matrix

l

,11

A-B= -12

Two matrices are said to be equal if and only if there orders are equal and their corresponding elements are equal. Example: Matrices B and C are equal.

C=[3

2

2] -1

What is the Sum of two matrices? If A and B are two matrices of the same order, the sum of A and B, denoted by A + B, is the matrix for which each of its elements is the sum of the corresponding elements of A and B.

A

=[-15

-2 6 ] 8 -3

~=AB- 1 8

J

Solution:

Example: Find DC.

2 -3] [ ~- ~1

c

=[-4-1 1 3] 2 5

A+B=[9 5 7] 2 8 -7

4(-4)+(-1)(-1) [ 1(-4)+5(-1)

2(-4)+(-3)(-1) 2(1)+(-3)(2) 2(3)+(-3)1 4(1)+(-1)(2) 1(1)+5(2)

-5 -4 -9] [

DC= -15

-9

L2

If matrix A is reflected in its main diagonal, so that ali rows become columns and all columns become rows without changing their relative order of entries in the rows and columns, the result is a transpose

2

7

11 28

matrix, AT. Example:

4(3)+(-1)( 1(3)+5(5

·

A_ [-4-1 21 53]

I -~ ~.

What

--4 -1

AT= [

~ ~

l

ts a Cofactor of an Enta:y of a

.~atrix?

A cofactor of an entry of a matrix is the same as the cofactor of the sam~ entry in the determinant of the matrix and thus, is defined only for square matrices. S1gn conventions:

[:

:]

+ - +] - + [+ - +

1 xample: Find the co-factor of 6 in the lollowing matrix.

A=[~-~~

2] 1

The determinant of A is:

~~

l=

(1)(1)- (2)(2) =- 3

Thus, the cofactor of6 is - 3:

What is a Cofactor Matrix?

What is a Transpose Matrix?

~

D is a 3 x 2 matrix and C is a 2 x 3 matrix. The product DC can be obtained because the number of columns of D (2 columns) is equal to the number of rows of C (2 rows). The product DC will be a 3 x 3 matrix_ DC=

If A and B are matrices having the same order, then the difference of A and B, denoted by A - B, is defined as A- B =A+ (-B)

4

B.

A + B = [ 5 + 4 -2 + 7 6+1] -1+3 8+0 -:-3-4

What Is the Difference of two matrices?

where 8' is called the 1nverse matrix of matrix B

Supposed that A is a matrix of order m x p and B is a matrix of order p x n, then the product of A and B, denoted by AB, is the n x n matrix for which the element in the ith row and the jth column is the sum of the products formed by multiplying each element in the ith row of A by the corresponding element in the jth column of-

D=

B =[4 7 1 7 0 4

A=+f1

det A:=

'

1

What is the Product of two matrices?

Example: Find A + B.

Division of matrices is done by multiplying the numerator by the inverse matrix of the denominator.

Let: A and B are matrices

-1 + -7 5 8 -8 -2

Diagonal Matrix

2J 2 -1

''',l,

A-B=[~ -3] [-4 -3] 0 -6] [-5 6

0 0 -2

B =[3

•=[~ ~]

.)~~

Solution:

6 0 0] C= 0 1 0 [

[: 1:

What is the Division of Matrices?

B.

A cofactor matrix is formed by replacing each element in' the given matrix by its cofactor. Example: Find cofactor matrix of A

A=[! ~J Cofactor matrix of A= [

~ ··~]

What is an lnvers~ Matrix? Steps required to find for the Inverse Matrix (say. A):

a. Form the cofactor matrix of matrix A b. Fom1 the transpose matrix of the cofactor matrix A c. Evaluate the determinant of matrix A d. Divide each element in the (matrix cofactorf · Example: Find the inverse matrix of A.

2 0 A=i2 1 4

4 2 6

-2

4

Cofactor matrix A= 1-12

6'

-4

8

I ht~ .:quivalent matrix is: (Cofactor matrix A)T

=I

0

-2 4 0

6

-3 -12

0

6

6

6

--3

418 .1001 Solved Problems in Engineering Mathematics (2"d Edition) by Tiong & Rojas

I

det A =

6. The value of the determinant is not changed if a column is replaced by the column plus a multiple of another column. Similarly for rows.

1 2 0 1 4 I =6 4 2 6

I2

[:

[ l ~II

~- .

Day 17- Advanced Engineering Math 419 -2

•;

:.

.•

'

det A=

I5

.

.·'. . . 1·

3

-'~ ','

-5

7 -5 I=

0 11

A=~;; -~

-6

''"c

.

f

Determinant of a 2 x 2 matrix: 1-2 Thus , A- 1 = _.!_ 4 6 0

f

-12 6 6

0 6 -3

det A= 1: :1

~,;

Determinant is a square matrix (i.e. the number of rows= the number of column). Every deierminant can be associated with a real number.

;.'i

Determinant of a 3 x 3 matrix:

2. If two columns (or rows) of a determinant are interchanged, the value of the resulting determinant is equal to the negative of the value of the given determinant.

3. If two columns (or rows) of a determinant are identical, the value of the determinant is zero. 4. 'If the elements of a column (or row) of a determinant are multiplied by k, the value of the determinant is multiplied by k. 5. If the elements of the jth column of a determinant 0 are the sum aij + bij, then 0 is the sum of the determinants 0' and 0" in which all the columns of 0, 0' and 0: are the same except the jth; furthermore, the jth column of 0' is aii, i = 1, 2, 3, ... , n, and the jth column of 0" is bij. I = 1, 2, 3, ... , n. Similarly for rows.

5

0 11

-5

5 0

-6

3 11

=1 d

e f g h i

d e g h

Determinant of a 4 x 4 matrix:

I

Using Pivotal Element method: Example: Find det A.

2 -4

3

-1

~1 CD~2 · · · 2

I 3

?

-4

-1

-2

5

1

4

11

Using Modification method:

~

-4 3 1 2 -4

-1

@0 -1

5

det A= 200

1

4

What is a Laplace Transform? The Laplace transform of a function f(t) denoted by l,. [ f(t) ) is defined as a function of a variable "s" by the integral:

Multiplying column 2 by 1 and add it to column 1 Multiplying column 2 by 2 and add it to column 3 Multiplying column 2 by -2 and add it to column 4

4- (5)(2)

j 2 for 2"d column

0

where: t > 0 and s is any number (real or complex) laplace transform of some elementary functions:

The new matrix becomes,

5 3

.........

.... ····•········•• ...

r----·;--;~·;--;~~--;~: and

="'

F(s) = L [ f(t) 1 Jf(t) e-"1dt

.

A=~~~ cb~: · ~· · ·

2- (-4)(-1) 3- (-4)(-2) -1-(-4)(2)1 2 3- (2)(-1) -4- (2)(-2) -1- (2)(2) (1):: 1- (5)( -2)

51(-1)2•1

-3

= 5 [(-5)(-3)- (11 )(5)) (-1)

detA =

1-2- (5)(-1)

detA=(5)~-5

det A= 200

Set the encircled numbers to ·zero by·

det A= (aei + bfg + cdh)- (gee+ hfa +idb)

i

2 for 2"d row 1 for 151 column

det A= 2(0)(-6) + (-5)(-5)(3) + 7(5)(11)[3(0)(7) + (11)(-5)(-2) + (-6)(5)(-5))

A=

a b c a b

A=

-2 -5

1,.·1

det A =I d e g h

a.

7

3

b.

det A

-5

_,,_,_,

a b c

What are the Properties of Determinants? 1. If the rows of one determinant.are the same as the columns of another, and in the same order, the two determinants are equal.

·t

detA =ad- be

What is a Determinant?

-2

f $

-2 -5 A=(1)j

0 11

-5

F(s)

1.

1

2.

t

3.

t"

4.

e±at

5.

t" e±at

6.

e±at,sin kt

-6

7

5 0@ 3 11

f(t)

i(-1)2• 2

-6

Set the encircled number to zero by multiplying column 1 by 1 and add it to column 3

1 s 1

-

52 n! sn+t s=t:a n! -·- )"+1 (s+ a k (s +a) 2 + k2

4ZG l 001 Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Roj!! e±at cos

7

s=t=a

kt

2

(s+a) +k

8. 9.

[II

a

sin at

s2 +~ s s2 + a2

cos at

10. sinh at

a s2 -a2

11. cosh at

s ~2 -a2

12. t sin at

2as (s2 + a2 )2

13. t cos at

sz- a2 (s2 + az )2

14. sin (wt + 8)

s sin8+kcos8 ~~

II

[ l ,111

2

15. cos (wt + 8)

~uote: "Obvious is the most dangerous word in mathematics." - Eric Temple Bell

.I

I \~;,;

x:

l~ tvlon

·N

} ·;:%:,

·t

i:'

Theory

!~

, Problems

0 0

--52

GOOD LUCK!

Solutions

Notes

I h

'

determinants dates back to the ancient Chinese who use barnboo rods in representing the coefficient of unknown quantities, and gain acceptance ·when introduced in Japan by the 1th century greatest Japanese mathematician Seki Kowa (1642- 1708)! Even if German Mathematician Gottfred Wilhelm von

Leigniz (1646 -1716) and Swiss mathematician Gabriel Cramer (17041752) gave their valuable contribution to the subjeGt, it was Alexandre-Theopl1ile Vandermonde (1735 -- 1796) ir1e one regarded as ti">e formal founder of determinant theory.

\Ved

Thu

~ Fri

0

Sat

7U: ECE Board April 1999 : ;1mplify the expression i1997 + i1999 , where i 11; an imaginary number.

r;,~-----~_,_,

Adva11ced Engineering Math Complex Numbers 1 Forrr1 of Complex Numbers Operation of Complex Numbers Matrices Operation of Matrices (Addition, Subtraction, and , Multiplication) Transpose Matrix Cofactor of Entry of a Matrix Cofactor Matrix Inverse Matrix Determinants Properties of Determinants Laplace Transforms Laplace Transforms of Functions

I

j

C. 0,

2i -1 1+ i

714: CE Board May 1:994

'O.Critiia: Did you know that. .. the theory of

0 0 0

Tue

s sin8-kcose +k2

Prm:eed to the next page for your 17th test Detach and use the answer sheet prCNlded at the last part of this book. Use pencil number 2 in shading your answer.

Topics __

I''

']

1\ ll

The expression 3 + 4i is a complex number. Compute its absolute value.

0

c

-i 1+i

I>

1- i

711:: EE Board April1997 : >unplify: i29 + i21 + i

A

4

B. C. D.

5 6 7

715: EE Boal!'d October 1993 /\

3i

ll

1- i

I.

1+ i :Ji

IJ

Write the polar form of the vector 3 + j4.

'/II J: u:~

Boat•d April 1'997 3217 form a + bi the expression i

W1 111 · 111 tlw 1·\.'i. I 1111

i\

1'-

II

I I

I

I

A

6L:53.1°

B,

10L:53.1 °

C,

5L:53.1°

D.

8L:53.1°

422 . 1oo·i Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas 1:t:t1 EE Board October 1997

71ft: ME Board April1997 Evaluate the value of

r-10

x

..J-7

[

B.

-JiO

C.

JjO Jfi

D.

II!

[ l

l~

71.'7: EE Board April 199ft 2 Simplify (3- i) - 7(3- i) + 10 A.

- (3 + i)

B.

3+i

C.

3-i -(3-i)

D.

718: EE Board April. 199ft 1200 If A= 40 ei , B = 20L- 40•, C = 26.46 + jO, solve for A + B + C.

A B.

35.1 L45° 30.8L45° 33.4L45°

71.9: EE Board October 1997 What is 4i cube times 2i square?

A.

-8i

B.

8i

C.

-8

D.

-8f

7:tO: EE Board April1997 What is the simplified complex expression of (4.33 + j2.5} square?

B.

c. D.

12.5 + j21.65 20 + j20 15 + j20 21.65 + j12.5

7:t1: ECE Board November 1998 Find the value of (1 + i) 5 , where i is an imaginary number.

A. B. C. D.

A.

B. C. D.

;I

B.

1- i -4(1+4) 1 +I 4(1 + i)

'i!

1.9 + j1.1 3.26- j2.1 2.87 + j2.1 2.25-j1.2

:~

C.

D.

221-91i

731.: EE Board April 1:999

169 21+52i

Evaluate cosh ( j~ ).

What is the quotient when 4 + 8i is divided byi 3 ? A. B.

C. D.

8-4i 8+4i -8 +4i -8-4i

If A = -2 -3i, and B ::;: 3 + 4i, what is

~?

A. B.

13 -90+220i

C. D.

169

I

What is the simplified expression of the 6+j2.5 complex number--- ? 3+j4

A.

- 0.32 + j 0.66

B.

1.12-j0.66 0.32 -.j0.66 -1.75+j1.03

C. D.

A.

B. C. D.

A. B. C.

25 -18+i 25

D.

18+i 25

square square square square

root root root root

of 3 of 3 of 3 + of 3 +

2i i i 2i

77.91 EE Board June 1:990

7:£51 EE Board October 1997

50 + j35 F.1nd t h e quot'1ent of 8+ j5

. . 4+3i Rat1ona11ze ..:i .

A.

2

A.

7.47 L3Q•

11 + 10i

2.47 L53°

5 5+2i

730: EE Board March 1998

--

5 2 + 2i

. l'fy (2 + 3i)(5- i) S Imp I (3- 2i)2 .

B. C. D.

1.34 + j0.32 2.54 + j0.866 2.23 + j0.21 1.28 + j 0.98

734: EE Board October 1997 Evaluate the terms of a Fourier series 2 ei10nt + 2 e-i1oxt at t = 1.

A. B. C.

2 +j 2 4

D.

2 + j2

.

s1n x = x-

follows:

!2B = 2

A.

u

c

D

5- j5 -10+j10 10-j10 1b+j15

x3

+

xs

31 51 x2

+ ......

x4

COS X= 1--+-+ ......

Three vectors A, B and C are related as at 180°, A+ C =- 5 + j15,

C = conjugate of B. Find A.

7:t&: EE Board October 1.997

A.

7351 EE Board March 1998

D.

C.

0.5+j1.732 j0.866 j1.732 0.5 + j0.86S

Given the following series:

C.

--

A. B. C. D.

6.47 L3° 4.47 L3•

1 + 2i

B.

D.

B.

3

Evaluate In (2 + j3).

Perform the operation: 4 (cos 60• + i sin 60°) divided by 2 (cos 30° + i sin 30°) in rectangular coordinates.

25 -18-i

Evaluate tanh ( j~ ).

7331 EE Board April1999

77.8: EE Board Aprilt.997 18-i

0.707 1.41+j0.866 0.5 + j0.707 j0.707

737.1 EE Board Aprill.999

77.7: EE Board Aprill.99&

1111

7:t4: EE Board October 1997

4

13 -7 + 17i

7:£31 ECE Board April1999

27.7L45•

C. D.

A.

.1:~·

A.

''I

A: i

Ill

Find the principal 5th root of (50( cos 150• + jsin 150°) ].

Day 17 -Advanced Engineering Math 423

2!

4!

x2

x3

ex = 1 + X + 2f + 3! + ...... What relation can you draw from these series?

A.

e,x = cosx + sinx

B.

eix =cosx+isinx

C.

e ix

·

·

=I COS X+ Sin X

D.

740: CE Board November :1996

iex =icosx+isinx

Compute the value of x by determinant.

'736: EE Board October 1997 One term of a Fourier series in cosir)e form is 10 cos 40nt. Write it in exponential form.

r:

[: l,

A.

B.

C. D.

5 6 i40nt 5 ei407tt + 5 e-i407tt

10

e-i 40lt!

A. B.

o

c.

10 6 i40rd

D.

c. D.

1 -2

2 -1

3 -2

3

1

4

A.

4 2 5 0

B. C. D.

738: ECE Board November 199:1 Evaluate the determinant:

D.

D.

452

-1

0

0

2

-1

3

3

5

-1

3

2

-3

-4

-3

-4

-1~ -~1

B.

~-~ -~,

D.

3 0

c.

(1, 2, -1) {-1,-2,1)

If A= 1-1

2

2

I,

what is the

I~ -~1

_,-20

the second row, third column element?

B.

-1~ ~I

c

··I~ ~I

D.

[

0

-1

1

3

is multiplied by

1

~ ] ;; eqoal to mo, then ~trix [ ~ ] ;,

A.

3

B.

1

c. 0 D.

-2

747: CE Board November 1997 Given the matrix equation, solve for

01 -1

A. B. C. D.

x and

--4,6 -4,2 ~4,-

2 -4,-6

7481 EE Board A.pril1996 If matrix [

0 1 0

0"0 0 0 0 0 0 0

-1 2]

[ ~ ~ J[;J=[ ~ J

1 0

~ ~

}s

multiplied by

[~ }s

J

equal to zero, then matrix [; is

I

q 0

0 0 1 C.

[

1 2

y.

1 0 0 A.

41 ,what is the cofactor of

' 0 5 7

120 531

.,...

B.

2 3 1

-2

A.

C.

1 -2

743: EE Board October 1997. 1

1

cofactor with the first row, second column element?

-28 16 52

(2, 1, -1) (2, -1 , -1 )

489

B. 389 c. 326

If A= 1-2

-32

A. B.

A. A.

5

multiplied together, write the product.

Evaluate the determinant:

3

.

=


If matrix

Solve the equations by Cramer's Rule:

D.

14

:'il

F;ngineering Math 425

746: EE Board April 199ft

7441 EE Board October 1997

2x-y+3z=-3 3x + 3y- z 10 -x-y +z=-4

'110 -101 101 -110

2

2

0 4

1

;I

I~

'7421 EE Board Aprlll997

0 5 3

c.

3 2

3

D.

745; EE Board October 1997 If a 3 x 3 matrix and its inverse are

1 6 0 4 2 7

A. B.

0

2

Given the equations: x+y+z=2 3x "'- y- 2z = 4 5x - 2y + 3z = -7 · Solve for y by determinants.

Evaluate the determinant:

,1111

2

10 14

-1

741: EE Board Aprll1997


A. B.

X=lI

4

Advanc~cj

Day 17 -

424 . 100 1'Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas

0 1 0 ' 0 0

A. B. C.

D.

8 1

-4 0

749: EE Board October 1997 450 100 If A 16 7 3 and B 0 1 what 1 2 5 0 0 1

=

=

is A times B equal to?

4 0 0 A.

lo

1

o

0, 0 5

oj ,

426 l 00 i Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas 0 B.

io

0

0

7

o

1 0 0

75::&: EE Board October :1997 3 1

2

Transpose the matrixl-2

0

0

6 7 0

c. is [

~'

l

'~

A.

2 3 5 4 5 0

~'

[,

-1

9 4

D.

16

7

2 3

3

B.

2 5

. I-12 31 I + 2 Matnx. 1-11 21 I=

Matnx

Matrix 1-2 2

B.

Matrix 1-1 1

c.

Matrix

~-~

Matrix

I~

D.

4 2

[ -11 19

:]

C. D.

[ -10 -19 [-11 -20

2 3

1

2

.~! ~1

~~~

f

-1

2 -1

·.:. iI

0

·';'~

A.

1

B. 2

0 -5

[!

]

~J

.C. D.

756: EE Board April :1997

[ 2 e( exp -t ) - 4 e( exp - 3! ) 1 [ e( exp -2! ) + e( exp - 3! ) ] [ e( exp -2t ) - e( exp - 3t ) 1

B. C.

[ 2 e( exp -t) ][ 1-2 e( exp- 3t) 1

D.

757: EE Board Marc:h :1998 Determine the inverse laplace transform of l(s) =

200 -

50s + 10625

3

2 0

A.

l(s)

2

2

-1

B. C.

l(s) = 2e-251 cos100t

D.

l(s) = 2t e-251 cos 1OOt

The inverse laplace transform of s I [ (s square) + (w square) 1is

-~ ) (~ ~) (~ ~) 5 c~ -1 J (-~

f :~

A.

sin wt

B.

w

C.

(e exponent wt) cos wt

D.

759: Find the inverse laplace transform of 2s- 18 . -

754: EE Board April :1997 k divided by [ (s square)+ (k square)] is the Inverse laplace transform of:

B. C. D.

as a funct1on of x.

2 cos x - sin 3x 2 cos 3x- 6 sin 3x 3 cos 2x- 2 sin 6x 6 cos x- 3 sin 2x

760: Determine the inverse laplace

cos kt sin kt ( e exponent kt)

9] . 6

D.

1.0

_:]

755: EE Board April :1996, EE Board April :1997

A

The. laplace transform of cos wt is

II

t·

-

s 2 +9

A. B. C.

s I I (s squam)

= 2e-251 sin100t l(s) = 2t e- 251 sin100t


A.

A

s2

-2

G~)

5

D.

1-1

3 5

w I [ (s square)+ (w square)] w I (s + w) s I (s + w)

B. C.

A. .£:,

753: Determine the inverse matrix of

Find the elements of the product of the two matrices, matrix BC.

B.

't"


Find the laplace transform of [ 21(s +1) ]- [ 41(s + 3) 1.

1

Elements of matrix C =

A.

0

2 -1

11

-1

r

2

Elements of matrix B = [

8] -5

2

-2 D.

75:1: CE Board Mzy :1996

[ 11 -20

C.

2

~I

0 -2 3

-1 0

11


A.

2 -1 1

I0

-1

I

t

1

t

f transormo, ~.·~.

4s'l 8s

(w square) ]

1 I . ·-·e s1nht

4

1 e·''I sinh! 7

1 4

C.

-e 1 cosht

D.

..!e21 cosht

2

:I

·~

jl ~

Day 17 -Advanced EMi.nE!~il:lg Math 429

•

i

-1

i2=

[Ill

D D D D D D

~II

[I l

Mon

,,

Tue

,1111

Theory

Problems

Solutions

Wed

Thu

~ Fri

D D Notes

Sat

ANSWER KEY

711. A 712.A 713.C 714. B 715.C 716. B 717. D 718.C 719. B 720.A 721. B 722.A 723. c

724. B 725.A 726.C 727. B 728.C 729. A 730. B 731. A 732.C 733. D 734.C 735. B 736. B

737.C 738. B 739.C 740. B 741. c 742.C 743. B 744. D 745.A 746.C 747.A 748. D 749. D

750. D 751. A 752. B 753. A 754. B 755. A 756. A 757.A 758. D 759. B 760.A

--

Advanced Engineering Math Complex Numbers Form of Complex Numbers Operation of Complex Numbers Matrices Operation of Matrices (Addition, Subtraction, and Multiplication) Transpose Matrix Cofactor of Entry of a Matrix Cofactor Matrix Inverse Matrix Determinants Properties of Determinants Laplace Transforms Laplace Transforms of Functions

. 3

c:J c:J

RATING 43-50 Topnotcher

c:J

If FAILED, repeat the test

.

=-I

Since

·1

j427:::-

i

16 is exactly divisible by 4

i2 = -1 i3 =- i i4 = 1

Note:

,,,··l; IIi

',

·'

'

'

j17:::

1

t

Since

1996 is exactly divisible by 4,

'' ,;,_'

:t

Substituting:

a

I

Since

f1996

= 1,

i1997

=i

i1998

=- 1

11999

=- i

.1997 + i1999 I

i

i16=-

If the exponent of" i " is exactly divisible by 4, then the simplified equivalent of the imaginary number is equal to 1.

1

Substituting:

•

j3217 - i427

Let r

+ i18 = i - (- i) + (- 1) = 2i - 1

= absolute value of the complex no. (a + bi)

r= -/a

2

+ b2

Substituting:

r=

= i + (- i) =0

j{3)2 +(4)2

r =5

Ill 20 is exactly divisible tiy 4, i20

= 1,

i21

=i

The polar form of the complex number, •a + jb" is given by: z = r L6 where:

r= -la

2

+ b 2 and

e = tan-1 E.

Since 28 is exactly divisible by 4, i28

= 1,

Substituting:

~

r

='1/32 + 4-

i29= i

Conditional

0-24 Failed

i426::: -

i16=

I~ ,

30-42 Passer

c:J 25-29

i

J

.. ::\

,'

i4 =1

·····················i······················

J

Topics

i

r=5 substituting:

•

i29 + i21 + i ::: i + i + i = 3i

e = tan-1 .i 3

e =53.1°

Since 32'16 is exa(.ily divisible by 4, 1,

i3216::::

I

i3217

=i

·;nu;e 424 is exactly divisible by 4 i424

== 1'

i4?5 "'

i

Thus, the complex number is 5 L53.1 o

a ..J-io =JfO .[:1 '..J-10 = .Jffii

a


430 I 00 I Solved Problems in Engineering Mathematics (2"d Edition) by Tiong & Rojas

•

H=.fi~

"--1 =.fii

2

( 4.33 + j2.5) = 4.33 2 + 2( 4.33)(j2.5)

+ l (2.5)

Substituting:

R ../-10 X R

../-10 IIH [

../-10

~Ill

[

1~11

Iii

l,

'~~

a

../-10

=

X

=

x.fl X R

(JiO i) {.fi i) .fi0i

,,,

--.3-

( 4.33 + j2.5) = 18.749 + j21.65 +(-1)(6.25)

I li

ltr.

=

..fi0(-1)

=

-.fiO

2

-

7 (3- i) + 10 = 9- 6i + i2

• -

21 + 7i

+10 2

(3-i) -7(3-i)+10'=19-6i+(-1)-21+7i 2

(3-i) -7(3-i)+10=-3+i 2

(3-i) -7(3-i)+10=-(3-i)

a Note: Convert all the complex number in form

rectang:.~lar

A= 40 ei 120' A= 40 .L120° A = - 20 + j 34.64 B = 20Z-40° B = 15.32- j12.855

c

= 26.46

A + B + C =- 20 + j 34.64 + 15.32 - j12.855 + 26.46

= 21.78 + j 21.785

•

Note:

= 30.8L45o

i2 i3

= -1 =- i

(2+3i)(5-i) = 13+13i(5+12i) (3-2i) 2 5-12i 5+12i

Rationalizing: -2-3i = -2-3i(3-4i) 3 + 4i 3 + 4i 3 - 4i

. 1 + i = 1.4142 .L45° (1 + i)5 = (1.4142 .L45° )5 (1 + i) 5 = (1.4142) 5 .L5(45°) (1 + i) 5 = 5.656 .L225° (1 + i)5 = - 4- 4i (1 + i)5 =- 4(1 + i)

•

50 (cos 150° + j sin 150°) = 50L150° ~50L15oo = (50)115 .L150°(1t5)

~50.L150° =2.1867.L30°

~5QL150° = 1.893 + j1.093

(2+3i)(5-i)- 65+156i+65i+156(-1) (3-2i) 2 - 25+60i-60i-144(-1)

-2-3i 3 +4i

(2 + 3i)(5- i) -7 + 17i (3-2i) 2 =-1-3-

-18-i 25

--=--

~50.L150° = 1.9 + j1.1

tm 4 + 8i

-i.....

4 + 8i · since i3 = - i

-~

~i3--

Rationalizing: Multiply the denominator with its conjugate.

4+3 8i = 4+ 8i(i) i

-i

4 +8i

4i+8i2

1

-1

i

4+3i 2-i

~

I

(2 + 3i)(5- i) _ 6s + 156i + 65i + 156i2 (3-2i)2 - 25+60i-60i-14M

-2- 3i -6 + 8i - 9i + 12i2 3 + 4i = 9 - 12i + 12i - 16i2 -2-3i -6+8i-9i+12(-1) 3+4i = 9-12i+12i.-16i(-")

liD

-.3-=--.2-

(4i 3)(2i 2)·= (4)(- i)(2)(-1) ( 4i 3 )(2i 2 ) = 8i

Rationalizing:

-2-3i 3+4i

2

(3-2il =5-12i (2+3i)(5-i) 13+13i (3-2i)2 = 5-12i

liJI

( 4.33 + j2.5) = 12.5 + j21.65

2

(3-2i) =9-12i+4i2

= -8 + 4i

I

Note: (r.Le)" = r" .Lne

(3- i)

4 +8i

2

2

2

4+8i- 4i+8(-1) i3 -(-1)

(2 + 3i)(5- i)- -91 + 221i (3- 2i)2 169

•

6 + j2.5 = 6.5.L22.619° 3+ j4 5.L53.13°

Rationalizing: 4+3i_4+3i(2+i) -2-i 2-i 2+i

~----

4 + 3i 8 + 4i + 6i + 3i 2 2=1= 4+2i-2i--,i 2

6 + i2 ·5 = 1.3.L- 30.5° 3+ j4 6 +i 2 ·5 =1.12-"0.66 3+ j4 J

liD

4 + 3i 8 + 4i + 6i + 3( -1) 2=1 = 4 + 2i- 2i-(-1)

4(cos60° + i sin60°) = 4L60° 2(cos30° + isin30°) 2L30°

--=--

4 + 3i 2-i

4(cos60° + i sin60°) = .L o 2 30 2(cos30° + isin30°)

4+3i 1 2" -=+ I 2 -i

4(cos60°+i sin60°) = 1.732 +i 2(cos30° + isin30°)

5 + 10i

5

liD (2 + 3i)(5

~~) = 10- 2i + 15i- 3i2

(2 + 3i)(5- i) = 13 + 13i

4(cos60° + i sin60°) = 2(cos30° + isin30°)

J3 + i

---------------·-::::D~a;:.!_..:1..:.7_ 7_Advanced Eng~ering_!\iath__.43;'!,_

43Z l 00 1Solved Problems in Engineering Mathematics (Z"d Edition) by Tiong & Rojas

•

180") eJ•., + e- f.f• = 2cos (1t x-;-

4

50+ j35 8 + j5

e1~ +e·i! =1.4142 Therefore,

61.03L35°

= 9:43L32° 50+ !35 =6.47 L3"

.• ) =~1.4142

cos h( 14

8+]5

cosh(H)

IIIII! [

Hklllli

l:l o [

Iill

l. 'llllllj

Ill

2

~=-2

tanh(.7t) J- = 3

B A=-28

10cos(40nt) =10 [ ~-~e

Substitute: t = 1

10cos(40nt) = 5 ei~Om + 5

=0.707

Note:. eJ9 + e-fo

c =-5+ j15

.,-2 (a + jb) + (a - jb)

=-5 + j15

-2a-2jb+a- jb = -5'+ j15

-a- jb =-5 + j15

e~- e -~• e'~ +e-~.

X

Euler'~ equat1ons

., +e-r,., = 2cos [1t x-7t180°] er,

3

eii +e-ig = 2cos60°

e~+~-~=1

e~ + e-~

By inspection:

:/,~

• ,

=j2sin60"

•

tanh(ii) = j1.732

'I!

cosn(H) =e • + e·r.

=2cos0 7

.:...ix3

x4

Euler's equation

let: x

ix5

= In (2 + j3)

x =In (3.6 eious)

x -ln3.6 + lne1ooo

0:.1

e

,, ·

x

21

~~ 1~)(····· . . ··~3 ···.···~~·················),! .. + 11 x-3!+5!+ ..... ! l

)

v i sin x

D

1

5

•D=le i

I

1 3

e 1'

-=

2~osn

1 :

-2 -4

-~ ~- · 2 -3

-3 -4

!li

eix = cos x + i sin x

rm , ~~~·

0

=-101

I

I hus,

0 5 3

0=6+0+0-0-35·-72

=:1--~+---·+ ... .

v

6 2

- (3)(4){6)]

X 6 .i

COS X

4

- (0)(2)(0) -- (5)(7)(1)

X4

2

1 6 0 4 2 7

D = [(1)(2)(3) + (6)(7){0) + (0)(4)(5)

................ ., .. .., ........................................

2 + j3 = 3.6 ejS6.J•x,:.

2 + j3 = 3.6ei0· 98 Note: ejll + e·jll

•

5

-')1(6

Convert (2 +j3) to polar form, then to · exponential form: 2 + j3 =3.6L56.3"

2

4

3

0:::5

+6! .......... .

111

=e• +e·• jg

- x2

.

1

2

- (4)(-2)(2)]

e =1+1X+-+-+-+2! 3! 4! 5! 3

1

D ::: - 4 - 12- 6 + 9 + 2 + 16

2

.1t)_j1.732 t an h( J- - - -

3

6 -i40.t

- (3)(-1)(3)- (1)(-2)(1)

) (ix ) (ix ) (ix ) =•1+. I X(ix+ --+-+--+-2! 3! 4! 5! (ix)6 +6! ........ ..

,.

A=-10+j10

e·i10•)

Let x = ix

Therefore,

coshx

=2 (ei10x .+

.2

_J'l

-j40l
D=l-2 -1 -2 -2 -1 3 1 4 3 1 D = [(1)(-1)(4) + (2)(··2)(3) + (3)(-2)(1)

x""4

e1

Therefore,

A =-2(5- j5)

7 Euler's

( 180")] x =~ [ 2cosl107t x -n-

e~ +e-~ =j1.732

a=5 -3b = 15 b=-5 a=5

=2 cos 0

equation

A+C = -5+ j15 ·-28 +

•

x =2 (ei1o~ + e-i10•)

Note: efo + e·j(l =2 cos a } efo _ e·iO = j 2 sin 0

= a - jb

j40.t·

Let: x = 2 el10.t + 2 e·i10.t

x = 2 ei10n(1) + 2 e-i1D•!11

tanhx = e• -e·• e• +e·• ;

B

X =1.28 + i0.98

•

121

~ == 2L180"

Let: B = a + jb; C

i•'

1./

Using Pivotal element method: (Use the

second row,

tirst column element as the

pivot number).

7 Euler's equation

D = (1) -2-1(5)

14 _ 2i5)

3 _ 2(-1) 2-1(-1)

1--4- 3(5)

:..3- 3(-1)

4

5

[

~lilt

l.li [

+'

-4- 3(3)

4 -7 -13 -19

D = (-1) l-7

3 -19 0

Ill!

1- 2(3) I -3-1(3) (-1)2

-5

5

..,.6

3 0

-2

-1

0 8 4

X=l

X=(1)1

31

-2

1

2

-6

-2

-1

5 -4

ll

oG)

-2 X=

1'111

t

- (0)(-6)(4)- (-13)(-7)(5)]

I

-1

-4 3 -2

8 4

2

-1

8

3

4

2

..,.6

•

I! I

X= [(-2)(3)(-6) + (-1 )(-2)(4) + (-4)(8)(2)- (4)(3)(-4)

D = 326

a X=l

4

-1

2

2

0

2

10

3

0

14

2

4

3

X=- 28

Ill

Multiply column 4 by -2 and add it to column 1: 3(- 2) + 4 =- 2

x+y+z=2 3x-y -2z =4 5x - 2y + 3z = - 7 By Cramer's rule; y =

1(-2)+2=0 1(- 2) + 10 = 8 5(-

1

D

1(-2)+2=0

1

=I 3

2) + 14 = 4

Multiply column 4 by -2 and add it to column 3: 3(- 2) + 2 =- 4

-1 5 -2

ci

•

1

1

-2 3

3 5

i

:f

-1

l

-2

'I~

,r

D = [(1)(-1)(3) + (1)(-2)(5) + (1)(3)(-2) - (5)(-1)(1)- (-2)(-2)(1)- (3)(3)(1)]

D = [- 3-10-6 +5-4- 9]

Dy

=I 3 5

2

1

1

2

4 '--7

-2 3

3 5

4 -7

Dy = [(1 )(4)(3) + (2)(-2)(5) + (1 )(3)(-7) - (5)(4)(1)- (-7)(-2)(1 )- (3)(3)(2)]

=12

D 12 y=-

~

-1 3

-1

~1

-1

1

3

2

-1

-3

-1

3

3

10

-1

-1

3 -1

Dz=l

I 32.

-1

·I -1

.- 1

Dz

3

2 3 -4 I -1

= [(2)(3)(-4) + (-1)(10)(-1) + (-3)(3)(-1) - (-1)(3)(-3)- (-1)(10)(2) - (-4)(3)(-1)]

D = ((2)(3)(1) + (-1)(-1)H) + (3)(3)(-1)

Dz

- (-1 )(3)(3)- (·1 )(-1 )(2)- (1 )(3)(-1 )] D

=[6 -

1 - 9 + 9- 2 + 3)

=6

= [- 24 + 10 + 9-9 + 20 -· 12]

Dz= -6 Dz

Z=o -3

-1

3

-3

-1

10

3

~1

10

3

Z=-

-1

z = -1

Dx=l

-4 Dx

-4

-1

= ((-3)(3)(1) + (·1)(-1)(-4) + (3)(10)(·1)

-6 6

Therefore, the answer is ( 1, 2 - 1).

- (-4)(3)(3)- (-1)(-1)(-3) "(1)(10)(•1)]

•

ox= (- 9- 4 .. 30 + 36 + 3 + 1OJ= 6 Ox x=o

X::

Dy=:

2

3~

4 0 5 7

A=l-12

Let: X.'= cofactor matrix of A

X=l~ ~~(-1)2+ 3

x=s 1

3 - 36 + 30 - 8 + 9]

= Dy

6

D = -27

5(- 2) + 4 =- 6

y

2x--y+3;t=-3 3x+3y-z= 10 -x-y+z:=-4

(.

1(-2)+0=-2

The new matrix becomes,

. l '

5

=[20 -

6

oJ

- (2)(-2)(-2)- (-6)(8)(-1)] X = [36 + 8 - 64 + 48 - 8 - 48]

Dy

y=2

D= (-1) [-156 + 570 + 0-285 + 0-455]

D = (-1)(-326)

- (-1 )(10)(3)- (-4)(-1 )(2)

y=-27 y=3

'jg

I

= • 81

-81

,Sf

.,

Dy

Dy = [(2)(10)(1) + (-3)(-1)(-1) + (3)(3)(-4)

D

I~

-2

= [12 ... 20-.21 • 20- 14 -18]

Y= Dy

·..~

( -1)2+4

3 -2 2 -6

8 4

Dy

- (1 )(3)(-3)]

.;,

I

D = (-1) [(4)(3}(-13) + (5)(-6)(-19) + (-5)(-7)(0)- (-19)(3)(-5)

'~~

0 3

-4

_____ . ____ J:>ay 17 -Advanced Engineering Math 435

~ -~

434. 1001'-Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas

1

2-3 10 3

-1

-4

X=-\~ ~I

312-3 ~1

3 -1

10 -4

Ill

CD

3 A =1-2

-1

2 0

0

2

-1

436 1001 Solved Problems in Engineering Mathematics (2"d Edition) by Tiong & Rojas

X= ll'lil [

~~11

[

1~1

•

lll

~~~~

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A (A

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I

X=-

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4

''[

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X=-1-20

0

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I

!!f

El

•

Note: T~e transpose of a given matrix is formed by interchanging the rows and columns.

'I'

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~

jl

~-

][;] = 0

A =j-2

1 -1

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2

3

.li

A = 3 x 3 matrix A" 1 = inverse of matrix A

Let:

'!ill

6- 3y + 2y = 0 y=6 x=2-6

= cofactor matrix of A

Let: x

Day 17-Advanced Engineering Math 437

)

=A ( ~)

)

= 1(unity or identity matrix)

A

Note: A unity matrix is a matrix whose elements in the main diagonal are all number 1.

I

1 0 0 1 0

•

1 0 0

6 7 31X

0 1 0 0 0 1

2 5

Unity matrix = 0 1

o o

El

[ 1 -1 2] [ z 2

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X

y

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l

• 1-12

;

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+2

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-7 Eq. 1

1

1

+

=0

-7 Eq.2

Substitute Eq. 1 in Eq 2: 3(2 - y) + 2y = 0

• •

I I. .

I

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-52]

= [ 0(3) + (-5)(4)

. =[ _;~

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··58

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rhus, A_,=

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_1_19 1-92

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J

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s

- -2 2

= cos wt 7 Formula

S +W

2s-18

2s

18

s2 + 9

= S2 + 9 -

s2 + 9

5 -1

s

Note:

} cos at = - 2- s +a 2 . a Formulas! sm at=-2- s +a 2 2s -18 . Thus, - 2 = 2 cos 3x-6sm 3x s +9

By completing the square of the denominator:

I

8s = 4 ( s 2

-

2s)

_4s2 - 8s = 4 ( s 2

-

2s + 1) - 4

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ml

. kt s1n

1m 1m

cos wt = - 2- -7 Formula 2

= - 2-k -2 s +k

-7 Formula

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1\ n "''

s-1 a

2 s

I

-1]

1

1[

12

]

Note: 7 Formula

4 I

2

--=4 (s-1) -1

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s

S

I hw;, ;> e

= 2e-251 sin100t ·

•

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~J

[!

..

. A = ac bd , 1ts 1nverse 1s g1ven by: rnatnx

·, lit

100 ) (s+25)2 +(100)2

2s-18 _ 2 [ - s J- 6 [-3-J 52 + 9 s2 + 9 s2 + 32

Note: For a 2 x 2 matrix, say,

~-~ ~I

= -1 +2

1(3) + 2(4}

3(x) + 2(y)

O=l; ~I

·1-.l

[~

-1

Solving the determinant of the given matrix:

21=12 11 1 -1 3

-1

=0

•

2

0

0 =-1

=I 0

[ ~ ; J[;]=[ ~ J

0

- 2 -1

-7 Formula

....,.__ 20_0~us, s2 -50s+ 10625 -

Ill

4 5 0 6 7 3 2 5

Th

k

(s+a) 2 +k 2

-2(

0=9-10

By inspection, since the resulting product is zero then, x = y = z = 0.

•

A transpose = j1 2

Since matrix B is a unity matrix, then A x B is equal to matrix A.

Note: e-•t sinkt =

0 -1

3

By inspection, since the resulting product is zero, x = y = 0.

4 5 0

2

•

<; I :~

e"1 sinh kt =

k2 (s-a) -k 2

1 1 "('h us -----, h\ =- eI s1n ' 4s~ 8s 4

+Formula

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Physics Vector and Scalar Quantities Classifications of Vectors Speed and Velocity Distance and Displacem~nt Acceleration Laws of Motion Forces Law of Universal Gravitation Work, Energy and Power Law of Conservation of Energy Momentum and Impulse Law of Conservation of Momentum Gas Laws Properties of Fluids Archimedes Principle

I ·~·

0

Wed

0

Thu

0 Fri

[g Sa!

What are Vector and Scalar Quantities? Vector quanities are quantities whose measurement is specified by magnitude ;md direction. The following are e~amples of vector quantities: Weight, momentum, torque, velocity, displacement, <~cceleration, electric field intensity, etc. Scalar quantities are those quantities which have only magnitudes. The following .~,,~examples of scalar quantities: Speed: 11\ass, volume, energy, length, I• "nperature, pressure, voltage, time etc. 1he term "scalar'' comes from Latin "scala" wl11ch means "a ladder". v.~ctor 1s the line whose length indicates ''' ·~•;aft' fhP maqnitude of the vector 'l'l<~rJtrty and whose direction indicates the "" ''"'"" nl ltw quant1ty The term "vector" l.ol111 "vt·h•·r"" wl11chrw~ans

'"'"'"·""'II 'It'

I ;lily''

Yl/hat are the_Ciassifications Vectors?

1.

2.

3.

91

Free vector- is one whose action is not confined to or associated with a unique line in space. Sliding vector- is one for which a unique 1ine in space must be maintained along which the quantity acts. Fixed vector- is one for which a unique point of application is specified and therefore the vector occupies a particular position in space.

44Z. 1001 Solved Problems in Engineering Mathematics (2"0 Edition) by Tiong & Rojas

t'lllllb~

lt1m1~~~~

What is the difference between a Speed and Velocity?

Newton's Second Law: (The law of acceleration)

Slug is the mass to which a force of one pound will give an acceleration of one foot per second per second.

Speed is defined as the distance per unit time. Speed is a scalar quantity.

"Whenever a net (resultant) force acts on a body, it produces an acceleration in the direction of the resultant force that is directly proportional to the resultant force and inversely proportional to the mass of the body."

Gram force is one-thousandth the pull of the earth upon a standard kilogram at a place where g has a value of 980.665 cm/s 2

Velocity is defined as the displacement per unit tim~. Velocity is a vector quantity.

t""

What is the difference between a Distance and Displacement?

l.,,

Distance is a length from one point to another usually measured in a straight line. It is a scalar quantity.

a-lii~ It

'~~

Dayl8- Physics 443

Dis!)lacement is the change in position, specified by a length and a direction. Displacement is a vector quantity.

What is a Frictional Force?

Instantaneous acceleration is the time rate of change of velocity. Uniformly accelerated motion is defined as the motion in a straight line in which the direction is always the same and the speed changes at a constant rate.

Frictional force a force acting on the body whenever it moves while in contact with another body. This force always opposes the direction of the motion. The frictional force is proportional to the normal force and is directed parallel to the surface.

Newton's Third Law: (The law of reaction) • For every force that acts on one body there is a second force equal in magnitude but opposite in direction that acts upon another body."

Newton's First Law: (The law of inertia) • There is no change in the motion of a body unless an unbalanced external force is acting upon it." Inertia is the property of the body by virtue of which a resultant force is required to change its motion.

~

Centrifugal Force

Let Fe = centripetal force. For Sl System:

F=J..i.N 2

Fe"' tnVr -r-.

What is a Force?

where: J..l. = coefficient of friction

Force is a push or a pull that one body exerts on another. This includes gravitational, electrostatic, magnetic and contact influences.

Coefficient of kinetic friction is the ratio of the frictional force to the perpendicular Ioree pressing the two surfaces together

F

Constant forces are forces that do not vary with time. External forces are those actions of other bodies on a rigid body while those forces that hold together parts of a rigid body are' called internal forces.

J..l.k

"''N

Coefficient of static friction is the ratio of the limiting frictional force to the normal Ioree.

Weight (of a body) is the resultant gravitational force acting on the body due to all other bodies in spac.e. It is always a vertical force acting downward.

Fmax

J..I.,=N

For English System:

J:n'Vr... Fe=~

Dyne (dyn) is the force that will give to a mass of one gram an acceleration of one centimeter per second per second. Poundal is the force that will give to a mass of one pound an acceleration of one foot per second per second.

~· ~:

·~

~' '

......I·

I

What is a Centripetal Force? · Centripetal force is the force (real force) on the body towards the center of rotation when the body is moving around a curved path. Centrifugal force is the force (apparent lor cc) on the body directed away from the ':• ~~~ll~r of rotation when the body is moving uorrrHI a curve path.

lbf

9c.

where: m = mass in Ibm Vr tangential velocity in ft/sec r = radius of curvature in ft

=

gc = 32.2 Newton (N) is the force that will give to a mass of one kilogram an acceleration of one meter per second per second.

Newtons

where. m =mass in kg Vr = tangential velocity in m/s .r = radius of curvature in m

2

What are the Laws of Motion? There are three Laws of Motion which are commonly called as "Newton's Laws of Motion".

.......

F=ma

What is an Acceleration? Acceleration is the change of velocity per unit time.

~

Ibm- ft sec 2 -lbf

Normally, the centrifugal force is equal in magnitude with the centripetal force.

What is the Law of Universal Gravitation? The Law of Universal Gravitation is also known as Newton's Universal Law of Gravitation is stated as follows:

i·!

444 100 i Solved Problems in Engineering Mathematics (2nd Edition) "Every particle in the universe attracts every other particle with a force that is directly proportional to the product of the masses of the two particles and inversely proportional to the square of the distance between their centers of mass."

Day 18- Physics 445

tiy Tiong & R~ What is the Law of Conservation of Energy?

Energy is also defined as the ability to do work. Energy is a scalar quantity.

The Law of Conservation of Energy is stated as follows:

The typical units for energy are joules, calories and BTU (British Thermal Unit). The unit BTU and calorie are used for thermal energy.

"Energy can neither be created nor destroyed; it merely changes from one form to another."

11.111111

(

Calorie is defined as the amount of heat required to raise the temperature of one gram of water 1°C.

lllillltlllf 1~1~111~

(

F

l'lllklli

l~~~~

'~~

Transformation of Kinetic & Potential Energy:

m1

mz Gm 1mz

F=~

British Thermal Unit (BTU) is defined as the amount of heat required to raise the temperature of one pound of water 1°1=.

What is Work? Work refers to the process of changing the energy of a particle, body or system. Work is a scalar quantity. The typical units of work are joules, foot-pound and inchpound.

fngh :=; ~tnv

EP = mgh

The unit "joule" is equivalent to the units of N • m and kg • m 2/s 2 . This unit was named in honor of the English Physicist, James Prescott Joule (1835- 1889).

where: W = weight of body m = mass of body g = gravitational acceleration g = 9.81 m/s 2 h = height of body

Mathematically, work is defined as the product of force and the displacement in the direction of the force.

Kinetic ener~:IY is the energy in motion.

W = Force x distance What is Energy? Energy is the property of the body or system of bodies by virtue of which work can be done.

2

or

v

= J2gh

Transformation of Work & Kinetic Energy: Work

= Kinetic Energy

Fs=imv

1 Ek =-mv 2

2

where: m = mass of body v = velocity of the body

"If there is no net external force acting upon a system of bodies, the momentum of the system does not change."

\11

,,

;~~

Impulse is the product of the force and the time during which it acts. Impulse is equal to the change in momentum. Impulse= FAt. Impulse·= p 2

-

P1

Impulse= mvrmat -mvinitiat

2

What is Power?

'·

EP =Wh

The law of conservation of momentum is stated as follows:

What is Impulse?

What are Two Types of-Energy? .

Potential energy is also known as the energy of position or configuration or gravitational energy. This type of energy is decreases as the elevation of the body decreases and increases as the elevation of the body increases. Normally, the lost in potential energy is converted into heat or kinetic energy.

What is the Law of Conservation of Momentum?

Potential Energy = Kinetic Energy

Energy is classified either a potential energy or a kinetic energy. where: G = gravitational constant G = 6.673 x 1o· 11 N. m 2fkg 2 G = 6.673 X 1o·B cm 3 /~. s 2 G = 3.436 X 10'8 lbf-ft /slug 2 G = 3.320 x 10'11 lbf-ft2/lbm 2 G = 3.436 X 1o·B ft 4/lbf-sec4

where: m = mass of the body v = velocity of the body

Power is the time rate of doing work or the amount of work done per unit time. Power is a scalar quantity.

P=Y!_

t

The typical units for power are watts, ftlbf/sec and horsepower.

1 watt = 1 joule per second 1 hp = 746 watts

where: F =force At = change in time P2 = final momentum P1 = initial momentum

What are the Two Types of Collisions? Elastic collision is a collision of two bodies in which kinetic energy as well as momentum is conserved. Inelastic collision is a collision of two bodies in which only the momentum is conserve but not the kinetic energy.

What is the Coefficient of Restitution? What is Mom~ntum? Momentum is the product of the mass and velocity of a body. Momentum is a vector quantity. ·

p

c~

nw

Coefficient of restitution is the negative ratio of the relative velocity after collision to the relative velocity before collision.

e'= __vz,..- '~~2s _ ~ -v2v1A-

'~~1e

'~~1a /vtA

446 100 1 Solved Problems in Engineering Mathematics (2"d Edition) If e = 1, the collision is perfectly elastic while if e = 0, the collision is completely Inelastic.

by Tiong & Rojas

Day 18- Physics 447

What are the Properties of Fluids?

w

Density is the mass per unit volume. This is expressed mathematically as

Fluid displaced

What are the Gas Laws?

m

IIIII ( 1111001111

[

1~11.!11

Boyle's Liiw: "If the temperature remains constant, the product of the pressure and volume is constant." This was named after l~ish physicist Robert Boyle (1627- H)91):

:11\l'~

'i~~i

Weight density is the weight per unit volume. The relationship between density and weight density is expressed in the following equation. ·

.PV.:;:k

l.,,,

,,,

P=v

D= W Charles' Laws: "If the volume of a confined gas is constant, the pressure is directly proportional to the absolute temperature."

.Pl ~fz.

T1 -:-r2 "If the pressure of a confined gas is constant, the volume is directly propl)rtional to the absolute temperature."

v

D=pg

.

. density

BF=pV.

where:

.,

"' . density ofwater 1

The values for the density of water are as follows:

v: ····~.

BF

Specific gravity is the density of the substance relative to that ofwater. This is also known as relative density. Water is considered the standard substance which has a maximum density at 4•c.

I IJ;

II

General Gas Law: The combination of the Boyle's Law and the Charles' Uiws may be regarded as the general gas law. This is expressed · mathematically as

PV::::nRT

where:

n= n

=

or

Pi\'1 P.Na

Tt

= Tt

BF"'W

62.4 pounds I ft3 1000 kg I m 3 9.81 kNim 3 9810 Nlrn 3 1 gram/cc

Note: The buoyant force is always acting at the centroid (center of gravity) of the submerged volume.

The specific gravity of water at densed condition (4"C) is exactly 1.0

Proceed to .the next page for your18th test. Detach and use the answer sheet provided at the last part of this book. Use pencil number 2 in shading your answer.

What is the Archimedes Principle?

GOOD LUCK I

The Archimedes' Principle is stated as follows:

m

w

mass .of the gas atomic mass of the gas R = universal gas constant

"When a body is immersed (partially or wholly) in a fluid, it is subjected to an upward force (buoyant force) which is equal to the weight of the fluid displaced,~ This was discovered by Greek mathematician and inventor, Archim~des · (287- 212 B.C.).

Also by equilibrium, the buoyant force is equal to the weight of the body, thus

j,

.....!=~

1'1 T'z

Vs = volume submerged p = density of the fluid

, :s_!

it!

~:t,¥ "i.!

•· \!rtibia: Did you know that. .. the most difficult problem in mathematics is the "Fermat's Last Theorem! The search for the proof of this theorem begun right after Fermat's death in 1665 and remained an unsolved theorem through centuries of hopeless search. A British-born professor in mathematics at Princeton University, Andrew Wiles brought an end to the search of the proof in 1995, i.e. 330 years later. Because of this, Fermat's Last Theorem was regarded as the Mount Everest of Mathematics.

~uote: "With me everything turns into mathematics." - Rene Descartes

Day 18- Physics 449.

•I

II

•'\J

Topics

(

lii!IMI

[]

~llllil~l

LOlli

Physics Vector and Scalar Quantities Classifications of Vectors Speed and Velocity Distance and Displacement Acceleration Laws of Motion Forces Law of Universal Gravitation Work, Energy and Power Law of Conservation of Energy Momentum and Impulse Law of Conservation of Momentum Gas Law Properties of Fluids Archimedes Principle

Mon

lli>'i~lijl

[

.(

0

~IIIWi~ll

Tue

0 0 [Q] 0 0 0

I

'""

Theory

Problems

Wed Thu

Solutions

Fri

[J ~ Notes

Sat

761: ME Board October 1994 The weight of a mass of 10 kg at a location

7&3: ME Board October 1994, MEBoardApril1998

where the acceleration of gravity is 9.77 m/s 2 is

The mass of air in a room 3 m x 5 m x 20 m is known to be 350 kg. Find its density.

A. B. C. D.

A. B. C. D.

79.7 N 77.9 N 97."7 N 9T7 N

7&:&: ME Board AprU :1998 How much does a 30 Ibm weigh on the moon? Gravitational accelerati.on in the 2 moon is 5.47 ft/s 2 and in earth is 32.2 ftls .

A. B.

c. D.

2.0 lbf 3.21bf 3.41bf 5.0961bf

;~

7651 ME Board October 1997, ME Board April t998 100 g of water is mixed to 150 g of alcohol

C. D.

(p = 790 kgfm 3 ). Calculate the specific vol.ume of the solution, assuming that it is mixed completely.

7701 EE Board October 1996 A 10 g block slides with a velocity of 20

3 0.82 cm /g 0.88 cm 3/g 3 063 cm /g 3 1.20 cm /g

A. B. C. D.

A.

100 g of water is mixed to 150 g of alcohol (p 790 kg/m 3 ). Calculate the specific gravity of the total mixture.

B. C. D.

A.

'771t ME Board April 1.997 A 60 ton rail car moving at 1 mile/hr is

B.

c. D.

1.862 0.963 0.286 0.862

767a ME Board ApriiJ.996 The specific gravity of mercury relative to water is -13.55. What is the specific weight of mercu~y? The specific weight of water is 62.4 lbf/ft . .

'

I ~

.: ,'.r

1.167 1.176 1.617 1.716

A.

B. C. D.

3

102.3 kN/m 3 132.9 kN/m 3 150.9 kN/m 3 82.2 kN/m

7&81 ECE Board November 1998

A 16 gram mass is moving at 30 em/sec while a 4 gram mass is moving in an opposite direction at 50 em/sec. They collide head on and stick together. Their velocity after collision is

A. B. C. D.

0.14 m/s 0.21 m/s 0.07 m/s 0.28 m/s

D.

coupled to a second stationary rail car. If the velocity of the two cars after coupling is lftls {in the original direction of motion) and the coupling is completed in 0.5 second, what is the average impulsive force on the 60 ton rail car?

A. B. C. D.

50!bf 3500 !bf 1200 lbf

60 lbf

7721 ME Board April :1:997 What momentum does a 40 Ibm projectile posses if the projectile is moving at 420 mph?

A. B. C. D.

24,640 lbf-sec 16,860 lbf-sec 765 lbf-sec 523.6 lbf-sec

7691 ME Board October 1996

.~~

A 60 ton rail car moving at 1 mile/hr is instantaneously coupled to a stationary 40 ton rail car. What is the speed of the coupled cars?

A. B. C.

I,,

I\

the density of the block?

988 kg/m 3 1255 kg/m 3 3 2550 kg/m BOO kg/m 3

15 cm/s 10 cm/s 25 cm/s 5cm/s

773; ME Board Aprii199S, ME Board April1998 A 10-kg block is raised vertically 3 meters. : What is the change in potential energy? Answer in Sl units closest to:

7&4: ME Board April1996 An iron block weighs 5 N and has a volume of 200 cubic centimeters. What is

A. B. C.

r;m/s on a smooth level surface and makes a collision with a 30 g block moving in the opposite direction with a velocity of 10 cm/s. If the collision is perfectly elastic, what is the velocity of the 30 g block after the collision?


=

.. I

0.6 mph 0.4 mph

ll

0 88 mph 1 mph

D.

320 J 350 kg-m 2/s2 294 J 350 N~m

4\50 · l 00 l Solved Problems in Engineering Mathematics (2"d Edition) by Tiong & Rojas

(

1111~1»1

~~ll!illlll

lif;illlll (

11~11~11

l~'""

'

,111

"174: ECE Board AptriiD 1.997

779: ME Board October l'.991J

An aircraft engine develops a forward thrust of i 5,000 N. If the gross mass of the aircraft is 100 tons, what horsepower does the engine develop if it is flying at 1000 l\ph?

What is the water pressure if manometer is 0.6 m Hg? Mercury is 13.6 times heavier than water.

A. B.

c. D.

150,000 5585 5400 3108

'il

A. B. C. D.

B.

27.4 kPa 47.2 kPa 79.97 kPa 72.4 kPa

3.81 kCal 32 BTU 2300 Cal 9.41 kJ

A.

7

'1.63 X 10 hp 3.16 X 107 hp 3.61 X 107 hp 1.36x107 hp

A force of 200 lbf acts on a block at an angle of 28° with respect to horizontal. The block is pushed 2 feet horizontally. What is the work done by this force?

C. D.

320 480 540 215

~:~

'~.

C. D.

78:1: EE. Board October 1994 Assuming the barometer reads 760 mm Hg, what is the absolute pressure for 900 mm Hg gauge?

A.

B. C. D.

74.213kPa 221.24 kPa 48 kPa 358 kPa

J J J J

7"1&1z What average force is necessary to stop a .45 caliber bullet of mass of 15 grams and speed of 300 m/s as it penetrates a block to a distance of 5 em?

78:&: CE Board May 1.994 A barometer reads 760 mm Hg and a pressure gage attached to a tank reads 850 cmof oil (sp.gr. 0.80). What is the absolute pressure in the tank in kPa?

A B. C. D.

168.1 kPa 186.1 kPa 118.6kPa 161.8kPa

783: EE Board April 199& A sealed tank contains oxygen at 27°C at a pressure of 2 atm. If the temperature increases to 100°C, what will be the pressure inside the tank?

A. B.

12.5 kN 13.0 kN

C.

13.5 kN

A.

D.

12.0 kN

B.

4.92 atm 4.29 atm

2.49 atm 9.24 atm

784: ME Board April :1996 A volume of 400 cc of air is measured at a pressure of 740 mm Hg abs and a temperature of 18°C. What will be the volume at 760 mm Hg and 0°C?

1\.

11

c. IJ

I ~

376 326 356 366

cc cc cc cc

34°C 45°C 60°C 90°C

78&: ME Board April1998 A transportation company specializes in tile shipment of pressurized gaseous rnaterials. An order is received for 100 l1ters of a particular gas at STP (32oF and I atm). What minimum volume tank is ru:cessary to transport the gas at 80°F and .r maximum pressure of 8 atm?

\~

~ "'i

788: ME Board Octol>er 1.997

A 10 Ibm object is acted upon by a 4.4 lbf force. What acceleration in ft!s 2 does the object possess?

A. B.

C. D.

A. B. C. D.

785: EE Board October :1995 The pressure of the nitrogen gas thermometer is 76 em at 0°C. What is the temperature of a liquid in which the bulb of the thermometer is immersed when the pressure is seen to be 87.7 em?

1,934.5 m. 3,508.4 m. 4,168:2 m. 2,647.7 m.

"Jt'li/1 ME Board Apnrifl lt998

A. B.

'l'li\'

A mercury barometer at the base of Mt. Makiling reads 654 mm and at the same time another barometer at the top of the mountain reads 480 mm. Assuming specific weight of air to be constant at 12 3 N/m , what is the approximate height of Mt. Makiling?

B.

C. D.

C. D.

780: EE Board October 199&

7'1&: ME Board October 1997 . A rocket is moving through a vacuum. It changes its velocity from 9020 fUsee to 5100 ft!sec in 48 seconds. How much power is required to accomplished this if the rocket's mass is 13,000 slugs?

A.

Day 18- Physics 451

-'{'

'1"1§: EE B«~J;u•d October :1994 if a i 0 kg piece of copper falls 100 m, how much heat might be produced? A. B. C. D.

,\\)\

A. ll

<: ll.

16 liters 141iters 10 liters 12 liters

'187x EE Board April1996 !\ 20 liter sample of gas exerts a pressure ',f 1 atm at 25°C. If it is expanded into a 40 111"' vessel that is held at 100°C, what will 1" • rts final pressure? 1\

II

ll

0.50 atm 1 0 atrn o :\15 atm 0 ti:~ atrn

12.4 10.0 14.2 13.0

789: A 50 kN truck traveling with a speed of 50 kph hits a lamp post and is brought to rest in 0.1 s. What is the average force of the truck?

A. B. C.

D.

-408 -508 -608 -708

kN kN kN kN

790: A tennis ball moving horizontally to the left at 40 m/s hits a racket and rebounds horizontally to the right at 30 m/s. If the mass of the ball is 100 grams, find the impulse of the force (in kg-m/s) exerted on the ball by the racket.

A B. C.

1 -1 7

D.

12

7911 Two steel balls of masses 500 kg and 50 kg, respectively are placed with their centers 0.5 m apart. The two balls attract with a force of

. A. B. C. D.

6.67 X 10"10 N 6.6'7 X 10"7 N 6.67 X 10-6 N 6.67 X 10"3 N

19:&: EE Board October 1.995 A 50 g mass hangs at the end of the spring. When 20 grams more are added to the end of the spring, it stretches 7 em more. Find the spring constant. A.

2.8

B.

2.9

C.

4.:3

D.

2.5

452 1001 Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas , !l


A.

Determine the submerged depth of a cube of steel 0.3 m on each side floating in mercury. The SJJecific gravities of steel and mercury are 7.8 and 13.6 respectively.

C. D.

B.

3min. 1 min. 4min. 2min.

::1

',,

~t: I

798: ME Board April199& A. 101!11 [

B. C.

IIMilliil~ll

D.

~~~~Iii

7941 EE Board October 1995

L.,,l,ll

'

,1111

A.

A block of wood floats in water with 5 em projecting above the water surface. When placed in glycerine of specific gravity of 1.35, the block projects 7.5 em above the liquid. Determine its specific gravity. A.

B. C. D.

0.514 0.704 0.836 0.658

7951 EE Board October 1996 A solid cube material is 0."75 em on each 3 side. If it floats in oil of density 800 kg/m with one-third ofthe block out of the oil. What is the density of the material of the cube? · A. B. C. D.

3

533 kg/m 3 523 kg/m 3 513 kg/m 3 543 kg/m

79&1 CE Board November 1993 A hollow cyJinder 1 m in diameter and 2. m high weighs 2825 N. How many kN of lead weighing 110 kN/m 3 must be fastened to the outside bottom of the cylinder to make it float with 1.5 m submerged in water?

A. B.

C. D.

0 0

Mon

'~1~1111

(

Topi4:S

What is the power required to transfer 97,000 couiombs of charge through a potential rise of 50 volts in one hour?

0.155 m. 0.165 m. 0.134 m. 0.172m.

8.5 kN 6.5 kN 10.5 kN 9.5 kN

7971 ME Board October 1995t ME Board Oetober 1996 How long must a current of 5 amperes pass through a 10-ohm resistor until a charge of 1200 coulombs passes through?

C.

0.5kW 1.3kW 0.9kW

D.

2.8 kW

B.

Tue

.,:·

799: EE Board October 199&

How much oil at 200°C must be added to ·~ 50 grams of the same oil at 20°C to heat it to 70°C? A. B. C. D.

B. C. D.

15.87·c 10.3o•c 8.65°C 23.5TC

Wed

0

0 0

Solutions

Thu

Fri

~ Sat

0

soo: EE Board October 19'96

A.

0

Theory

Problems

12.39 grams 29.12grams 19.23 grams 23.91 grams

The temperature of three different liquids are maintained at 15°C, 2o•c and 25•c respectively. When equal masses of the ,[1. first two liquids are mixed, the final ··r: 1 temperature is 18·c and when equal .· · masses of the last two are mixed, the final ': temperature is 24°C. What temperature will be achieved by mixing equal masses of the first and the last liquid?

0

Notes

RATING

ANSWER KEY

c

761. 762. D 763.A 764.C 765. D 766. D 767. B 768.A 769.C 770. D

771. B 781. B 772. c 782. A 773. 783. c 774. B 784. D 775.C 785.A 776. D 786: B 777. B 787. D 778. c 788. 789. D 779. c 780. A .,790. C

c

c

Physics Vector and Scalar Quantities Classifications of Vectors Speed and Velocity Distance and Displacement Acceleration Laws of Motion Forces Law of Universal Gravitation Work, Energy and Power Law of Conservation of Energy Momentum and Impulse Law of Conservation of Momentum Gas Law Properties of Fluids Archimedes Principle

791. c 792.A 793. D 794. D 795. A. 796.A 797.C 798. B 799:C 800.D

c:J c:J

34-40 Topnotcher 26-33 Passer

c:J 20-25

0

Conditional

0-19 Failed If FAILED, repeat the test.

454 1001 Solved Problenis in Engineering Mathematics (2nd Edition) by Tiong & Rojas

"

rm [

1111~~"~1

ljlil(l'~*ijl

(

li;li~llll

•

"

IIDal

W=mg

mt = m1 + mz

W = (10 kg)(9.77 m/s 2 )

mt = 100 + 150

W= 97.7 N

mt =250

l·r.J.i I

g

•

Day 18- Physics 455 Initial momentum = Final momentum

Note: Specific weight of water is 62.4 3 lbs/ft3 or 9.81 kN/m (j)

Vt

w

30 32.2

(!)water

= mt.+ mz

Mass in earth = Mass in moon

Pt

P2

Note: Density of water (Pt) is 1000 kg/m v 0.100 0.150 t = 1000 + 790

5.47

w = 5.096 lbf.

LOIIIIII

4

3

Vt = 2.8987 x 10" m x

3

C°~m J 0

1

Vt = 289.87 cm 3

~IIIII Volume of air is the same as the volume of the room ·

v = 3(5)(20)

•

V = 300 rn 3 m

p=-

v

- 350 kg

P- 300 rn3 p

vt

289.87

m1

250

·~.· I

..

...

..

''i~ '

•-

mt = m1 + m2

m = 0.5096 kg.

1 = v2 ,_ v1, Vt-V2

v1 -

= v2 ,_ v1, 20- (-10) = V2·- (-v1 ')

,,,2

m p=-

v

0.5096 kg p=---0.0002 m 3 p = 2548 kg/m 3 Note: From the choices, the nearest 3 answer is 2550 kg/m

V = 0.14 m/s

3

Initial momentum= Final momentum

+ mN2 = (mt + m2)V

v

= 0.100 + 0.150

mN1

t Vt

1000 = 2.8987

60(1) + 40(0) = (60 + 40)V

'790 X

10"4 m 3

m

p=-

0.250 kg P- 2.8987x10c4 m3 p = 862.45 kg/m

=

3

•

-- -

v2

m1

m2 ~--·--·

____,_

V2'

~).Q't.t.r.I)Nl l\'l'!>l'l'!l~ I --II I'

taeJ

t

~

}

IV

,

:

=m(V1 -

V1 ')

F(0.5) = _?0( 2 000) (1.4667 -1)

Vt

V1'

V =~X~X5280 ft 1 hr 3600 s 1mi v1 = 1.4667 ft/s F(t)

V = 0.6 mph Note: Since the second car is stationary, its velocity (Vz) is zero.

v

sp. gr. = 862.45 1000 sp. gr. 0.862

-1oo = -10(3o..:.v2 ')+30V2 • v2' = 5 cm/s

P2

Pwater

-> Eq. 1

-100 = -300 + 10V2 '+ 30V2 '

mN1 + m2V2 = (mt + m2)

+!':lz..

sp. gr. = ·_e__

•

Substitute Eq. 2 in Eq. 1

3

V = 0.0002 m3

v2

V1 • = 30- V2

16(0.3) + 4(-0.5) = (16 + 4)V

Note: Density of water (pt) is 1000 kg/m

5 = m(9.81)

. v2 ·-v1· e=-V1-V2

Initial momentum'= Final momentum

mt = 0.100 + 0.150

=10(-V1') + 30W

Note: For a perfectly elastic collision, coefficient of cestitution (e) is equal to 1.

v2 -+-----

Vt == Vt + V2

W=mg

(~) 100 em

132.9 kN/m 3

m1 +m2

Pt

3

COm=

-v

= 1.2 cm 3/g

Vt::: mt

V = 200 cm x

13.55 (9.81)

mt

mt = 0.250 kg.

= 1. 167 kg/m 3

10(20) + 30(-10)

- 100 =- iOV1' + 30V2' 7 Eq. 1

COm=

Vt

v=--=--

v

m1V1 + m2V2 = mtV1 + m2V2

sp.gr. ::. __ m_

Vt= Vt +V2

'l,~!~llill

'

~

32.2 F := 3478.509 lbf

Note: From the choices, the nearest answer is 3500 lbf.

V = 420 mix~x5280 ft hr 3600 s imi V=616ft/s Momentum

=mV

I'

!lf..,

•

456 lOO 1 Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas

Momentum= 40(616) Momentum

=24640 lbm-ft/s

24640 . Momentum=-- = 765 lbf-sec 32.2

• • '

1,, l!r Jl,

b.KE =

~m(V0 2 - V 2 )

[

l.ili~~:~l

PE =mgh

t.KE =

= 10(9.81)(3) PE =294 J

~(13000)[(9020}

'11~1~1~11

t.KE

L""'l'

PE

~H1111

V

-(5100n

!t/

=3.597x1 011 lb-ft

F(s) =!mv2

.:~

1 ~

p = t.KE

2

hr

)i

1hp P = 7 .49x1 09 lb-ftls x 550 lb-ftls

=15000(277.778)

P = 4166670 watt 1hp ) p = 4166670 watt ( 746 watts

p = 5585 hp

•

F=13.5 kN

•

=(Sp.gr.Hg )( (I)Waler )h

p = 13.6{9.8)(0.6) P = 79.97kPa

•

Paouom =Prop + roAi,h

w =rcose(s) w

Ass~:~ming

no losses, the total heat equals the initial PE of the copper. cal 4.2 J Q = 2335.714 cal

x--

1 BTU ' Q =2335 714 calx - - · · 252 cal Q =9.268 BTU

roHghb = roHgh, + ro Airh

2

h = 1934.5m

x9.81 ft/s

p2

= 100 + 273 P2 = 2.49 atm

Ill

PtVt

P2V2

740(400) 760V2 18 + 273 = 0 + 273 V2 =365.38 cc

Note: From the choices, the nearest answer is 366 cc.

•

m + palm

W = 479.55 N-m

I 'ub•

pgngo

w~480J

I ',~>·•

13.6(9.81)(0.9) + 13.6(9.81)(0.76)

"""'

221.47 kPa

iiii'IW<'I I'>

/74 /4 kPa

PtV1 = P2V2

~

T2

Note: V1 =V2 78 87.7 0-273 =--,=;

T2 =306.95° K T2 = 306.95-273

13.6(9810)(0.654) = 13.6(9810)(0.48) + 12h 1kg 1m . -W = 3.53.179 lb-ft X 2.202 lb x 3.281 ft

T2

27 - 273

(sp.gr.)Hg(I)Walerhb = ( Sp.gr.)Hg (I)Waterhl + roAirh

=353.1791b- ft

Ntltu I rom the choices. the nearest

Note: From the choices, the nearest answer is 2300 cal

T1 Note: V, = V2

Pressure (P) =Specific weight (ro) x Height (h)

p

W = 200cos28° (2)

let: Q =heat

P1V1 = P2 V2

~=--,=;

Let: h = height of Mt. Makiling

PE = 9810 J

=168.1 kPa

P = {wH9 }h

=10(9.81)(100)

0=9810 J

~(0.015)(300)2

Pabs

F=13500N

P=1.36x10 7 hp

PE ::::mgh PE

F(0.05) =

'.

;I;,

11 P = 3.597x10 48 P = 7 .49x1 0 9 lb-ft/s

1 km

...

·~·.

at

= 1000 kmx~x!OOO m

P=FV

•

Work done by retarding force = initial kinetic energy of the bullet

'·

3600s V = 277.778 m/s

p

J

= Pgage + Palm

Pabs = 0.8(9.81)(8.5)+13.6{9.81)(0.76)

"-+ s 2

• • Pabs

v-

~t

llll!(!:iljljil

F-

8

p = b.K_§_

101!111

[

•

Dayl8- Phyf:!iCs 457

•

T2 = 34°C

P1Vt

P2V2

~=--,=; 1(100) 8V2 32+460 = 80+460. V2 =13.7 liters

v2 "'141iters

--···----------------------------------~ !8- Physic~ 459

458 .1 001 Solved Problems in Engineering Mathematics (2"d Edition) by Tiong & Rojas

•

Ill = P2V2

P1V1

T1

(

•lllll!limll

(

111!1~111~1

T2

'll~li~~ll

L_IIIMI

~"""

4.4 =

•

= ks

50= ks

k

W=BF

F = k(s + 7)

~-

( sp.gr)wood (Ah)

50+ 20 = k(s + 7)

:r

Impulse = P2 - P1 Impulse= 3- ( -4)

Eq. 1

( sp.gr. )\11/ood Pwater VTotal =

'i

Impulse = change in momentum

70

Impulse= 7 kg-m/s

•

-~

50 S= -

P2 = 3 kg-m/s

2

5 em

=

P2 = mVz p2 = (0.1)(30)

(__!Q_}a 32.2

A

-7 Hooke's Law

I" = tension in the spring due to the load attached to it k = spring constant s elongation of the spring due to the load attached to it

where

p1 = (0.1)(-40)

Newton's second law of motion

a= 14.2 fUs

F

P1 = -A kg-m/s

I!'P.9!tl lliiia ~

~:~

P1 =mV1

P2 (40) 25 + 273 1()0 + 273 P2 = 0.63 atm

F = ma

II

Momentum = mass x velocity

___!t?_QL = ~1111~~~~

j

= ks + 7k

(sp.gr.)Wood

-) Eq. 2

Pwater VDisplaced

= A(h- 5) h-5 =h-

--+ Eq. 1

Substitute Eq. ·1 in Eq. 2:

A

(50)

7() cckj-- + 7k

\ k

F

m1

7 em

k=28

m2

i!ll

s

Glycerine sp.gr. = 1.35

Using the relationship between impulse and momentum: Impulse = change in momentum F6t

=mV1 - mV0

50000 N m = ----=~9.81 m = 5096.84 kg

Using the formula for universal gravitation:

W=BF

.F = Gm1m2. sz

(sp.gr. )VII:Jod rwater VT

where : F = force of attraction in N. m1 and m2 = respective masses of two particles in kg. s = distance between the centers· of the two particles in m. G = gravitational constant

G = 6.67 x 10'11 N-m V. = 50 km 0

x -~x 1000 m 3600 s km

hr V0 = 13.89 m/s Substituting: F(0.1)

=5096.84(0 -13.89)

F = -707951 N F = -707.951 kN

2

kg2

"i:'

0

= !.:_35(h- 7.5)

F

=

.

(0.5)

F = 6.67 x 10-6 N

h

W=BF

( sp.gr. )Steel Pwater \/Total

=(sp.gr. )Hg

3

Equations 1 and 2: 2

7 81(9 81)(0 3) = l3.6(9.81)(0.3) d

d=0.172m

Substituting:

(6.67x10- 11 )(500)(50)

(spgr.),i\I'A:d (Atl) = (1.35)A(h-7)

BF

2fv""

Pwater VD

i .ci I\ ::':.: base area h ·:-:· hf~ioht the V·Jood

-~=-_!?_ = ~~~-~5(h - 7.5)

h h 11- 5 = 1.35h -- '10.125 1'1 = 14.64 em Substitute in

1:

2

14.64-5 sp.gr.vvoor! = -·-14'.64. • . :·!S.l.LJi.,!l/ood

cc

0.658

-~ Eq. 2

Day 18- Physics 461

_.60 100 1 Solved Problems in Engineering. Mathematics (2nd Edition) by Tiong & Rojas

• [

•01111~1

(

illll~lll~il

BF2 = Pwat.,;vLead BF2 = 9.81Vtaad

w

=

I

wlead = 11 ovlead

BF1 + BF2 = Wcy,inder + Wtoad vlead = 0.0772 fT13

11~:~11~11

Heat gained = heat loss mc 2 (24- 20) = mc3 (25- 24) c 2 = 0.25c 3 ~ Eq. 2

wlead = 110(0.0772)

L" "~"·'

W=BF Pcube Vrotal = POit Voisplaced

l".,

Pcube(x

3 )

=

800(~x ){x

2 )

Peube(0.0075) = 8oo( ~ }o.oo75) 3

•

Peube = 533.33 kg/m3

3

WLaad = 8.5 kN

•

Substitute Eq. 2 in Eq. 1: Q =It

3c1 =2(0.25c3 )

1200 = 5t t = 240 sec

c 1 =0.1667c3

1 min t = 240 sec x 60 sec

·

Heat gained = heat loss mc1(t -15) = mc3 (25- t)

t=4 min

We

0.1667c3 (t -15) = c 3 {25- t)

Note: ampere = coulomb/sec.

0.1667t- 2.5 =25 -t

Ill

t =23.57°C

Q::lt 97000 = 1(3600) I =26.944 A

BF1 = Pwater Vo;splaced

* *

2 BF1 = Pwatar( )d Y 2

BF1 = 9.81( )<1) (1.5)

BF1 =11.56 kN

•

Heat gained = heat loss mc1(18 -15) = mc 2 (20 -18) 3c1 = 2c 2 ~ Eq. 1

11.56 + 9.81Vlead = 3.825 + 11 ovlead 1!1:!~1~11

Heat gained hE!at loss 50c(70- 20) = mc(200- 70) m = 19.23 grams

P=EI p = 50(26.944) p = 1347.2 watts · 1 kW p = 1347.2 watts x 1000 watts P=1.3472 kW

•

Heat = mc(at)

where: m = mass c = specific heat at change in temperature

=

I

~

fl 484 ·lOOI Solved Py>blems in Engineering Mathematics (aM. Edition) by Tiong & Roj..-

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~hat .!iL~!!_eerin.!l Mec~anics?

C.

Mechanics is the branch of physics that considers the action of forces on bodies or fluids that are both at rest and in motion.

What .11re the

Engineering Mechanics is the branch of that applies the principles of mechanics to any design that must take 111to account the effect of forces.

·~ngineering

:

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I he branches of mechanics are the l••llnwing:

condition, the forces or vectors are transformed into a force polygon. For equilibrium, the force polygon must close.

F1 I\

,,

Statics. It deals with bodies in the ;;tate of rest.

F2 Dy11arnir.s or Kinetics. It deals with l11>
"

fo!

G1·aphical condition: Under this

4:

,<;;, !0\ ;,""

.•·'

are the Branches of Mechanics?

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CQ~s

A body is in equilibrium if it satisfies the following conditions.

1. Wh_~t

lio'~····~·.)l..;t~~* ............... .
Kinematics. It refers to the study of motion without reference to the forces which caus1~s the motion,

~guilibr_!!!m1

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Engineering Mechanics Branches of Mechanic..s Conditions for Equilibrium Friction Parabolic Cable Catenary Centrolda! Moment of Inertia Moment of Inertia with respect to the base Transfer Formula Mass Moment of Inertia

--.....1

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466 · 1001 S0lved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas

N~

6

F2

Force Polygon

2.

where:

Directional condition: If three or more non-parallel fqrces or vectors are in equilibrium, then they must be concurrent F1

c) Length of parabolic cable, S:

F=~N tan~=

F

Day 19- Engineering Mec;:hanics (Statics) 467

S =L + 8d2 32d4 3L - 5L3

f.1

= length of parabolic cable =sag = span or distance between supports w = unit weight or load per unit length T = maximum tension (usually at the support) H = minimum tension (usually at the lowest point of the cable)

where: S d L

A cable is analyzed as a parabolic cable when the loading is uniformly distributed horizontally throughout the cable.

111 1 ~~~

F3

3.

Analytical condition: If forces or vectors are in equilibrium, then 'it must satisfy the three static equations; namely

}:F, =0

- i

LFy=O

LMx =0

l

=

0=0

w

JTIITffiUTl.!!J ·

S =length L =span

Analyzing half of the

A cable is analyzed as a catenary when the loading is distributed along and throughout the cable. The word "catenary" comes from the latin word which means "chain". It is a graph of the equation y:::: cosh x.

H

What is Friction? !......

Friction is defined as the limited amount of resistance to sliding between the surfaces of two bodies in contact

w

p

r

y ltU'Jio

U2

X

---

a) Tension at the lowest point, H:

Span, L L=2x

·X= c.ln

E.

A.

H= (j)t:

X

x-axis

Distance from the x-axis to a point on the catenary, y:

Y2 = 8 2 +c2

ad

S =L + 8d2 32d4 3L - 5L3

If the cable has uneven supports, the formulas to be used are the same, only that all unequal dimensions will now contain subscripts 1 and 2. For example, the distance from the origin to the left support is now taken as x1 rather than x, and xz for its distance to the right support, and so on.

What are the Centroidal Moment of Inertia of Common Figures? Moment of inertia is also called the second moment of area. Centroidal moment of inertia is the moment of inertia of the figure with respect to an axis passing through the centroid.

A. II

Rectangle

y

Maximum Tension, T: Like a parabolic cable, the maximum tension in a catenary occurs at the support

Centroid

hll

01

T=-my

R

c

Length, S

b) Tension at the support, T:

N

S+ Y

length= 2S

y-axis

T

D.

What is a Catenary?

cabl~;:

_..,.._

Minimum Tension, H: Like a parabolic cable the minimum tension in a catenary occurs at the lowest point H=wc

N = normal force F = frictional force R = resultant force ll = coefficient of friction cp = angle of friction

What is a Parabolic Cable? Fz

C.

~

b

-

t

~I

1 ~

468. 100l'Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas

.

What are the Moment of Inertia of Common Figures with respect to the Base?

,

. b~h . ........... •. Y -.1'2·.

i)h3·

ilx=f2

'·~

Day 19- Engineering Mechanics (Statics) 469

~.

where: m = mass of sphere r = radius of sphere

i ~

I

j

lit

B.

Spherical shell

·.•

A. B. llii!IWI [

[

Triangle

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Rectangle

' JD. ·-,.

.-b-aXIS .

....._:

~--.,

bh3

t=bh$

B.

Circle

3 .

--~----~---1---- X

3

,:':"

t ).

TOaA .

36

1=_g_mr 2

.I

4· 64

•

\

lte

,rab3

'4···

a.,,:,

X

A

mrtbia:

= distance from the centroid

Did you know that ... the integral sign

to the axis area of the figure

=

I'· _'lrt!~b

r--;r

where: a = length of semi-major axis b = length of semi-minor axis

A.

J

and elongated S denoting sum (Latin for "summa"), was introduced by Gottfried Wilhelm Leibniz, who named integral calculus "calculus summatorious" ! ... the definite integral which is defined as the integral between two values of an independent variables is also known as "Riemann Integral" after the German mathematician Georg Friedrich Bernhard Riemann (1826- 1866)!

What are the M~ss Moment of Inertia of Common Solids?

,1

~

GOOD LUCK I

L=l8 +Ad2

~

>


For composite figures and for axis not at the centroid nor at the base, moment of inertia may be calculated using the transfer formula, which is as follows;

y

....

~· 1::: 1 mr·

2

.

where: m = mass of sphere r = radius of sphere

1tr•

where: d

I

rt(O·.· ............ '' ••,::/1-·.}~A

1rt)" Jx··.=-.::~

Ellipse

Cylinder

b-axis

where: r = radius of circle D = diameter of circle

D.

C.

... ....

b

•'X "'·'')'

where: m = mass of sptrere r = radius of sphere

Sphere

..

,l:~:~~yf

l


~·

:i

A.

~~I

B. C. D.

[

Engineering Mechanics Branches of Mechanics Conditions for Equilibrium Friction Parabolic Cable Catenary Centroidal Moment of Inertia Moment of Inertia with respect to the base Transfer Formula Mass Moment of Inertia

~~~~II

Mon

tlln~~~~~

,.,qi<~il

D

14!~1~~~~

Tue

D D

L.""""''

l:. ,

Theory

~

Wed

D Thu

o· o

Problems

A. B.

c. D.

104.48• 105.58• 106.69° 107.96°

A.

B. C. D.

A.

B. C.

c. D.

so:sz What is the magnitude of the resultant force of the two forces 200 N at 20° and 400 N at 144"?

332.5 323.5 313.5 233.5

N N N N

8041 ECE Board November :1998 A load of 100 lb is hung from the middle of a rope, which is stretched between two rigid walls 30 ft. apart Due to the load, the rope sags 4 feet in the middle. Determine the tension in the rope. A.

B.

I

D.

sos:

1651bs 1731bs 1941bs 1491bs

A boat moving at 12 kph is crossing a river 500 m wide in which a current is flowing at 4 kph. In what direction should the boat head ifit is to reach a point on tho other side of the river directly opposite its starting point?

D.

121 kN 265 kN 211kN 450 kN

612.38 N 628.38 N 648.16N 654.12 N

809: EE Board October 1:99:! A man can exert a maximum pull of 1000 N but wishes to lift a new stone door for his cave weighing 20,000 N. If he uses a lever, how much closer must the fulcrum be to the stone than to his hand? 10 20 10 20

times times times times

nearer farther farther nearer

m m m

m

length. It carries a uniformly distributed load including its own weight of 200 N/m and a concentrated load of 100 N, 2 meters from the left end. Find the reactions if reaction A is at the left end and reaction B at the right end.

B. C. D.

RA = 81 0 N & RA = 700 N & RA =810 N & RA = 700 N &

on: A beam

Rs Rs Rs Rs

= 700 = 800 = 780 = 810

N N

N N

of span "x" meters with uniform loading of "w" kilograms per meter is supported at one end (A) and a distance of 2m from the other end (B). Find the reaction at support A.

A.

wx 2 2(x-2) kg

B.

wx(x-4) 2(x-2) kg

C.

wx(x-2) 2(x-2) kg

D.

A. B. C. D.

2.48 3.24 3.43 4.21

8U: EE Board October 1:991 A simply supported beam is 5 meters in

A.

808: CE Board November 1:994 A 200 kg crate impends to slide down a ramp inclined at an angle of 19.29° with the horizontaL What is the frictional resistance?

A. B. C. D.

810: A beam rests on a fuicrum, 1.2 m from one end. A weight of 350 kg is suspended from this end causing the beam to balance. If the weight of 350 kg is suspended on the opposite end of the beam, it is necessary to suspend a 1000 kg weight on the first end in order to effect an even balance. Find the length of the beam. ·

A. B. C.

88.67 kg 100 kg 70.71 kg 50 kg

807: ECE Board November 1998 A block weighing 500 kN rest on a ramp inclined at 25° with the horizontaL The force tending to move the block down the ramp is _ _ _ .

D.

80:&1 ME Board October 1:996 Two forces of 20 units and 30 units act at right a!)gle. What is the magnitude of the resultant force? 36 42 40 44

I

C.

Sat

80J:r Three forces, 20 N, 30 N and 40 N are in equilibrium. Find the largest angle they make with each other.

I

B. C. D.

B.

D D Notes

A.

A.

Fri

Solutions

downstream upstream downstream upstream

80&: EE Board October 1997 A 100 kg weight rest on a 30° incline plane. Neglecting friction, how much pull must one exert to bring the weight up the plane?

Topics

[

19.47° 19.47" 18.43° 18.43°

wx 2(x-2) kg

47_? 100 r Solved Problems in Engine~~ing Mathematic_;U?n<~.~Tiong & Rojas

II

(

(

oi
"ffi iilll~~l

Sll!U When one boy is sitting 12 m from the center of a see-saw, another boy rnust to sit on the other side 1.5 m from the center to maintain an even balance. However, when the first boy carries a(1 additional weight of 14 kg and sit 1.8 m from the c,enter, the second boy must move to 3 m from the center to balance. Neglecting the weight of the see-saw, find the weight of the heavier boy.

~~~.~ 11~~1JJI .. ~~~~~1:!11

L.

.,lllMII

~~"'

A.

30 kg

B. C. D.

42 kg 34 kg 45 kg

8:14: CE Board November 199ft A 40 kg block is resting on an inclined plane making an angle of 20° from the horizontal. If the coefficient of friction is 0.60, determine the force parallel to the incline that must be applied to cause impending motion down the plane.

A. B. C. D.

82 77 87 72

815: EE Board Oc:tober :1997 A 250 ib block is initially at rest on a tlat surface that js inclined at 30°. If the coefficient of kinetic friction 0.30 and the coefficient of static friction is 0.40, find the force required to start the block moving up the plane.

A. B. C. D.

1901b 2121b 125 lb 751b

8161 A 600 N block rests in a surface inclined at 30°. Determine the horizontal force P required to prevent the block from sliding down. Angle of friction between the block and the inclined plane is 15°.

A. B.

160.75 N 198.55 N

C. D.

164.60 N 190.45 N


tU7: ME Board March 1998

821: EE Board October 1991.


Assume the three force vectors intersect at a single point. F1 ::: 4i + 2j + 5k F2 = -2i + 7j -3k F3 = 2i- j + 6k What is the magnitude of the resultant force vector, R?

A certain cable is suspended between two supports at the same elevation and 500 ft apart, the load is 500 lbs per horizontal foot including the weight of the cable. The sag of the cable is 30 ft. Calculate the total length of the cable.

A copper cable is suspended between two supports on the same level, spaced 600 m apart. The cable hangs under the influence of its own weight only. Under these conditions, it is desired to calculate the maximum sag (at the center of the span) when the maximum stress in the material 2 is 1000 kg/cm . The cross-section of the cable is 1. 77 sq. em. Weight of cable 1.6 kg/m. Use parabolic equation.

I

A. A. B. C. D.

B. C. D.

"14 12 13 15

503.76 ft. 502.76 ft 504.76 ft 501.76ft

822: EE Board April1994 1111:111: EE Board March :l998 Given the 3-dimensional vectors: A = i (xy) + j (2yz) + k (3zx) B = i (yz) + j (2zx) + k (3xy) Determine the magnitude of the vector

sum lA + A. B.

C. D.

Bl

at coordinates (3,2, 1)

32.92 29.92 27.20 24.73

A. B.

C.

819: At what angle does the force F = 6.23i- 2.38j + 4.92 k N makes with the xaxis? A. B.

C. D.

:39.2° 40.2° 41.3° 42.2°

= =

14.73

D.

16.16

82:J: A cable carries a horizontal load of 20 kg/m. Neglecting its own weight, find the maximum tension on the cable if the distance between the supports is 100m and the sag is 5 m. A.

Assume the three force vectors intersect at a single point. F1 = i + 3j + 4k F2 2i + 7j- k F3 -i + 4j + 2k What is the magnitude of the resultant force vector, R? 15 13.23

D.

64.02 m 66.37 m 67.76 m 69.28m

A. B. C. D.

B. C. D.

5099 kg 50591
824: CE Board May 1993 Determine the sag of a flexible wire cable weighing 60 N/m over two frictionless pulleys 100 m apart and carrying one 10 kN weight at each end. Assume the weight of the cable to be uniformly distributed horizontally. The cable extends 5 m beyond each pulley to the point they are attached to the weights. 1\ II

7.2 rn

c

I G rn

ll

/Am

r 11 rn

42.26 m 43.26m 44.26m 45.26 m

826s A cable weighing 0.4 kg/m and 800 m long is to be suspended with a sag of 80 · m. Determine the maximum tension.

A. B.

420 kg

D.

416 kg 400 kg

c.

414 kg

827: A cable weighing 60 N/m is suspended between two supports on the same level at 300 m apart. The sag is 60 m. Compute the distance of the lowest point of the cable from the ground level. A.

B.

s:zo: ME Board Oc::tober ll99&

A. B. C.

A ·cable supported at two points of same level has a unit weight , ro of 0.02 kg per meter of horizontal distance. The allowed sag is 0.02 m and a maximum tension at the lowest point of 1200 kg and a factor of safety of 2. Calculate the allowable spacing of the poles assuming a parabolic cable.

=

C. D,

205.5m 196.8m 200.5m 188.2 m

8:18: Find the location of the centroid of the composite area consisting of a 10-inch square surmounted by a semi-circle. The centroid of a semicircle is located 4r/37t above the base (diameter) of the semicircle of radius r. A.

B. C. D.

6.0 6.2 6.4 7.0

inches from the inches from the inches from the inches from the

bottom bottom bottom bottom

47 4. 1001 ·solved Problems in Engineering Mathematics (2"d Edition) by Tiong & Rojas

~ 'Ill

8%9: EE Board March 1998

il

Load

~I

(

(

d~ijj!I~I[IJ

' ltl~i~~l!ll 1111'~11~~~11

coordinate

ycoordinate

Kilowatt load

1 2

0

2

1

1

3 4

1 2 2

3

100 180 200 120 150 200 180 100

111~11111~~111

5

L.,,,l""'"

6 7

~"'"""'

D.

Electrical loads are arranged on horizontal x, y axes as follows: X-

3 3 4

8

0 4 1 3

2

--

A. B. C. D.

X= 2.000, y = 2.049 x=2.163,y=2.195 X= 1.854, y = 2.211 X= 2.146, y = 1.902

2304 in

4

833: EE Board March 1998 An isosceles triangle has a 10 em base and a 10 em altitude. Determine the moment of inertia of the triangular area relative to a line parallel to the base and through the upper vertex in cm 4 .

A. B. C. D.

Topics

[QJ tvlon

2750 3025 2500 2273

0

Tue

0

834: ECE Board Apri11999 What is the moment of inertia of a cylinder of radius 5 m and mass of 5 Kg? A.

B. C. D.

Theory

0

120 kg-m 2 80 kg-m 2 62.5 kg-m 2 . 2 72.5 kg-m

Problems.

830: A rectangle has a base of 3 em and a height of 6 em. What is its second moment of area (in cm 4 ) about an axis through the center of gravity and parallel to the base? A.

B. C. D.

64 34 44 54

831: EE Board March 1998 A circle has a diameter of 20 ern. Determine the moment of inertia of the circular area relative to the axis perpendicular to the area through the center of the circle in cm 4 .

A.

B. C. D.

14,280 15,708 17,279 19,007

83Z: ME Boa,.d October 1993 The moment of inertia of a section 2" wide x 2' 0" high about an axis 1'0" above the bottom edge of the section is:

A. B. C.

1834 in 4 384 in 4 9214 in 4

835: ECE Board April :1998

Solutions

What is the inertia of a bowling ball (mass = 0.5 kg) of radius 15 em rotating at an angular speed of 10 rpm for 6 seconds?

A. B. C. D.

0

0 0 0 0

Wed Thu

Engineering Mechanics Branches of Mechanics Conditions for Equilibrium Friction Parabolic Cable Catenary Centroidal Moment of Inertia ' Moment of Inertia with respect to the base Transfer Formula Mass Moment of Inertia '

'

Fri

Sat

Notes

2

0.001 kg-m 0.002 kg-m 2 0.005 kg-m 2 0.0045 kg-m 2

ANSWER KEY 801. A 802.A 803.A 804. 805. B 806. D 807. 808. 809. D 810.C

c

c c

811. A 812. B 813. B 814. 815. B 816. A 817. B 818. B 819. 820.A

c

c

c

821. 822. D 823.A 824. 825. A 826. 827. B 828.D 829. A 830. D

c c

RATING

831. B 832. D 833. 834. 835. D

c c

c:J c:J c:J

0

30-35 Topnotcher 21-29 Passer 17-20 Conditional 0-16 Failed


iI

476 · 100 1 Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas

~,I

II

'II

For equilibrium, the force polygon must close.

il

l

30ft

T L


- ..

F=P

P = Wsine

F = 200(9.81)sin19.29°

P = 100sin30° p =50 kg

w

I

(

:E Finclined = 0

P =648.15 N

1m

Ill

II!IUI~~llill

tl'~ ~~~k4~11

40N

("l~lli11*1141il

a..~~~~~lli~~M

2Tcos9=100 2

L""'."'"

\.,.

,,,

1001bs

:EFv = 0

By cosine law: 2

( 40) = (20) + (30) e 104.48

2

-

T=~

2(20)(30)cose

=

1!1

2cose

T=

-~0 • '

100 2cos75.068° T=1941bs

p

-25°! 500 kg

.G

t)

4 kph

.

LMFulcrum = 0

.

........·

30

.

20000(x 2 ) = 1000(x 1 )

·.

_20_0_0_0x~2

x1 = 1000

The component, P along the plane of the force 500 kg is the force that tends to move the block down the ramp.

R = J(20)2 + (30)2 R = 36 units sinB=12 .e = 19.47"

II

P=Wsine P = 500sin25° P=211 kN

a

=0

II

100 kg

x1 = 20x 2 Thus, the fulcrum must be placed 20 times nearer his hand than the stone door.

:E Finclined

4

Ill

.-..-,

350 kg

We

X

F1=200N By cosine law:

Let: We= weight of beam

R~ = 200 2 + 400 2 - 2(200)(400)cos56° R=332.5N

LMFulcrum =0

n

Ill "15 tane=-

4

9=75.068°

W8 (0.5x -1 .2) =350(1.2)

· 3o•....--: 1OOsm \ 30"!1 00 kg

~

Let:

P = component of the weight of crate along the inclined

.p

Lf~ndlnod =0

.

..... -············

F = frictional resistance

.

\

1

We =

420 7 Eq. 1 O.Sx-1,2

~

I

11


478 · 100 l·Sclved Problems in Engineermg Mathematics (2"a Edition) by Tiong & Ro~. 1000 kg

W8

'I

0.5x

·~

,

ll

[

411111~!1~~

l,llil.l'lllll

M: L

~

=0

i

wx

:

Fwenm

x-2

2

4 ~~li~~jl~

L,,,.,.,.,

~'""'

0.5x -1.2

WX (X ;

RA (x- 2) =

Equattng equations 1 and 2:

Substitute the value of N in Eq. 1:

p = 125 + 0.4(216.506)

··

y-axis 0.6N =40(9.81)sin20° + P

4)

P=0.6N-134.208

I

We

-

w,

!12m i

::J p = 0.6(368.735)-134.208

F

L~=O

f . :: RA

Re

a

=0

WA +14

11.8

2:Fx = 0 Pcos6+F=Wsin8

P = 87 Newtons

WB =0.8WA

::5 ~::;:::::•:: :~ 8

I

y-axis

Substituting the value of N in Eq. 1:

.3

"'{8 (1.5)- WA (1.2)

A

e =~oo

N = 368.735

t

EDI

W= 600 N

N = 40(9.81)cos20°

1.5m : :: :~;; ~:;; t: ::::::;; ::; ·

x=3.43 m

1m

~Eq.1

N=Wcos20°

~

0.5x -1.2 0.5x -1.2 420 = 1200 - 350x + 440 350x =1200

P=212!b

LFY =0

wx(x- 4) kg RA = 2(x-2

~= 1200-350(x-1.2)

N=Wcose N = 216.506 lb

Ra

We= 1200-350(x-1.2) 7 Eq. 2

2:Fy = 0 N = 250cos30°

! (

I

1

RA i

350(x-; 1.2) + W8 (0.5x -1.2) = 1000(1.2)

fl

e = 20°

,t

A?:! U:U~ !1. lt

1

---····,·

I diM~~~

H:

~

x-1.2

w = 40 kg.

X

2- 2

350 kg

Pcos30° + tan15° (N) = 600sin30° W=250 lbs.

e =30°

We

3.0m

m

Pcos6+f.1N=WsinEl

N=

!

300-0.866P 0.268

~Eq1

L;Fy = 0

t;;;;:~;;;:·~~7~-;;;-:;w~d

2:MA =0 100(2) + 1500(2.5)- R8 (5) = 0 R8 =700 N

2:M8 =0

N = Wcos8+Psine

F

N = 600cos30°+ Psin30°

LMF=O

y-axis

W8 (3)- (WA + 14)(1.8) = 0

RA(5) -100(3) -1500(2.5) = 0 RA =810 N

2.4WA -1.8WA -25.2 = 0

Ill

RA(x -2) =

wx(~-2)

P = WsinEl+f.1N Pcc125+0.4N

2:Fx =0 F = WsinEl+P f.1N=WsinEl+P

Eq.2

P =Wsin6+F P = 250sin30° +0.4N

LMs=O

~

Equating equations 1 and 2:

(0.8WA)(3)-(WA +14)(1.8)=0 WA =42kg

Ill

N = 519.6 + 0.5P

'-'F L~ X = 0

~

Eq.1

Note: Since in the condition of the problem, the block is to start moving, then ''~":tile cooftrcient of static friction.

300-0.866P =519.6+0.5P 0.268 P = 160.75 N!3wtons

480 100 l Solved Problems in Engineering Mathematics (2"d Edition) by Tiong & Rojas

11

(

r = ../6.232 +2.382 + 4.922 r=8.287

'Iii'

R=F1 +F2 +F3

I~

R = (4i+2j+5k)+(-2i+ 7j -3k)+(2i- j+6k)

II

R =4i+8j+8k

,111

(

a

~fUtlll~lllili~

~ lU~~ili~l\ lt:II~~Wil!l~l!

1-il~l ~~~il~ I

L. "~'" "

l"~'"'

2

+(Ak)

Ell

roe H=--·· 8d H= 20(100) 8(5)

2

IRI =J42 +82 +62

By cosine law:

5.465

1!11!11

2

·= 8.287

2

Substitute the values two vectors:

of x,y and z to the

A= i(xy)+ j(2yz)+k(3zx) A= K3)(2)+ j(2)(2)(1) + k(3)(1)(3) A=6i+4j+9k

•

(:) = 41.3°

R=F1 +F2 +F3

B = K2)(1) + j(2)(1)(3) + k(3)(3)(2) IRI = A +B = (6+2)i +{4 +6)j+(9+18)k

IA+BI =J(A;t +(Ai +(AS

Ja

2

2

+10 +27

J22 +142 +52

2

Ell

3(~00)

11

'J·········~~axis

•

y

T2 =( ~!: +H2 l2) 12

.

17702"" [.1.6(:00lj +H 2

T=5099 kg

H =1703.67 kg ro=60N/m

11

,..roe 8d

1703.67 "' .~~~00)

2

8d-

d"' 42.26

m

s = 500 + 8(30)2 -

J

+ 50002

T =10300 N

4

8d . 32d S = L + - - -33L 5L

lA + Bl =29.88 units

4.92

J( 20(~00) J

T '= 10000 + !30N (5m)

Let: S = length of the cable

2

T=

m

IRI = 15 units

•

A+B =8i+10j+'27k

Note: This force is equivalent to the tension (i) at each support.

~s~

2

2

IRI= J(A;) +(AJ) +(AS

B=2i+6j+18k

x 1.77 cm 2

F = 1'770 i
ii!

R =(i+3j+4k)+(2i+ 7j-k)+(-i+4j+2k) R=2i+14j+5k

B = i(yz) + j(2zx) + k(3xy)

F=SA . 1000 kg F =--err?

2

T=~Lr +~

+ 6.23 2

- 2(8.287)(6.23)cose cose = 0.7517339

liMill

F

S=A

H=5000 kg

d2 =r 2 + x 2 - 2rxcose

IRI = 12 units

•

L=69.28 m

d = J2.38 2 + 4.922 d-5.465

IRI=F~r~(Aj)

lA +BI =

----------------------------~D~a~y~i~9~-~E=n~gtn~·~e~e~nn~·~M~~tat~~

•

~···tao c

4

32(30) 5(500) 3

s = 504.76 ft

T2 =

(mL)

2

~

+H2

\ 2

2

103002 =[60(100)] --2-. +H2 H=9853.42 N

T=roy

T = 0.4(130 +c)

2

roL H=8d

e =H[~J L2 =' 1200[8(0.02)]· 2 0.02 L2 = 4800

H= ool2 8d 9853.42

2S =800

~ ~0(100)

d= 7.6

2

8dm

S =400 m y2= 8 2 +c2 (80 + c)2 = 400 2 + c2 6400 +'2G + c 2 "'400 2 + c 2 Co,960 m

~

Eq. 1

48~ 100 l Solved Problems in Engineering Mathen1!ticsJ.?..::=..~iticm) by 'f!ong & ~5?1!!.

1

Substitute the value c in Eq. 1:

-~~

T = 0.4(80 + 960)

il

T=416 kg

I .

x=150

2

Ar = 10

(

·1 )( n: \1

+l:Z l4)10)

2

Substitute values:

Ar = 139.27 in 2

. •

'---~

x=150

.

rm

'J2==10+

3n 12.12 in.

I=

let: J = polar moment of inertia nd 4 J=-

Substitute Y1 and Y2 in Eq. 1: 139.27y = 100(!)) +

J"' n(20)

4

32

y = 7 inches from the bottom

, $

J = 15708 cm 4

S+y

X=Cin-C

X

S + (60 +C)

150=cln-~---

= _..!.._..!____.--"._e______, __ , _________.!!._.!.!.

c

bh 3

Ll +L2 +

->Eq. ·1

lx=--

12

x~~O):_!OO(~+;;m:1)+1ax2'J+'~!~+;m;3)"':100:3~~~

lx =~(24[ 12 lx =2304 in 4

1C0+100+2Xl+ 1:tfJ+ ito+ilJD+1lD+'!CO

Yz = 8 2 +c2

X=2

(60+c) 2 =S 2 +c 2 3600 + 120c + c 2 = S 2 + c 2

y =-

s == FGoo .; 120c

---'*' Eq. 2

'-p 1

:~.::..~~-~~---~-:.:;:.L~.:_:~~·-:.:.:._r_:._-:~:_~l L1 + L2 + L 3 ......L,

- 1~) + 10C(1l::.?Q:5~+E()\0)+ ~4)+20Q£1l_+'l80(3) +.!_C0:2) ®+®+2Xl+~+~+2Xl+®+~

....~,.·································:Jr.····················~.········ X1

Y-

Substitute Eq. 2 in Eq. 1:

150

y =2.049

= cln ../3600+120c + (60 +c}

By trial and error:

2 d =-h

h = 10 em

c

3

c = 196.8 m

l'l!'R ~

h=6

b = '10 em )(

4r

10 in.

J':

ivn + Ari

,

br1 3

4

~lb

r.

_..

lx =i Xo +

oh

.

L-1'·~-~--+-bh

Using transfer-axis moment formula:

':'_!'_

!Xl =co

2

-1-

3

.L\.cf

\ 1"-,

j

2:' ;uul:lil'

3.mr2 5

I= 3_(0.5)(0.15)2 5 I= 0.0045 kg-m 2

32

-~~ri!)

~)li-~~--JWA&~~ft!!t.,$1 L

I= 62.5 kg-m 2

1Xo =54m 4

5 in.

Y2::::

c

2

ix., = 3{6)3 3 - 3(6)(3)2

·7 Eq. 1

wher-e:

60

1 2 1=-mr 2 I= _!(5)(5) 2

ArY =A1Y1 +

)>,

•

bh 3 1x.. =--bhd2 4

+ Asomi-cil'"...im

Ar =

Day 19- Engineering Mechanics (Statics) 4~3

-------

(2

,=,

-hi

3 )

(4 )l

1 2(10)(10) 3(10)

r

~

t

486 ·1001 Solved Problems in Engineering Mathematics (2nd Edition) by Tiong:_ & Rojas

I

'~I

I'

I

Topics

0

Mon

~ Tue ''"~~~~~~

0

Wed

Theory

Rectilinear Translation - Horizontal Translation -Vertical Translation -Free Falling Body Curvilinear Translation - Projectile -Rotation D'Aiembert's Principle Centrifugal Force Banking of Highway Curve

Probien1s

Soluttons

[] []

Notes

Sat

Fri

:cctil!near translation is a motion in a !llane or one dimension.

r

i

are l:h<"

i l1.e 11 ;m::,lalion:

Not this path

ca::::·.::·.-..:. . . . . ,....,:::::::}:::::::·····. .+::::~.·. This path (straight line)

of rectilinear Top View

1.

2. 3.

Hori2:ontGI transiation Vertical translation C;Jiviline;;;,r

Vo

a=O .............)!»-

11 ., u nnl;1l tr·;;n~;laiion (motion) is a straight ~~l~•illl)l :llui'ifJ ~~

w1,.,,., dllH•

t

..

I

r

·i

•jj

iliG

[ :,,,

l·y'

'9jd;l'l/)

H 'f 1 1\111

hcni.;-:ont;]j ~··

···j,·,· {, '!

~;\a;J;;;_

ri:

s

rr:~5·'- :1nc,~tv

·"

Elevation View

v ~

,, 'ii

488 1001 ·solved Problems in Enginee~~g Mathematics (2nd E@;ion) by Tiong & Rojas

1J

(

where:

••IIWI~~II1

....

IIIIIN~~~~~~~~~

[

,~,

llll~ij/·~11!

111~~111~1~

~'"'""

·-~Vo

S=Vt

I

Vo = initial velocity (time = 0) V = final velocity (time = t) S ::: distance traveled after elapsed time = t

.............

-~

,~

____:__J s The following formulas

~g

will be used:

x =V0 coset

: a

'f

y

So the time of flight, t is:

-~v

·-~v v

a

Since the ho'rizontal component of the velocity is constant throughout the projectile, then the Value Of X iS calculated as follows:

Vo=O

y

If acceleration is constant, the initial velocity, Vo is not equal to the final velocity, V.

Vo

• t

as it goes down and decreases as it goes up.

V=Vr:;

II,,

Day 20- Engineering Mechanics (Dynamics) 489

t=~ V0 cos6

' 1 '/=2gt2 In the figure, the value y is obtained using the formula:

The following formulas will be used:

~~~lil' V""Vo:t:'ati

'""~~"

What Is a Curvilinear Translation? Curvilinear translation (motion) is a motion along a curve path either on the vertical plane or on horizontal.plane. This type of motion includes a trajectory or projectile, which is a curve path on the vertical plane and rotation on horizontal plane.

v ~J • ""';.; 1:;~ ·, ""'':VIfSII'I:<>•'""-·l;l~· ,

. . '

,2

Substituting the value of the time of flight "t" in the above equation: 2

V 7 =V0 ~ ;:J:2a$ V = V0 :tat

1 S ""V0 t±:o-at2 '

where:

.2

Vo = initial velocity (time = 0) V = final velocity (time = t) S = distance traveled after elapsed time = t a = acceleration

Note: Use + if accelerating and - if decelerating.

where:

Vo =initial velocity (time= 0) V = final velocity (time = t) S = distance traveled after elapsed time = t a = acceleration

Note: Use + if going down and - if going up. If the acceleration of the body is not given, then it is presumed that the acceleration to be used is the acceleration due to gravity (gravity on earth). g = 9.81 m/s2 g = 32.2 ftls2

What Is a Free Falllnp Body? yvhat Is a Vertical Translation? Vertical translation (motion) is a straight motion along a vertical plane. In this type of motion, acceleration cannot be equal to zero since the velocity of a body rnoving in vertical motion, increases

A body is said to be free falling if the initial vertical velocity is zero. The acceleration of a free falling body is due to gravity which is equal to 9.81 m/s2 or 32.2 ftls 2 .

A projectile is a body which after being given an initial velocity with an initial angle of release is allowed to travel under the action of gravity only. A projectile is a trajectory which is a graph of a parabola.

v

(V cose )-· 2 (V cose J 0

0

Simplifying the above equation will produce the "General Equation of Projectile".

....................

"··-.. ' (x,y)

(O,O).x.···· . . . . .•

y

. O -X- +1 -X= "v,0 Sin 9

X

""'!!!lll!lllll!llii!I-.,4~W.flll\'!i~Y 1

j;·,'·., ..····.....

J.!ii!i:t'#/Z;;;;pi

@ ~¥k

~

v.

Let: and Vy be the horizontal component and the vertical component, respectively of the initial velocity, Vo.

kVo

tJilfrv.

v. = V

0

Vv

cose

=V0 sine

Since the variable "x" is in second degree while the other variable "y" is in the first degree, then the graph is a parabola. Note: The vertical component of the velocity decreases as it goes up and is zero at maximum point of the projectile and increases as it goes down, while the horizontal component is constant. If the initial point of the projectile is on the same elevation as the point of impact, the following formulas may be used:


490 ·1 Ob l Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas

I

I(

r; (

diM1 1 I!~IIII

Wh;;it is a Rotation?

Highest Point

11

Rotation is a motion along a circular path.

...····················•·······················... Vo . •••

Ymax

/'\e

•·••••

F".J,p.~~'<'Wb~!\'t~v«;'"'1J,'i~'"i11,-W('"&!~ '!:f,-"?'ry..'<'"""!-:;,.">~''7~"-"J1<_'-,,Y~

""-:oW>->"'

Initial Point

Maximum height, Ymax:

L. , . ., .,

~REF

.'\ "( "·

s

Let: S V

or R:

''·

g If the initial point of the projectile is not on the same elevation as the point of impact, the following formulas for ran!=)e, R may be used: Highest Point

...···············:··""········ Vo

=linear distance in meters

or

REF::: W a ~~

._,L. _t.lt,. .,.,_,., Initial Point

R _ 2V0 2 sinecos(e + ~) gcos2 p Highest Point

Fe =F g

f:C'lrC~0

v,,no

is the force thRt will movinq in a circular o.8th

1''1

fmm U1e center. ~"'=ret

where: F

= force due to friction

Why is Banking of necessary?

Hl!it~l_"!\'

go. !D'_~

In order for the weight of the carlo help the frictional force resist l.he force, banking of highway curve is necessary. Once banked, a car can move at higher speed due to bigger resistance of the centrifugal force.

For variable anqulat aceeleration, the following formulas must be used.

ui "'w02 ±2a6 "I

ro"" ro 0 ±at

9 ""ro0 t ±

1

rap view ot" c1rcular track

D'Aiembert's principle is stated as follows:

gcos2 13

acceleration

::-c

s,.. wf!

~Q~eJ§.mbe!f~

R "'~V,/ si~eoos(e-~)

If there is no force other than the force due to friction between the tires of the car and the pavement of the road to resist the centrifugal force, then:

·-

Impact Point

·\e -

Elevation view

REF ==ma

e = angular distance in radians m = angular distance in rad/s a ::: angular acceleration in rad/s2

2

F

!,\i.;;.-~themat.~cally,

= linear velocity in m/s a ::: linear acceleration in m/s2

R"' Vo 9in2El

Fe~-

reverse effective force is equal to the ,,,nduGt of mass and accelertion.

i-

Range, Xmox

=weight of body

V = tangential velocity r = radius of curvature g = gravitational acceleration

..

=Range = R

where: W

. ...... ;,.. a

'T'~~ -"'"(~": -'>4\'

f•jllii!III\WI

~ ....:

,. ;r:fe:
-....

···...

rJ1i~~~~lfil!l

f... ,~lll~~~~~lll

"1"' ,'~

'{7

Impact Point Xmax

i

force mentioned in D'Aiembert's nrinciple is known as the reverse

"When the body is subjected to an acceleration, there exists a force opposite the direction of !he mo!ion and equal to lhA product of mass and acceler~:.~tim;"

I

·II'

The ideal angle of banking, calculated l!Sing:

centr if11gal force 2

I,

1/V\11

'JI

tane"' v gt

e may be

492

100 l Solved Problems in Engineering Mathematics (2"d Edition) by Tiorig & Rojas

1\,

For greater velocity without skidding,

~

tan(e + cl>), V gr

I,

2

Topics tancjl

dillllllll~lm

(

Utliljl11ill\lllll

,111 ..~1~~111 (

=ll

D

1

where: cjl = angle of friction ll = coefficient of friction

Mon

IIIIIIIIO!I

~ •.,.,1


[QJ

I

<4 ~~~~~1111 L'l

L.,..

I

I

Tue

D D

'QI;ribia:

~uote: "He who loves practice without theory is like the sailor who boards ship without a rudder and compass and never knows where he may be cast"

Theory

Wed

Problems

~

Thu

Solutions

Fri

D D D

GOOD LUCK I

Did you know that ... about 500 B.C., th~ Pythagorean Brotherhood was originally aware of the four regular polyhedrons and considered them to represent the four basic elements namely tetrahedron- "fire", octah3dron- "air", hexahedron- "earth", icosahedron - "water"! When the Pythagoreans learned the existence of the fifth regular polyhedron, dodecagon, they considered it to represent the fifth element - "universe".

D D Notes

Sat

s:s«t: ME Board April 1.996 What is the acceleration of a body that 111creases in velocity from 20 m/s tp 40 m/s 111 3 seconds? 1\. ll. ('

ll.

5.00 m/s2 6.67 m/s2 ·1.00 m/s2 8.00 m/s2

•H7: ECE November 1.998

- Leonardo da Vinci

Rectilinear Translation - Horizontal Translation -Vertical Translation - Free Falling Body Curvilinear Translation -Projectile -Rotation D'Aiembert's Principle Centrifugal Force Banking of Highway Curve

I low far does an automobile move while its ·.peed increases uniformly from 15 kph to •1', kph in 20 seconds?

I\ II I.

ll

185m 167 rn 200m 1l2m

838: CE Board November J.99fl A train passing point A at a speed of 72 kph accelerates at 0. 75 m/s 2 for one minute along a straight path then decelerates at 1.0 m/s2 . How far in km from point A will it be 2 minutes after passing point A?

A B. C. D.

3.60 km 4.65 km 6.49km 7.30 km

839: CE Board May J.99fl From a speed of 75 kph, a car decelerates at the rate of 500 m/min 2 along a straight path. How far in meters will it travel in 45 seconds?

A. B. C. D.

790.293 791.357 793.238 796.875

m m m m

'

494 I 00 I Solved Pmblems in Engineering Mathe~ti~ (2"' Edition) by Tiong & -~ ,~,

'I, 1('

«;

840: CE Board November 1997 A train starting at initial velocity of 30 kph travels a distance of 21 km in 18 minutes. Determine the acceleration of the train at this instant. A. B. C.

0.0043 0.0206 0.0865 0.3820

D.

m/s 2 m/s 2 m/s 2 m/s 2


-~'itJII~

An automobile moving at a constant velocity of a 15 m/sec passes a gasoline station. Two seconds later, another automobile leaves the gasoline station and accelerates at a constant rate of 2 m/sec2 . How soon will the second automobile overtake the first?

A. B.

=

1.02 m/s 2 . rn/s 2 3.4 m/s 2 18.1 m/s 2

C. D. 84~:

B. C. D.

building 40 meters tall will hit the-ground with a velocity of:

50 m/sec 28m/sec 19.8 m/sec 30m/sec

847: ME Board April J:'JC):! Using a powerful air gun, a steel ball is shot vertically upward with a velocity of 80 meters per second, followed by another shot after 5 seconds. Find the initial velocity of the second ball in order to meet the first ball 150 meters from the ground. 65.3 m/sec 45. 1 m/sec 56.2 m/sec 61.3m/sec

ME Board October 1!995

36 54 24 20

844: CE Board May 1998 Determine the velocity of progress with the

given equation: D = 20t + seconds. A.

84&: ME Board AprU :1995

A ball is dropped fro, n the roof of a

A B. C. D_.

The distance a body travels is a function of time and is given by x(t) = 18! + 9f. Find its velocity at t = 2. J.\.

4.52 s 4.42 s 5.61 s 2.45s

18.6-m/s

•.ii 'J~

'

'il'

8491 lEE ha~rd October 199& A ball is dropped from a height of 60 meters above the ground. How long does it take to hit the ground? A. B. C D.

2.1 3.5 5.5 1.3

sec sec sec sec

ilil:!OO: ECE Board April 1998 A baseball is thrown from a horizontal plane following a parabolic path with an initial velocity of 100 m/s at an angle of 30° above the horizontal. How far from the throwing point will the ball attain its original level?

A. ll

c

n.

890m 883m 880 m 875m

~'ill!

If a particle position is given by the expression x(t) = 3.4e- 5.4! meters, what is the acceleration of the particle after t 5 seconds?

B

A. B. C. D.

C. D.

842: EE Board October :1996

A.

19.8 m/s 21.2 m/s 22.4 m/s

845: ECE Board April 1999 ·A ball is dropped from a building 100 m high. If the mass of the ball is 10 gm after what time will the ball strike the earth?

A. B.

15.3 sec 16.8 sec 13.5sec 18.6 sec

C. D.

B. C. D.

Day 20 - Engineering Mechanics (Dynamics) 495

~ t+1

when t = 4

848: EE Board O<.rto~r 1!99:1> A ball is thrown vertically upward from the ground and a student gazing out of the window sees it moving upward pass him
15.25 m

B. C. D.

14.87 rn 9.97 m 11.30 m

ME Board April :11:991 /\ plane dropped a bomb at an elevation of 1000 meters from the ground intended to hit the target at an elevation of 200 meters from the ground. If the plane was flying at :1 velocity of 300 km/hr, at hat distance Irom the target must the bomb be dropped lo hit the target Wind velocity and o~tmospheric pressure to be disregarded. .'\ 11.

(;

u

1024.2 rn 1055.6 m 1075.5 m "1064.2 m

8§3: ME ik,lard April :1995, ME Board October 199& A shot is fired at an angle of 45° with the honzontal and a velocity of 300 fps. Calculate, to the nearest value, the range of the projectile. A. B. C. D.

932 yards 1200 yards 3500 yards 4000 yards

854: CE Board May :1.995 A projectile leaves a velocity of 50 m/s at an angle of 30° with the horizontal. Find the maximum height that it could reach.

A. B. C.

D.

31.86m 31.28m 30.63 m 30.12 m

1!155: ME Board October 1997 A shot is fired with an angle of.45o with the horizontal with a velocity of 300 ft/s. Find the maximum height and range that the projectile can cover, respectively.

A B. C. D.

800 ft, 1600 ft 923 ft, 3500 ft 700 ft, 2800 ft 1800 ft, 3000 ft

S$6: CE Board November :1.996> A ball is thrown from a tower 30 m high above the ground with a velocity of 300 m/s directed at 20° from the horizontal. How long will the ball tlit the ground?

1.1',-;::e,: ME &11!1rd: O~"l:ooor :ll'Jil'\ll1 ll!o muzzle velocity of a projectile is 1500 Ips and the distance of the target is 10

''Illes. The angle of elevation of ihe gun ""1st be: /\ II

21"59' 2:~ tl i'

Jl

/h"Hl'

0

?!lo:r:r

A. . B. C. D.

s:n:

2'1.2s 22.2 s 23.2 s 24.2 s

In the last 2 seconds of NBA finals featunng Chicago Bulls VS Utah Jazz, with the latter ahead by 2 points with the former at 94-92 count Bulls Michaei.Jordan decides to snoot from a certain point on the rainbow territory wt1ich counts 3 points if converted. During the play, i(Jorcjan releases the ball at 7 m from tne basket a11d 2. i 5 m above the ground and an

I Day 20 -

496 1001, Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas

~II

I.

~ (

*MIIV\~1111~

r•ii.

ifi~IIII~Wl·

inclination of 40" with the horizontal and assuming no block was made by the opponents, at what velocity will the ball be given to cast the winning basket? The basket is 10 feet from the ground. A. B. C. D.

8.57 m/s 8.86 m/s 9.03 m/s 9.27 m/s

~UIIIIill

~ .. ,.ur~l~~l.il~

L. ..,,,IJ

l ..,.:!

8581 CE Board May 1995 A projectile is fired with a muzzle velocity of 300 m/s from a gun aimed upward at an angle of 20" with the horizontal, from the top of a building 30 m high above a level ground. With what velocity will it hit the ground in m/s?

A. B. C. D.

298 299 300 301

m/s m/s m/s m/s

8591 CE Board May :1.995 A stone is thrown upward at an angle of 30" with the horizontal. It lands 60 m measured horizontally and 2 m below measured vertically from its point of release. Determine the initial velocity of the stone in m/s.

A. B. C. D.

22.35 m/s 23.35 m/s 24.35 m/s 25.35 m/s

8601 CE Board November :1.99Z A wooden block having a weight of 50 N is placed at a distance of 1.5 m from the center of a circular platform rotating at a speed of 2 radians per second. Determine the minimum coefficient of friction of the blocks so that it will not slide. Radius of the circular platform is 3 m.

A. B.

c. D.

0.55 0.58 0.61 0.65

86!.1 ME Board October :199:1. The flywheel of a puncher is to be brought to a complete stop in 8 seconds from a speed of 60 revolutions per minute. Compute the number of turns the flywheel will still make if its deceleration is uniform.

A. B. C. D.

5 turns 3 turns 4 turns 6 turns

8621 ECE Board April :1998 What is the speeq .of a synchronous earth's satellite-situated 4.5 x 10 7 m from "the earth?

A. B. C. D.

11,070.0 kph 12,000.0 kph 11,777.4 kph 12,070.2 kph

86~1

ECE Board November :1998 A rotating wheel has a radius of 2 feet and 6 inches. A point on the rim of the wheel moves 30 feet in 2 sec. Find the angular velocity ofthe wheel.

A. B. .C. D.

2 4 5 6

rad/s rad/s rad/s rad/s

8641 CE Bqard Novembeio :1997 A turbine started from rest to 180 rpm in 6 minutes at a constant acceleration. Find the number of revolutions that it makes within the elapsed time.

A. B. C. D.

500 540 550 630

8651 A flywheel is 15 em in diameter accelerates uniformly from rest to 500 rpm in 20 seconds. What is its angular acceleration?

A. B. C. D.

2.62 3.45 3.95 4.42

rad/s2 rad/s2 rad/s2 rad/s2

.

~W1

:~1

;!)

866: ME Board April1991 A boy tied a 80 grams stone to a string which he rotated to form a circular motion with a diameter of 1000 mm. Compute for the pull exerted on the string by the stone if it got loose :eaving at a velocity of 25 m/sec. A. B.

120 N 100 N

C.

150 N

D.

135 N

867: EE Board April :1997 A man keeps a 1kg toy airplane flying horizontally in acircle by holding onto a. 1.5 m long string attached to its wing tip. The string is always in the plane of the circular path. If the plane flies at 10 m/sec, find the tension in the string.

A. B. C.

28N 15N 67N

D.

18 N

868t ME Board October :1.996 An automobile travels on a perfectly horizontal, unbanked circular track of radius R. The coefficient of friction between the tires and track is 0.3. If the car's velocity is ·15 m/s, what is the smallest radius it may travel without skidding?

A. B.

c. D.

68.4 m 69.4 m 71.6 m 7-6.5 m

8691 CE Board November :1998 A hi-way curve has a super elevation of 7degrees. What is the radius of the curve such that there will be no lateral pressure between the tires and the roadway at a speed of 40 mph?

A. B. C. D.

265.71 m 438.34 m 345.34 m 330.78 m.

EngiJl~(!ring

Mechanics (Dynamics) 497

8'701 ME Board April 1998 Traffic travels at 65 mi/hr around a banked highway curve with a radius of 3000 feet. · What banking angle is necessary such that friction will not be required to resist the centrifugal force?

A. B.

c.

D.

3.2• 2.5° 5.4°

1Po

8711 ECE Board April1999 Determine the arigle of the super elevation for a 200 m hi-way curve so that there will be no side thrust at a speed of 90 kph.

A. B. C.

19.17" 17.67• 18.32•

D.

20.11"

8'721 ECE Board April :1998 The inclination of ascend of a road having a 8.25% grade is _ _ _.

A. B.

4.72° 4.27°

c.

5.12"

D.

1.86°

8731 ME Board April :1.996 A cyclist on a circular track of radius r 800 feet is traveling at 27 fps. His speed in the tangential direction increases at the 2 rate of 3 fps • What is the cyclist's total acceleration?

=

·A. B. C. D.

2.8 fps2 3.1 fps2 3.8 fps2 4.2 fps 2

8'74r ME Board October 1997 A concrete hi--way curve with a radius of 500 ft is bar;~ked to give lateral pressure equivalent to f = 0.15. For what coefficient of friction will skidding impend for a speed of60mph? ·

A B.

c.

1-1. ~

0.360 Jl < 0.310 Jl > 0.310

~.

·

498_ 1001 Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas

,,, ~~II

,('

~~

,.~.,,.J ',! lllllllllliii!M~

D.

ll < 0.360

87'>: EE Board April1993 What force is· necessary to accelerate a . 30,000 pounds railway electric car a.t the rate of 1.25 fUsec2 , if the force required to overcome frictional resistance is 400 pounds? A. B. C. D

1565 pounds 1585 pounds 1595 pounds 1575 pounds

87&: ME Board October 1995 A car weighing 40 tons is switched to a 2 percent upgrade with a velocity of 30 mph. If the train resistance is 10 lb/ton, how far up the grade will it go?

A. B. C. D.

1124 ft 2014 ft 1204 ft 1402 ft

on on on on

A. B. C. D.

19.63 feet 19.33 feet 18.33 feet 19.99 feet

•...

•,

( '.1·1

',

880: ME Board April1997 A pick-up truck is traveling forward at 25 m/s. The bed is loaded with boxes whose coefficient of friction with the bed is 0.4. What is the shortest time that the truck can be brought to a stop such that the boxes do not shift? A. B. C. D.

2.35 4.75 5.45 6.37

Topics

D Mon

~ Tue

s s s s

D D D D [I] D

slope slope slope slope

B. C. D.

4250 N 0.68 N 680 N 42.5 N

878: ME Board April1998 An elevator weighing 2,000 lb attains an upward velocity of 16 fps in 4 seconds with uniform acceleration. What is the tension in the supporting cables?

A. B. C. D.

2,150 lb 2,4951b 1,950 lb 2,250 lb

8791 ME Board April1998 A body w.eighing 40 lb starts from rest and slides down a plane at an angle of 30• with the horizontal for which the coefficient of friction ll = 0.30.' How far will it move during the third second?

Wed

Problems

Thu

Solutions

Fri

D D

877: EE Board April199& A car moving at 70 km/hr has a mass of 1700 kg. What force is necessa~ to decelerate it at a rate of 40 cm/s ?

A.

Theory

Rectilinear Translation - Horizontal Translation -Vertical Translation -Free Falling Body Curvilinear Translation -Projectile -Rotation D'Aiembert's Principle Centrifugal Force Banking of Highway Curve

Sat

Notes

RATING

ANSWER KEY 836.B 837.B 838.B 839. D 840.B 841. B 842.B 843.B 844. B 845. A 846.B 847. D

848.D 849. B 850. B 851. D 852.C 853.A 854. A 855.C 856.A 857.C 858.D 859. D

860.C 861. 862.C 863.D 864.B 865.A 866. B 867. 868. D 869.A 870.C 871. B

c

c

872. A 873.8 874.B 875.A 876. 877. 878.D 879.B 880.D

c c

c:J 3!-45 Topnotcher c:J 27-37 Passer c:J.22-2b Conditional

0

0-21 failed


'l 500 .100 LSolved Problems in Engineering Mathematics (2"d Edition) by Tiong & Rojas

,,,

Ill

1~1

~

""''"""~!

s,

V = V0 +at 40 = 20 +a(3)

rm

S1 = 2.55 km

lj•~lllll:~

L.MIIIIIIJ

l. .~~

Overtaking point ~! t,= t2+2 -v,=15 a,= 0 ;,

s,

V1 = V0 +at1 v, = 20 + 0. 75(60)

V = V0 + at -7 Eq. 1

V. = 15 km x 1000 m 0 hr 1 km V0 = 4.167 m/s

x~ 3600 s

~

1

S2 = Vot2 + 2at2

V = V0 +at 12.5 = 4.167 + a(20) a = 0.41665 m/s2

2

s = 4.167(20) + ~(0.416665)(20) 2

2

S =167m

! .ALNAA
S2 = 2100 m

1 ;:::

~~~

1 2 =V2 t 2 .+ -at 2 2

si = o(t 2) + i<2)(t2)

s = 2.55+2.1 S =4.65 km

s2 =

-7

Eq. 2

s, =S2 15t2 +30 = t/

60 min

t2 2

-

15t2 - 30 = 0

By quadratic formula:

l!!l!tl liiiiiJI

1 2 S = V0 t+ --at

2

Vo-

v, -

s = 1250(

45

60

2

)+

45 60

t - 15±J15 -4(1)(-30) 2 2(1)

~(500)( ) 2

2

t - 15±18.574

S =796.875 m

tz

t 1 =1 min. =1 min . ~-----------;·~~------~

s,

s2

V. = 72 km x 1000 m x ~ 0 hr 1 km 3600 s V0 = 20 m/s

2-

•

Taking + sign:

V. = 30 km X 1000 m X _!.!!!: 0 hr 1 km 3600 sec V0 = 8.333 m/s

2

2 t 2 = 16.8 seconds

.1

s = V t+-ae 2 0

1m

21000 =8.333(18){60) +ia[(18)(60)]

a =0.0206 m/s

ge

dx v = - = 18+18t dt

V=18+18(2) V =54 m/s

1m 5 D=20t+t+1

dO

-5

V=-=20+-dt (t + 1)2

.5

V=20--2 ( 4 + 1)

V = 19.8 m/s

•

Note: Since the ball was dropped, initial velocity of the ball is zero.

t2 = 15+18.574

2

1. 2 S1 = V0 t 1 + at1

2

=1st+

2

Equating equations 1. and 2:

x~

~->Gx

At timet= 2:

S=S 1 +S2

V. = 75 km x 1000m 0 hr 1 km V0 = 1250 m/min

dV a=·--= 20.4t dt a= 20.4(5) a= 102 m/s2

'

15t + 30____.---2 . -7 Eq. 1

t/

sw1

i

S 2 = 2.1 km . S2

v = dx = 3(3.4)e- 5.4 dt v = 1o.2e- 5.4 dV = 20.4t dt

i

2

s, = V.t11 s, = 15(t2 +2)2 s

Ill

s = V0 t+-~ae

;;.;

s2

Solving for total distance:

Substituting values to Eq. 1:

;

-v2=0

:ul;.;w{~n';; .m.tMittiMM¥.: =;;~.

2

s2 = 65(60) + i(1)(60)2 V= 45 km X 1000m X~ hr 1 km 3600 s V = 12.5 m/s •

:l!!~Z.,X

t2

V1 = 65 m/s

I"'I!Wf!il~(l

""'""1

•

= (20)(60) + i<0.75)(60f

S1 = 2550 m

a= 6.67 m/s2


.l

2

1oo = o + i<9.81)e X

2

1 2 h=V0 t+-gt

= 3.4t 3

t = 4.52 seconds -

5.4t

d

l,,i

• •


502 I 001 Solved Problems in Engineering Mathematics (2"d Edition) by Tiong & Rojas

-IIi I

I.!I

S:

.liM

1

h3 = V3t3 - 2gt3

V 2 = V0 2 +2gh 2

V

o

h1

••

···········;;0,;···

.

''

•,; t 2

.

= range of the projectile

g

V3=? .

10

t

R=

m

•••••••••

0 =52 - 2(9.81)h h=1.3 m

0 =802 - 2(9.81)h h1 =326.2 m

1 2 h2 = v,t2 + 2gt2 326/2-150 = O(t2) + ~(9.81)t/

t2

=6 seconds

t3=~+t2-5

t3 =8.155+6-5 t3 =9.155 s

.

.. _ . lt

L

v

~·~"'!'!'l'~~~~

V/ = V0 2 -2gh

=8.1555 s

r~

•

i#~

Let: H = maximum height

;;

0 = 80-9.81t,

/.

~~=30o

'Vo=?

• • Vo

gx2 Y = xtane- 2Vo2cosze (9.81)X 2 -800 = xtanoo - 2{83.33)2 cos2 oo

2

R = Vo sin2e 2 10(5280) = 1500 sin[2e] 32.2 e = 24.540

-.. ·,

.£1. I 1&.¥4

R

2 sin 2 e H=-"-o- 2g

v

H = 699 feet

g

~9=3~

•

H=31.86m

2

t = 3.5 seconds

...

t

2(9.81)

2 300 (sin 45° H = ---'-----''2(32.2)

Ill

60 = O(t) + ~(9.81)t 2

...

2 50 (sin30

H=-~---'--

t

X= 1064.2

1 2 h = V0 t +-gt

=100

.. Am

1000 3600 V0 = 83.33 m/s

V0 = 300 X

H=10+h H=10+1.3 H =11.3 m

a·

2 2 sin_ H=-"-o_ 2g

V2 = V/ :...2gh

t,

J

32.2

BD v, =50/

8001

t'

V1 = V0 -gt1

300 sin [ 2( 45°)

-----:"~-'""'

R = 2795 ft x 1 yard 3 feet R = 932 yards

H

Yl=

g 2

R = 2795 ft

Vo = 83.33 ; -

6 .1 ..

h3

J

.

•··4"':.

.:v1. = 5 ...·i ;0·;:

l

1002 sin[ 2(30°)

9.81 R =833 m

V=O

•

R = V/sin2e

R = V/sin2e

R=

h 1

h2

T : · · · · ...·• t3 .~

!Vo=80

.•1

Let: R

v3 = 61.3 m/s

V1 1= o

. t1 ~ .

il

150 = V3 (9.155)-~(9.81)(9.155)2

= + 2(9.81)(40) V = 28m/s

.: :

2

·.

·,

iii. liiJ$$ q;;;

~

~ I

0 = 24032.4'

2

R = V0 sin2e g 2

300 sin [ 2( 45°) R=--32.2 R = 2795 feet

J

Note: l'he closest answer from the choices is 700 feet and 2800 ft.

•

504 .1 00·1 Solved· Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas

1,1 1.11

S:

·~ij

...

1111111~11

V=O

Let: t

.-·

t1

·.r

1:··'vo

7

v.

~

...,.

E 1 q.

0

·,

~·-~~111U~

l~. :~

v.

H= o

t 1 = 10.46 seconds

v.y -2gH

0=(300sin20") -2(9.81)H 1 2 H = --gt 2 2

=-i(9.81)t2

t 2 = 10.747 seconds Total time = 11 + 12

•

,· .·

t1

h = H-0.9

2

. t2 •.

V2x =Vox

~ Eq. 3 ·

0

v2x = 300cos20°

Vo

ti/v' (}\e

.

-

0...9

~2

2

2.15m

9.14- 0.066V02

4

x= 7 m

~i

83.54 -1.206V/ + 0.004V0 2 = 0.2025V (0.021V 0

0

2

-

0.9)

j

V=rro = 1.5(2)

N

!

v

V=3m/s

wv2

Fe=gr

IJ

'II II

Fe= 30.58 N

:II

LFtt.=O

I

•

l 2 =0.45v0.021V 0 -0.9

Squaring both sides:

:-----

~i

v2 = ~(v2S + (v2S

v2 3.05lm =10ft.

Fe

v2y

Fe= 50(3)2 9.81(1.5)

V2 =

9.14 = 0.066V0 + 0.45v0.021V~ - 0.9

o = 40"

...

v2x = 281.9 m/s t = t1 + 12

. 02 -9.14 = 0.066V0 + 0.45.j0.021V

H

•

·,

V2 y2 = 0- 2(9.81)(30 + 536.59) V2y = 105.43 m/s

"

Vo = 25.35 m/s

...

V2 y2 = V/- 2g(30 +H)

,·

'

·. .

0 = (300sin2o•f - 2(9.81)1-t

t2 = 0.45v'h

....................... ;,;·.·t-tj·.·:.;··

·,

H =536.59 m

h = 0.021V0 2 - 0.9 1 = 0.45J0.021V 2 -0.9

V=O

X

-2 = 60tan30o _ (9.81)(60)2 . 2(Vo )2 cos2 30•

,12

V/ = V0 / -2gH

h = -i(9.81)t/

Total time= 10.46 + 10.747 Total time= 21.2 sec

~

!

2

1 2 h = -gt2 2

2

•.;··:.,

'*' ... :·· ..

i

..·..

30

300 sin 40" H=----2(9.81) H = 0.021V0

H= 536.595 m

. ·,

. !<~.=-~~:. . . ..

e-

2

2

30 + 536.595

sin 2g

2

T .·, H

,.·v

Eq. 2

1

2

_..

Vay / ~

t 1 = 0.066V0

o = 300sin2o• - 9.81t1

V2 =

v1.=o

. .•· t1/

0 = V0 sin40°- 9.81t1

v =vov -gt

V0 =.9.03

V = VoY -gt1 V = V0 sin 0 - gt1

30

.l.

2

liiiiilll

·

(

.. . . . ¥-~~-~. :. ~. :. ~. ~->::: l Vo=?

83.54 -1.206V02 = -0.1 E

=V0 cos4o•t

..

4

=0.004V04 -0.

t = 9.14

h

.....():-. .~. = 2~·

L~,J

83.54 -1.206V0 2 + 0.004"

lf- = V0 coset H

r:::::1

= total time (t) of the flight

Day ZO- Engineering Mechanics (Dynamics) 505

~(281.9) 2 + (v2S

=301 m/s

:1 'I

F -Fe =0

!

P- 30.58 = 0 F=30.58 N F 30.58 Jl=-=-- N 50 J.1=0.61

:.II

I

il 'I

~ .

506 l 00 1 ·Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas

Ill

!l,.:ll 1~111 ,(',j

o> = 60 rev x 1 min 0

.ll!llll

rev m0 = 1 sec

min

.,ill,~

(

"""'"""''Ill lill ..

lllllii~J~

····"''"I

l

V=rm 15= 2.5m m =6 rad/s

60 sec

,'·f

:f I

Ill

Ill Note: The force exerted by the string is equal to the centrifugal force

wv2

Fe=-m =m0 -at

0 = 1-a(8)

180=0-a(6)

re:s

1<1o•1Ujll!ll'l~~l

a=0.125

,.~.,. ",1

m2 = mo 2 -2a9

a=30 - . 2

....

mtn

Fe

I

0 = (1)

"Uii·M~II!

•

2 -

m2 = mo2 +2a9 2

1802 = (1) + 2(30)()

2(0.125)9

9 = 4 rev or 4 turns

•

V=rm 1 rev 2n rad 24 hr ., 1 rev -5 rad

hr

-

3600 s

V = (4.5 X 107 )(7.27 X 10'5 ) V =3271 m/s V=3271m x ~ s 1000 m V =1.1777.4 kph

e =540 rev

X

3600 S -;,:-

9.61r r =265.41 m

=66.67 N

"' 67 N

Ill w

V = 65 mi X 5260 ft X ~ hr mile 3600 s V = 95.33 ft/s

=

N

m =m0 -at

•

9= 5.40

l:FH :.Q

re:s

a =2.62

Fe= F = tJN

Ill V= 90'k"'

l:Fv =0

Fe V

= 25 m/s

F.e---- wv2 f.LW=

X

1000m X~ km 3600 s

V=25 m/s

v2

gr

s V=-

hr

N=W Fe= tJW

r =0.5"m

Ell

tane=g; (25)2 tan9 = 9.6:1(200)

9 = 17.67,0

wv2 gr

t

0.3 = 152

V=30 2 V = 15 ft/s

6 r=. 2 +12

r = 2.5 ft

\

2

tan7° = 17.88

1.5

llii1l

52.36 =0- a(20)

s

gr

500 rev 1 min 2n rad m=---x--x-min 60s rev m 52.36 r~d/s

m=--X--X--

m=7.27 x 10

tan~=~

Fe= mV2 r Fe = 1(10)2

rev

V= 40mi x.5280ft X~ X~ hr mile 3.218 ft 3600 s V =17.88 m/s '2

gr

m =m0 -at

•

Day 20- Engine4!rin9 Mechanics (Dynamics) . 507 .

Fe= WV2 ~r

9.81r r = 76.5 m

Ell 'i i I

I

I '

508 ·100.1 Solved Problems in Engineering Mathematics (2nd Edition) slope= tan9

1"11

tan9

~~~

~

2 8.53° + cjl =tan·, , {88} 32.2(500)

=0.0825 e = 4.71°

Day 20- Engineering ~-~cha~tics (Dynamics) 509

by Tiong & Rojas V. = 50 mi X 5280 ft 0 1 mi _ hr V0 = 44 ft/s

I

I

X

....!...!!!:_

V = V0 +at

3600 s

16 = 0 +a(4)

Ill

L !'=inclined =0

J.l = tancjl

LFv =0

J.l = tan17.16°

REF =;:·F + Wsin9

J.l = 0.309

..•.,.,,I

I'

a

•

Fe= wv2 gr Fe= mV2 r Fe =ma

40(2000) a= 10(40) + 40(2000)sin1.146° a = 0.8049 ftls2

w

kz .............,..+a 'WJ-REF

-

LFH =0

v2

V2

i.JL

i&k&JL

0 = (44)

2

..,.

2(0,08049)8

Note: The nearest value from the choices is 1,204 ft.

Wa F· p = ---+ g

T=W+ Wa g T = 2000 + 2000{4) . 32.2 T = 2248.41b

•

+a

............

Ill

p = REF+F

T=W+REF

Note: The nearest value from the choices is 2250 lbs.

=V02 -2aS

s = 1203 ft

F = ~00 lbs.

a=' r 27 2 a=----' 800 a, =0.911 ft/s

.............,..-a

\

30000 ( 1.25) + 100 p = 32.2

N

p = 1556Jbs

2 a =a/ +a/

LFH =0

2

Note: REF means revers~ effective force by d'Aiemberts principle.

x 5280 ft x _h_r_

•

F=REF F=ma F = 1700(0.4)

a2 = (0.911)2 + (3) 2

Ell V = 60 mi hr V =88 ft/s

Wa =F+Wsine g

The coefficient of friction is less than 0.310

at

a= 3.1 f/s

a= 4 ft/s 2

cjl = 17.160

mi

..........

3600 s

-a

V=O

'\kf_Position of the body after 3 s LFinclined =0 Wsin9 =REF+ F Wsin9

F=680N

Ill.

T

tane = 0.25

sin30° = ~ + 0.3cos30° 32.2 a= 7.734 ft/s 2

a

2

Qf

tane = 0.02 9=1.146°

1

w

g

Wsine = Wa + J.1(Wcos9) g

e"' 8.53° tan (9 + $)::;: V

= Wa + J.LN

!

~EF

Let: V

=velocity after 2 seconds

S =distance traveled iii the third secol)c:Ywith reference from the position at the end of 2 seconds.

,I!

~

, I~ jl.'

.

510 1001 Solved Probtema in Engineering Mathematics (2'"1 Edition) by Tiong & Rojas

i

'[

''"~l,,l

v v

~I

.11~11

[ r····' "1 ~ ~~~~~u~ ....

l.~,.,,l

l . . :!

·:

V = V0 +at = 0 + 7.734(2) = 15.468

'l

',';: j

ftls

1 2 S =. Vt+-at 2 •

•

s = 15.468(1)+~(7~734)(11 s = 19.33 ft

s w REF

F N

LFH=O

I!

F=REF

w

J1N=-a g

w

J1W=-a

g

a= J19 a = 0.4(9.81)

R,~•·'<·~~O'~'

•(.'·O~+

~<~

-'•'~<'(••X~~'

I

~.~x.

I

a = 3.924 m/s2

~·•11"~:~''•><;v•;<;'

'

V=V0 -at 0 = 25- 3.924t t =6.37 s

"<<•'

(~''

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512 :1001 SolVeclProblems in ~S Mathematica(a"'i Edition) by Tions & Roiaa

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is Simple Stress?

~>imple stress is defined as the force per 1111it area. The typical units of stress are 2 MPa, lbf/in and ksi (thousand of pounds per square inch). The MPa is equivalent to MN/m?. or N/mm 2

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perpendicular to the area resisting the forces. Normal stress is also called as bearing stress. Shear stress, r, is the type of stress which is caused by forces acting along or parallel to the area resisting the forces. Shear stress is a!so called as tangential stress.

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:;trength of Ma'terials is the branch of :;cience that deals with the elastic behavior of load bearing engineering materials. :;tren9th of Materials is also known as the im1Ghanics of materials.

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Wl1at is_ Stre!]!Lth of Materi~l!!.1.

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Thu

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Strength of Materials Simple Stress Types of Normal Stress Simple Strain Hooke's Law $tress .. strain Diagram Thermal Stress Thin-Walled Cylinders

Area

1he two basic types of stress depending the orientation of the loaded area are •<~11ll:1l :,;iTess and shear stress .

~

'lll

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Nonnal ~;ilxl.'s~. o, IS the type of stres::; wh1< .II 1:. "·" l';~:d I lV forct"S actin<)

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p fl""A

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. .,,I,!, ,.lol!ll!ll~ll

I

Ultimate stress (or ultimate strength) refers to the highest ordinate in the stressstrain diagram.

where: G

't"""A

What are the Two Types of Normal Stress?

Strain,

What

0

E=l

Is Stress-Strain Diagram?

I

Allowable stress is the maximum safe stress which the material can carry.

II 1:,. ]•

Factor of safety is the ratio of the ultimate stress to allowable stress.

!

F

'I

D •••e

..••·"'

Mathematically,

E

0" OCE

Shearing strain is the angular change between two perpendicular faces of a differential element..

I,

Modulus of rigidity (G) -refers to the

I I

modulus of elasticity in shear.

What .is..Tit~!!~~ Thermal stress is the stress on the ~ material caused by the intern<;~! forces due

cr=EE

Tensile Stress

Compressive Stress

What is a Simple Strain? Strain refers to the elongation of the material with respect to its original length when subjected to a load. The typical units of strain are mm/mm, inch/inch or no unit at all. Strain may be expressed as a percentage of the original length.

The constant of proportionality, E is the modulus of elasticity of the material. It is also known as the Young's modulus. The Young's modulus was named after Thomas Young who introduced this constant of proportionality in 1807. Also,

P

Eo

A ""'L When the material is subjected to a shearing stress, as shown in the figure on the next page, ·

I

Stress

"Within elastic limit, the stress is proportional to strain." Area

I

material when loaded.

between the stress and the strain is known as the stress-strain diagram. The diagram below is the stress-strain diagram.

Hooke's Law formulated by Robert Hooke in 1678 is stated as follows:

i

Working str·ess is the actual stress of the

= shear modulus

A diagram that shows the relationship

What is Hooke's Law?

I'

Rapture strength is sometimes known as the stress ·at failure.

1:i=GS

+

p

The two types of normal stress are the tensile stress (forces away from the area) and the compressive stress (forces towards the area).

the materia!' even without any corresponding increase of load.

to change in temperature. The change in temperature can cause a change in length, area and volume of the materia!. The temperature deformation (linear) may be Strain

A -7 proportionality limit B -7 elastic limit C -7 yield point D -7 ultimate strength E -7 rapture strength F -7 actual rapture strength

calculated using

cr:::.~ .•. ·,:::=J, -······· S. ubje.·c.· ed. .t.

.,.r..·---~--

~.:.·t.•~mp. change .E

...

li

~--------~~

.

L

o1t

I

Elastic limit refers to the stress beyond which the material will not return to its original shape when the load is removed. fhe permanent deformation caused by excessive stress is called permanent set. Ylold point refers to the point where there Ill an appreciable elongation or yielding of

'•1!,

Br == aLt6.1') where:

a = coefficient of linear expansion

:1~ : I'

1111'

I

~~~~ 1,• 516 : 100 1 Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas

'"1~1

1.11 ~

•:~1

'Miat is a Thin-Walled Cylinder?

What is Torsion?

A thin-walled cylinder is a cylinder under

Torsion refers to the twisting of solid or hollow circular shafts.

pressure and has a thickness, t :::; inner radius.

r

1~ of its

Day 21 -Strength of Materials 517 where: T = torque applied L =length J = polar moment of inertia cross-section G = modulus of rigidity

I I

I

A. Shearing stress:

t

GOODLUCKI

-r:p

(-""'~1

'{ "'j

:::::·,,

P =27tf T

where: T = torque applied p = radial distance from the center of cross-section J = polar moment of inertia of the cross-section

,. ..1111111

.,.,.,J,, ',In'

B. Maximum shearing stress:

'I!IJI!II"Ii!lfll'lll

The thin-walled cylinders will experience tWI> different stresses, namely, tangential stress and longitudinal stress.

Max.f::;~$f·. 'J

where: P =power in watts T = torque in N-m f = frequency or' speed rps

What is a Helical Spring? A spring which forms a helix is called a helical spring.

A.

Maximum shearing stress:

where: r = radius of the cross-section T = torque applied J = polar moment of inerti~ of the cross-section

A. Tangential Stress: pD

16PR (1 3

t""

rcd

C. Maximum shearing stress of: where: p = pressure in N/m3 D = inside diameter in mm t = thickness in !Tim

or

+..!!_) 4R

pO

3

Another term for tangential stress is ciR:umferential stress, or hoop stress or girth stress. Note that the longitudinal stress is one-half the value of th~ tangential stress.

"Mathematics is often defined as the science of space and number ... not until the recent resonance of computers and mathematics that a more apt definition became fully evident: mathematics is the science of patterns!"

m

Maxit.~ 1&1··

where: P = axial load R = mean radius of helical spring d = diameter of rod/wire of spring m = ratio of the mean diameter of the spring to the mean diameter of the spring rod or wire

•>''"'\'

;_.'.~·'"

B. Hollow shaft

where: p pressure in N/m D = inside diameter in mm t = thickness in mm

4m-4

Did you know that... QED, the abbreviation of Quod Erat Demonstrandum, Latin for "which was to be demonstrated" was commonly used by · mathematicians to indicate that a conclusion has been reached, was first introduced by Euclid using its Greek equivalent in the 3'd century B. C. !.

- Lynn Arthur Steen

/c

O'L=M

16PR(4m-1 + 0.615) rcd 3

A. Solid shaft

·. . · ·.:-:;;r

B. Longitudinal Stress:

p:

·~tibia:

~uote:

:;:2t

=

Proceed to the next page for your 21st test. Detach and use the answer sheet provided at the last part of this book. Use pencil number 2 in shading your answer.

E. Transmit power, P

~~

CJT

of

Max. t·= . ~~:;;P '"'i .• t #''

2R

D

d

d

il,l!

m= - = where: d = inner diameter of shaft D = outer diameter of shaft D. Angular deformation,

'!I',II

B.

Spring deformation:

9:

. e....;r:~· -Ji·

1111111'111111

3

0 = 64PR4 n Gd

where: n = number of turns G = modulus of rigidity

Day 21 -Strength of Materials 519

·~

,,,a

~,

1.1

(' It

'st

Topics

0 0

·-'' '1

Mon

:::::·1, ""'!iilllli

Strength of Materials Simple Stress Types of Normal Stress. Simple Strain Hooke's Law Stress-Strain Diagram Thermal Stress Thin-Walled Cylinders Torsion He.lical Spring

I

. . .I,

Tue

0

,f'

Theory

-·"""

Problems

0 0

Solutions·

Notes

~ Wed

0 0 0

Thu

Fri

885: ME Board April 199ft If the ultimate shear strength of a steel plate is 42,000 psi, what force is necessary to punch a 0.75-inch diameter hole in a 0.625 inch thick plate? A.

B. C. D.

What force is required to punch a 1/2-inch hole on a 3/8 thick plate if the ultimate shearing strength of the plate is 42,000 psi? A. B. C. D.

A.

B. C. D.

12.75 mm 12.. 57 mm 17.75 rnm

15.75 mm

88:&: ME Board April1998 A force of 10 N is applied to one end of a

10 inches diameter circular rod. Calculate the stress.

A. B.

0.20 kPa 0. '15 I
C. D.

0.051{Pa 0.10 kPa

24.~0

24,620 24,960 24,740

B. C. D.

11.77 mm 13.18 mm 10.25 mm ·12.6 mm

1!:

I

891: EE Board October I990 A water reservoir of 24m high and 12m in diameter is to be completely filled with water. Find the minimum thickness of the reservoir plating if the stress is limited to 50 MPa.

887: A single bolt is used to lap joint two

A

steel bars together. Tensile force on the bar is 20,000 N. Determine the diameter of the bolt required if the allowable shearing stress is 70 MPa?

B. C. D.

!\. ll.

has an outer diameter of 200 mm and is subjected to a force of 74 kN. Find the thickness of the wall if the allowable compressive stress is 10 MPa.

890: EE Board Apri11996 A cylindrical water tank is 8 m in diameter and 12 m high. If the tank is to be completely filled, determine the minimum thickness of the tank plating if the stress is limited to .40 MPa.

A. 88&: ME Board October 1995

4.09 in~hes 3.96 inches

24.5 mm 28 mm 21 mm 26mm

89:t: EE Board April 1995

Sat

88lU IECE Board November 1998 An iron column of annular cross-section

63,000 68,080 61,850 66,800

C. D.

883: ME Board April 199ft A steel tie rod on bridge must be made to withstand a pull of 5000 lbs. Find the diameter of the rod assuming a factor of safety of 5 and ultimate stress of 64,000 psi.

A. B.

C. D.

0.75 0.71

0.84 0.79

884: EE Board October 1996 Determine the outside diameter of a hollow steel tube that will carry a tensile load of 500 kN at a stress of 140 MPa. Assume the wall thickness to be one-tenth of the outside diameter. 111.3 mm B. 109.7 mm C. '113.7mm D. 112.4 mm

A.

c. ll

17 mm 18 mm 19 mm 20mm

The stress in a 90-cm diameter pipe having a wall thickness of 9.5 em and under a static head of 70 m of water is

A 888: EE Board October 199& What is the stress in a thin-walled :;pherical shell of diameter D and a wall thickness t when subjected to internal pressure p? .

!\

S = D/pt

II.

s = 4D/pt

c:

S = pD/4t S = pD/t

()

889: ME Board April I998 ornpute the safe wall thickness of a 76.2 diameter steel tank. The tank is ·.cci>wr.ted to 7.33 MPa pressure and the ·.1•·•:1 cn
c

•Ill

!\ II

I I 1/ inches I W) 111dws

B. C. D.

ill:

325 kPa 32.5 kPa 32.5 MPa 3.25 MPa

II [,

893: ME Board October 1994 A cylindrical tank with 10 inches inside diameter contains oxygen gas at 2,500 psi. Calculate the required thickness in mm under a stress o.f 28,000 psi. A. B. C. D.

11.44 11.34 10.60 10.30

894: ME Board April 1:995 A solid shaft 48.2 em long is used for a transmi,ssion of mechanical power at a rate of 37 kW running at 1760 rpm. The stress is 8.13 MPa. Calculate the diameter.

I

II

'I~! ll;

520 :100 l Solved Problems in Engineering Mathematics ~2nd Editio~:l_!~fl~q & Roj~

l~~l

IJil,

g .....'"'J

" . '1 UHJIIill

A. B. C.

D.

r ---

A.

B. C. D.

1.512 x 10-s m4 1.215x10-am4 1.152x10-em4 1.125 x 1o-e m4

A. B. C.

D.

200 GPa 180.32 GPa 148.9 GPa 106.48 GPa


9001 ME Beard April 1:997

90S: EE Board Aprii19C)6

What is the modulus of elasticity if the stress is 44,000 psi and unit strain of 0.00105?

What power would a spindle 55 mm in diameter transmit at 480 rpm. Stress allowed for short shaft is 59 N/mm 2

A cylinder of diameter 1.0 em at 30°C is to be slid into a hole on a steel plate. The hole has a diameter of 0.99970 em at 30°C. To what temperature the plate must be heated? Coefficient of linear expansion for steel is 1.2 X 10" 5 em/" C.

A.

B. C.

....,,J,

30mm 35mm 40mm 50mm


D.

41.905 X 106 42.300 x 106 41.202x106 43.101 X "106

8961 ME Board October :1995 A 2-inch solid shaft is driven by a 36-inch gear and transmits power at i 20 rpm. If the allowable shearing stress is 12 ksi, what horsepower can be transmitted? A. 8. C. D.

29.89 35.89 38.89 34.89

sen: ME BOard October :199/i A hollow shaft has an inner diameter of 0.0:-15 m and an outer diameter" of 0.06 m. Compute for the torque in N-m, if the stress is not to exceed "120 MPa. A.

B. C. D.

4500 4100 4300 4150

8<~JI!I$ ME Board October 1996 Compute the nominal shear stress at the surface in MPa for a 40-mm diameter shalt that transmits 750 kW at 1500 rpm. Axial and bending loads are assumed negligible.

A. B.

c. 0.

218 312 232 380

A. B. C.

D.

42.12 kW 50.61 kW 96.88 kW 39.21 kW

A.

B. C. D.

90•: A 30-m long aluminum bar is subjected to a tensile stress of 172 MPa. Find the elongation if E = 69,116 MPa?

62°C 65°C 48°C 55°C

906: EE Board April :t99S A. B. C. D.

An iron steam pipe is 200 ft long at 0°C. What wi.ll its increase in length when heated to 100°C? Coefficient of linear expansion is 10 X 1o-s ft/"C.

0.746 m 0.007 m 6.270 mm 7.46 em

90:&: EE Board October 1'996 A steel wire is 4.0 m long and 2 mm in diameter. How much is it elongated by a suspended body of mass 20 kg? Young's modulus for~steel is 196,000 MPa.

A. B.

c. D.

0.18 ft Q. 12ft 0.28 ft 0.20 ft

9071 ECE Board November 1996 A.

1.123 mm

B.

1.385 mm 1.374 mm 1.274 mm

C. D.

A simple beam 10 m long carries a concentrated load of 200 kN at the midspan. What is the maximum moment of the beam?

903: A steel wire is t} m long, hanging

A. B. C. D.

vertically supports a load of 2000 N. Neglecting the weight of the wire, determine the required diameter if the stress is not to exceed 140 MPa and the total elongation is not to exceed 4 mm. E = 200,000 MPa.

908: ME Board Oc:tober 1993 A beam supported at both ends and carrying a uniformly distributed load:

A.

3.4 mm

B.

4. 4 mm

A.

C.

4.26 mm 5.4 mm

B.

D.

899: ME Boa!l'd O~tobe~ 1lHS

904: A copper rolled wire 10 m long and

A hollow shaft has an inner diameter of 0.035 mandan oute.r diameter of 0.06 rn. Determine the polar moment of inertia of the hollow shaft.

1.5 mm diameter when supporting ,,. weight of 350 N elongates 18.6 rnrn. Compute the value of the Youn!ir:·. .modulus of this wire.

250 kN-m 500 kN-m 400 kN-m 100 kN-m

C. D.

J

has its maximum bending moment at the supports has its maximum shear at the center of the beam has its maximum shear at the supports has uniform shear throughout the length of the beam

909: A simply supported beam, 10 m long carries a uniformly distributed load of 20 kN/m. What is the value of the maximum shear of the beam due to this load? A. B. C. D.

250 kN 100 kN 1000 kN 500 kN

910: A simply supported beam, 10 m long carries a uniformly distributed load of 20 kN/m. What is the value of the maximum moment of the beam due to this load? A.

B: C. D.

10,000 kN-m 5,000 kN-m 2,000 kN-m 250 kN-m

11

....

.

•.• H l~~ll

D

1!!!111 liiiiiill

= 0.2 m

I

p cr =kA where: k = factor of safety

,.

Topics

D D

·-"""

,......w~ll'~~

lvlon

...111l" .,.,.,1116

f:


liiilll

'll".

r,\: .

I

Tue

_.,,J,,

r

(g

D

_ ...Ill

Wed

Theory

D D D D D

64000 = (5)5000

Strength of Materials Simple Stress Types of Normal Stress Simple Strain Hooke's Law Stress-Strain Diagram Thermal Stress · Thin-Walled Cylinders Torsion Helical Spring

A A= 0.3906 in2

A=~d 4

A

=75000

10 X 106

....

A= 0.0075 m2 A= ~0 2 -~d 2

4 4

2

-

~d 4

4 d = 0.71 in

liiiill

4

0.0075 = ~(0.2)

2

0.3906 =~d 2

A

Thu

Problems

Uli :·

··!1

cr=~

D

2

d=0.1745 m

Fri

Solutions

Notes

Solving fort: D = d+2t

Sat

0.2=0.1745+2t

881. A 882. A 883. 6 884. D 885.C 886. D 887. 888. 889. 6 890.A

c c

891. 6 892. D 893. 6 894. D 895.A 896.6 897. A 898. D 899. D 900.C

901. D 902.0 903. 6 904. D 905. D 906. D 907.6 908.C 909. 6 910. D

.....

RATING

ANSWER KEY

t = 12.75 mm

[:} 18-24 Passer

140 X 106 = 500000

d= 10

15-17 Conditiona.l

. In

1ft X 1m - . . x 12 in 3.281 ft

d=0.254

[:} 0-14 Failed

m


7t 2 7t ( A= -:0 - - D- 2t )2

=A

cr=

4

p

~d2

4

4

A =·0.282702 10

cr=---

0.'00357 = 0.2-8270 2

~(0.254)2 4

D = 0.1124 m

cr

J

4

7t( A= -D 2 - D·-2(0.10) )2 7t

4

I

A A =0.00357 m2 Note: t = 0.1 D

p

cr

cr=~ A

liiilill

[:} 25-30 Topnotcher

0

t = 0.01275 m

= 197.35 Pa

"' 0.2 kPa

D;, 112.4 mm

lill, 1'\11' 1

Day 21 - Strength of Materials SZS


'•••"•'

t = 3.89 mm

·-·1'' .....,,,,

A

Ill

p = 61850 lb

l!tiB

liiiiilll

p = 9810(70)

cr,

p = yh

2cr1 p

2(40

A

X

.10

6

)

p

t = 0.4464 in x 2.54. em X10 · ·mm --

ln

where d = diame~er of the bolt 70 X 106 = 20000

~(d2) d=0.019m d=-19mm

~

Ill pD

pD crL = 7 Formula 4t

p = 2nfT 60

Note: The biggest pressure occurs at the bottom of the tank

37000 = 2n(1760)T 60 T = 200.75 N-m

p = yh

7td3

8.13 X 10& = 16(200.75)

where: k = factor of safety

120 x 106 =

em

d=0.050 m

6

d=50 mm

2(.50 X 10 )

t = 0.028 m t= 28 mm

16TD n(D4 - d") 16T(0.06) 7t[(0.06)4 - (0.035)4

1!9!!11 liii6ia p

= 27tfT 60

750000 .

27t(1500)T 60

16T t=-

7td3

7td3

t = _23,...-54_4_0(,_12_,_}

crt=kpD . 2t

t-

16T t=-

2cr1

liiiiilll

•

33000(12) p = 35.89 hp

T =4774.648 N-m

p = 235440 Pa t = pD

lf!I!R

2nfT. 33000(12)

T =4500 N-m

1m

2t

p = 9810(24)

liiiiilll

=

t = 11.34 em CJr =

I!

p = _2n~(1_2-'0)'-'--(1_8_84_9_.55.....!._)

28800- 2500(10) 2t

t = 11.77 mm

~(d2)

16 T 3 n(2)

T = 18849.551b-in

pD

t = 0.01177 m

p CJ=--

686700(0.9} 2(0.095}

CJr = 2t

t = _1,.-17_72_0_,_(8-::'-;) CJ = -

=

12000 =

Ill

t = pD

J!IP.II!W Uilll

2t

cr1 = 3.25 MPa

p = 117720 Pa

p = 24740 lb

nd3

cr1 = 3252789.474 Pa

p = 9810(12}

p 42000 = n(0.5)(0.375)

E=41.905x106 Pa

= pD I

Note: The biggest pressure occurs at the bottom of the tank

A

II; I:

16T t=-

cr

p CJ=-

1::

liiiill

p = 686700 Pa

cr, = pD 2t

p 42000 = n(0.75)(0.625)

1.:"

E = 44000 0.00105

..,.

p = yh

1m

p

CJ = -

1111

E=~

Note: The biggest pressure occurs at the bottom

t = 0.0389 m -~

'

]I' :'

&

2t

Cl: --~~~~~~-

=21

215.4 X 106 = 3(7.33 X 10 )(0.762)

'~~-:

.rill

Ill

pD

CJr 6

~,

•t~~JOII·II

11.1

p A

CJ=k-

111

I I

~

t=

16(4774.648) 3

7t(0.04) t=380 MPa

J

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SZ& :100 1 Solved Problems in Engineering Mathematics (2"d Edition) by Tiong & Roip

'"'"

'" H

Ill

4

4

2

.

I)=

16T 7td3

t=-

59=~3 n(55)

=1927391.637 N-mm

T = 1927.39N- m

p

•

15 = 1.274 x 10-3m 15 = 1.274 mm

140

ID

X

d =0.00429 m d=4.26mm

AE p aS=A

~

AE where: P = weight of the body P mg 20(9.81) P 196.2 N

= = =

2000(6)

d = 0.0044 m d= 4.4 mm

15 = aL

E 8 . I)= (172 X 10 )(30) 69,116x106 8 =0.0746 m &:7.46 m

=

Note: To be safe for both stress and elongation, use d = 4.4 mm.

•

E = 106.48 MPa

ill

:':ll

10m

HI:

R2

L:Fv =0 R1 + R2 = 200 2R1 = 200 R1=100kN

a

Maximum shear= 100 kN

Refer to solution in Problem # 909: Consider the half of the beam:

I

~

Consider the half of the beam:

max. moment= R1(2.5)

1:11111 .,

Iiiii

R1=100 kN

max. r:noment = (100)(2.5} max. moment = 250 kN-m

:;- ::fc 1

max. moment= LMc max. moment= R1(5) max. moment== {100)(5)

4

E = 1.0648 x 1011

R1

max. moment= LMc

~::::::::.·:

~(0.0015)2

~

Note: Maximum'tnoment.of a symmetrically loaded system is at midspan.

200kN

l>= PL AE 350(10) .....-0.0186 = E

]I!

20(10)=200 kN

20(5)=100 kN

R1 + R:;i = 200 2R1 = 200 R1=100kN

~d2 (200000 X 108)

Substitute Eq. 2 in Eq. 1:

15 = PL

LFv =0

AE

0.004

. ;r~~ ; :=;;~;t

Note: Since the load is at midspan, then the reaction at both supports must be equal.

15 = PL

Eq.2

~

,I'!"1 'I

fi~J:P!Jj

t 1)

200kN

t'~:::.

4

-

Strength of Materials 527

Maximum shear of a symmetrically· ioaded system is equal to the reaction at the supports.

01 = 10 X 10-6(200)(100° -0°) 81 = 0.20 ft

106 = 2000 !:d2

-+ Eq. 1.

Or= aL(t 2

Ill

Considering limitations of elongation:

l5 = PL

t 1)

t 2 = 55°C

•

196.2(4) 3.1416 X 10--6(196,000 X 108 )

A

=~ltfr

=aL(t2 -

0.0003 = 1.2 X 10'5 (1)(t 2 -30°)

Considering limitations of stress: p y=-

60 p = 21t(480)(1927.39) 60 P=96.88 kW

a

81

A=3.1416x10.s m2

J=1.125x10-6 m4

T

ST = 1- 0.99970

81 = 0.0003 em

A = *(0.002)

)

• •

Day 21

A=~d 2 4

J = _!t_(D4 - d4) 32 J= ;2[(0.06) -(0,035}

•

~~

Pa

I

J

max. moment

500 kN-m l

I 11

1

'f··'l 530 100 f Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas

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E~:onomics is a sc1ence which deals with !he attainment of the maximum fulfilment of society's unlimited demands for goods and service. ·

V¥.1:!J!1Js Engineering

Economics?

..,~~·-··

•i .. I>.;-_... < '- ,, .\ " ' ' ;.,

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Engineering Economics is the branch of t!conomics which deals with the .1pplication of economics laws and theories IIJVolving engineering and technical r 11 ojects or equiprnents. IIVIJ
~

,.

'·, _.

!i

II

IIi

Compound Interest Continuous Cornpounding Nominal and Effective Rate of Interest

··-~ <1 ... "

.w.!•at is E_@.!lOmics? ~

Engineering Economics Consurner & Producer Goods and Services Necessity and Luxury Different Market Situations Demand· Supply law of ~upply and Demand Simple -Ordinary Simple interest - Exact Sirnple Interest

consumer goods and services refer to 1/w 1>1oducts or services that are directly 11·:.·d l1y pt~opl" to satisfy their wants. I ·.. JIIlpl, .. ; :lit' food. dotltlrH), shc!ter or lit liiH' l"IJ

Producer goods and services are those that are used to produce the consumer goods and services.

What.l§.l.b!L.Q1fi~I!LIJCe betwemJ.

!'!feces*illY...¥!nr!_Lu~.!Y1

'

Goods and services are divided into two types, namely necessity and luxury. Necessity refers to the goods and services that are required to support human life. needs and activities.

Necessity product or staple product is defined as any product that has an income-elasticity of demand less than one. This means that as income rises, proportionately less income is 'spent on such products. Examples qf necessity products include basicfoodstuffs like bread and rice, clothing, etc.

I, ' 'I~

"

I

;I

If!

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Day_2_g- Engtneering Economics (Simple and_ Compoun
532 l 00 i Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas

l,illt,,, 1.11.,1111

~.,,',,

'

jljll

Luxuries are those goods and services that are desired by human and will be acquired only after all the necessities have been satisfied.

The price of any commodity or product will depend largely on the market situation. The following is a tabulation of the different market situations:

r,,.

_.,.J

Luxury product is defined as any product that has an income-elasticity of demand greater than one. This means that as income rises, proportionately more income is spent on such products. Examples of luxury products includes consumer durables like electric appliances, expensive cars, holidays and entertainment, etc. Necessities and luxuries are relative terms because there are some goods and services which may be considered by one person as necessity but luxury to another person. For example, a man living in the· Metropolis finds a car as an absolute necessity for him to be able to go to his workplace and back to his home. If the same person lived and worked in another city, less populated with adequate means of public transportation available, then a car will become a luxury for him.

A.

Few seller and many buyers: The bulk of market supply is in the hands of a relatively few sellers who sell to many small buyers.

B.

Homogeneous or differentiated products: The products offered by the suppliers may be identical or more commonly, differentiated from each other in one or more respects. These differences may be of a physical nature, involving functional features, or may be purely "imaginary" in the sense that artificial differences are created through advertising ahd sales promotion.

C.

Difficult market entry: High barriers of entry which make it difficult for new sellers to enter the market.

Monopoly is the opposite of perfect competition. This market situation is characterized by the following: Perfect competition (also known as atomistic competition) refers to the market situation in which any given product is supplied by a very large number of vendors and there is no restriction against additional vendors from entering the market. Perfect competition is a type of market situation characterized by the following:

A.

What are the Different Market. Situations? The term "market" refers to the exchange mechanism that brings together the sellers and the buyers of a product, factor of production or financial security. It may also refer to the place or area in which buyers and sellers exchange a weli-defined commodity.

B.

Buyer or consumer is defined as the basic consuming or demanding unit of a commodity. It may be an individual purchaser of a good or service, a · household (a group of individuals who make joint purchasing decisions), or a government.

C.

Seller is defined as an entity which makes product, good or service available to buyer or consumer in exchange of monetary consideration.

Absence of all economic friction: There is a total absence of economic friction including transport cost from one part of the market to another.

This market situation provides an assurance of complete freedom on the part of both the vendors and the buyers though the latter benefits more from the reduced prices brought about by the competition while more and better services are afforded by the vendors or players in the industry .

1~11

..... 1111111

E.

D.

Many sellers and many buyers: Since there is a large number of selleos and a large number of buyers, each seller and buyer will become sufficiently small to be unable to influence the price of the product transacted. Homogeneous prodm:ts: The products offered by the competing sellers are identical not only in physical attributes but are aiso regarded as identical by the buyers who have no preference between the products of various producers. Free ma1·ket-entry and exit: There are no barriers to entry or impediments to the exit of the existing sellers.

Perfect information: All buye!s am! ail sellers have complete inforrn''il.ion · on the prices being asked and otlemd in all other parts of the mmlwt ·

A

One seller and many buyers: A market comprised of a single supplier selling to a multitude of small, independently-acting buyers.

B.

Lack of substitute products: There are no close substitutes for the monopolist's product.

C.

Blockaded entry: Barriers to entry are so severe that it is impossible for other sellers to enter the market.

There exists a perfect monopoly if the single vendor can prevent the entry of all other vendors into the market. The monopolist is in the position to set the market price.

What is a Demand? Demand is the need, want or desire for a product backed by the money to purchase it. In economic analysis, demand is always based on "willingness and ability to pay" for a product, not merely want or need for the product. The demand for a product is inversely proportional to its selling price, i.e. as the selling price is increased, there will be less demand for the product; and as the selling price is decreased, the demand will increases. Figure 1 below illustrates the relationship between price and demand.

A natural monopoly is a market situation where economies of scale are so significant that costs are only minimized when the entire output of an industry is - supplied by a single producer so that supply costs are lower under monopoly than under perfect competition and oligopoly. Oligopoly exists when there are so few suppliers of a product or service that the act1on of one will inevitably result in a ~•imilar action by the other suppliers. This tyrt~ of mark0t sJtuation is characterized by till' loiiOWIII
Price

. .: jp1

L

·········~··

2

........................ !

·······6~

!

!P2 !

j

Figure 1

Demand

I':,I '-~

§.!1..Jgo·-I Solved Problems in .Engineeri;l!~athematics (2r.d Editionl.PY 'I'iong & ~jas

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rwh

Assuming a linear function for the relationship between price and demand, it shows that at point 1, the selling price, P, is high, thus there is less demand for. the product as compared to point 2, where 'the demand, Dz is great because of lower selling price, Pz. What is a Supply? Supply is the amount of a product made available for sale. If the selling price for a product is high, more producers will be willing to work harder and risk more capital in order to reap more profit. However if the seiling price for a product declines, capitalists will not produce as much because of the smaller profit they can obtain for their labor and risk. Therefore, the relationship between price and supply is that they ,are directly proportional, i.e. the bigger the selli1·1g price., the more the supply; and the smaller the selling press, the less is the supply. Figure 2 below illustrates the price-supply relationship.

Price

. . . . . . . . . . . . . .Sz···························f/ .. / l

1

/

[

·······~·1···

p2

"Under conditions of perfect competition, the price at which any given product will be supplied and purchased is the price that will result in the supply and !he demand . being equal." The relationship between price-supplydemand may be illustrated by combinirlfj figures 1 and 2. For general application, the curves for supply and demand are no longer represented in linear function.

_ _

j

What is the Law of S!ill.P_tl and D_emand? The law of supply and demand is stated a~ follows:

When there is additional demand without an additional supply, a new and higher price is established as shown in Figure 5. Price

~~'!.I

i

~

d n=-

360

.....

The simple interest is:

···•····...

I

!

I

tl....---;::-------

···· ······

uUOOIV

.

=Demand

•

l "' Demand ,.,·! ----·--·Units

Figure 3 illustrates the price-supplydemand relationship, showing equal supply and demand at a given price. When there is additional supply without an additional demand, a new and lower price is established as shown in Figure 4.

Previous Supply

New Supply

~~JJ

._ _ _ _ ,.

t---- · -------·Figure 4

Figure 5

-

What is an Interest?

Figure 3

Price

•·····... Previous Demand

Units

!'""'

r,t··/···············~

Units

Ordinary simple interest is based on one banker's year. One banker's year is equivalent to 12 months of 30 days each. Also, 1 banker's year = 360 days. The value of n is:

+

Price I

~Supply

Figure 2

Day 22 - Engineering_J::<.:O_I\OIDi<:§._@J!lj)l_e and Compound Interest) 535

Price

/ !pj

::f

Suppose a debtor loans money from a creditor. The debtor must pay the creditor the original amount loaned plus an additional sum called interest. For the debtor, interest is the payment for the use of the borrowed capital while for the creditor, it is the income from invested capital. What is Simple Interest? Simple interest (I) is defined as the interest on a loan or principal that is based only on the original amount of the loan or principal. This means that the interest charges grow in a linear function 'bver a period of time. It can be calculated using the formula I= Pin where: P = principal i = interest per period n number of interest period

=

!lil

',·_p·· d - 1-

'I

360

Exact simple interest is based on the exact number of days in a given year. A normal year has 365 days while a leap year (which occurs once every 4 years) has 366 days. Unlike the ordinary simple interest where each month has 30 days, in the type of simple interest, the number of days in a month is based'on the actual number of days each month contains in our Gregorian calendar.

1

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''I To determine the year whether leap year or not, one has to divide the year by 4. If it is exactly divisible by 4, the year·is said to be leap year otherwise it will be considered just a normal year with 365 days. However, if the year is a century year (ending with two zeros, e.g. 1700, 1800 ... ), the ye;3r must be divided by 400 instead of 4 to determine the year whether or not a leap year. Hence, year 1600 and year 2000 are leap years. Under this method of computation of interest, it must be noted that under normal year, the month of February has 28 days while during leap years it has 29 days. Again, the values of n to be used in the preceding formulas are as follows:

il

'II !I ,,

:'1' I

I

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For ordinary or normal years: There are two types of simple interest, namely ordinary simple interest and exact ~;irnple interest.

d n = 365

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Day 22- Engineering_~~nomics (Simple and Compound·Interest) 537

536 , 100 l Solved Problems i~ E!!,qineering Mathematics (2nd Edition). by Tiong &: Rojas

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d n= 366

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What is a Discount? Consider the following case:; wnere discount is involved:

Compound interest is defined as the interest of loan or principal which is based not only on the original amount of the loan or principal but the amount of loan or principal plus the previous accumulated interest. This means that aside from the principal, the interest now earns interest as well. Thus, the interest charges grow exponentially over a period of time.

I

Using the formula for nth term of a G.P. an= atfn-1 an = P(1 + i)(1 + i)n-1

CASE 2:

Discount "" Future worth~ present worth

' The rate of discount is the discount on one unit of principal per unit time.

Principal

Interest

1

p

Pi

P(1 + i)

P(1 + i)i

-2

P(1 + i) 2

P(1 + i)2 i

,,,,, ''.

Total Amount P+ Pi= Pl1 + i) P(1 + i)(1 + i) = p (1 + i)2 P(1 + i)2 (1 + i) =P(} +

let x =..!!!..

NR

p( +

F=

"

1 .;rNR)N

I tt,l'''' ,,

F=P[(1+.;JrR)N

but

Lim (1 + x~oo

.!)x =e x

therefore,

B.

The tabulation above shows that the future amount (total amount} is just the value P(1 + i) with an exponent which is numerically equal to the period.

PRESENT WORTH, P

.... I 0

It is also observed that compound interest is based on the principles of geometric progression and using such method, the total amount after each period are as follows: a,= P(1 + i) First term, 2 Second term, a2 P(1 + i) 3 Third term, a3 P(1 + i)

p

1

2

3

.......................................... :

. F ,.,p~<"~ :~;~;/,f

n

f:t:r

=principal NR =nominal rate

where: P

e =2.71828 ...

= number of years = ccintinuous compounding

N e
compound amount factor

P-

=

F

- (1 + i)n

=

a ,

NR)mN

F=P ( 1+m

''~

where: P = Principal i = interest per period (in decimal) n = number of interest periods (1 + i)n = single payment compound amount factor

J2

r = -1. a,

but for !Tl periods per year

I

w~ere:

Solving for the common ratio,

d '1=:::-'1-:-d

..,,, ,, ,

F = P(1 +·i)"

F =P(1+it

and so on.

.

n

3

Cash Flow

j..

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1

t:;i~,:~:J .......................................» ~'1f};~:

Let d = rate of discount

1 d=1-- ·thus 1+ i '

I... . 0

The future amount of the principal may be derived by the following tabulation:

3

In a bank loan, a person borrows P 100 for 1 year with an interest of 10%. The interest as computed is P 10. The bank deducts the interest, which is P10 from P 100 and gives the borrower only P 90. The borrower then agreed to repay P100 at the end of the year. The P1 0 that was deducted represent the interest paid in 'advance. In this case, discount also represent the difference between the present worth (i.e. P90} and the future worth (i.e. P 100}.

FUTURE AMOUNT, F:.

Compound interest is frequently used in commercial practice than simple interest, more especially if it is a longer period which spans for more than a year.

Period

The basic equatioh for future worth of compound interest is

an =P(1+i)n

A.

that is not due yet but payable in some future date, desires to exchange it into an immediate cash. In the process, he will accept an amount in cash smaller that the face value of the bond. The difference between the amount he receives in cash (present worth) and the face value of the bond or financial security (future worth) is known as discount. The process of converting a claim on a future amount of money in the present is called discounting.

The concept of continuous compounding rs based on the assumption that cash payments occur once per year but compounding is continuous throughout the year.

r = 1+i

CASE 1:

A person has a bond or financial security

What is Continuous Compounding?

P(1 + i)2 r=-P(1 + i)

What is Compound Interest~

For leap years:

j

( : i)" = single payment present 1 worth tactor

The present worth of continuous compounding is

F

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1 :j 538 · 1001 Solved Problems ii). Engineering Mathematics (2nd Edition) by Tiong & Rojas

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What is the Difference Between Nominal and Effective Rate of Interest?.

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Rate of interest is the cost of borrowing money. It also refers to the amount -earned. by a unit principal per unit time.

nominal rate are equal if the mode of compounding is per annum or annually. I

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There are two types of rates of interest, namely the nominal rate of interest and the effective rate of interest. Nominal rate of interest is defined as the basic annual rate of interest while effective rate of interest is defined as the actual or the exact rate of interest earned on the principal during a one-year period. For example: A principal is invested at 5% compounded quarterly, In this statement, the nominal rate is 5% while the effective is greater than 5% because of the compounding which occurs four times a year. The following formula is used to determine the e.ffective rate of interest:

ER = [1+it-1 where: m = number of interest period per year i = interest per period . NR 1=-

m

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·.. i

Proceed to the next page for your 22nd test. Detach and use the answer sheet provided at the last pa.rt of this book. Use pencil number 2 in shading your answer. GOOD LUCK I

Mon

'Urribia:

0

I

Theory

ER=[1+NmRr-1

Substituting the values of m and i:

ER =[1+

ER =0.0509 ER =5.09% So, the actual interest rate is not just 5% but 5.09%. However the effective rate and·

0 I 0 Sat

'I'

I

0 0

Fri

II

Ii 'I

II! l

i

ME Board April 1.995 P 4,000 is borrowed for 75 days at 16%

C}U:

per annum simple interest How much will be due at the end of 75 days?

c

P4,133.33 P 4,333.33 p 4,166.67 P 4,150.00

· 9I::t: CE Board May 1.997 fl deposit of P 110,000 was made for 31 days. The net interest after deducting 20% withholding tax is P 890.36. Find the rate of return <:mnually.

0 5

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1

i

D. Note: i = NR if the mode of compounding is annually

Consumer & Producer Goods and Services Necessity and Luxury Different Market Situations Demand Supply Law of Supply and Demand Simple Interest - Ordinary Simple Interest -Exact Simple Interest Discount Compound Interest 1 Continuous Compounding . Nominal and Effective Rate of I Interest

~ Thu

Notes

-Goethe

I Engineering Econ~mics

,l•li

Problems

Solutions

~uote: "Mathematicians are like Frenchmen; whatever you say to them they translate it into their own language and forthwith, it is something entirely different."

Wed

,_

A. B.

or

0 0 0

Tue

Did you know that ... the Gregorean Calendar we are using was named after a former teacher of law at the University of Bologna, Ugo Buoncompagni who became Pope Gregory XIII in 1572! In February 24, 1582, he issued a Papal edict directing the former Julian Calendar be allowed to catch up with the Lord's Time and that aside from leap year every four years, leap year be once in every four centennial years, i.e. every 400 years.

!1111

Topics

c

11.95% 12.75% 11.75 o;,,

IJ.

1/.?5 'X,

fl. U.

913: ME Board Aprill993 Agnes Abanilla was granted a loan of P 20,000 by her employer CPM Industrial Fabricator and Construction Corporation With an interest of 6 % for 180 days on the principal collected in advance. The corporation would accept a promissory note for P 20,000 non-interest for 180 days. If discounted at once, find the proceeds of the note. A. B. C. D.

P 18,600 P 18,800 P 19,000 19.200

II

IIIII

,,,:I

91.4: ECE Board November 1.998 What will be the.future worth of money after 12 months, if the sum of P 25,000 is invested today at simple interest rate of 1% per month? A.

p 30,000

11:

!


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B.

C. D.

"'~"'

P 29,000 P 28,000 P 27,859

money was borrowed and the loan is payable at the end of one year. How much is the actual rate of interest.

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9S5: ECE Board November S999

A.

If you borrowed money from your friend with simple interest of 12%, find the present worth of P 50,000, which is due at the end of 7 months.

B.

C. D.

~.~~~:,.

. ;li.

12% 14% 10% 19%

C. D.

P46,200 P 44,893 P 46,730 P 45,789

9Sft: Annie buys a television set from a merchant who ask P 1,250 at the end of 60 days. Annie wishes to pay immediately and the merchant offers to compute the cash price on the assumption that money is worth 8% simple interest. What is the cash price? A. B.

c. D.

P 1,233.55 P 1,244.66 p 1,323.66 P 1,392.67

917: ME Board April1998 It is the practice of almost all banks in the Philippines that when they grant a loan, the interest for one year is automatically deducted from the principal amount upon release of money to a borrower. Let us therefore assume that you applied for a loan with a bank and the P 80,000 was approved at an interest rate of 14 % of which P 11 ,200 was deducted and you were given a check of P 68,800. Since you have to pay the amount of P 80,000 one year after, what then will be the effective interest rate? A. B.

c. D.

15.90% 16.28% 16.30% 16.20%

p 408.00

B.

P415.00 P 551.00 P 450.00

C. D.

D.

A. B.

C. D.

14.49% 12.36% 14.94% 14.88%

c.

D.

The amount of P 50,000 was deposited in the bank earning at interest of 7.5% per annum. Determine the total amount at the end of 5 years, if. the principal and interest were not withdrawn during the period? A. B.

930: ME Board April199~

c. D.

7.71% 7.22 15.7a %, 21.81 %.

°/:

92&: ECE Board April1998

The amount of P 12,800 in 4 years at 5 % compounded quarterly is A.

19.25% 19.48% 18.46% 18.95%

P 71,781.47 P 72,475.23 P 70,374.90 P 78,536.34

An interest rate is quoted as being 7.5% compounded quarterly. What is the effective annu!'ll interest rate?

B.

B. C. D.

!II: I,

I

P 14,785.34 P 15,614.59 P16,311.26 P 15,847.33

Alexander Michael owes P 25,000.00 due in 1 year and P 75,000 due in 4 years. He agrees to pay P 50,000.00 today and the balance it) 2 years. How much must he pay at th'ti"end of two years if money is worth 5% compounded semi-annually? A. B. C. D.

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P 38,025.28 P 35,021.25 P 30,500.55 P 39,021.28

93S: ECE Board November 1998 At an interest rate of 10% compoundec:l annually, how much will a deposit of P · 1,500 be in 15 years?

927: ECE Board April1999 Find the present worth of a future payment of P 100,000 tobe made in 10 years with an interest of 1.2% compounded quarterly.

Mandarin Bank advertises 9.5 %account that yields 9.84 % annually. Find how often the, interest is compounded.

·B.

P 6,500 P 8,600 P 5,500 P 7,500

D.

A.

3% 13.2% 12% 12.55%

922:ME Board October 1995, EE Board October 1997

A.

A. B. C. D.

C.

What is the corresponding effective rate of 18% compounded semi-quarterly?

D.

=

925: ME Board October 199&

Proble~

C. 918: EE Board April 199ft

The effective rate of 14% compounded semi-annually is

921: ECE Board April!.999

A. B.

1

inheritance as of the boy's 6 h birthday, if the interest is compounded annually? Assume i 4%.

929: ECE Board April!.999


B. C. D.

4.06% 1.00% 2.04% 3.36%


A man borrowed P 100,000 at the interest rate of 12% per annum, compounded quarterly. What is the effective rate?

A.

A bank pays one percent interest on savings accounts four times a year. The effective annual interest rate is

c.

A bank charges 12 % simple interest on a P 300.00 loan: How much will be repaid if the loan is paid back in one lump sum after three years?

A.


A. B.

919: ME Board April1998 A. B.

Day 22- EnginE!E!ring Economics (Simple and Compound Interest) 541

d~

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1111

','? •. .. '·

A. B.

C. D.

Daily Monthly Bi-monthly Quarterly

A

P 30,444.44

B. C.

p 33,000.00 P 30,655.68 P 30,546.01

D.

928: EE Board June 1990 On his Gu' birthday a boy is left an

A man borrowed P 20,000 from a local commercial bank which has a simple interest of 1So/o but the interest is to be deducted from the loan at the time that the

11\herit.mce The inheritance will be paid in 51 ;t lump swn of P 10,000 on his 21 lllrlhd<~y Wh
~J

P6,100.0U P 6,234.09 P 6,265.87 P 6,437.90

932: CE Board May !.995 How long (in year's) will it take money to quadruple if it earns 7 % compounded semi-annually?

A. B.

c. D.

20.15 26.30 33.15 40.30

II

I

542 l 00 1 Solved Problems in Engineering Mathematics {2"d Edition) by Tkmg & Rojas

l

Day 22 - Engin~~~!l9. Economics (Simple artd Compound Interest) 543

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9331 ECE Board April :1999

A.

P 1,925.00

In how many ye.ars is required for P 2,000 to increase by P 3,000 if interest at 12% compounded semi-annually?

B. C.

P 1,860.00 P 2,345.00

A.

8

B.

9

D.

7

c.

10

D.. P 2,160.00 938: CE Board May 2:996 P 200,000 was deposited on January 1, 1988 at an interest rate of 24 % compounded semi-annually. How much would the sum be on January 1, 1993?

934& ME Board April :1.996

Consider a deposit of P 600.00 to be paid back in one year by P 700:00. What are the conditions on the rate of interest, i% per year cOmpounded annually such that the. net present worth of the investment is positive? Assume i 2 0.

A. B. C. D.

0 :=; i < 14.3 % 0 :=; i < 16.7% 12.5% :=; i < 14.3%

A. B.

P 401,170 p 421,170

C.

P 521,170 P 621,170

D.

939: CE Board November :1996 If P 500,000 is deposited at a rate of 11.25 % compounded monthly, determine the compounded interest after 7 years and 9 months.

B.

P 660,550 P 670,650

A company invests P 10,000 today to be repaid iq 5 years in one lump sum at 12 % compounded annually. How much profit in present day pesos is realized?

C. D.

P 680,750 P 690,850

p 7,632 P7,236 P 7,326 P 7,362

936a ME Board April :1996 A firm borrows P 2,000 for 6 years at 8 %. At the end of 6 years, it renews the loan for the amount due plus P 2,000 more for 2 years at 8 %. What is the lump sum due?

A. B. C. D.

p P P P

5,355.00 5,892.00 6,035.00 6,135.00

9371 ME Board October :1996 A deposit of P 1, 000 is made in a bank account that pays 8 % interest compounded annually. Approximately how much money wilL be in the account after 10 years?

for eight years, at which time the principal is withdrawn. The interest has accrued is left for another eight years. If the effective annual interest rate is 5 %, what will be the 1 withdrawal amount at the end of the 16 h year?

at an interest rate of 7% compounded semi-annually. How much is the sum now?

A. C. D.

p 706.00 P 500.00 P 774.00 P 799.00

943: ME Board April :1998 P 1,500.00 was deposited in a bank account, 20 years ago. Today it is worth P 3,000.00. Interest is paid semi-anr1Ually. Determine the interest rate paid on this account.

B.

C. D.

3% 2.9% 3.5% 4%

944: ME Board April :1998 Fifteen years ago P 1,000.00 was deposited in a bank account, and today it is worth P 2,370.00. The bank pays interest semi-anf1ually. What was the interest rate paid in this account?

A merchant puts in his P 2,000.00 to a small business for a period of six years. With a given interest rate on the investment of 15 % per year, compounded annually, how much will he collect at the end of the sixth year?

A. B. C. D.

A. B. C. D.

9401 ME Beard October :1996

3.8% 4.9% 5.0% 5.8%

94:11 ME Board April :1998 If P 5,000.00 shall accumulate for 10 years at 8 % compounded quarterly, find the compounded interest at the end of 10 years. A. B. C. D.

P 6,005.30 P 6,000.00 P 6,040.20 P 6,010.20

A. B. C. D.

P 4,400.00 P 4,390.15 P 4,200.00 P 4,626.00

945: ECE November 1998 . A man expects to receive P 25,000 in 8 years. How much is that money worth now considering interest at 8% compounded quarterly?

A. B.

c. D.

p P p P

13,859.12 13,958.33 13,675.23 13,265.83

p 2,000,000 P 2,000,150 P 2,000,300 P 2,000,500

947: ME Board October 1995 In year zero, you invest P 10,000.00 in a 15% security for 5 years: During thattime, the average annual inflation is.6 %. How much, in terms of year zero pesos will be in the account at maturity? A. B.

C. D.

A.

16.7% :=; i < 100% A.

B. C. D.

946: CE Board November 1994 P 500,000 was deposited 20.15 years ago

B.

9351 ME Board October :1995

A.

942: ME Boa&-d April :1998 A sum of P 1,000 is invested now and left

P 12,020 P 13,030 P 14,040 P 15,030

948: ECE Board April1998 By the condition of a will, the sum of P 20,000 is left to a girl to be held in trust fund,by her guardian until it amounts toP 50,000. When will the girl receive the money if the fund is invested at 8 % ' compounded quarterly?

A.

B. C. D.

7.98 years 10.34 years 11.57 years 10.45 years

949: ME ~ard October 1996 You borrow P 3,500.00 for one year from a friend at an interest rate of 1.5 % per month instead of taking a !oan from a bank at a rate of 18% per year. Compare how much money you will save or lose on the transaction.

A. B. C. D.

You will pay P 155.00 more if you borrowed .from the bank. You will save P 55.00 by borrowing from your friend. You will pay P 85.00 more if you borrowed from the bank. You will pay P 55.00 less if you boHowed from the bank

" 544 lOQl,Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas

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950: ME Board April %996 What is the present worth of two P 100 payments at the end of the third year and fourth year? The annual interest rate is 8%.

A. B. C. D.

'1"opics

P 153 P 160 P 162 P 127

0 L.:]

Engineertng ·Economics Consumer & Producer Goods and Services Necessity and Luxury Different Market Situations Demand Supply Law of Supply and Demand Simple Interest ·· Ordinary Simple Interest - Exact SimplfJ Interest Dist::ount 1 Compound Interest Continuous Compounding Nominal and Effective Rate of Interest

Mon

iiUMf

Tue

(,,,!

0 [_j ~01 D lQ [] .~ l__; D

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Theory

VIed

Problems

.,

Thu

Solutions

Fri

r~otes

Sat

j

!

ANSWER KEY 91 ·1. 912. 913. 914. 915. 916. 917. 918. 919. 920.

A

c B c c

A 8 D A D

921. 8 922. D 923. B 924.A 925.A 926. 8 927. 928. 929.A 930. D

~'Uk~4&~A«d
c c

93'1. c 932.A 933.A 934.8 935.A 936.·C 937.0 938. D 939. 0 940. D

941.

c

942.A 943. 944. D 945. D 946. B 947. D 948.C 949. D 950. A

·i't

c

w

0 c:J 0

0

RATING 34·-40 Topnotcher 26-33 Passer 20-25 Conditional

0··19 Failed

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il


546

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,,,,,,

1m !

11

I

...,till ~!dij

~-·

~~~~~

~ 4,000[1+0.16(;6~)]

•

I~

a

i ~ 0.1628 i ~ 16.28%

i=0.1175 i = 11.75%

a

Interest= 0.06(20,000) Interest= 1,200

F = P(1 +in) F = 25,000[ 1+ 0.01(12)] F = 28,000

a F=P(1+in)

• •

(1+~r -1

ER = 0.1255 ER= 12.55%

= P[1 +

F = 1,233.55

o.os( °

6 )] 360

• •

ER = 0.1948 ER = 19.48%

=(1+ it ·- 1

ER=(1+~r-1 ( 0 0951'' 1.0984 = 1+-·-····j ' n .

ER=(1+ 0 ~ 1 J -1

F =P(1 + i)" 10,000 = P(1 + 0.04)

21

'p = 4,388.336

F1 = 4,388.336(1 + o.o4t

= (1+

F1 = 5,552.645

·~ J -1

0 4

Note: From the choices, the nearest answer is 5,500

•

ER =0.1449 ER = 14.49%

ER = (1+ir -1 ER = ( 1+

0.~75

r

F =50,000(1 + 0.075)

p

50,000

t

(

F = 12,800 1+

0

5

a

-1

=o:on1 =7.71%

F = P(1 +i)"

i~llllll i

F = P(1+i)" F = 71,781.47

0

·~ r4) 5

t

1

l

3

4

2

25,000

1

75,000

.

Solving for the effective rate per year:

F = 15,614.59

ER = (1+it -1 F = P(1 + i)" 100,000 p

I

F

F, = P(1+ i)"

ER ER

ER = (1+it -1

ER

F = P(1+in)

•

ER = (1+it -1

l

p!

ER = (1+it -1

ER

F = 408

21

• F1

ER = (1 + i)m -1

F=P(1+in)

mJ

[

1m

i = 0.19 i=19%

ER =

6

ER =0.01 ER =1.00%

ER = (1+iJ -1

p = 46,728.97 Note: From the choices, the nearest answer is 46,730

1.2so

•

0

:. the mode of interest is quarterly

F = 300[ 1+ 0.12(3)]

7 50,000 = p[ 1+ 0.12( 12)]

a

I

3,200 = (20,000- 3,200)(i)(1)

a

Proceeds= 20,000-1,200 Proceeds= 18,800

a

I

I=Pin

1,112.95 = 110,000(i{3~~)

•

By trial and error: N=4

I~ 0.16(20,000) I= 3,200

I= Pin

-~J

I

a 0.801 ~ 890.36 I= 1,112.95

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Pin

Da! 22 -Engineering Economics @imEle and ComEound Interest} 547

11,200 ~ 68,800(i)(1)

F ~ 4,133.33

·-r· =~

F~P(1+in)

F

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i 001 Solved Problems in Engineering: Mathematics ~znct Edition~ by Tiong: & Rojas

=p (

1+

=30,655.68

0

·~

2

rO)

'ER=(1+

0

-~~J -1 1

ER = 0.050625

1

I

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:

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4

3

l

2

25,000

2

75,000

2.5 = (1.06)

F = P(1 + i)"

2

F = 200,000(1 +0.12t

"

F = 621,170

log2.5 = log(1.06t

~

= 25,000 + 75,000 50 OOO + _P_ 4 2 ' (1+i) (1+if (1+i)

+

p = 39,201.28

DJI

ID

F = P(1 +i)"

ot

Interest= F- P

•

4P = 3.5oo(1+

0.~7r

4 = (1.035t Take log on both sides: log4 = log(1.035) log4

2 "

5

Note: From the choices, the nearest answer is 690,850

•

=2nlog1.035

n = 20.15 years

6

F=P{1+i) +P(1+i)

2

8

F = 1,000(1 + 0.08)

10

F = 2,158.92 Note: From the choices, the nearest answer is 2, 160.

.F = P(1 + i)" F = 5,000( 1 +

F=11,~0

= 1,500 ( 1 + 4

40

NR =3.5%

ml

NR = 5.8%

•

NR)20(2)

3,000

NR = 0.035

30

0 8

·~ r

F= P(1 + i)" F = 2,'000(1+0.15)

Note: NR = nominal rate F = P(1 + i)"

F = P(1 + i)"

2 = (1 + 0.5NR)

2

NR = 0.058

F = 6,034.66

•

NR)15(2l

2.37 = (1 + 0.5NR) 2

Note: From the choices, the nearest answer is 706

F = P(1 + i)"

2,370 = 1,000 ( 1 +

F = 2,000(1.08) + 2,000(1.08)

•

F16 = 705.42

Interest = 690,848.73

F = P(1 + i)"

Profit= 7,623.42

1m

8

Interest= 1, 190,848.73-500,000

Profit= 17,623.42-10,000

F = P(1 + i)"

F16 = P(1 + i)8 F16 = 477.455{1 + 0.05)

Profit= F -P

F = 6,265.87

Let: F1s = total amount after the end of 161h year

F = 1,190,848.73

F = 17,623.42

F = 3,5oo(1 + o.1

= 1,000{1 + 0.05) 8

Money left= 1,477.455-1000

F =50 ooo(1 + 0.1125)93 ' 12

F = 10,000(1 + 0.12}

F = P(1 + i)"

F

Money left= 477.455

i = 0.1667 i = 16.67%

75,000 (1 + 0.050625t

= P(1 + i)"

F = P(1 + i)"

700 = 600(1 + i)'

&l 000 +p = 25,000 1 2 (1 + 0.050625) (1 + 0.050625) '

F

After the principal is withdrawn: n =93

n"" 8 years

rm

•

Interest= 6,040.20

n = 12(7)+9

log2.5 = 2nlog(1.06) n = 7.86 years

50,000 + p1 = p2 + p3

Interest= 11,040.20-5,000

F = 1,477.455

ID

:

. p3

=10

"

Take log on both sides:

r··············; ................................................................:

Interest = F- P n = 2(1993 -1988) n

2

0 12) 2, 000 + 3, 000 = 2, 000 ( 1+ ~

1

~p

1m F = P(1 + i)"

t t i' l i

Day 22_- Engineering Economics (Simple and Compound Interest) 549

•

6

F = 4,626

F = P(1 + i)" 25,000

0

8(4)

8

= p ( 1 + ·~

)

p = 13,265.83

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550 ··1001 Solved Problems in Engineering Mathematics (2"d Edition) by Tiong & Rojas

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F = 500,000(1+

0~7r.tsc2J

F = 200,166

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F = P(1+i)"

•

Computing for the amount due after one year.

a. Borrow money from a friend

F = 3,500(1+ 0.015)

0 0 0

12

P'

=value of the account after 5 years considering there was no inflation. value of the account in today's peso due to inflation

b. Borrow money from a bank

5

20,113.57 = P'(1 + 0.06)

5

p• = 15,030

•

p

o~ar

Thu

Solutions

0 0 Fri

Sat

Notes

"'_5_ + _li_4 (1+it

50,000 = 20,ooo(1+

Problems

P =P1 +P2

p

=P(1 + i)"

(QJ

0 0

Therefore, you will pay P55 less by borrowing the money from the bank.

F= P'(1 +i)"

Wed

Theory

F =4,130

F = 20,113.57

F

0

1

F =3,500(1 + 0.018)

F = P(1 + i)" F =10,000(1 + 0.15)

Tue

F =P(1 +i)"

=

Engineering Economics Consumer & Producer Goods and Services Necessity and Luxury Different Market Situations Demand Supply Law of Supply and Demand. Simple Interest - Ordinary Simple Interest -Exact Simple Interest l;)iscount Compound Interest Continuous Compounding Nominal and Effective Rate of Interest

Mon

F =4.185

let: F

a

Topics

F =P(1 + i)"

Note: From the choices, tt)e nearest answer is 200,150

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(1+o.oat

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P=153

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2.5 =(1.04t" Take log on both sides: log2.5 = log(1.o2t" log2.5 = 4nlog1.02 n = 11.57 years

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What is Annuity? '

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Annuity is· defined as a series of equal payments occurring at equal interval of time. When an annuity has a fixed time span, it is known as annuity certain.

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Types of annuity:

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1. 2. 3. 4.

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Ordinary Annuity Annuity Due Deferred Annuity Perpetuity

Annuity - Ordinary Annuity - Annuity Due - Deferred Annuity - Perpetuity Capitalized Cost Annual Cost Bond Depreciation ··Straight Line Method -Sinking Fund Method -Declining Balance Method - Sum-of--Years Digit Method Break-Even Analysis Legal Forms of Business Organization

Let A be the periodic or uniform paynent and assuming only four payments:

t-rll

0

1

2

3

4

A

A

A

A

.

~ ........,... A(1+ i)

: ................... ->

..

1. Ordinary annuity is a type of annuity where the payments are made at the end of each period beginning from the first period. Derivation of formula for the sum of ordinary annuity:

~'

1

A(1 + i)2

~ ............. ,. ...............,._ A(1+ i) 3

Let: a1 =A

a2 = A(1+i)

a3 = A(1+il a4

= A{1+i) 3

F

554 .1001 Solved Problems in Engineering Mathematics (2"d Edition) by Tio~g & Rojas

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Annuity is based on the principles of compound interest. Hence computation of the sum of annuity may be done using the formulas for geometric progression.

. . .I I I I 0

PRESENT WORTH OF ORDINARY ANNUITY:

2

0

3 .. .

A'

.lIII

Solving for common ratio:

r = a2

a,

A

r = A(1+i) A r =1+i

A

A

Solving for the sum:

But: F = ---=-----""" i Substituting ~he value ofF:

i

3 ...

2

0

p = A[(1+i)" i(1+ i)"

A

A

[<1 + i)" -1] i(1 + i)"

0

1

2

3···0·1

n

l Il II.

where: i = interest per period n = number of periods A = uniform payment

A

A

A

A

A

A

A

«!•• .. •················· .. •• ................... ;

where: i = interest per period A = uniform payment

3.

Capitalized cost of any structure or property is the sum of its first cost and the present worth of all costs for replacement, operation, and maintenance for a long time or forever, or ' Capitalized cost

=First cost + Cost of perpetual maintenance

Annual cost

=ann1.1al depreciation cost + interest of first cost

+ annuaLoperating cost + maintenance cost

Deferred annuity is a type of annuity where the first payment does not begin until some later date in the cash flow.

Bond is a financial security note issued by businesses or corporation and by the · government as·a means of borrowing longterm fund. It may also be defined as a long-term note issued by the lender by the borrower stipulating the terms of repayment and other conditions. Bonds do not represent ownership of a business or corporation and therefore not entitled to share of the profits. The bondholder has no voice in the affair of the business. However the bondholder has a more stable and secured investment than does holder of common or preferred stock of the business or corporation. Bonds are issued in certain amounts known as the face value or par value of the bond. When the face value has been repaid, normally a~ maturity, the bond is said to be redeemed or retired. Bond rate is the interest rate quoted in the bond.

I

The following illustrates the normal life cycle of a bond.

Jtc!>~~

Lender

A

Cash Flow of Annuity Due

[<1+i)"-1]

amount factor

A

What is a Capitalized Cost? Annuity due is a type of annuity where the payments are made at the beginning of each period starting from the first period.

.

I

4···0CJ

worth factor

i

-.- - = uniform series compound

2.3

i

=-----""" = uniform series present

. . Af(1+.it,.,-1].

l .

p

where:

2.

interest of the first cost and the annual operating and maintenance costs, or

A

P=~

A

··························~ f''

F,;:;:.

-1]

n

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1'

A

SUM OF ORDINARY ANNUITY:

A

A

!Tlll

A[(1+i)" -1]

S= A[(1+i) -1]

5--·n

PRESENT WORTH OF PERPETUITY:

0

4

4

What is a Bond?

P=-F(1 +i)"

1+i-1

3

4. When an annuity does not have a fixed time span but continues indefinitely, then it is referred to as a perpetuity. The sum of a perpetuity is an infinite value.

Using compound interest formula:

S= A[(1+it -1]

2

Cash Flow of Deterred Annuity

A

-«····· ...............................:

S=a,(r"-1) r-1

1

A

..........................~ F

p

Day 23 -Engineering Economics (Annuity, Depreciation, Bonds, etc.) 555

Borrower

What is an Annual Cost? Annual cost of any structure or property is the sum of the annual dFmr<>,..,i::~tion cost, i

)II '

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Day 23- Engineering Economics (Annuity, Depreciation, Bonds, etc.) 557

· 100 1 Solved Problems in Engineering Mathematics (2nd Edition) by Tiong &Rojas

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Lender

f ~ Lender

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2

3

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Zr

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l

Zr

Zr

Types of Depreciation:

Annual depreciation charge, d

:~ ~-·: : : : : :.·.·:.·.·:·.~·: ·.~:·:~·-~: : : : : ~ Let V n = value of bond n years before redemption

A.

Physical Depreciation is' due to the reduction of the physical ability of equipment or asset to produce results.

B.

Functional Depreciation is due to the reduction in the demand for the function that the equipment or asset was designed to render. This type of depreciation is often called obsolescence.

,,..,..,..i,fi,...-.+,.,

f4=&~

0 =*' l§,[\ {:J, \ ~

7

Eq. 1

Using formula for present worth of annuity:

p _ A[(1+i)" -1] 1

i(1+i)"

-

Borrower Value of a bond is the present worth of all the amounts the bondholder will receive through his possession of the bond. The two payments that the bondholder will receive are the following: A. Periodic payments as interest of the bond until it is redeemed. B.. Single payment upon maturity of bond. This payment is usually equal to the par value of the bond. Derivation of the formula for the value of a bond:

Methods of Computing Depreciation: 1. STRAIGHT LINE METHOD In this method of computing depreciation, it .is assumed that the loss in value is directly proportional to the age of the equipment or asset.

zr[(1+i)" -1]

P1 =

Cn = cost after "n" years (salvage/scrap value)

= life of the property

n

Book value at the end of "m" years of using, Cm

Cm =Co-Om~ where: Dm

= total depreciation after "m" years

Dm =

·dCu ~en ---~

n

p =-F2 (1 + i)"

c

where: Co

p =-2 (1 + i)"

= first cost

Cn = cost after "n" years (salvage/scrap value) n

Substituting in Eq. 1

d[(1+it -1]

3. DECLINING BALANCE METHOD In this method of computing depreciation, it is assumed that the annual cost of depreciation is a fixed percentage of the book value at the beginning of t~e year. This method is sometimes known as constant percentage method or the Matheson Formula.

= life of the property Matheson Formula:

_ zr[(1+i)"

Vn -

i(1+ i)n

-1]+~

Book value at the end of "m" years of using, Cm

(1 + 1)1'1

k=1--t£:.

Crn=C0 -Dm where: Z = par value of the bond r = rate of interest on the bond per period C = redemption price of bond i = interest rate per period n = nuf11ber of years before redemption

where: Dm = total depreciation after "m" years Dm d(m)

=

VCo'

or

!

i' '

Annual depreciation charge, d

Using the formula for present worth of compound interest:

fl

= first cost

where: Co

i

i(1+ i)"

;;;==-

l !i

an

Borrower

4. Borrower redeems bond after n years, p~ys principal& gets back

v!.~

4

! l ! l Zr

Vn = p1 + p2

lend«

Depreciation is the reduction of fall in the value of an asset or physical property during the course of its working life and due to the passage of time.

2. SINKING FUND METHOD In this method of computing depreciation, it is assumed that a sinking fund is established in which funds will accumulate for replacement purposes.

What is Depreciation?

Vn

vc;

.k=1--&.m ·

The value k is the constant percentage. Hence k must be decimal and a value less than 1. In this method, the salvage or scrap value must not be zero.

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1 558 .1001 Solved Problems in Engineering .Mathematics (2"d Edition) by Tiong & Roj~

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revenue Respective depreciation charges:

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'arrtbta: Did you know that. .. there are three problems in Geometry, which attracted the interest of mathematicians in the ancient times, considered as "impossible problems". They are the following:

n-1

~~ ~(Co .,. Cn) ~)ears

Second year:

n-2

. daf"T(Co -C")kY~iilrs

Third year:

production

.and so on ... Book value at the end of ".tJ:~ .. years of using, Cm

¢ 11,

I.

GOOD LUCK!

.· o1 .~{C0 ~C"}'!~e~rs

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1

Proceed to the next page for your last test. Detach and use the answer sheet provided at the last part of this book. Use pencil number 2 in shading your answer.

4. SUM-oF-YEARS' DIGIT (SYD) METHOD

First year:

I I

,;


=C0 .i(d1 +d2 ·-i-,,.:~t.d"') .

Sum of year's digit,

2: years

What are the Legal Forms of Business Organization? The legal forms of business organizations are the following:

1.

.~yea($ =!i(n + 1)

Sole proprietorship.- considered as t~e simplest type of business organization wherein the firm is owned and controlled by a single person.

2

2. What is a Break-Even? Break-even refers to the situation where the sales generated (income) is just enough to cover the fixed and variable cost (expenses). The level of production where the total income is equal to the total expenses is known as break-even point. Break-even chart is a diagram which shows relationship between volume and fixed costs, variable costs, and income. The following is an example of a breakeven chart.

3.

Partnership- is a firm owned and controlled by two or more persons who are bind to a partnership agreement.

1. Duplication of a cube -: to construct a cube whose volume shall be twice that of a given cube .

2. Squaring a circle - to construct a ·square whose area shall be equal to the area of a given circle.

3. Trisection of an arbitrary angle- to construct an angle that is exactly onethird of a given angle.

Corporation - is a firm owned by a group of ordinary shareholders and the capital of which is divided up to the number of shares. It is also defined as a distinct legal entity separate from the individuals who owns it and can engage in any business transaction which a real person could do. This is sometimes known as joint-stock company or a cooperative. i

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______Day 23- Eng-ineering Economics (Jinnuity, Deprecia!~!~.:?o~:!E~,~~l~~~ D.

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P 40,544.29

9551 ME Board April :1998

I

D D [] D D D ~ D []

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Tue

Wed

Problems

Thu

Solutions

Fri

Notes

_

Topics

Annuity -Ordinary Annuity - Annuity Due - Deferred Annuity - Perpetuity Capitalized Cost Annual Cost Bond Depreciation -Straight Line Method - Sinking Fund Method - Declining Balance Method -Sum-of-Years Digit Method Break-Even Analysis Legal Forms of Business Organization

Man

Theory

....

Sat


953: ME Board Oetober :1.996

A man purchased on monthly installment a P 100,000 worth of land. The interest rate is 12 % nominal and payable in 20 years. What is the monthly amortization?

You need P 4,000 per year for four years to go to college. Your father invested P 5,000 in 7 % account for your education when you were born. If you withdraw P 4,000 at the end of your 1ir., 181h, 191h and 201h birthday, how much will be left in the account at the end of the 21st year?

A.

P 1,101.08

B. C. D.

P 1,121.01 P 1,152.15 P 1,128.12

A. B.

95Z: ECE Board April :1998

C.

Money Qorrowed today is to be paid in 6 equal payments at the end of 6 quarters. If the interest is 12% compounded quarterly. How much was initially borrowed if quarterly payment is P 2000.00?

D.

A.

B. C. D.

P 10,834.38 P 10,382.90 P 10,586.99 P 10,200 . 56

P 1,700 P 2,500 P 3,400 P 4,000

954: ECE Board November 1998 What is the accumulated amount of five year annuity paying P 6,000 at the end of each year, with interest at 15 % compounded annually?

A. B. C.

P 40,454.29 . P41,114.29 P 41,454.29

How much must be deposited at 6% each year beginning on January 1, year 1, in order to accumulate P 5,000 on the date of the last deposit, January 1, year 6?

A. B.

c. D.

P 751.00 P 717.00 p 715.00 P 725.00

956: ECE Board November 1.~98 A debt of P 10,000 with 10 % interest compounded semi-annually is to be amortized by semi-annual payment over the next 5 years. The first due in 6 months. Determine the semi-annual payment

A. B. C.

P 1,200.00 P ·t,295.05 P1,193.90

D.

P 1,400.45

9571 EE Board October •997 A young engineer borrowed P 10,000 at 12% interest and paid P 2,000 per annum for the last 4 years. What does he have to pay at the end of the fifth year in order to pay off his loan?

9S9s ME Board O~t(llbell' 1!:.9'94 If you obtain a loan of P 1M at the rate of 12% compounded anrually in order to build a house, how much must you p.ay monthly t\) amortize the loan within a period of ten years?

A. 8. G. D. 9~0~

Ecm: ha1·d Ap~·!i~ l[9'91i How rnuch mljst you invest today in order to withdraw P 2,000 annually for 10 years

if the interest rate is 9°/c.?

B. C. D.

P 6,919.28 P 5,674.00

P 6,074.00 P 3,296.00

9S8s EE Board April 199'7 Mr. Cruz plans to deposit for the education of his 5 years old son, P 500 a.t the end of each month for 10 years at 12% annual interest compounded monthly. The amount that wili be available in two years is

12,853.~12

A

P

B. C. o.

P12,f581.37 P 12,::>65.32 P 12,Ba5.32

9fl>:K.f ECE Bom!rd ApwH fi."~

A person buys a piece of !ot for P 'IOC,OOO dow.1payment and ·i 0 deferred s~~mi annual payments of P 8,000 each, startmg three years from now. What is the 1-m~sent value of the investment if the rate of interest is 12% compounded semi-

{1,

A.

P13,994.f7 p P 15,855.45 P ·t2,900.25

B.

c. D.

p 134,866.80 P 143,999.08 p 154,696.80 P 164,969.80

9<&:2: ClE B~ilard l!llay :!998 A man loans P '187,400 from a bank with interest at 5% compounded annually. He agrees to pay his obligations by paying 8 equal annual rayments, the first being due at the end of 10 years. Find the annu<:~i payments. ·

A. B.

P 13,000 p 14,500

A. B.

P5G,143.03

C.

P 13,500

D.

P 14,000

C. D.

P 62,334.62 P 38,236.04

P 44,982.04

,. Day 23- EngineeringEconomics (Annuity, Depreciation, Bond, etc.) 563

562, 100 l Solved Problems in Engineering Mathematic:; (2"d Edition) by Tiong & Rojas

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~ A housewife bought a brand new washing machine costing P 12,000 if paid in cash. However, she can purchase it on installment bas1s to be paid within 5 years. If money is worth 8% compounded annually, what is ner yearly amortization if all payments are to be made at the beginning of each year?

A.

B. C. D.

p 2,782.85 P 2,872.58 P 2,400.00 P 2,827.58



97Sz A fund donated by a wealthy person

A man inherited a regular endownment of P 100,000 every end of 3 months for 10 years. However, he may choose to get a single lump sum payment at the end of 4 years. How much is this lump sum if the cost of money is 14% compounded quarterly?

A small machine has an initial cost of P 20,000, a salvage value of P 2,000 and a life of 10 years. If your cost of operation per year is P 3,500 and your revenues per year is P 9,000, what is the approximate rate of return (ROR) on the investment?

to I lEE to provide annual scholarships to deserving EE students. The fund will grant P 5,000 for each of the first five years, P 8,000 for the next 5 years and p 10,000 for each year thereafter. The scholarship will start one year after the fund is established. If the fund earns 8% interest, what is the amount of the donation?

A B. C. D.

P P P P

3,802,862 3,702,939 3,502,546 3,602,431

A. B. C.

D.

25.0% 22.5% 23.9% 24.8%

97:2.: CE Board November 1996 964: ME Board October 1996 Mr. Ayala borrows P 100,000 at 10% effective annual interest. He must pay back the loan over 30 years with uniform monthly payments due on the first day of each month. What does Mr. Ayala pay eoch month? A.

B. C. D.

P 870.00 P 846.00 P 878.00 P 839.00

968: ME Board April 1998 A parent on the day the child is born wishes to determine what lump sum would have to be paid into an account bearing interest at 5 % compounded annually, in order to withdraw P 20,000 each on the child's 18th, 19th. 20th and 21•t birthdays. How much is the lump sum amount? A. B.

C.

D.

P35,941.73 P33,941.73 P 30,941.73 P 25,941.73

965: ME Board April 1998 A house and lot can be acquired by a downpayment of P 500,000 and a yearly payment of P 100,000 at the end of each year for a period of 10 years, starting at the end of 5 years from the date of purchase. If money is worth 14% compounded annually, what is the cash price of the property?

969: ME Board April :1998 An instructor plans to retire in exactly one year and want an account that will pay him P 25,000 a year for the next 15 years. Assuming a 6 % annual effective interest rate, what is the amount he would need to deposit now? (The fund will be depleted after 15 years).

A man paid 10% down payment of P 200,000 for a house and lot and agreed to pay the balance on monthly installments for "x" years at an interest rate of 15% compounded monthly. If the monthly installment wasP 42,821.87, find the value of)(? A. B. C. D.

80 employees a holiday bonus. How much is needed to invest monthly for a year at 12% nominal interest rate compounded monthly, so that each employee will receive a P 2,000 bonus?

A.

B. C. D

P810,100 P 808,811 P 801,900 P 805,902

'!i!Wo: ME Board April 1998 A piece of machinery can be bought for P 10,000 cash or for P 2,000 down and payments of P 750 per year for 15 years. What is the annual interest rate for the time payments?

A. B. C. D.

4.61% 3.81% 5.71% 11 0 '%

A.

p 249,000

C.

B. C.

P 242,806 P 248,500 P 250,400

D.

D.

970: EE Board October :1997 An investment of P 350,000 is made to be followed by payments of P 200,000 each year for 3 years. What is the annual rate of return on investment for the project? A. B

C. D.

41.7% 32.7% 51.1% 15%

P P P P

D.

12,608 12,610 12,600 12,300

P 101,605.71 P 10i,505.21 P 100,506.21 P 99,601.71

976: ME Board April 1998 A company issued 50 bonds of P 1,000.00 face value each, redeemable at par at the enq of 15 years to accumulate the funds required for redemption. The firm established a sinking fund consisting of annual deposits, the interest rate of the fund being 4 %. What was the principal in the fund at the eno of the 1'2th year? A. B.

97~: ME Board April 1998 A manufacturing firm wishes to give each

B.

A.

11 9 5 7

A. B. C.

C. D.

p 35,983.00 P 38,378.00 P 41,453.00 P 37,519.00

977: ME Board Aprii199:Z A unit of welding machine cost P 45,000 with an estimated life of 5 years. Its salvage value is P 2,500. Find its depreciation rate by straight-line method. A.

B. C. D.

17.75% 19.88%. 18.89% 15.56%

974: CE Board November 1995 Find the present value in pesos, of a perpetuity of P15,000 payable semiannually if money is worth 8% compounded quarterly.

A.

P 372,537

B

P 374,977

C. D.

P 373,767 P 371,287

978: EE Board A.pril :1997 A machine has an initial cost of P 50,000 and a salvage value of P 10,000 after 10 years. Find the book value after 5 years using straight-line depreciation.

A. B. C.

p 12,500 P 30,000 P 16,400

D.

P 22,300

'

'•l. ')

Day 23 - Engineering Economics (Annuity, Depreciation, Bond, etc.) · 565


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98~: An equipment costs P 10,000 with a

The initial cost of a paint sand mill, including its installation, is P 800,000. The BIR approved life of this machine is 10 years for depreciation. The estimat~d salvage value of the mill is P 50,000 and the cost of dismantling is estimated to be P 15,000. Using straight-line depreciation, what is the annual depreciation charge and what is the book value of the machine at the end of six years?

salvage value of P 500 at the end of 10 years. Calculate the annual depreciation cost by sinking fund method.at 4% interest.

A.

B. C. D.

p 74,500 P 76,500 P 76,500 P 77,500

; p 340,250 ; P 341,000 ; P 342,500 ; P 343,250

980: CE Board November :1997 The cost of equipment is P 500,000 and the cost of installation is P 30,000. If the salvage value is 10% of the cost of equipment at the end of 5 years, determine the book value at the end of the fourth year. Use straight-line method. A.

B.

c. D.

P 155,000 P 140,000 p 146,000 P 132,600

98:1: ME Board Apriil1998 An asset is purchased for P 500,000.00. The salvage value in 25 years is P 100,000.00. What is the total depreciation in the first three years using straight-line method? A. B. C.

D.:

p P P p

48,000 24,000 32,000 i6,000

98:&: ME Board April 1:998 A machine has an initial cost of P 50,000 and a salvage value of P1 0,000.00 after 10 years. What is the book value after 5 years using straight line depreciation? A. B. C. D.

p P P P

35,000.00 25,000.00 15,500.00 30,000 00

A.

P 791.26

B.

P 950.00 P 971.12 P 845.32

C. D.

984: CE Board November :1995 A machine costing P 720,000 is estimated to have a book value of P 40,545.73 when retired at the end of 10 years. Depreciation cost is computed using a constant percentage of the declining book value. What is the annual rate of depreciation in %? A.

B. C. D.

28 25 16 30

985: CE Board May :1996 A machine costing P45,000 is estimated to have a book value of P 4,350 when retired at the end of 6 years. Depreci~:~tion cost is computed using a constant percentage of the declining book value. What is the annual rate of depreciation in %? A. B. C.

D.

33.25% 32.25% 35.25% 34.25%

986: ECE Board November ll.998. ABC Corporation makes it a policy that for any new equipment purchased, the annual depreciation cost should not exceed 20% · of the first cost at any time with no salvage value. Determine the length of service life necessary if the depreciation used is the SYD method. A.

B. C. D.

9 years 10 years 12 years 19 years

I • I

987: ME Board April :1998

991.: CE BOard November :1996

A company purchases an asset for P 10,000.00 and plans to keep it for 20 years. If the salvage value is zero at the end of 20th year, what is the depreciation in the third year? Use SYD method.

At 6%, find the capitalized cost of a bridge whose cost is P 250M and life is 20 years, if the bridge must be partially rebuilt at a cost of P 1OOM at the end of each 20 years.

A.

A.

P 275.3M

B. C. D.

P 265.5M P 295.3M P 282.1M

B. C.

D.

p 1,000.00 P 857.00 P 937.00 P 747.00

988: ECE Board Aprilt.999

992: CE Board May :1997

A Telephone company purchased a microwave radio equipment for P 6 million, freight and installation charges amounted to 4% of the purchased price. If the · equipment will be depreciated over a period of 10 years with a salvage value of 8%, determine the depreciation cost during the 5th year using SYD.

A corporation uses a type of motor truck which costs P 5,000 with life of 2 years and final salvage value of P 800. How much could the corporation afford to pay for another type of truck of the same purpose whose life is 3 years with a final salvage value of P 1,000. Money is worth 4%.

A.

B. C. D.

P 626,269.10 P 642,786.07 P 638,272.08 P 627,989.90

989: ME Boaa·d April :1998 An asset is purchased for P 9,000.00. Its estimated life is 10 years after which it will be sold for P 1,000.00. Find the book value during the first year if sum-of-years' ·digit (SYD) depreciation is used. A. B.

c. D.

P P p P

8,000.00 6,500.00 7,545.00 6,000.00

990: EE Board April :1997 The maintenance cost for a sewing machine this year is expected to be P 500. The cost will increase P 50 each year for the subsequent 9 years. The interest is 8 % compounded annually. What is the approximate present worth of maintenance for the machine over the full 10-year period? A.

13. C. D.

P 4,700 p 5,300 P 4,300 P 5,500

A. B. C. D.

P 8,450.66 P7,164.37 P 6,398.24 P 9,034.56

99~: ME Board October :1995 A company must relocate one of its

factories in three years. Equipment for the loading dock is being considered for purchase. The original cost is P 20,000, the salvage value of the equipment after three years is P 8,000. The company's rate of return on the money is 10%. Determine the capital recovery rate per year. A. B. C. D.

P 5,115 P 4,946 P 5,625 P 4,805

9941 EE Board October 1998 The annual maintenance cost of a machine shop is P 69,994. If the cost of making a forging is P 56 per unit and its selling price is P 135 per forged unit, find the number of units to be forged to breakeven. A. B.

886 units 885 units

Day 23- Engineering Economics (Annuity, Depreciation, Bond, etc.) 567

566 :1001 Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas

,,,,,

""'a:a (/l~

I

C.

D.

688 units 668 units

995: CE Board May 1998 A manufacturer produces certain items at a labor cost of P 115 each, material cost of P 76 each and variable cost of P2.32 each. If the item has a unit price of P 600, how many number of units must be manufactured each month for the manufacturer to break even if the monthly overhead is P 428,000.


A company which manufactures electric motors has a production capacity of 200 motors a month. The variable costs are P 150.00 per motor. The average selling price of the motors is P 275.00. Fixed costs of the company amount to P 20,000 per month which includes taxes. The number of motors that must be sold each month to break even is closest to:

A.

A. B.

B.

A. B. C. D.

1053 1138 946 1232

9961 ME Board April1996 Steel drum manufacturer incurs a yearly fixed operating cost of$ 200,000. Each drum manufactured cost $160 to produce and sells $ 200. What is the manufacturer's break-even sales volume in drums per year?

A. B. C. D.

1250 2500 5000 1000

9971 JRT Industries manufactures automatic voltage regulators at a labor cost of P 85.00 per unit and material cost of P 350.00 per unit. The fixed charges on the business are P 15,000 per month and the variable costs are P 20.00 per unit. If the automatic voltage regulators are sold to retailers at P 580.00 each, how many units must be produced and sold per month to breakeven? A. B. C. D.

..~

104 200 120 150

10011 ME Board April1998

Compute for the number of locks that an ice plant must oe able to sell per month to break even based on the tbllowing data: Cost of electricity per block- P 20.00 Tax to be paid per block- P 2.00 Real Estate Tax- P 3,500.00 per month Salaries and Wages- P 25,000.00/month Others- P 12,000.00 per month Selling price of ice - P 55.00 per block

c. D.

1228 1285 1373 1312

9991 EE Board October 1997 The annual maintenance cost of a machine is P 70,000. If the cost of making a forging is P 56 and its selling price is P 125 per forged unit. Find the number of units to be forged to break even. A. B. C. D.

1015 units 985 units 1100 units 1000 unit!;

1000: ME Board Aprill998 XYZ Corporation manufactures bookcases that sells for P 65.00 each. It costs XYZ Corporation P 35,000 per year to operate its'l)lant. This sum includes rent, depreciation charges on equipment, and salary payments. If the cost to produce one bookcase is P 50.00, how many cases must be sold each year "tor XYZ to avoid taking a loss? A. B. C. D.

2334 539 750 2333

C.

o.

40 150 80 160

~'

;I

Day 23- Engineering Economics (Annuity, Depreciation, Bonds, etc.) 569.

1&.1

K':ll' I

I

ID

4.1

F1 =P1(1 + i)"

n

012345

e '] rnuity •rlr·:··' .f!~

)

;~A ~~

'

..

:•\

t_

!.111on

AAAAA

Ordinary Annuity Annuity Due Deferred Annuity Perpetuity Capitalized Cost !! i!._nnual Cost 1 Bond Dc1preciation -Straight Line Method -Sinking Fund Method - Declining Balance Method - Sum-of.. Years Digit Method Break-Even Analysis Legal Forms of Business Organization 1

--~ Theory

\Ned

[~

r-·--~

LJ

[I]

~ Fri

Problems

Solutions

Thu r~-·o

[J (._J I Sat

Notes

A

~...1. . 1. . L. . L. .L. . . . . . . . . . . . . . ..i

p

F2 =

2 -

P= A[(1+i)"-1J

(1 + i)" i F3

Alr(1 +0.12.) 12(20)_1]·

•

12 . 0.12)12(20) i ( 1+-12 A=1,101.08

I 951' A 952. J\ I 953. P..

I

954.A 955. B . 956. B 957.A 958.C 959.A 960. D 96'1. B 962. A 963.C

964. 965. 966. 967. 968. 969.

D B A B

c B

970. B 971. D

9l2.C

973. B

977. c 978. B 979. 8 980.C 981. A 9!32. D 983.A 984. B 985. B 986.A

974. D 987. B 975. D 988.A 976. D

989.

c

990. A 991. C 992. B 993.C 994.A 995.A 996. c'

997. c 998.A 999.A

1000.

~~

1001.D

c:J·43-51 Topnotcher

[~

c:J

0

33-42 Passer 25-32 Conditional 0-25 failed


-:-1] ----=-.:..-o.=o1--'---=

F3

(

=19,002.95

1 0.12)

+4

6000[ (1 + 0.15) 5 -1] 0.15 F = 40,454.29

6 .

I

p1 ............................................................Jo-jF1 ·

16 17 18 19 20

2~

llll

A A A A

L.. .L.. .L..~Fl

l2 F= A[(1+i)"-1] i

F = A[(1+i)"-1] i F = ----'=-'-------'-------=!

p = 10,834.38

0

=F3 (1 + i)1

Money left = F1 - F3

111

(1+i)"i

•t

4

Money left= 20,702.81-19,002.95

P= A[(1+i)"-1]

RATING

t

I

Money left= 1,699.86"' 1, 700

·---·----· ·

.

F3 =17,759.772(1+0.07)

100,000 =

6

ANSWER KEY

A((1+i)"-1]

F2 = H, 759.772

P-= 2.ooo[(1+T) -1J ~---·---··

21

F. - 4000[(1+0.07)

0 Il"'l L_J Tue

F1 = 500(1 + 0.07) F1 =20,702.81

lllll"'""""""""'' "'""""'""'! Topics

II

L..,..:F3

• •

F = A((1+i)"-1] i

5,000 = A[(1 +0.06) . 0.06 A·= 717

6

-1]

P= A[(1+i)"-1] (1+i)"i

A((1

J

+ o.;o)2(5)_ 1 10, 000 = --:-"-';.._..__~-!,...------"" (1+ o.~or5) (0.05)

A= 1,295.05

570 · 1001 Solved Problems in Engineering Mathematics (2"d Edition) by Tiong & Rojas

k.l (J.l~

a·a '···

•

ER = {1+it 012345

ER={1-+:i)

l l l l 1 ~ ~ ~ ~ ~ ~ L....L. .L. .J ............... 1

. ! Pz

.•.•

58,880.69 = p2 (1 + 0.06)

0.12 = {1 + if -1

p2 = 43,999.078

n

4

2000((1+0.12) -1] _

--=-.:.._---,-4_

+

(1+0.12} (0.12)

Total amount= 100,000+43,999.078

mJ

(1+i)"i

F + - - = 10,000 (1+i)" F <

_ -10,000

(1+0.12} F = 6,917.72

Note: From the choices, the nearest answer is 6,919.28

•

A({1 + 0.009488) 1,000,000 =

•

120

-1]

120

{1 + 0.009488)

(0.009488)

A= 13,994.17

10

A[ (1 + i) "-1]

p1 = ----'=-------"8

A[(1 + 0.05) -1]

pl = ----'=---~ 8

(1 + 0.05) (0.05)

10 payments

5 6 7 8

100,000

F

=sao[( + ~) 1

2

M)_ 1]

0.12 12 F = 13,486.70

p

P1 = 6.643A

15 16

A A A P1

A

!

6.643A = 187,400(1 + Q.o5l

A

~•.L. . L....L.. . . . . :l... J

2 ~...................... J

P1 = P2 (1 + i)"

A= 44,982 04

•

Solving for the interest rate per month,!:·

-

I

= (1+if 2 2

+ 1 = 1.1 0 = ( 1 + if

i = 0.007974

Borrowed, money = A + P A[(1+0 007974) 359 -1] 100,000 = A + _ ____e_:.___~__:_---"--359 (1 + 0.007974) (0.0 007974)

•

A= 839

10 payments

!". . . . . . . . . . lll . . . . .

!l l l l l

500,000

~.. J. . .l. . . l_J.. . !

p2

4 5

p 0

10

pl

AAAAAA

(1 + i)" i (1 + 0.06) (0.06)

p1 = 58,880.69

1+i

A

6

7

A A

A

n

14 15

01234$

_ 8ooo[(1+0.06f -1]

p1-

Solving for the interest rate per month:

ml

J

1

A

~., ..L. .L. . L. . . . . . . . . . . L. .!

0 1

p __ A,_(( 1_+_i)_"-_,1 Note: From the choices, the nearest answer is 13,500

p

\

1718

AA

(1+i)"i

F= A((1+i)"-1]

i

n

359 360

Total number of payments = 12(30) = 360

(1 + 0.09} (0.09)

0 1

3

AAA.A

r. L. J. . ..t. .............L. . ~

-1]

10

r. . . . . . . Tn. . . . . n

=

A A

2

Pz ~.................. .

p = 12,835.32

i.= 0.12/12 = 0.01 n 12(2) 24

9101112

P1

(1 + i)" i

3 years

0 1

. . . . . . . . . . lll . . . . . n (

0 1

P= A((1+i)"-1]

- 2000[(1 + 0.09)

I'll

r rrr . . . . . . . . .

8 payments ~--A··

A

P-

=

5

-1] (1 + 0.08)5 (0.0.08)

ml

Total amount= 143,999.078

p = _A,_(( 1_+_i)_"---"'1

p1 + p2 = 10,000

(1+i)"i

A~1 + 0.08)

A= 2,400

Total amount= 100,000+P2

=12(10) =120

J

.J

A[(1+i)"-1]

12,0,00 =A+

5

i =0.009488

•

__h..;.__:.~

p1 = p2 (1 + i)"

-1

F

P1

=1

12 2

i

I

Day 23- Engineering Economics (Annuity, Depreciation, Bonds, etc.) 571

Cash price = A + 1-'

1 1 12,000 =A+ ---'A[~(_+_:_i)_"-...:!.] (1 + i)" i

~

A A

r.L . . .l. ....L. . . . . . l.... J

...................... :

"-1]

_ A[(1 + i) Pl-----. (1+i)"i

I. I

57Z : l 00 l Solved Problems in Engineering Mathematics (2"d Edition) by Tiong & Rojas - 100,000[(1 + 0.14)

"' ,,,...

p1-

10

-1]

10

(1+0.14)

p,;, A[(1+i)"-1]

(0.14)

(1+i)"i

P1 = 521,611.56

(/'Ill

Q.'l

0 4 24

P=~ 2

I

p=

100,000[(1+ 24

(1+i)"

0.~4) (0~4)

(1+

p - 521..611.56 2 4 - (1+0.14)

·~ ) -1]

p = 1,605,836.76

p2 = 308,835.90 F = _A[=-(1_+_i)_"--=-1] Cash price = 500,

ooo + P2

i

Cash price= 500,000 + 308,835.90 Cash price= 808,835.90 Note: From the choices, the nearest answer is 808,811

•

100,000[(1 + 0.035) . 0.035 F = 2,097,102.97

16

-1]


•

J

p = _A[h.:.(1_+...:...i)_"---=1 (1-+ i)" i

i'l

A= 5,500 11111·11111'1!1

Substitute the value of A in Eq. 1:

p = 242,806 0 5500[(1 + if -1] - - - = - - . .10 :---"'

{1+i) i

350,000

i

I, IIIII

2000 +-10 = 20' 000 (1+i) '

•

By trial and error, i = 24.8%

0

2

3

! l-1

F=

A

A

A

:

:

:

Down payment= 10% of cost of house and lot

p~ ... L.....L .....l

Lump sum price = P + F

i

A= 9,000-3,500

15 - 25,000[{1+0.06} -1] p15 (1 + 0.06) (0.06)

•

I

Note: A = revenue/year- operating co.sVyear

200,000 = 0.10(cost) cost= 2,000,000

Lump sum price= 1,605,836.76

il

+ 2,097,102.97

Cash price = Down payment + Present worth of the Annuity

.....,. J

1 1 10 000 = 2 000 + _A.=.:[('--+_i:_)"--....=. ' ' (1+i)"i 10,000 = 2,000 +

750[ (1 + i) 15

15

(1 + i)

-1]

A

A

A

r. .L ...L...l. ...!

1-

24 payments

•

p = __,A[:.:_(1_+i)_"----=!1] (1+i)"i

p2 ~.......................... ......................sv ~

~T

15 16 17 18

A A

A

A

=

(1 + 0.05} (0.05}

39 40

A A

4

P1 = 70,919

. . .1. . .1_. . . . . . . . . r. . l A

iii

ii

0

1

!

2

3

A

9 10

·················-

20,000

70,919 (1+0.05t

p = 30,941.72

(1.0125)

12

=

42,821.87[(1 +0.0125) 12•-1] --=~12:;::----'--

(1+0.0125) x(0.0125) 2

-1 = 0.52543(1.0125f x 2 (1.0125f x =2.1072 x

Taking log on both sides: 12xlog1.0125 = log2.1072

il! I

x = 5 years

P=__!l_ (1 + i)" p =

~ 800,000

:

A A A 4

...i

p1

(1 +I")".I 2,000[(1 + 0.05) -1]

p1

= 0.0125


p1

J

i = 0.15/12

(1 + i) (i)

17 18 19 20 21

0

p~·······················

-n. . . lll r. . . n LJ.._...Lltl. L_.. J. A A

. . . . . . . . . . . . l' ! ! !

- 200,000[(1+i) -1] 350,0003

p __ A['=-(1_+;..;..,i}:--n --=-1

16 payments

Balance= 1,800,000

3

p1

Ill 0 1 2

liiltiiiill

A


Balance = 2, 000,000- 200,000

(1 + i)" i

i

10.667 = (1 + if5 -1 15 (1+i} i

Balance= cost- downpayment

p = A[(1+i)"-1]

Lump sum price= 3,702,939.73

p1 + p2 = 20,000 A[(1+i)"

- 1 L_~~

-·(1+ i)"i

(1 + 1)

n

=20,000

~Eq.1

·I

:j

Day 23- Engineering Economics (Annuity, Depreciation, E!onds, etc.) 575

574 1001 Solved Problems in Engineermg Mathematics (2"d Edition) by Tiong & Rojas

u.l lll 'Ill'

al

•

I

•

. . . . . . . . . . . . !!-!!1

0

F= A[(1+i)n-1]

i A[(1 + 0.12) 80(2,000) = ' 12 0.12 12 A = 12,615.80

12

-

p2

r. .L. .L. .L. .L..J

p3 ....:................... .

J

= (1 + i)2

A2 [ ( 1 + i)n - 1]

--'~----='

( 1+

i)" i

2

A

3

(1 + 0.08)

A

A

A

J ..~ F

50(1,000)= A[(1+0.04t' -1] 0.04 A= 2,497

5

l ! !!1..

F~-

I

Ps ,.;

1

p - 10,00.Q

-1]

4

-

0.08

p4 = 125,000 p 5-

p 5

-

12

-1]

C 5 = 50,000- 4,000(5)

•

C5 =.30,000

-

(Cr,- cost to dismantle)

n

d = 80,000- (50, 000- 15, 000) 10 d = 76,500

Let:, Ce = book value at the end of 6 years C6 =C 0 -d(m)

•

C6 = 341,000

C0 = 500, 000 + 30, 000 C0 = 530,000 en= 0.10(500,000)

d =Co- Cn n d = 45,000-2,500 5 d = 8,500

p4 =-:-

5ooo[(1+0.08} p1 = - - " - - . , .5 - - - (1 + 0.08) (0.08)

I

1m

A

(1+i)"i

.

- 2,497[(1+0.04) F1. 0.04 F1 = 37,519

I I II

p4 •

A A A

5

_ A[(1+i)" -1]

l !!1

A

C5 = C 0 -d(m)

C6 = 800,000- 76,500(6) oc

A A A A

1 [(1+i)n -1] ---"'-~--"'

n d = 50,000-10,000 10 d = 4,000

Let: Cs = book value at the end of 5 years

Let: F1 =value of the given annuity when n = 12 years

11

012345

P1...:....: ..... .:. ..... :..... .:. .... :

d =Co -Cn

d = C0

F= A[(1+i)n _ 1]

p3 = 21,738.97 0

. .. . . . .. . ....

•

14 15

L .....i ...... i........................L. ..

p3 = 31941.68

p = 15,000 . 0.0404 p = 371.287.13

p1 = 19,963.55

1

A

p p2 3 - (1 + i)5

i

-

• lll .........n i

P=~

p1 -

Total= 99,601.71

p2 =31,941.68

i = 0.0404

A

Total= 19,963.55 + 21,738.97 + 57,899.186

8ooo[(1+0.o8t -1] p2 = ' (1 + 0.08) 5 (0.08)

0.~8J =(1+i)2

A

Total= P1 + P3 + P5

0

-

( 1+ N4R

7 8 9 10

A A A A A

1}

p2 -

(1+

•

6

en= 50,000

d =Co -Cn

n d = 530, 000 - 50, 000

5

p4 {1+ifo

depreciation rate =

_i_

d = 96,000

Co 125,000 (1+0.08)

10

P5 =57,899.186

depreciation rate =

8 500 · x1 00% 45,000

depreciation rate = 18.89%

. Let: C4 = book value at the end of 4 years

C4

=C 0 -

d(m)

C4 = 53o,ooo- 96,000(4) C4

...

146,ooo

l,.


4 .1

u.l (Jll a:~

I

•

_______ .!_)_~ 23- Eng!~ng Eco!!2._mic~!!\nnuitL.~nreciati
k d =Co

720;000

-c.

k

n 25

=

d 16,000 Let: D = total depreciation after- "m" years

Lyears =.!'_l(n + 1) -----

=0.25

,.._..., years= 55

l~ years "' 3~(20 + 1)

vco-

I

Substituting:

=16,000(3)

Substituting:

Book value at the end of 6 years, C6: C1o = 4,350

0=48,000

a

- "(10 d3 " ' 000--

\ d, "'(9,000 -1,000)1(_!Q).

&iii~!

1!'-R

d=Co-Cn

n

......

= book value at the end of 5 years C5

=C0 -d(m)

Solving for C1:

/

4

I

Using_SYD~

the largest charge of depreciation is the first year.

years "''

~)ears

d1

=(Co -en {

n(~_2-+ 1)

C0

LY!'!ars = n 0

d

Using formula for sum of an A. P.

=791.26

p == ----~-..._________ _).

....

5n = n(n +1)

Book value at the end of 10 years, C1o: C1o 40,545.73

=

Substituting in the formula:

d5

2 n

=9 years

a

Capitalized cost = C 0 + Co-- Cn

C1 =Co - d1

n-2

I

d3 =(Co- c.) ) years)

Note: From the choice!i, the nearest answer is 4,700

= 626,269.10

~

r

+50[tf.~"!~i.'rj ________1~-------.! (0.08){1+0.08t p =4,653.88

d5 = (6,240,000- 499,200)( 5~)

2

k=1-~:

= 6,240,000

Substituting:

ryears =·n(n + 1)

Ill

i(1+ i)"

5oo[(1+o.oat -1l

-

c, ~' 0.08(6,240,000) c, = 499,200

d = .!...(1_...;.0,0_0_0-_5_00-d-).!....(0_.04...!..) (1+0.04f -1

·1!

G[-'------------- -- ----.11. ------J

C0 == 6, 000, 000 + 0. 04( 6, 000, 000) ' (1+i)"-1

'1,545.46

' )10 (008). (1+fl08

0.20C 0 = Co ( LY:ars)

d =(Co- Cn) ~

,,

lll'(.1·-·'"-•] ro(1+1)·" ---1 _ '"' IIJ ' p o' ·-"-"-. ____..__ + (1+ i)" i - f (1+ i)"

= 1Oi:!2_:~ 1}

Lyears = 55

L :ears)

•

C1

Using uniforrn gradient formula:

2

C 5 = 30,000

C 1 ~, 9,ooo- 1,454.54

l

I -.. n_-_---··j d,, =(Co ... '"'"\L:rears P

Using formula for sum of an AP.

C 5 = 50,000- 4,000(5)

•

d1 ""t454.54

k =32.25%

llifiM

Let: Cs

\55

18 \1 0) ('···-·-) 210

d3 = 857

Substituting in the formula:

d = 50,000-10,,000 10 d = 4,000

lj

k=1-_rs;

D = d(m) D

i I

2

?

"'"' 2-L, years= 210

El1

,I

2)ean.; = 10(10 + 1)

2

k=25%

d"' 500,000-100,000

L years "' !~+ 1) '2--

=1-19/40,545.73

d1 = (Co -

1111

,

C~ {~):ars

)

(1 + i)" -1

Capitalized cost = 250 + ---

100

--

(1+0.06)20 -1

Capitalized cost = 295.3 million

u.l u.J

K·~ I

Let AC 1 = annual cost of the old motor truck AC 2 = annual cost of the new motor truck AC ;=(C )i+ (Co1-Cn1)i 1 01

(1 +

ir -1

135x

= 2,258.82

= 56x + 69,994

79x = 69,994

• Let: x

AC ~ (5,000)(0.04) + (5,000- ~00)(0.04) 1 2 (1 + 0.04) -1 AC1

To breakeven: Income = Expenses

x = 886 units

=number of units to be manufactured per month

Income

= Expenses

X=

•

= (C 02 )(0.04) +(Co2

2,258.82

= 0.04C 02

C02

= Expenses

40x = 200, 000 x = 5, 000 units

Income

= Expenses

125x = 15,000

CH = (Co)i+ (Co -Cn)i

(1 + i)" -1 CR

= 20,000(0.10)+ ( 2 0,000-8,000)(0.10) 3

(1 + 0.1 0) -1

" . 1·

.R = 5,625

• Let: x

x = 120 units

= number of blocks to be sold per

month Income = Expenses 55x = 20x-t2x +3,500+ 25,000+12,000 33x =40,500 x = 1,227.3 blocks

i f~t: X ::::

10er of units to be forged

:'Ill

Income = Expenses

1il

65x = 50x + 35,000

'I. :I

-

x "' 1,228 blocks

1

'l!lllli,l i

x "' 2,334 cases

I'

I

,11,1111 1.11,,11

Let: x = number of motors to be sold each month

I

!;

Income = Expenses 125x

580x = 85x + 350 + 20x + 15,000 CR = Annual Cost

Let: x = number of cases to be sold each year

275x = 150x + 20,000

ml Let: x = number of units to be produced per month

Let CR =Capital recovery rate per year

x"' 1,015 units

x = 2, 333.33 cases

-1,000)(0.04) 3 (1+0.04) -1

= 7,164.36

-

\.

15x = 35,000

200x = 200,000 + 160x

+ 0.32C 02 - 320.35

125x = 56x + 70,000

x"' 1,053 units

Income

AC 1 = AC 2 2,258.82

Income= Expenses

1,052.42

Let: x = number of units to be sold out per year

Equating:

Let: x = number of units to be forged

x = 1,014.49 units

406.68x = 428,000

AC = (C )(0.04) +(Co2 -1,000)(0.04) 2 02 3 (1 + 0.04) -1

•

69x = 70,000

600x =c 115x + 76x + 2.32x + 428,000

AC = (C )i +(Co2- Cn2)i 2 02 (1+if-1


=20,000

x = 160 motors

:I!

582 I Op i Solved Problems. in En¢!1eermgMathexnatics~2nd Edition)·by Tiong·& Rojas

+

Appendix A:

Appendix A - Glossary 583

Glossary

' .r ' " " ~ '

' • .;,

"" ">-'

. ~ ·~

<'

v

···>! ,., ••.. """"" ........ ~.-

·-~.-·

•0·

-~·

••

.

A

' ~' < ' •'

~

A

_;«,

,..

""

abscissa the position of any point on a plane from the y-axis. Also known as 'the x-coordinate. The plural form of abscissa is abscissae.

" < .. ~ < :

·> i

-~

.(• ·~,

M"'

..

.<

•0

,0

'f~

....

~ --~

•-<

" : • •• ·< "'

>·•?:·' "'"""'""'

~--. -~--· ~ ~ ~ .:~ •

~ ... " ' ' ~ ~ "j/>-., ·~'

I

h

~

... ~.-_...

'!>~~-"~-~

\ ,..·.!..«.-'I'S '>~';?, f'~<'<

.. '·

'"-

1

~

N

,.

0

. ...

V

.h

'

h

', , •. ' ·•., ,. <' " •.

h" 'f " ... 1" ' ':: '

,.,,,

~-··"•'

··;- ,..,., -~.. 4

-~·

.....

,·,;.~~~"""""' ~ 'II: .'

'1)"'

' i• .....

,, .. "''* .,

....

_,,..~~--

0

~ ~-~

'

~

'

x

)' :'<"'

vrdinate

,.,;-.~;.

-~~

~

',(.<

.,

,

•!

•

N 1 "'

0

•

0

°o"O

~

·:

-~"

~ < -/.:,,> ••

. .; ,, ~

..

~,

. ..,

<:'

.. , • ,. -;;.·"' ' .,_ ,· •

< ~ ~ 1

,,

~

.•..

-., "

'" ·~ •« ;.

~

..

,.~~,.-,_~·'-··~·~··

~

..

4

"

~ ~ 'J· ~ .,

v

,

•••

'

,Y

~

~

. , .• ·.,,

,

...., ;

:

v

Y,

<

'

'

:..,

'l, ~ ~

.• i ~

• ·" ,

... · "'

•

~

...

,.

'

'

'

h

•••

'

~

•

~

-axis

accuracy a measure of precision of a numerical value !!f some quantity.

>-I abscissa

'

~

'

~

:·

Rectangular coordinate system showing abscissa and ordinate. The abscissa of point Pis x.

the

absolute term (syn. constant term) a term in an expression which does not have a variable . absolute value the number written arithmetically omitting the sign that prefixed it. The absolute · value of the number x is denoted as I x I

<

' ' .., -.'

!Y

, On'gin(O,O) I<

absolute error the difference between approximate value and the exact value:·

"·);':'

',x

...

* -~ .... ~ '·

'

,.... ,

~ ~

..

''· ...

t " .. '

('•·-"·""'"'~""""...;,.·.,~~

.• -~ ~

•

P(x,y) ·-·~

~."' .... '"""' "; ....... - ....~ <

-~ ~ "•""" ..... :' ,... • " h ( ' .,. :• t' "'

··\\"''

"· v

acceleration the rate of change of velocity per unit time. Acceleration is a vector quantity. The standard unit is meter per second per second (m/s2). The gravitational acceleration on earth is denoted as g and is equal to 32.2 feet per second2 or 9.81 meters per second2

v-axis

*

•< r/,;!J'->1.:>

~..,•• ,.,..""" ,.~ "-"'""-~-~ ~ .,, l<- ~"'."1

~

~-

·+·

.............

~ ~ ~ "'"' .. ' •• >

abundant number a natural number where the sum of its distinct factors exceeds the number. For example, the factors of 12 are 1,2,3,4 and 6. The sum ol the factors is 16. Since 16 is greater than 12, hence 12 is an abund.ant number.

,..,.,_·' *!.....,

~

, ~.:-.. 'I ,,. ' ~-. " ,• .., fi,<" " ~-" ' • _,; :·' ._,

"'''

:

~ " • <

",. "" *

..... ,...·,,..... !:' "

t '• '<," \

the number 10 in the hexadecimal number system

, , . , ..... .,. •.•

i'i=-~~,~·-·,,,

<

abstract algebra the part of algebra that deals with study of groups, semiiJroups, rings, modules, fields and similar structures.

'

.,_,..

-~

'

, ~ :'. '

acre British unit of area which is equal to 4840 square yards. In metric equivalent, 1 acre = 4046.8 square meters or 0.40468 hectares. acute refer to something less than 90° It comes from Latin 'acus" which means "needle" acute angle an angle less than a right angle or 90 degrees.

~<90° The angle {1 is an acute angle.

absolute value of complex number the distance of the complex number from the origin when the complex number is represented as the point with rectangular coordinates (a,b). Absolute

acute triangle a triangle having all angles acute angles.

Ja

add to combine numbers of quantities by getting the toti:il number of units contained in them.

value of a+ bi is

2

+ b2

absolute value of a vector the numerical value of the length of a directed line segment representing the vector. Absolute value of the vector ai + bj + ck is

Ja

2

+ b2 + c2

addend the number added to a certain number to produce a sum. Example: In -the equation 5 + 2 = 7, the number 2 is the addend whi,e the nu~ber 5 is the augend and 7 is called the sum.

.584 _1001 Sol~d Pro~el!!!'! in Engi_neering Mathematics (2nd E<,!~tionLey Tiong:_& Rojas. addition the process of combining numbers or quantities. The result of addition is called sum.

""'K{ ld

I

addition formulae formulae which express a trigonometric functions of the sum or difference of ·two angles in terms of the function of the individual angle. The foilowing are the addition formulae: sin (A +B) " sin A cos B +cos A sin 8 sin {A - 8) :: sin A cos B - cos A sin 8 cos (A +B) = cosA cos B -sin A sin B cos (A - B) = cos A cos B +sin A sin B tan A ·+-tan B tan(A +B)= ·--------1-tan A tanS tanA-tanB lan(A- B)= - - - - - - 1+tan A tan B additive identity an identity clement of an additive operation. The additive identity is the integer zero. additive inverse the negative of the numbHr. The additive inverse of 5 is -5. ad infinitum (latin term) continuing with no end.

algebra the study of operation and relation among numbers through the use of variables or literal symbols instead of just constants. The Latin term "algebra" comes from the Arabic "al-jabr" which means "to set or consolidate'.

algebraic equation an equation of the form f(x) = 0 where f is a polynomiai of dr;.'gree n with coefficients in a given base field, usually rationals, n is the degree· of the algebraic equation. For example, x2 + 3x + 4 = 0 is a second degree (n = 2) algebraic equation. algebraic expression any combination of symbols and numbers related to !he fundamental operation of algebra.

algebraic number a real number that is a root of a polynomial equation with integer coefficient algorithm a step by step procedure by which operation can be carried out.

an

aliquot part refers to any divisor of a number that is not equal to the number itself. This is

adjacent angles two angles that have !he same point or vertex and a common side between them.

Ai·jabr wa'l muqabalah Rough Translations, Balancing Equations, an Arab texbook written in the early 800s by AI-Khowzrizmi (from whose name the word algorithm was coined). This is where !he name algebra came from and from this text Europe came out of the dark ages and learned algebra.

Q{o_ 8, and f3 are adjacent angles agonic another !enn for skew as in agonic line or skew line.(see skew)

aleph the first letter in the Hebrew alphabet, ~. This letter was first used in mathematics by Georg Cantor to denote the various orders, or sizes, of infinity

e

also known as proper divisor.

alphanumeric (syn. alphameric character) combination of alphabet, numerical symbols, punctuation marks and other symbols used in computer works. alternate angles either pair of angles contained between two given lines and a transversal and lying on opposite sides of l.he transversal. These angles are equal if the given lines are parallel.

284, while the numb_er 284 have proper factors of 1, 2, 4, 71, and 142 which sums up to 220. ampere an Sl unit of electric current

algebraic curve a curve !hat describes an algebraic equation

adjacent lying next to. An adjacent angle (side) is an angle (side) that lies next to anothei angie (side).

)p


(a)

(b) (}and f3 are alternate angles. Figure 5b shows equal alternate angles since the lines are parallel. alternating series an infinite series whose successive terms are alternately positive and negative, such as 1 -1/2 + 1/3- 1/4 + 1/5 ..... altitude a line segment drawn from a vertex perpendicular to the opposite side (called base). altitude of a trapezoid a perpendicular distance between the bases or parallel sides of a trapezoid.

amplitude the maximum displacement between in either positive or negative direction from a reference level. amortization as applied to the capitalized cost the distribution of !he initial cost by periodic charges to operation as in depreciation or the reduction of a debt by either periodic or irregular prearranged program. anallagmatic curve a curve that is invariant under inversion. This curve includes the Cassinian ovals, Lirnacon of Pascal, Strophoid, Cardioid, etc. Analytic Geometry the branch of Mathematics which deals with the properties, behaviors and solutions of points, lines, curves, angles, surfaces and solids by means of algebraic method in relation lo the coordinate system. This was invented by Rene Descartes.

altitude of a triangle the perpendicular segment from a vertex of the triangle to the line • containing the opposite side. It also refers to angle the basic figure formed by two line segments the perpendicular distance from the apex of a or rays with a common end point. The symbol triangle to the base. for angle is".!". Angles are measured in different units, namely degrees, radians, grads ambiguous case the case of a solution of a plane and mils. triangle where the given data. lead to two 1 revolution = 360 degrees solution:;. c = 2n radian = 400 grads = 6400 mils

A~B 8 - .............:..... · .

Figure shows that

if sides a, band angle

A are given, each of the two triangles ABC and AB'C satisfies the given

angle, of depression or elevation the angle between the horizontal and the line of sight to the observed point. If the observed point is below the horizontal from the observer, it is an angle of depression. It is an angle of elevation if the observed point is above the observer.

conditions. Object

amicable numbers two integers such that each is equal to the surn of the ·distinct proper factors of the other. An example of amicable numbers are the numbers 220 and 284. The number 220 has a proper factors of 1, 2, 4, 5, 10, 11, 20, 22, 44, 55 and 110 which have the sum of

Lineof~-x horizontal

Angle of elevation

,,,....,.

honzontal

~\f~"x f Obiect Line of sight __

Angle of depression

~

586 100·1 Solved Problems in Engineering Mathematics (2"d Edition) .by Tiong & Rojas

w.

~I

angle of inclination the smallest positive angle that the straight line makes with the positive x-axis.

apotome a number that has the form of

)(JA- JB).

This

is

a

number

angular relative to or in terms of angles such as angular acceleration, angular velocity etc.

categorized by Euclid as one of the irrational numbers

annulus (syn. circu/auing) a plane figure which contains an area of a ring-shaped region lying between two concentric circles. Another term for annulus is ring. Area of annulus is 1t(R2 r2).

appreciation (ant. depreciation) increase in !tie value of an asset. approximation a number which is a close estimate of another number. The symbol of approximation is "'· arabic numerals the symbols 0,1 ,2,3,4,5,6, ... that represent the counting system in the decimal number system (base 10). (see numerals)

Annulus or Circular ring

annuity a series of equal payments occurring at equal interval of time. Type of annuity are, ordinary annuity, annuity due, deferred annuity and perpetuity. annuity due a type of annuity where the payment is made at the beginning of each period starting from the first period. antecedent the first term of the ratio. In the 2:3, 2 is the antecedent and 3 is consequent. (See consequent). antecedent is equal to the product of the and the consequence.

ratio the The ratio

antilogarithm the inverse function of a logarithm. antiprism a semi-regular polyhedron from two nsided polygon and 2n triangle. apex the highest point of a figure with respect to the base or plane of the base. 0

E~C

arbitrary constant a non-numerical symbol· holding a place for an unspecified constant. For example, in the general linear equation y = mx + b, m and b are arbitrary constants while x and y are variables. arc a part of a circle between two points on the circle. A portion of a curve between two of its points A and B. Archimedes Principle Any body immersed in a fluid is subjected to a buoyant force which is equal to the Weight of the fluid displaced. This was discovered by Greek Mathematician, Archimedes( 287- 212 B.C). Archimedean solid a solid made from regular polygonal· sides of two or more types that meet in a uniform pattern around each corner. It is a convex semi-regular polyhedron. Archimedes' spiral (syn. Archimedean spiral) spiral with the polar equation r = ae, with its graph as shown in the figure. The radius vector, increases with polar angle, e, a is the constant of proportionality.

y

A

8 Dis the apex of the polygon '

apothem the radi~s of the circle inscribed in a regular polygon .. A line from the center of a polygon perpendicular to one of its sides.

----1--t~------X

Archimedes' spiral

IIIII

Appendix A ~Ql()f!l!~ 587 ·are a larger unit of area in the metric system. It is equivalent to 100 square meters. area a numerical measure of a two dimensional geometric figure enclosed within a specific boundary Argand diagram (syn. Gaussian Plane) the rectangular coordinate system used for the representation of a complex number. The xaxis and the y-axis are known as the axis of reals and axis of imaginaries, respectively. Named after the Parisian bookkeeper, Jean RobertArgand (1768 -1822): argument (1) the angle between OZ, where Z is the point representing a complex numbE;r on an Argand diagram and 0 is the origin, and the real axis (2) theinput of a fUnction (3) an informal mathematical proof arithmetic the branch of mathematics which deals with calculation of integers using the fundamental operations of addition, subtraction, multiplication, division and the extraction of roots. arithmetic mean (syn. mean, average) the sum of 'n" numbers divided by n. For example, the mean of 2,9 and 7 is 6. arithmetic progression is a sequence in which the difference between any two successive terms is a constant and is called the common difference. arithmetic series the sum of the terms of an arithmetic progression. arm one of the lines forming the a:1gle. array an arrangement of numbers in rows and columns (see matrix) assets refers to everything a company or corporation owns and has a money value. Assets are as current assets (cash, bank account and other items that can be converted into cash), trade investment (investment in associated companies), fixed assets (land, building, etc) and intangible assets (goodwill, patent, etc.)

astroid (syn. ~tar curve) a hypocycloid with four cusps and with parametric equation of x = cos3 t , y = sin3 t. The rolling circle of this hypocyloid has a diameter one-fourth that of the fixed circle.

, I

Astroid

astronomical unit (AU) the mean distance between the earth and the sun. It is about 1.495 x 108 km. asymmetric not symmetric asymptote a straight line that approaches the curve more and more closely but never really touches it except as a limiting position at infinity. The word asymptote was coined by Thomas Hobbes (1588- 1679), using various latin stems meaning roughly "to fall together but not touch". atmospheric pressure the pressure caused by the weight of air at a given point. Standard value is 14.7 pounds per square inch or 760 mm of mercury. It is also equal to the weight of a column of water about 30 feet high and in metric system, it is equal to 100 kPa. Aubel's theorem "Given a quadrilateral and a square drawn on each side of it, the two lines connecting the centers of the squares on opposite sides are perpendicular and of, equal length."

II, II, I

!I

augend the number or quantity to which the addend is,added. Example: 5 + 2 = 7. The number 5 is the augend while 2 is the addend. automorphic number a number n whose square ends in n. Example: 5 is an automorphic number because square of 5 is 25 and 25 ends with the number 5. auxiliary ~ircle a circle with radius equal to half the major (transverse) axis and its center is at the center of the ellipse (hyperbola).

'"!!

588 l 001 Solved Problems in Engineering Mathematics (2nd Editi
w.

Ellipse

Bang's theorem 'If all the faces of a tetrahedron have the same perimeter, then the faces are all congruent triangles."

l&J

K~

I

bar graph a chart or diagram consisting of horizontal or vertical rectangles or bars, each of which represents an interval of values of a variable and has height proportional to the quantities.

auxiliary_____fi arde

Auxiliary circle Avagadro's constant the number of atoms of n grams in an element with atomic weight n. The value of this constant is 6.02214199 x 1023 This was named after the Italian phys1cist Amedio Avagadro (1776 -1856) average the usual term used for arithmetic mean. average acceleration the change in velocity of a body divided by the time interval during which the change occurs.

barycenter (see center of gravity) base a 'side of a polygon wh1cn 1s at the bottom of the orientation. · Bertrand's conjecture 'If n is an integer greater than 3, then there is at least one prime number between n and 2n - 2." This postulate is named after the French mathematician Joseph Bertrand (1822- 1900)

------------------------=AP:.cP:.:::e::.n:::d=ix A- Glossary 589 possible outcomes of an event (i.e. success and failure) and the possibilities if the outcomes are independent and constant binomial expansion expansion of a binomial in the form of (x + y)" in accordance with the binomial theorem. binomial theorem the theorem that gives the form of the expansion of any positive integral power of a binomial (x + y)". Its general equation is (x+y)"=x"+nxn·1y +n(n-:!.)_xn-2Y2+ ... 2! ... +n x yn-1 +yn bisect to divide a geometric figure into two equal parts. bisector (syn.bisectrix) a line or plane that bisects a given angle or line or any geometric figure.

bicorn any of a collection of quartic curves that has a rectangular equation of y2( a2 - x2) = (x2 + 2ay- a2)2. This is also known as "cocked-hat".

bit (abbreviation of binary digit) the digits 0 and 1 in the binary number system.

average velocity the displacement of a body divided by the time interval during which the displacement occurs.

bicuspid curve a quartic curve with an equation (x2 - a2)(x- a)2 + (y2- a2)2 = 0

book value the recorded current value of an asset. The value of an asset that is recorded in the book of records of the corporation.

axiom a statement of truth which is admitted without proof.

bilateral having two sides .or relating to the right side and left side of an object

axiom of induction 'Any property that belongs to zero, and also to the immediate successor of any natural number to which it belongs, belongs to all natural numbers.'

billion refers to one million million (1012) in the United Kingdom and Germany and one thousand mi~ion,(109) in the United States and France. In US and France, 1 million = 1,000,000.

axis the fixed reference line used in a coordinate system.

binary relating to the binary notation or binary code.

axis of symmetry a line around which a geometric figure is symmetrical.

binary logarithm logarithm of a number to the base 2

axis of the conic the line through the foc;us and perpendicular to the directrix.

binary number system (syn. Dyadic number system) a system of notation for real numbers that uses the place value method with 2 as the base. Only two digits are considered, 0 and 1 sometimes called as 'bit' (abbreviation of binary digits).

B B the number 11 · in the hexadecimal number system.

binol1Jjal a mathematical expression of two terms. Example: 5x + 4y binomial distribution (syn. binomial probability) the di:;trillrllion of p10hahilil1P:. wlmrn lhnm are two

Boolean Algebra an algebra which deals with the operation of complementation, union and intersection. It is devised by the British mathematician George Boole (1815 - 1864) who is best known for his innovatory work in formal logic. Boyle's Law At constant temperature, the pressure is inversely proportional to the volume.

bundle a family of lines or planes which all passes through a single point. butterfly theorem Let M be !he midpoint of a chord PQ of a circle, through which two other chords AB and CD are drawn. If AD intersects PQ at X and CB intersects PQ at Y, then M is also the midpoint of XY. The resulting figure of this theorem forms a butterfly.

,I!! I ,rll

byte a sequence of bits; a unit of information equivalent to a single character; a unit capacity of a computer.

!

c C the number 12 in the hexadecimal number system. The number 100 in Roman numerals calorie the amount of heat required to raise the temperature of one gram of water 1°C. calculate to determine the value of a given mathematical procedure; to compute. Calculus the branch of mathematics created in the seventeenth century by Isaac Newton(16421727) and German mathematician, Gottfreid Wilhelm von Leibniz (1646-17'16) which rest on the basic principles of limits. Calculus is divided into two subjects, namely, Differential Calculus and Integral Calculus. canonical form a form of any given polyhedron distorted so that every edge is tangent to the unit sphere and the center of gravity of· the tangent points is the origin

Brianchon's theorem 'Given a conic section, if we circumscribe a hexagon about it, then the major diagonal of the hexagon are concurrent.'

cap the symbol n, which is used to denote the union between two sets

Briggsian logarithm (see common logarithm)

cardinal numbers numbers used to count objects. Example: one, two, three, ... twenty...

l:lritish thermal unit (BTU) the amount of heat required to raise the. temperature of one pound of water 1" F. 1 BTU = 252 calories •

cardioid a heart-shaped curve generated by a fixed point on a circle as it rolls roung another circle of equal radius. Its equation is = a(1 - cos).

r

bulk modulus the ratio of the volume stress to the volume stress.

II

il II

"!' .1

590 · 100 1 Sblved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas Carmichael number a number n that is a Fermat pseudoprime to any base, that is, it divides (a" - a) by any a. This number is also known as "absolute pseudoprime' Cartesian coordinates •(syn. rectangular coordinates) a method of locating a point by pair of numbers denoting the distances from two fixed reference intersecting lines. The first number is called abscissa which is the distance from the y-axis while the second is called ordinate, Which is the distance from the x-axis. The two intersecting lines are called coordinates axes. r.ash flow the flowback of profit plus depreciation from a given project. Catalan solid a polyhedron that is a dual of an Archimedean solid. These solids are named after the Belgian mathematician Eugene Catalan (1814- 1894)

ceiling function the function of a number x that is the smallest integer that is not smaller than x. center of gravity (syn. barycenter ) the resuliant of. the passes regardless body in soace.

centroid, center of mass, point through which the distributed gravity force of the orientation of the

center of mass (see center of gravity) centesimal degree (see gon or grad) centillion a number 10 raised to a power of 600 or 10600 . central angle an angle whose vertex is at the center of a given circle. central conic a conic with a center. Examples are ellipse and hyperbola. The only non-central . conic is the parabola.

Am>enciix A- Glossary 591 center. Its equation is (x - h)2 + (y - k)2 = r2 where center IS at (h, k) and radiU$ is r. The parametric equation is x = r cos 8 or y = r sin 8. The word circle comes from Latin "circus" which means 'large round" circular cone a cone whose base is a circle. circular cylinder a cylinder with a circular right section. circular prime a prime number that remains prime on any cyclic rotation of its digits.

circumcenter the point of concurrency of the perpendicular bisectors of the sides of a triangle. circumcircle a circle that circumscribed a given polygon. circumference the boundary of geometric figure, especially a circle

centrifugal force a force acting outward on a body as it traverses a curvilinear path ·

circumscribe to draw a geometric figure around another geometric figure in such a way that they are in contact but does not intersect.

catenoid the surface generated when a catenary is rotated about its directrix.

centroid (see center of gravity)

cissoid a curve with a rectangular equation of

cathetus a line that is perpendicular to another line. This usually refers to one of the lines in a right triangle that is not the hypotenuse

century a period of 100 years. The word century comes from Latin 'centuria' whfch means 'one hundred"

caustic the envelope of rays of light reflected or refracted by a given curve from a given point source of light

.chain rule a rule of differentiating a function of a function , f [u(x)]:dj/dx = (dj/du) · (du/dx) characteristic the integer part of the logarithm.

Cavalieri's Principle Given two solids and a plane. Suppose that every plane parallel to the given plane, intersecting one of the two solids, also intersects the other and gives a cross-sections with the same area, then the two solids have the same volume. Named after Bonaventura Cavalieri (1598- 1647). Cayley's sextic a sinusoidal spiral curve with a rectangular equation of 4(x 2 + y2 -ax)3 = 27a2(x2 + y2)2. This curve is named after Arthur Cayley

chord a segment whose end points lie on the circle. chord of contrast the chord joining the points of tangency of the two tangent lines from a point P outside the circle. · cipher an old name for zero. lt is derived from Ara.bic "sifr", Latin "cephirum" and Italian 'zevero". circle a close plane figure every point of which is equidistant from a fixed point called the

coefficient of restitution the ratio of the total momentum after collision to that of before collision. It is denoted as 'e'. coefficient of static friction the ratio of the limiting frictional force (maximum) to the normal force. The coefficient of static friction force. The coefficient of the static friction is always greater than the coefficient of kinetic friction. coefficient of volume expansion (see bulk modulus)

circular ring (syn. annulus) see annulus

central tendency a central value between the upper and lower limits of a distribution around which the scores are distributed.

· catenary a plane curve described by a heavy uniform, flexible cable ha'lging freely between two points. The term 'catenary" comes from the latin word which means chain. Its equation is y =a cos h (x/a).

coefficient of kinetic friction the ratio of the frictional force to the perpendicular force. It is denoted as mu (J.!).

cofunction (syn. complementary function) In trigonometry, the function of a pair have equal values for complementary angles. Example: Sin 30° = Cos (90° - 30°) or Cos (60°). Hence, sine and cosine are complementary functions. coincident having all points in common. collinear points points that lie on the same line. cologarithm the logarithm of the reciprocal of a number. columns the numbers in order which appears vertically in a matrix.

x3

y2 = - - (2a- x) clelia the locus of a point P that moves on the surface of a sphere in such a way that /8 is constant, where 4> and 8 are the longitude and colatitude, respectively cochleoid a spiral curve that has the rectangular 1 2 2 equation of ( x + y ) tan-

U)

=

ay and a

. asine polar equat1on of r = - - . .

e

coefficient In algebra, it refers to the numerical factor of a term. For example: In the term 5x, 5 is the coefficient of the term.

combination an arrangement of a set of objects in no specific order. The combination of 'n' different things taken 'r' at a time is given as n! C(n,r)=~. \n- rI'11 r, If taken all, C1n.n1 = 1 combinatorics the study of the ways of choosing and arranging objects from given collections and the study of other kinds of problems relating to counting the number of ways to do something. commensurable refers to a ratio of two values that results to a rational number. Example: If two distances have a ratio which is a rational

Appendix A- Cl?ssary

59Z 100 l Solved Problems in Engineering Mathematics (Z"d Edition) by Tiong & R~

"-'

~I

number, then the commensurable.

two

distances

are

common denominator an integer or polynomial ·that is exactly divisible by each denol111nator. common difference the difference of two successive terms of an arithmetic progression or sequence. common factor (syn. common divisor) Of two or more integers or ·polynomials, an integer of polynomial which is a factor of each. For example: 5 is a common factor of 20 and 25. common logarithm (syn. Briggs logarithm) logarithm to the base 10. This was created by the geometry professor of Gresham College in Lo11don, Henry BrirJgs (1561-1630) as an improvement of the natural logarithm. common multiple an in!eger or polynomial that is multiple of each in a given set. For example: 90 is a multiple of 15 and 18. common tangent a line that is tangent to two or more curves. commutative law law stating that the sum or product is unaffected by the order of the terms. In addition, a + b = b + a. In multiplication, a times b =b times a. complementary angles two angles whose sum is 90" or right angle completing the square the process of modifying a quadratic polynomial to obtain a perfect square (trinomial). complex fraction any fraction which contains one or more fractions in either numerator · or denominator. complex number a number of the form a + bi with a and b real constants and i =the square root of ~.

.

composite numbers an integer which is the product of two integers, both different from 1 and -1. The integer 15 is a composite number since 15 "' 5 limes 3.

conjugate angles (see explemenlary angles)

compound interest the interest charges under !he condition that interest is charged on any previous interest earned in any period of time, as well as the principal. Fomula for future amount of a principle in compound interest is F =P(1 + i)", with n as the number of periods and i is the interest per period.

conjugate arcs two circular arcs together make a full circle conic (see conic section) conic section the locus of a point which moves such that its distance from a fixed point (focus) is in constant ratio, e (eccentricity) to its distance to a fixed straight line (directrix)

compound number a quantity expressed in different but related units. Example. 5 hours and 32 minutes. computable number a real number for which there is an alqorilhm !hi, given n, calculates the nth digit concentric circles circles having the same center with unequal radii. conchoid a shell-shaped curve that has a . o{ rectangular equation a(x- a)(x 2 + y2 ) = k2 x2 conclusion a oart of the theorem which is to be proved. ·

coordinates the abscissa and ordinate together, Also known as rectangular coordinates or Cartesian coordinates.

conoid a surface or solid formed by rotating a conic section about one of its axes.

corollary a statement of truth which follows with little of proof trorn the theorem

consequent the SflCOnd term of a ratio. See antecedent

corporation (type of business mganiza!ion) a

continued fraction a fraction in the form of: 1 x = ao +1 a1 + - -,----1-a2 +--· --1

conditional equation an equation which is satisfied by some, but not all, of the values of the variables .for which the members of the equation are defined.

convex polygon a polygon with no side extended will pass through the cenler of ihe polygon. Also it contains no interior angle gr
coplanar r)oints set of points that lie on the same plane

constant term a term in a polynomial thai does not contain a variable

concurrent having a common point.

conclusion and the conclusion becomes the hypothesis

conical surface a surface generated by a moving straight line (generator) which always intersects a fixed plane curve (directrix) and which always passes through a fixed point (vertex) not in the plane of the curve

constant of integration an arbitrary constant term in the expression of indefinite integral of a function.

concrete number a number that counts a physical quantity. For example~5 apples, 8 cats

59~

coprime refers lo two or more numbers thai have no factors in common other than 1

no

distinc;t legal entity separate from the indi~iduals who own it and which can engage in any business transaction which a real person can do. cosine law a law which relates the sides and angle of an oblique triangle. I! is used in solving the parts of an oblique triangle. cotermina! angles angles of rotation same initial side and terminal coterminal angle of 75• is 435 and difference of coterminal angles

which the side. The -285•. The is always

360".

a+--

cone a solid bounded by a conical surface (lateral surface) whose directrix is a clbsed curve and a plane (base) which .cuts all the elements.

3

confocal conics two conics which share the same . focus. A confocal . ellipse and hyperbola intersect at right angle. congruent geometric figures having identical shape and size.

a4 ...

continuity a property in mathematics that refers to smoothness of a function or curve

couple a pair of forces equal in magnitude, opposite in direction and not in !he same line

continuum any set that can be bought into one-toone correspondence with the set of real numbers

Cramer's rule a method of solving linear equations of several unknowns simultaneously using determinants or matrices. II is named after the S•,yiss mathematician and physicist Gabriel Cramer (1704 -1752).

convergent A convergent sequence has a finite limit. A convergent series has a finite sum Opposite of convergent is divergent.

conjecture a mathematical statement which has neither been proved nor dehied by counterexamples. Some of the famous conjectures are Fermat's Last Theorem and Goldbach Conjecture.

converse of a theorem* another theorem wherein the hypothesis and conclusion'of the first are reversed; that is, the hypothesis becomes the

J

counting numbers (see natural numbers)

critical point (syn. Stationary point) a point at which a function has a first deril,talive of infinity, thus havipg a slope which is vertical. cross product (see vector product)

!I!~

Appendix A- Glossary 595

594 1001 Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas.

~

I

de Moivre's' theorem A theorem that links complex numbers and trigonometry. It states that for any real number x and any integer n,

crunode a point where a curve intersects itself so that two branches of the curve ,have distinct tangeo1t lines

cyclic number a number of n digits, which, whem multiplied by 1, 2, 3, 4, ... , n, produces the same digits in a different order. ·

cube a polyhedron whose six faces are all squares. A cube is a regular hexahedron.

cyclic polygon a polygon whose vertices lie on a circle.

cubic curve an algebraic curve described by a polynomical equation of the general form ax3 + bx2y + cxy2 +dy3 + ex2 + fxy + gy2 + hx + iy + j =0

cycloid the plane curve traces out by a fixed point P on the circle as the circle rolls along a line, the base of the cycloid

decahedron a polygon with ten faces. There is no such thing as a regular decahedron.

cubic equation a polynomial equation of the third degree.

cylinder a solid bounded by a .closed cylindrical surface and two parallel planes (see circular cylinder)

decimal fraction number that consists of an integer part (which may be zero) and a decimal part (less the one) that follows the decimal marker, which may be a point or a comma.

cubit a measure of length which is approximately equal to the length of a person's forearm (from the elbow to the fingers). A Roman cubit is equal to 17.4 inches while the Egyptian cubit is equal to 20.64 inches

cylindrical surface a surface generated by a moving straight line (generator) which is always parallel to a fixed line, and · which always intersects a fixed plane curve (directrix) not in the plane with the fixed line

cuboctahedron a polygon obtained by cutting the corners oft a cube or an octahedron. A cuboctahedron has eight faces. that are equilateral triangles and six faces that are squares .. cuboid hexahedron will all faces rectangles Cunningham chain a sequence of prime numbers in which each rnember is twice the previous one plus one. cup the symbol v, which is used to denote the union of two sets curvature a measure of the amount by· which a curve deviates frorn a straight line curve (see locus) cusp a point on the curve where two branches coming frorn different directions, meet and have a common tangent. The word cusp comes from Greek "kera" which means "horn" and from Latin "cuspis" which means "sharp" cute number a number such that a square can be cut into n squares of, at most. two different sizes. cyclic quadrilateral a quadrilateral with all vertices lie on a circle

cylindroid a cylinder with elliptical cross-section.

D 0 the number 13 in the hexadecimal number system; the number 500 in the roman numerals .. Oandelin spheres the two spheres that just fit inside the cone, one on each side of the plane and both tangent to it and touching the cone, when the cone is 'sliced through by a plane

(cosx+isinx)" =cos(nx)+isin(nx) decagon a polygon of ten sides

decimal number system a system of notation for real number that uses the place value method with 10 as the base. decimal point a dot place between the integral and fractional parts of a number. declining balance method a method of computing depreciation in which the annual charge is a fixed percentage of the depreciated book value at the beginning of the year to which the depreciation applies. This method is also known as percent on diminishing value. decrement the negative.of an increment. defective equation any equation which, because of some mathematical process, has fewer roots than its original defective number (see deficient number)

0' Alemberts principle When a body is subjected to an acceleration, it is reacted upon be a reverse effective force opposite in the direction of the acceleration to balance. The value of the reverse effective force is equal to the mass time acceleration. Named after the French mathematician and physicist, Jean Le Rond d' Alember (1717 - 1783). de Malves' theorem "Given a tetrahedron in which the edges meeting at one vertex, X, form three right angles (i.e. the tetrahedron is the result of chopping off the corner of a cuboid), the square of the face opposite X is equal to the sum of the squares of the other three faces."

deferred annuity a type of annuity where the payment of the first amount .is deferred a certain number of periods after the first. deflection angle (syn. exterior angle) deficient number (syn. defective number) any natura! number the sum of its proper divisors. All prime numbers are deficient numbers. definite integral an expression of integrating an integrand between two limits of integration. Integrals with limits are definite integrals. degenerate conic conic obtained when the cutting plane passes through the vertex of the cone.

II

,,

Example of degenerate conics are the pointellipse, two coincident lines and two intersecting lines degree 1. A measure of an angle which is equal to 1/360 of a revolution.1degrce = 60 minutes and 1 minute = 60 seconds 2. The highest exponent or sum of exponents in any given term of a polynomial. The polynomial 7x5y2 + 8x4ys:2x2y is 9th degrees polynomial. delta curve a curve that can be turned inside an equilateral triangle while continuously making contact with all three sides. deltahedron a polyhedron whose faces are all equilateral triangles that are all of the same sizes deltoid a non-convex quadrilateral with two pairs of adjacent equal sides.

, I~

. il

denominator the number written below in a given fraction. The nurnber written above is called the numerator. For example, 3/4 has a denominator of 4. '

•ri'

density (syn. specific weight) the mass per unit volume of a substance.

l·l·i

depletion the lessening of the value of an asset due to the decrease in the quantity available. This refers to the natural resources such as coal, oil and timber in the forest

'!

I

lj I

depreciated book value t~e first cost of the capitalized asset minus the accumulation of annual depreciation cost charges depreciation the decline in the value of an asset due to the passage of time and constant use. determinant a number which is expressed as a square matrix (with the number of rows equals the number of columns) deviation in statistics, it refers to the differenGP. between any one of the sequence of observed value of a variable to some value such as the mean.

!,II

.~!Ii

J1 596 '1 001 Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas

'ID"

I

devil's curve a curve with a rectangular equation of y4 - a2 y2 = x4 - b2 x2 and the polar equation r2 ( sin 2 0- cos 2 0) = a2 sin 2 0- b2 cos 2 0 diagonal a segment joining two non consecutive vertices. It may be calculated using the formula, n/2 (n-3), where n is the number of sides of the polygon digit any specific symbol use to denote a number whether singly or in combination.

Diophantine approximation the approximation of a real number by a rational number Diophantine equations If there exist more unknowns than the number of equation but still can be solve bec(luse the values of the unknown are integers, these equations are called Diophantine equations (named after ~ Diophantus of Alexandria, a Greek mathematician" in the 3rd century BC). Diophantus in his book 'Arithmetica' carried out his extensive study to the solution to indeterminate equations.

diameter a chord containing the center of the circle difference the result of subtraction. differential calculus a branch of calculus that deals with the evaluation and use of derivatives and differernials. · differential eq<~ation an equation to be solved for an unknown function which involves the first or higher derivatives of the function ; an iquation that contains one or more terms involving derivatives of one variable with respect to another variable. Ordinary differential equation - one that involves function of a Gingle variable and some of its derivatives. Partial differential equation - one that involves functions to two or more variables and some of !heir partial derivatives Order ot a differential equation - refers to the order of the highest derivaiives that is present in the equation. Degree of a. differential equation - the highestyo.wer of the highest-order derivatives. differentiation derivative

the operation of finding the

directrix a fixed line opposite the focus of a conic section which the eccentricity of the conic section is defined. Dirichlet's theorem 'For any two positive coprime integers, a and b, there are infinitely many prime numbers of the form an + b, where n > 0.' discontinuity a point at which a function is not continuous. This is also kn?wn as 'jump" discriminant' the quanti!¥ that discriminates among · the pvssibilities of a quadratic equation. lt is expressed as b2 - 4ac. The discriminant • determines the nature of the roots. If b2- 4ac =0, the roots are real and equal. If b2- 4ac > 0, ttie roots are real and unequal if b2...: 4ac < 0, the roots are imaginary and unequal

a

displacement vector quantity which represent the charge in position of a point. It is equal to the product of the velocity and lime. distance the length of the shortest line segment between two points. distance formula the formula used lo measure the distance between two points. This formula was derived by the. use of the Pythagorean theorem.

d=~Vc2 -x~J +(y2 -y~J

dihedral angle the angle between two planes intercepted by a plane perpendicular to the common edge

divergent not having a finite limit

dimensions of the matrix the number of rows a'nd columns of a matrix

dividend in the expression a I b, a is iha dividend and b is the divisor

Appendix ~~-~Qlossary 597

I

division the process of obtaining the quotient; the inverse of multiplication divisor (see dividend) the number that divides the dividend

eccentricity the ratio of the distance from point to the focus (focal distance) to the distance from the point to the directrix. Eccentricity of parabola, ellipse and hyperbola are 1.0, <1.0 and >1.0, respectively.

dodecagon a polygon with 12 sides

ecenter (see. excenter)

dodecahedron a polyhedron of 12 faces. Each face of a regular dodecahedron is a regular pentagon.

ecircle (see escribed circle)

domain the set of all first elements of a relation

,' li I ' I

II

economic return the profit derived froni a project or business enterprise without consideration of obligations to financial contributors and claims of others based on profit

hi

dot product (see scalar product) dozen a term use to denote the number 12. This is derived from French "douze" which means twelve. 1 dozen = 12 items. duodecimal number system a number system using 12 as a place value. This number system still survive today as in 1 year = 12 months, a clock dia! has 12 hours, 1 foot= 12 inches, 1 dozen = 12 items and 1 gross = 12 dozens. dyadic number system (see binary number system) dynar:nic (syn. kinetic) branch of mechanics that deals with the forces that produced a motion. dyne a unit of force in the cgs metric system and is equivalent to 1 gram-cm/s2. 1 dynes= 10·5 N.

E

economical number a number that has no more digits than there are digits in its prime factorization effective interest the true value of interest rate computed by the equations for composed interest for a 1 year period. Effective interest can be computed using the formula, ER = (I + i)m- 1 where: m = number of interest period per year i = interest per period II!

Egyptian fraction a fraction in which the numerator is 1. This is also known as "unit fraction" Egyptian triangle a right triangle with sides 3,4 and 5 units eigenvalue a complex number, A, the satisfies the equation Ax = AX, where A is an n x n matrix and x is some vector. eight curve a curve with an appearance of a figure of eight and has a rectangular equation of 4

2

2

2

E the number 14 in the hexadecimal notation.

x = a (x

e il transcendental number which is approximately equal to 2.71828 ... 1t is commonly called as the Euler Number because it was introduced by Leonhard Euler(1707 - 1783) in the 18th century. It has the equation

a Lemniscate of Gerono

eccentric geometric figures not l:aving the same center.

-

y

) •

This curve is also known

I

II ill

j, Iii

,,'

,'I

i'

I'' '

'~II' ~

'

I ,

': II

element 1.member of the set 2. part of a geometric figure such as point, line or plane). eliminant (see resultant) ellipse a locus of a point which moves such that the sum of distance from two. fixed points (foci) is constant and is equal to the length of the major axis (2a).

I

,I

il,


598 100 l Solved Problems in Engineering Mathematics (2"" Edition) by Tiong & Rojas Eccentricity of ellipse is always le$S than 1. Standard equation of the ellipse is

(x-11)2 + {y-k)2 = 1 a2

y =(a+ b)sin(t) -csin[ ( +

ellipsoid (syn. spheroid) solid of revolution of an ellipse when rotated about one of its axes. When rotated about its longer axis, it is called as prolate ellipsoid. When rotated about its shorter axis, it is called oblate ellipsoid. The general equation of an eiHpsoid· in rectangular

yz

zz

coordinates is 2 + 2 + 2 = 1 a b c elliptic curve t[le type of cubic equations whose solutions takes the form of y2 +axy+by= x3 +cx 2 +dx+e emirp a prime number that becomes a different prime number when its digits are reversed. The word 'emirp' is 'prime' spelled backwards. Example is the number 13 which will become 31 when the digits are reversed but still a prime number empirical based only on observations and experimental evidences empty set a set the contains no element at all Engineering Economy the application of engineering or mathematical analysis and synthesis to economic decisions ennea • (syn. nona) a prefix which means nine. A polygon of nine sides is called a nonagon or enneagon. envelope an envelope of a family of plane curves is a curve that is tangent to every member of the family· epicycloid the plane curve traced by a fixed point on a circle as it rolls along the outside of a fixed circle. epitrochoid a curve traced out by a point that is a distance c from the center of a circle of radius b, where c < b, that is rolling around the outside of another circle of radius a. This curve is described by the followin!J parametric equntion~

error the difference between an approximate value and the true value which it approximates.

-ccos[(~ -1}] ~

b2

xz

x =(a +b)cos(t)

1}]

escribed circle (syn. excircle, ecircle) a circle tangent to one.side and to the extension of the two other sides of a triangle.

equal having the same value. The symbol of equal (=)was introduced in 1557 by Robert Recorda in his algebra textbook 'The Whetstone of Witte'. equally likely a term equiprobable events.

used

to

describe

equation a mat[lematical statement showing that two expressions have the same value. equiangular having all angles equal. An equilateral triangle is also equiangular.

Euler line a line that connects the centroid aml the circumcenter of a triangle Euler number the number e which is equal to 2,71828 ... To obtain this number, get the antinatural logarithm of 1. Or In e = 1. The formula fore is, · e =Lim n~a:

(1+ .!)n n

The symbol 'e' was introduced by the Swiss Mathematician, Leonard Euler (1707-1783). even function a function f(x) such that f(x) for all x

equidistant being of equal distance to any given points or lines.

even number numbers which are exactly divisible by two. All even numbers has for its last digit an even number.

equilateral triangle a triangle having all sides equal in length. An equilateral triangle is also equiangular. Each interior angle of an equilateral triangle is equal to 60°. equilibrium the condition when a body is acted on by no forces of several forces such that their vector sum (resultant) is zero. Forces are said to be in equilibrium if they are-concurrent (having a common point for the forces' line of action). equiprobable having the same probability. equivalent having some properties (geometric) in common. equivalent numbers a number such that the sums oftheir aliquot parts (proper ciivisors) are the same eradius (see exradiusj

'!I I

extract to find the value of a root.

equichordal point a point inside a closed convex curve in the ~lane, all the chords through which have the same length

equilateral hyperbola (see hyperbola)

exterior angle (syn deflectign angle) is the aAgle formed by the prolongation of one side and the adjacent side of the polygon. The sum of all exterior angles in any polygon is always equal to 360°.

=f(-x)

event in probability, it refers to the possible outcome of a trial. evolution the operation of root extraction. The opposite of involution. exact·accurate, precise excenter (syn. ecircle) the center of the ascribed circle.

extrapolate to estimate the value of a quantitr or measurement beyond the values which are already known. Opposite of extrapolatiOft is interpolation.

'"'"""''11.1

I: '!:·1:

extremes the first and the fourth terms in lle proportion of the four quantities

F F the number 15 in hexadecimal notation. face a plane surface of a geometric figure. factor (syn. divisor) each of two or more numbers which is multiplied together to form a product. factorial the product of the n consecutive positive integers from 1 to n. Example: n! = 1x 2 x 3 x ... x (n- 1) x n.· By definition, 0! = 1. The symbol factorial ( ! ) was introduce by Christian Kramp (1760 -1826) in 1808. Recursion formula states that: (n!)(n + 1) = (n + 1)! Substituting 0 for n, results to 0! = 1

excircle (see ascribed circle)

factorion a natural number that equals to the sums of the factorials of its digits in a ~iven base

explementary angles (syn. conjugate angles) two angle whose sum is 360"

factoring the process of converting expressions as sums into equivalent expressions as products.

exponent (syn. index) number (usually written superscript to another number) that is used to simplify repeated product. Example xJ = X·X·X

fallacy a contradictory or false proposition; an invalid argument of form of argument.

·exradius (syn .. eradius) the radius of the ascribed circle.

family ( syn. family of curves) a collection of related geometric configurations; a set of related curves or surfaces whose equations vary only in the values of the constants. III ~II ill :.

II~

~

600 1001 Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas Fermat's Last Theorem ( syn. Fermats' Great Theorem) a famous conjecture of mathema\ics which stales that X'' + yn = zn, .where n is 3 or greater x, y and z are all positive integers. This is regarded as the the least proved theorem in Mathematics. • Fiere de Fermat(1601,1665) claimed !hat he founded a marvellous root to this theorem, but the margin is too narrow to contain it. This is also known as the hardest problem in Mathematics or the "Mount Everest of Mathematics'.

Oblong numbers are numbers which can be drawn as dots and arranged in rectangular figure. Example: 2, 6, 12, 20, ... Pentagonal numbers are numbers which can be drawn as dots and arranged in pentagonal 'figure. Example: 1, 5, 12, 22, 35, ... Gnomon numbers are numbers which can be drawn as dots on equally long legs of a right angle. Example: 1, 3, 5, 7, 9, ...

Fermat number a number defined by the formula

F.:=22"+1 Fermat's little theorem "If P is a prime number !hen for ooy number a, (aP - a) must be divisible by P." Fibonacci numbers ( syn. Fibonacci sequence) the unending sequence integers formed according to ihe rule that each integers is the sum of the preceding two. The Fibonacci sequence is 1,1 ,2,3,5,8, 13,34,55,89 ....... Name after italian merchant and mathematician, Leonardo Fibonacci (1170 -c. 1250).

of

figure 1. any arrangement of points, lines, curves forming a geometric shape 2. The symbol for an integer, such as '8' or '69'. figurate numbers numbers that are represented by arrangements of dots as geometric figures. In plane figures, the following are examples of figurate numbers: Triangular numbers, Square numbers, Oblong numbers, Pentagonal numbers, Gnomon numbers. In solid tigues are the Cubic numbers, Tetrahedral numbers and Square Pyramidal numbers. Triangular numbers are numbers which can be drawn as dots and arranged in triangular figure. Example: 1, 3, 6, 10, 15, 20, ... Square numbers are numbers which can be drawn as dots and arranged in square figure. Example: 1, 4, 9, 16, 25, ..'.

cu'bic numbers are numbers which can be drawn as dots and arranged as a cube. Example: 1, 8, 27, 64, ... Tetrahedral numbers are numbers which can be ·drawn as dots and arranged as a tetrahedron. Example: 1, 4, 10, 20, ... Square pyramidal numbers are numbers which can be drawn as dots and arranged as a pyramid with square base. Example: 1, 5, 14, 30, ... finite can completely be counted off from 1 to the last whole number. first cost the initial cost of a capitalized property, · including transportation, installation, preparation for service, taxes, and other related initial !)xpenditure in order to make the property functional first derivative the derivative of a function. Normally the first derivative of the function is the slope of the function. First Proposition of Pappus the area of a surface generated by rotating.any plane curve about non intersecting axis in its plane is equal to the product of the length L of the curve and the distance traveled by its centroid. Or . expressed as

a

Area = length of arc x circumference described by the centroid of arc

Appendix A- Glossary 601 flat angle (syn. straight angle) an angle whose measure is 180". flow chart a sequence of logical computations often represented with rectangles, parallelograms and arrows. focal measurement from the focus to a certain point. focal chord a line segment joining two of its points and passing through a focus of a conic focal radius is a line SE!!lment from a focus to one point of the conic. focus a fixed point on the concave side of a conic section. Folium of Descartes a plane curve which forms a · loop on one side and intersect itself at a node. Its standard equation is xJ + yJ = 3axy where x + y + a = 0 is the equation of the line. formula a symbolic statement of mathematical expression which is syntactically correct. Fortune's conjecture 'If q is the smallest prime greater than P + 1, where P is the product of the first n primes, then q - P is prime.· fourth proportion In the proportion a:b the fourth proportion.

=c:d, d is

fractal a geometric shape that can be subdivided at any scale into parts· that are, " at least · approximately, reduced-sized copies of the whole. The term 'fractal' comes from Latin "tractus' which means a broken surface

fr3e body diagram the diagram of an isolated body with the representation of all external forces acting on it

frame a structure with at least one of its individual member is a multiforce member.

I I

1.1' ' I i

frequency the number of times an event occurs within a given period. frequency polygon a graph on which the frequencies of classes are plotted at the class mark and the class marks are connected by straig hi lines

,II

i

I

friction the limited amount of resistance to sliding between the surfaces of two bodies in contact frustum a part of cone or pyramid lying between the base and a plane parallel to the base. frustum of a regular pyramid the portion of a regular pyramid included between the base and a section parallel to the base frustum of a right circular cone the portion of a right circular cone include between the base and a section parallel to the base. function a relation in which every ordered pair (x, y) has one and only one value of y corresponding to the value of x fundamental operations of arithmetic referring .to the four operations - addition, subtraction, multiplication and division. In algebra, fundamental operation follows the sequence of 'My Dear Aunt Sally' which means that Multiplication and Division must be calculated first before Addition and Subtraction.

,II

I

il

,,li 1:'1

I:

fundamental theorem of algebra •The result that any polynomial with real or complex coefficient has a root in the complex plane.'

fraction a ratio of two integers such as a/b , with a not a multiple of b and b is not zero or one. fundamental theorem of arithmetic "Every The value 'a' is called the numerator and the positive integer greater than 1 is a prime value 'b' is called the denominator. When the number or can be expressed as a unique numerator is less than the denominator, it is a . product of primes and powers of primes.' common or vulgar or proper fraction otherwise it will be called improper fraction. An improper future worth the equivalent value at a designated fraction is a:ways written as a mixed number, future based on time value of money. that is an integer and a proper fraction such as

3Y,. This was formulated by Pappus (c.300-{;.350) of Alexandria.

\~i:!il il l!l

.,

',! I I

602 · 100 1 Solved Problems in· Engineering Mathematics (2nd Edition) by Tiong & RojilS

~

G g notation for gravitational constant. g =32.2 feet per second per second or 9.81 meters per second per second.• G notation for gravitational constant. G =6.67 x 10-11 N·m2/kg2.

..!

Gabriel's hom a surface of revolution of y =

X

generat"r (syn. generatrix) a point. line or plane whose motion forms a geometric figure. geometric mean (syn. geometric average) The geometric mean of n numbers is the nth root of the product of the numbers. For example, the geometric mean of a and b is M . The geometric mean is always less than the arithmetic mean except if all numbers are equal. geometric progression a sequence of numbers such that the ratio of any term· to the preceding term is constant.

centesimal minutes or centigon and 1 centesimal minute is divided into 100 centesimal seconds or milligan.

Hankel matrix a matrix in which all the elements are the same along any diagonal that slopes from northeast to southeast

googol the number represented by 1 followed by a hundred zeros or 10100. The name was coined by a 9 year old nephew of the American mathematician Edward Kasner in 1930s.

harmonic analysis the method of expressing periodic functions as sums of sines and cosines

googolplex the number represented by 1 followed by a google of zeros. 1 googolplex = 10 google = 1010'00 gr'ad (syn. gon or centesimal degree) unit of angle measurement with one revolution equivalent to 400 grads. gradient slope of line or the ratio of vertical change 1o horizontal change. graph a. plot of x values against y values for a given function -Gravitational . acceleration the constant acceleration of a free falling body. Denoted as 'g' and is equal to 9.81 meter per second per second or 32.2 feet per second per second on earth's surface. On moon, g =1.62 m/s2 and on sun, g = 274 m/s2.

geometric series a series whose terms form a geometric progression.

gravity the force of attracting between two bodies due to their mass.

Geometry the branch of mathematics which deals with the properties and relations of constructible plane and solid figures.

great circle the intersection of the sphere and the plane through the center of the sphere.

gnomon a geometric figure which is formed by cutting a parallelogram from one corner of another but larger parallelogram.

greatest common divisor the largesi integer that divides each. of a sequence of integers exactly. This is also known as 'greatest common factor" ·

gnomon numbers (see figurate numbers) Goldbach conjecture the conjecture that every even number (except 2) equals the sum of two · prime numbers. This was named after the Prussian-born number theorist and analyst, Christian Goldbach {1690-1764). gon {syn. centesimal degree, grad) 1/400 of the full angle. 1 revolution = 400 gons and 1 right angle= 100 gons. 1 gon is divided into 100


harmonic division the division of line segment by two points such that it is divided externally and internally in the same ratio harmonic mean a term in between two harmonic terms of a harmonic progression. It is the reciprocal of arithmetic mean. The harmonic . 2ab mean of a an db 1s - a+b • harmonic progression a sequence of numbers whose reciprocals form an arithmetic progression. Harshad number a number that is divisible by the sum of its own digits. This is also known as Niven number. height the measure of an altitude of any polygon. helix a curve in space which lies on a cylinder and crosses its elements at a constant angle. hemisphere a portion of sphere cut off by a plane through its center..A hemisphere is one half of a sphere. heptagon a polygon with seven sides

greatest lower bound the largest real number that is smaller than each of the numbers in a set of real numbers gross twelye dozen. This is equivalent to 144 items.

H Half-angle formulas formulas that express n trigonometric function of hnlf nn anglo.

Heron's formula the formula for the area qf a triangle with all sides given. This was named after the first century Greek mathematician Heron (c.A.D.75) of Alexandria. The formula is A= ~s(s-aXs-bXs-c) where a,b and c are the length of the sides and s= (P.+b+c) 2 Heronlan triangle a triangle with integer sides and integer area

hexadecimal number system a number system which uses a place value of 16. This number system uses 16 symbols (10 basic numbers and the letters A, 8, C, D, E. and F). hexagon a polygon of six sides.

I

hexahedron a polyhedron with six faces. A cube is a regular hexahedron. hippo pede a quartic curve which has a rectangular equation of 2

(x +y

2

t +4b(b-a)(x

2

i

I il 1

' 11

2

+y }-4b 2 x2 =0

where a and b are positive constants histogram a vertical bar graph that shows the frequencies of scores or classes of scores by the height of the bars holism the idea that the whole is greater than the sum of its parts homogeneous For polynomials, it is the one having all terms the same degree. For example, x3 + 3x2y + y3. II

Hooke's Law Within the elastic limit, the stress is proportional to strain. The constant of proportionality is called the modulus of elasticity, E or sometimes known as the Young's modulus. Named after the English mathematician and scientist, Robert Hooke (1635-1703).

[I'll,, I• li: 1.,',

ll':,j. :1,

II' lr1

Hydraulics the branch of Physics that deals with the properties and characteristics of fluids.

1 1

,·r1ii

hydrodynamic~

is the branch of hydraulics th!lt deals with l'fuids in motion.

11!11

hydrostatics is the branch of hydraulics that deals. with the fluids at rest. . hyperbola the locus of a point which moves such that the difference of the distances from two fixed points (foci) is constant and is equal to the length of the trilnsverse axis (2a). Eccentricity of hyperbola . is always greater than 1. Hyperbolic spiral a curve with a polar equation of re =a

1·1

II! 1

1:

l'i' il l 1

II1!111h. 1

,,I il

I


604 ·100 1 Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas hyperbolic functions functions of angle expressed in exponential functions such as sinhx = (ex- e-x) 2

inertia property of a body resisting any change in its state of rest or of uniform motion in a straight line.

icosahedron a polyhedron with twenty faces. A regular icosahedron has faces equilateral triangles.

inferential statistics the methods used to describe a population (universe) by studying a random sample of that population

identity referring to the property of being the same. idempotent the element x in some algebraic functions such that x times x =x.

coshx = (ex+ e-x) 2

infinite not finite, having value which are extremely large amount.

imaginary number number in the form of ~ and denoted as I

tanhx = ex -e-x ex+e-x hyperboloid a surface of revolution produced by rotating a hyperbola in space about one of its axes, called its axis of revolution

improper fraction a ratio of positive integer in which the value of the numerator is greater than.that of the denominator.

hypocycloid the plane curve tr;:~ced by a fixed point on a circle as it roll along the inside of a lixed circle. The parametric equations are:

impulse the product of the force and the time during which it acts

x =(a -b)cos(t)+ bcos[

(~ -1}]

y =(a- b)sin(t) -.bsin[(

~-1}]

infinitesimal a number that is greater than zero yet smaller than nay positive real number infinity the concept of being beyond the finite value. The symbol of infinity (oo) was introduced by John Wallis (1616-1703) in his Arithmetia in fini torum in 1655. inflection a point on a plane curve where the curve has a stationary tangent, at which the tangent is changing from rotating in one direction to rotating in the opposite direction

incenter the point of concurrency of the angle bisectors of a triangle incircle the center of the circle inscribed in a triangle.

inscribe to draw a geometric figure inside another geometric figure in such a way that the two figures havillJ common but not intersecting points.

inclination (see angle of inclination) hypotenuse the longest side of a right triangle or the side opposite to the right angle. hypothesis the part of the theorem which is assumed to be true. hypotrochoid a curve formed by the path of a point attached to a point c, which is not on the circumference of a circle of radius b that rolls around the inside of a larger circle or radius a. The parametric equations are: x = (a-b)cos(t)+ccos[

(~ -1}] ~

y =(a- b)sin(t)- csin[ ( -1}]

Incomputable number a real number with an infinite decimal expansion that cannot be enumerated by any universal computer

inscribed angle an angle whose vertex is a point on the circle and whose sides are chords

increment small change in the value of the variable. The increment of x from x = a to x = b is the difference, .b - a.

instantaneous power the limit of the rate of work done as time approaches zero. instantaneous velocity the limit of average velocity of the body as the . time interval approaches zero

indeterminate an expression with no direct meaning \IS a number. Example are 0/0, oo/oo, 0/oo, 00...

Integers are counting numbers (natural numbers). and th~ negative of the counting numbers and the number 0.

index (syn. exponent) refers to the number n of a railical

rr

Integrand the function to be integrated

I

induction a method of reasoning by which one infers a generalization from a series of instances

Integral the result of an integration. The integral sign

inelastic not elastic; having elasticity tess than one.

i notation for imagi[lary number, ~· I roman numeral for 1.

J by

Leibneiz. The word integral

eomes from a Latin origin which means 'making up a whole".

inequality a statement that one mathematical expression is 1greater than or less than another

Indefinite integral (syn. Primitive integral or antidiverivative) an integral with no restrictions imposed on its .independent variable.

j

Definite i'ntegral an integral defined by the limit values of the independent variable. Double integral an integral in which the integrand is integrated twice. Triple integral an integral in which the integrand is integrated thrice. Integral Calculus the branch of calculus which deals with evaluation of integrals and their applications. integral part the biggest integer not greater than the given number. In the number 5.12, the int!'Qral part is 5.. In the number -5.12, the integral part is -6. integration the operation of transforming a function to its definite or indefinite integrals. intercept refers to the intersection of two geometric figures. X and y intercepts are the intersection of the curve with the x and y axes respectively. interest the periodic compensation for the use of money. Sometimes referred to as the time value of money. interest rate the ratio of the interest payment to the principal for a given unit of time and is usually expressed as a percentage of the principal. International System (SI) the metric system,orunit defined by the General Conference of Weights and Measurements in 1960. Sl stands for the French equivalent, Systeme lntemational interpolation the procedure for estimating intermediate value that are not listed in a table of numerical values. The . simplest form of interpolation is linear interpolation, which has for its variation, which has for its variation of the functional described by a straight line. If the function does not satisfy the condition of linearity of variation, graphical interpolation is used. intersection a point where the. curve crosses the · coordinate axis. This also refers to the set consisting of the elements that are common to the original set.

I' IIIII


606 · 100 1 Solved Problems in Engineering Mathematics (2"ct Edition) by Tiong & Rojas intersection of two sets the set of all objects common to both sets invariant a value lhat is not changed when a particular function is applied involute of a circ)e a curve which is the path of the end of a taut string as it is unwound from a circle. involution the operation of raising to an integral power, X". This is opposite to evolution. irrational equation an equation in which a variable appears under the radical sign irrational number any number which cannot be expressed as a quotient of two integers: Examples are n, e, ../2 , etc. isochrone a set of points with the property that a given process or trajectory will take the same length of time to complete starting from ay of the points isodiametric having all diameters of equal length. .isogon an equilateral polygon. isomorphism a transformation in geometry tht does not change the lengths of sides and the measure of angles of the figure involved isoperimetric figures same perimeters

are figures that have the

isosceles having two sides of equal length. An isosceles triangle has two sides and two angles equal. The term 'isosceles" comes from Greek 'iso' (same or equal) and 'skelos" (legs). isosceles trapezoid a trapezoid having non parallel sides equal in length isosceles triangle a triangle having two side equal in length

kilowatt a unit of power equivalent to 1000 Watts;

J notation for imaginary number , ~ engineering and physics applications.

.

1 kN-m

eqUivalent to - - -

s

for

Johnson's theorem 'If three congruent circles all intersect in a single point, then the other three points of intersection will lie on another circle of the same radius. Jordan matrix a matrix whose diagonal elements are all equal and non-zero and whose elements above the principal diagonal are equal to 1 but all other elem~nts are 0 joule (pronounced 'jewel') a unit of work in the Sl units, named in honor of English physicist, James Prescott Joule (1835 - 1889). 1 joule = (1 Newton)(1meter).

kilowatt-hour (kWh) the usual commercial unit of electrical energy. Kilowatt-hour is a unit of work or energy, not power. 1 kWh= 3.6 MJ kinematics the study of motion without reference to the forces which causes the motion kinematic viscosity the ratio of viscosity to the density of the body. kinetic another name for dynamic. kinetic energy In Physics, the quantity equivalent to 1/2 mv2, where m is mass and v is the velocity. kite a quadrilateral with two pairs of congruent adjacent sides and forms like that toy kite.

Kepler's Laws the laws which described the motion of stars, planets and comets, formulated by Johannes Kepler (1571 - 1630). Kepler's three laws of planetary motion: 1. All the planets · of the solar system describe elliptical orbits, having the sun as one of the foci. 2. A radius vector joining any planet to the sun sweeps out equal areas in equal periods oftime. 3. The squares of the periods of revolution of the planets about the sun are directly proportional to the cubes of their mean distances from the sun (the major semiaxes of the elliptical orbits). kilogram (kg) mass of a particular cylinder of platinum-iridium alloy which is now kept at the International Bureau of Weights and Measures at Sevres, near Paris. 1kg = 1000 grams.

leading coefficient the coefficient of the term of highest degree in a polynomial of one variable. In the polynomial, 4x3 + 3x2 - 8x - 10 , 4 is the leading coefficient. leg Of a right triangle, any of the two sides other than the hypotenuse. lemma a proved proposition which is used mainly as a preliminary to the proof of a theorem lemniscate (syn. Lemniscate of Bernoulli) a curve whose equation is r2 = a2 cos 28, where (r,8) are polar coordinates. This was conceived by Jakob Bernoulli (1654- 1705) in 1694. length measure of a line segment.

K kappa curve a curve that resembles the Greek letter, K and has a rectangular equation of y2 (x2 +y2) = a2x2

Law of Universal Gravitation "Every particle in the universe attracts every other particle with a force that is directly proportional to the masses of the two particles and inversely proportional to the square of the distance between the centers of masses".

L

least common multiple the smallest integer that is an exact multiple of every number in a set of integers

L the number 50 in the roman numerals. Lame curve any family of curves related· to the . ellipse with a general equation of 1

1;1" +ltl" =

lamina a thin sheet of uniform thickness and density lateral area area of the surface exclusive of bases.

least upper bound the smallest number that is larger than every member of a set of numbers Leibniz harmonic triangle a triangle of fraction which is closely related to the famous Pascal's triangle and takes the form of 1/1 1/2 1/2 1/3 1/6 1/3 1/4 1/12 1/12 114

lateral edge the intersection of the lateral faces. lateral face any side of the polyhedron other than the base. lateral surface the union of the lateral faces of a prism latus rectum a line through the focus, parallel to the directrix and intersecting the curve

Lemniscate of Bernoulli a curve which looks like a bow of a ribbon and has a rectangular 2 equation of ( x2 + y2 ) = a2 ( x2 - y2 ) L'Hospital Rule (pronounced as Lopital's Rule) a method in calculus in evaluating indeterminate quantities such as 0/0 and oo/oo. Name in honor of Johann Bernoull's pupil, Guillaume Francois de L'Hospital, (1661-1704).

II

.. , I! .II

608 . 100 l S0lved Problems in Engineering Mathematics (2nd Edition) by Tiong & Roj~s

life the period of after which a machi#le or facility

long radius the distance between a· center and a vertex of a regular polygon.

should be discarded or replaced because of its excessive costs or reduced profitability. Also refers to the period of time after which a ·machine or facility can no longer be repaired in ord'er to perform its design function properly.

Ludolph's number a name used in Germany for the number pi (rr)

Limacon of Pascal a snail-shaped curve with a rectangular equation of

June the portion of a sphere lying between two semi-circles of great circles.

( x2 + y2

-

2rx

t

2

=k

2 (x

lozenge a rhombus with a 6Qo angle

+ y2 ) , where r is

the radius of the rolling circle or the rolled circle and k is a constant limping triangle a right triangle with two shorter sides differ in length by one unit line the shortest distance between any two points line segment a, portion of a line bounded by two points. linear equation an equation in which the variable or unknown appears only the first power and only in the numerator of any fractions literal equation an equation in which some or all of the known quantities are represented by letters. lituus a plane curve with equation r2 = ~ with (r,e) as polar coordinates; It resembles a trumpet shaped spiral and originated with the English mathematician, Roger Cortes (1682- 1716). The word lituus, is of Latin origin which means 'a crooked staff. Plural form of lituus is litui. locus the path of a point which moves according to a given law or equation. Plural form is loci. logarithm an exponent when a number, N is represented as a power of a fixed number called based. logarithmic spiral a type of spiral with a polar equation of r = ab6 logic the branch of mathematics which concern with how one statement can imply others, or how set of statements can be connected by chains of implications

M M the number 1000 in the roman numerals. Macclaurin trisectrix an anallagmatic curve that intersects itself at the origin, It has a rectangular equation of y2 (a + x) = x2 (3a- x) magnitude the absolute value of a vector quantity. mantissa the decimal part of a logarithm . In the expression log 25 = 1.39794.. the value 0.39794 is called the mantissa while the value 1 is the characteristic mass a measure of inertia of a body, which is its resistance to a change in ve~ocity. mathematical expectation expected value.

another term for

mathematics the group of subjects (Algebra, Trigonometry, Calculus, Geometry, etc.) used in investigation of numbers, space and the many generalization of these concepts. It is also defined as the science of patterns, real or ' imaginary. The word mathematics comes from Greet 'mathema' which means 'knowledge' Matheson Formula the formula used for Declining Balance Method depreciation, k =

_{CO vcn

where Co and Cn are the firSt and last cost. respectively. With this method d computing depreciation, the last cost, Cn should not be equal to zero.

Appendix A - Glossary _209 matrix a rectangular arrays of numbers forming m rows and n columns. Types of matrices: 1. Square matrix -· a matrix where the number f columns equals the number of rows. 2 Row matrix - a matrix which only one row. 3. Column matrix - a matrix which has only one column. 4. Lower triangular matrix - a matrix where all entries above the main diagonal are zero 5. Upper triangular matrix - a matrix where all entries below the main diaqonal are zero. 6. Scalar matrix - a diagonal matrix where a11 = an =a33 =.. . = k. where .k is a constant. 7. Unit matrix or identify matrix- a scalar matrix where k =1 8. Null matrix - a matrix in which all entries are zero 9. Complex m&trix - a matrix with at least one of the entries a complex number.

meter defined in 1960 as 1,650,763,73 wavelength of !he orange,red light emi!IE)d by atoms of krypton 86(BSY·) in a glow discharge tube. Redefined in 1983 as the distance !hal light travels (in vacuum) in 1/299,792,458 second. The latest definition is more accurate than the first.

mean (syn. average) 1. The arithmetic average of all the scores in the distribution 2. the average of two quantities.

method of exhaustion a method of finding an area by approximating it by the areas of a sequence of polygons

mean proportional the second and the third term~ of a proportion with the second term equals the third term.

mil a unit of angle measure with one revolution equivalent to 6400 mils

median 1. a line drawn from the vertex of a triangle to the midpoint of the base 2. (in stalistic) a point in the distribution of scores at which 50 percent of the scores falls below and tile 50 percent of the scores fall above median of a trapezoid a line joining the midpoints of the non parallel sides mediator the perpendicular segment.

measure a way of determining how large something in terms of weight, volume, mass, length, etc. measure theory the part of mathematics which investigates the conditions under ·which integration can be carried out. mechanics a branch of physical science which deals with state of rest or motion of bodies under the action of forces medial triangle the triangle .whose vertices are midpoints of the sides of a given triangle

of a line

Mersenne number a positive integer of "the form 2" - 1, where n is a prime number. This was named after a French mathematician and Franciscan Priest. Marin Mersenne (15581648).

mile means the secood and the third terms in the proportional ,of four quantities

bisecto~

a unit of length. Statue mile = 5280 feet Nautical mile " 1 minute angle on a great circle = 6280 feet. The nautical mile is 800 feet longer than the statue mile. The word mile comes from Latin 'mille passes" which means "one thousand paces"

milliard In United Kingdom, a thousand million. Billion (instead of milliard) is used ir1 the United States of America. million a thousand thousand. The word comes from Latin "miile" which means one thousand and the suffix 'ion' means great; so million iiterally means a great thousand minimal prime a prime number that is a substring of another prime when written in ba~ ~ 10

610 1001 Solved Problems in EngineeriJ;Ig Mathematics (2nd Edition) by Tiong & Rojas minuend In an expression x- y, x is the minuend minute a measure of both time and angle Time: 1 minute = _!_ hour = 60 seconds . 60

_!_ degree = 60 seconds

monotonic property of a function that is always strictly decreasing or strictly increasing but never both motion any change in the position of a body.

60

multinomial the algebraic expression consisting of a sum of any number of terms.

Mirifici Logarithmorum Canonis Description a book published by John Napier in 1614 which means • A Description of the Marvelous Rule of Logarithm'.

multiple any number of polynomial that is a product of a given number or polynomial and a multiplier. For example; 15 is a multiple of 3.

mixed number a number formed of an integer and a proper fraction. Example 51h

multiplicand the number being multiplied by another. In an expression 5 x 2 = 10, 5 is the multiplicand and 2 is the multiplier. Both 5 and 2 are factors of 10.

Angle: 1 minute =

Mobius strip a twisted surface in space formed by turning one side of a rectangle through 180° (relative to the opposite side) and join it to the opposite side. This was named after a German mathematician Agustus Mobius (1790-1868). . mode the most frequently appearing score or group of scores appearing in the distribution model theory The study of mathematical structures that satisfy a particular set of axioms, especially in the field of logic modulus of common logarithm in converting natural logarithms into common logarithms, the following formula is applied; log N = 0.43429 in N. The number 0.43429 is the modulus of common logarithm modulus of elasticity the ratio of the tensile stress to tensile strain. This is sometimes called Youg's Modulus, in honor of Thomas Young. moment of force another name for torque.

multiplicative inverse the reciprocal of the number. The multiplicative inverse of 5 is 1/5. multiplier (see multiplicand) mutually exclusive a condition where two events cannot happen at the same time, or when one occur, the other one will not occur and vice versa. myriad very large number. II comes from Greek 'murios' meaning "uncountable'

'N nadir the point on the celestial sphere directly below the observer. Nagel point a point in a triangle where the lines from the vertices to the points of contact of the opposite sides with the excirc!es to those sides meet ·

moment of inertia the quantity equivalent to the area times the square of the distance from the centroid to the axis considered. It has a unit to the 4th power.

Napierian logarithm logarithm to the base e = 2.718281828 ... This is also known as Natural logarithm. Denoted as In or log•.

momentum the product of !he rnass and velocity of the body

Napier's rules rules used in solving spherical right triangles.

monomial an algebraic expression of only one term

Rule 1 (Taro-ad rule) The sine o/uny middle part h; equal to the product of llln[lenl of the adjacent paris.

-------------------------------· Rule 2 (Co-op rule) The sine of any middle pari is equal to the product of the cosine of the ppposite paris. nappe either of the two parts into which a cone is divided by the vertex. narcisssictic number an n-digi! number equal to the sum of its digits raised to the nth power. Another term for this is "Plus perfect number" or 'Armstrong number". natural logarithm (see Napierian Logarithm) natural number are numbers considered as counting numbers. Example: 1, 2, 3, ... : Zero and negative numbers are not considered as natural numbers.

_____ A~dix A- Glossary 61.1

null equal to zero; empty null hypothesis a hypothesis that is being tested for rejection number theory the study of the whole numbers and their properties and relationships numeral · symbol or combination of symbols representing a number such as. Arabic and Roman numerals. Arabic numerals (e.g. 0, 1, 2, 3, 4, .. ) are the modification of the HinduArabic Numbers. The Roman numerals are certain letters of the Latin alphabet. Roman numeral Arabic equivalent I 1

v

5

X L

10 50 100 500 1000

c

negative less than zero. negative angle angle counterclockwise rotation.

measured

in

newton the amount of force thai gives an acceleration of one meter per second squared to a body with mass of one kilogram. Denoted as N. 1 N =1 kg-m/s2 Newton's First Law Every body continues in its state of rest. or of uniform motion in a straight line, unless it is compelled to change that state by forces impre~sed on it. This is also knowh as the law of inertia. Newton's Second Law The rate of change of linear momentum of a particle is C;Jqual to the total applied force.

D M

numerator In the fraction ~ , x is the numerator, y y is the denominator. numerical integration (syn. approximate integration) the process of finding an approximate value of a defmite integral without carrying out the process of evaluating the indefinite integral.

0

Newton's Third Law For every action, there is always an equal and opposite reaction.

oblate spheroid (syn. oblate ellipsoid) prOdUGed by rotating an ellipse through a complete revolution about its minor axis.

nominal interest the number employed loosely to described the annual interest rate

oblique angle consist of all angles except right and· straight angles

nonagon a polygon of 9 sides

oblique cal'tesian coordinate system a cartesian COOrdinate system in Which the X and y axes are not perpendicular

normal another orthogonal.

name

of

perpendicular

or

normal number a number in which digits sequences of the same length occur with the same frequency

oblique circular cone a circular cone whose axis is not IJerpendicular to the base of the cone. oblique triangle a triangle with no right angle.


612 1001 S~lved Problems in Engineering Mathematics (2"d Edition) by Tiong & Rojas oblong another term for rectangle. This comes from Latin "ob' meaning excessive and "longus' meaning long oblong numbers (see figurate numbers) obtuse angle an angle that is greater thc'm 90 degrees but less than 180 degrees. obtuse triangle angle.

a triangle having one· obtuse

octagon a polygon of eight sides octahedron a polyhedron with eight faces octal number system a number notation which uses base 8 as a place value. It uses the digits 0, 1, 2, 3, 4, 5, 6, and 7. octants referring to the eight compartments 9f the rectangular coordinate systems in space odd not divisible by two. odd function a function f(x) with the property that f(-x) = -f(x) for any value of X. ogive a graph of cumulative frequency distribution plotted at the class marks and connected by straight lines ordinal numbers numbers which state their position in a sequence. Example: First, second, third, ... ordinary annuity an annuity where the payments are made the end of each period starting from the first period. ordinate the position of any point on a plane from the x-axis. Also known as the y-coordinate. origin the intersection of the x and y axes of the cartesian or rectangular coordinate systems. The origin has coordinates of (0,0). orthic triangle the triangle whose vertices are lhe feet of the altitudes of a given triangle orthocenter the point of intersection of all the altitudes of a triangle.

orthogonal normal or perpendicular. Its symbol is

p

2

1

pandigital number a number that contains eacli of the digits from zero to nine exactly once and whose leading digit is non-zero. Example 1,432,576,098 Pappus theorems theorems that determine tha! surface area and volume of a figure generated about an axis. This is named after Pappus of Alexandria. First proposition of Pappus: "If an arc is rotated about an axis, it generates as surface are which is equal to the length of arc times the circumference described by the centroid of the arc.' Second proposition of Pappus: "If an area is rotated about an axis, it generates a solid of revolution, whose volume is equal to the area times the circumference described by the centroid of the area'. "parabola a locus of a point which moves so that it is always equidistant from a fixed point (focus) and to a fixed line (directrix). Ecr..entricity of parabola = 1. parabolic segment the plane region bounded by a chord a parabola perpendicular to the al{is and the arc of the parabola cut off by the chord. The area of the parabolic segment is 213 of the circumscribing rectangle.

~bh 3

parallel line or plane that will never intersect no matter how far they are extended. parallelepiped a prism parallelogram region.

3

4

Paraboloid

palindronic number a number such as 1,234,321 that reads the same forward and backward

A=

$

1

whose

base

is

a

B

Rectangular parallelepiped

parallelogram a quadrilateral in which both pairs of opposite sides are parallel.

3

6

4

The number can be found by adding the two numbers above it. For example, 6 in the triangle was obtained by adding 3 an 3, the numbers above it. Named (;lfter French mathematician, philosopher and physicist, Blaised Pascal (1623 -1662). pearls of Sluze curves that are generated by the rectangular equation y" = k(a- x)P xm where n,m and p are integers pedal curve the locus of the feet of the perpendiculars from a given point to the tangents to a given figure pedal triangle a triangle inscribed in a given triangle whose vertices are the feet of the three perpendiculars to the sides from some point inside the given triangle. Pelf equation a equation of the form y2 = ax 2 + 1 where a is any positive whole number except a square number

Parallelogram

parameter generally an arbitrary constant. partial fraction the parts of an algebraic expression which contain a polynomial in a single variable in the denominator, or in the denominator and numerator, when split partnership (type of business organization) an association of two or more persons for the purpose of engaging in a business for profit. pascal (Pa) a ~nit of pressure. 1 pascal = 1 newton per square meter.

percent on diminishing value (see declining balance method) pencil a collection of lines that passes fixed point or a given point.

th~ough

a

pentagon a polygon. of five sides pentagonal numbers (see figurate numbers) pentagram (syn. pentangle, pentacle) a starshaped figure formed by extending the sides of a regular pertagon and meet at vertices. pentedecagon a polygon of fifteen sides

paraboloid a solid of a revolution of a parabola. Volume of the paraboloid is always equal to one-half of the volume of the circumscribing cylinder.

Pascal's Law "If any external pressure is applied to a confined fluid, the pressure will be increased at every point in the fluid by the amount of the external pressure.' Pascal's Triangle a triangular array of numbers which is made up of the binomial coefficient of the binomial expansions:

per cent a word of latin origin which means every hundred. Its symbol is %. percentage a ratio by which the denominator is 100.

I

,'1 Appendix A- Glossary 615

614 1001 Sblved Proble~s in Engineering Mathematics~d Edition) by Tiong & Rojas perfect number a number the sum ·of whose factors including one but excluding itself is exactly equal to the number. Example: The factors of 6 are 3, 2 and 1. Adding the factors .will yield the number itself such as

1 + 2 + 3 =6. Hence, 6 is a perfect number. perfed power an integer of the form m" where m and n are integers and n > 1 perfect square a number that is the product of two equal whole numbers. Example 4 =2 x 2. perigon an angle equal to one revolution (360°) perimeter the sum of the sides of a polygon. This is known as circumference for a circle. permutable prime a prime number with at least two distinct digits, which remains prime on every rearrangement of the digits. permutation an arrangement of a set of objects in a particular order. The permutation of 'n' different things taken 'r' at a time is given as n! P(n,r)=~

1[1- r I'1

Mathematically, A= i+(%)-1, where i is the number of interior lattice points and b is the number of boundary lattice points

polyiamond a shape made from identical equilateral triangles that have been joined at their edges

place-value system a number system in which the value of a number symbols depends not only on the symbols itself but also on the position where it occurs

polynomial an expression of several terms. It may include any number of terms.

Planck time the shortest meaning period of time in quantum m.echanics

planimetry the measurement of plane areas.

postulate in Geometry, the construction or drawing of lines and figures the possibility of which is admitted without proof.

point a dimensionless geometric object having no properties except location and place.

pound a unit of force in the English (British) system. tf is equivalent to 1 slug-ft/s2. Also 1 pound

point of inflection !he point on the graph where the curve changes from concave up to concave down and vice versa

perpetuity an annuity where the payment periods extends forever or in which the periodic payments continue indefinitely. ·

Poisson ratio The ratio of the unit deformations or strains in a transverse directions is constant for stresses within the proportional limit.

circumference to the diameter of a circle. Its value is 3.14159 ... and has for its symbol 'Jt. This symbol for pi {n) was introduced in 1706 by William Jones (1675 -1749). Pick's theorem 'The area of a polygon can be found simply by cotlnting the lattice points on the interior and hoiJndary of the polygon.'

population (syn. universe) in statistics, it refers to all the members of a particular group of items or individuals. positive having values greater than zero.

plane a surtace such that a straight line joining any two points in it lies wholly in the surface

point of tangency is the point of contact of the tangent and the curve

pi an irrational number which the ratio of ttie

polyhedron a solid bounded by planes

pie chart a circular diagram divided into sectors of which the area are in proportion to the magnitude of the represented values.

perpendicular (syn. normal, orthogonal) forming a right angle.

Philosoph late Naturalis Principia Mathematica ('Mathematical Principle of Natural Science')a book publish by Sir Isaac Newton in 1686. This book clearly slates the fundamental law~ of nature which is now referred to as the Newton's Law, the cornerstone of mechanics.

polygonal region is the plane figure formed by fitting together a finite number of triangular regions

polar angle (syn. vectorial angle, the argument, the amplitude of the azimuth of the point) the angle the vector makes with the polar axis. polar coordinates coordinates in the form of (r,6) used to locate a point .in the rectangular coordinates system. To convert polar to rectangular, use the following r'!lations: X ::: r COS 6, y = r sin 6

=4.448221615260 N. power 1. the rate at which work is done or energy is transferred 2. (syn.. exponent) the number of times the number is multiplied by itself. power law a type of mathematical pattern in which the frequency of an occurrence of a given size is inversely proportional to some power of its size. power series an infinite series in which successive terms are of the form of constants times successive integral power of the variable. It takes the form of ao+a1x+a2x2 +a:JX3 x ..... . power set the set of all subsets of. a given set, containing the original set as well as the empty set. powerful number a positive whole number n such that for every prime number p dividing n, p2 also divides n. ·

pole the origin of the polar coordinates system polygon a closed figure bounded by line segments.

practical number a number n such that every positive integer less than n is either a divisor or a sum of distinct divisors of n

precision the accuracy in which a calculation is pertormed. present worth the equivalent value at the present , base on time value of money pressure the force per unit area. It has a unit of pascals in the metric system. Since pascal is a small unit, the unit bar or MPa are used instead. 1 pascal = 1 newton per square meter. 1 bar = 105 Pascals. The gauge pressure can be calculated using the formula: p = roh, where ro = density (specific weight) of the fluid and h is the pressure head. The absolute pressure is the sum of the gauge pressure and the atmospheric pressure. Standard atmospheric pressure = 14.7 psi = 1.01 x 105 Pa = 760 mm of Hg primitive integral (see integral) prime number an integer which has no other factors except 1 and itself principal In economics, it is the amount invested. prism a polyhedron of which two faces are equal polygons in parallel planes, and the order faces are parallelograms prismatoid a polyhedron having bases two polygons in parallel planes and for lateral faces triangles or trapezoids with one side lying one base and the opposite vertex or side lying on the other base of !h9 polyhedron. prismoid a prismatoid in which the two bases are polygons of equal number of sides and the lateral faces are quadrilaterals. prismoidal formula formula used in finding the volume of a prismatoid such as follows,

V=~(A1 +4Am +A2). where, L is the distance betw.een end areas, A, and A2 are end areas and Am is the area at the mid-section.

!

616 1001 Solved Problems in Engineering Mathematics (2"d Edition) by Tiong & Rojas probability the ratio of the successful outcome of an event to the total possible outcome of an even!. The value of the probability is always less than 1. product the result of multiplication. progression a sequence of numbers in which one is· designated as first. another as second , another as third and so on. Types of progression are Arithmetic progression, Geometric progression, Harmonic progression, Infinite Geometric progression, etc. projectile initial velocity of a body and then follows a path determined entirely by the effects of gravitational acceleration and air resistance. prolate spheroid {syn. prolate ellipsoid) an ellipsoid produced by rotating an ellipse through a coMplete revolution about its major axis

1 ll

App~!lciixA-

Glossary 617

Q

R

rational expression any algebraic expression that is a quotient . of two other algebraic expressions

QED latin word quod erat demonstrandum, which .means that a proposition has been proven. The Greek equivalent was used by Euclid in the 3rct century B.C.

radian the angle between two radii with an intercepted arc equal to the radius of the circle. 1 revolution is equal to 27t radians.

rational number any number which can be expressed as a .quotient of two integers (division by zero excluded).

quadrantal spherical triangle a spherical triangle with at least one side a quarter of a great circle. quadrants referring to the four divisions of the rectangular coordinates system. quadratic equation· an equation in which the maximum power of the unknown or variable is 2. Standard quadratic equation is in the form ax2 + bx + c = 0.

radical the symbol . [ This , symbol was introduced by Christoff Rudolff in 1525.

rationalizing the denominator the process of removing the radicals (or fractional. exponents) from the denominator

radical axis the locus of points of equal power with respect to two circles

ray a straight path of points that begins at one point and continues in one direction

radicand the quantity inside the radical (square root sign)

real number a non imaginary number. It includes the .rational numbers as well as the imaginary numbers all integers and natural numbers.

radius a segment from the center to a point of the circle

reciprocal multiplicative inverse of a number. For example, the reciprocal of 5 is 115.

radius of gyration the distance from a given axis that a particle of the same mass as a rigid body must be placed in order to have the same moment of inertia.

rectangle a parallelogram all of wthose angles are right angle

radius vector the distance of any point P from the origin in the polar coordinate system

rectangular hyperbola a hyperbola with length of semi-transverse axis, "a' equals the length of the semi-conjugate axis, 'b'. Eccentricity of this hyperbola is square root of 2.

a polygon offcur sides

radix the base of a number system. For example, 2 is the radix of a binary number system and 10 is the radix of the. decimal number system

rectangular parallelepiped a polyhedron whose six faces are all rectangles

quantity something with a magnitude or numerical value.

random variable a numeric quantity which can be measured in a random experiment.

quarternary having four variables.

range the set of all second elements of a relation

reduction the process of converting a fraction into a decimal form.

pyramid a polyhedron of which one face, called the base, is a polygon of any riumber of sides and the other faces are. triangles which have a common vertex

quartic a polynomial or polynomial equation that contains the fourth power of the variable, but no higher power

rate of return the interest rate at which the present work of the cash flow on a project is zero, or the interest earned by an investment

redundant equation any equation which, because of some maihematical process, has acquired an extra root

pyramidal numbers (see figurate numbers)

quinary number system which pertains to place value notation of base 5.

ratio the quotient of two numerical measure of two magnitudes of the same kind. The word ratio comes from the Latin verb 'ratus' which means 'to estimate'.

reentrant angle an inward-pointing angle of a concave polygon

proper fraction a ratio of positive integers in which the value of the numerator is less than that of the denominator .Ptolemy's Theorem In cyclic qua9rilateral, the sum of the product of two opposite Sides is equal to the product of the diagonals. Named after Ptolemy of Alexandria or Claudius Ptolemaeus (c.100- c. 168). puro quadratic a quadratic equation of the form ax2 + c =0, that is , the coefficient of the first degree term, b is equal to zero

Pythagorean theorem The sum of the squares of the sides of a right triangle is equal to the square of the hypotenuse; in equation, a2 + b2 = c2 with a and b are legs while c is the hypotenuse. This is named after the Greek· philosopher and mathematician, Pythagoras (c. 580- c. 500 B.C.) of Samos.

Quadratrix of Hippias the first curve recorded in . history that is not a part of a line or a circle. This curve has a rectangular equation of

y = xcot(;:) quadrature formulas- refers to the formulas used in numerical intregration. quad~ilateral

quintic a polynomial ~r polynomial equation that contains the..fifth power of the variable, but no higher power

ratio of similitude the common ratio of the · corresponding sides of two similar polygons

quotient the result of division. '

I

rational equation an equation which is satisfied by all value of the variables for which the members of the equation are defined

rectilinear pertaining to straight line.

reflex angle any angle greater than 180 degrees but less than 360 degrees regular polygon a polygon with all sides equal and all angles equal. A regular polygon is equiangular and equilateral. Also, a regular polygon is convex.

r 618 I 00 I Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas regular pyramid a pyramid whose base is a regular polygon and whose center coincides with the root of the perpendicular dropped from the vertex to the base relation is any set of ordered pairs (x,y) relative ~ensity (see specific grav1ty) relative error a measure of the difference between a number and an estimate. Relativity theory a theory formulated by Albert Einstein the revolutionized of man's understanding about time, space and gravity. Mathematically, it is expressed as E = mc2, where E, m and c are energy, mass and speed of light, respectively remainder the amount left when a quantity cannot be divided exactly by a divisor. resultant the single vetor quantity which is the sum of two or more vector quantities. rhombus (syn. diamond, lozenge) a parallelogram all of whose sides are congruent right angle angle equivalent to 90 degrees right triangle a triangle having one right angle rigid body a body which will not be· affected or deformed when an extremely large or extremely small load or temperature is applied. Rolle's theorem "Suppose a continuous function crosses the x-axis at two points a and b and is differentiable at all points between a and b; that is, it has a tangent at all points on the curve between a and b. Then there is at least one point between a and b where the derivative is 0 and the tangent is parallel to the x-axis Roman numerals (see numerals) root value that satisfy a given equation. rose curve a curve that has the shape of a flower with petals had has a polar equation of r asin(rrO)

A~endi:~tl\_~_Giossary

rounding (of numbers) replacing it with another number to produce fewer significant decimal digit. For an integer, fewer value carrying (non-zero) digit. Example: 3.14159 is rounded off to four decimal places as 3.1416. For numbers ... 5, the rule states that it should be rounded off to the .nearest even rounding boundary to minimize the systematic rounding error.

score another term used for the number twenty (20)

rows the numbers in order which appears horizontally in a matrix

second originally defined in 1889 as the fraction 1/86400 of the mean solar day, and redefined in 1967 as the duration of 9,192,631,770 periods of the radiation of a certain state of the cesium-133 atom

s sample any subset of a population sample space the set of all possible outcomes of an experiment. salient angle an outward-pointing angle of a polygon salient point the point where two branches of a curve meet and stop and have different tangents .salvage value the cost recovered or which could be recovered form a used property when removed, sold or scrapped. It is sometimes referred to as second hand value scalar quantity a physical quantity that is described by a single number only, the magnitude. It does not have a direction in space. ·

scrap value the value of an equipment if disposed as junk. This is sometimes referred to as junk value.

619

shear stress a stress that is caused by forces acting along or parallel to the area resisting the force ,Ill

short radius the shortest distance from the center of a regular polygon to any of its sides.

•I

!i:

secant a line which intersects the circle in two points. The reciprocal of the trigonometric function, tangent.

Second proposition of Pappus The volume of any solid of revolution is equal to the generating area times the circumference of the circle described by the centroid of the area. section of a solid the plane figure cut from the solid rv passing aplane through it sector a part of a circle bounded by the radii and an arc.

111:

Figure shows R, the short radius

Siegel's paradox "If a fixed fraction x of a given amount of money P is lost, then the same fraction x of the remaining ~ nount is gained, the result is less than t!:w v11ytnal and equal to the final amount if a fraction x is first gained, then lost.' Sierpinski number a positive, odd integer k such that k times 2n + 1 is never a prime number for any value of n

1 1

'1'111'11:: 11 ·1'

.,:11,,

segment part of a circle bounded by an arc and a chord.

significant figures fdigits the meaningful digits in a number. A number is considered significant unless it is used to the place a decimal point.

septagon a polygon of seven sides sequence (syn. progression) a succession of numbers in which one number id designed as first, another a~ second, another as third and soon

similar having the same shape but not necessarily . the same size simple interest the interest charges under the condition that interest in any time is only charged on the principal

series sum of a finite of infinite sequence.

scalar product of A and B is denoted as AB. Because of this notation, scalar product is also called as dot product

serpentine a curve which nas a rectangular abx . ( ) equat1on y x = -()

scalene triangle a triangle having all sides of unequal lengths

set (syn. class) a collection of objects.

x2 -a2

scattergram the relation between two variables is shown by a series of dots plotted on a graph

sexagesimal pertaining to the number 60.

scientific notation a number ·represent using powers-of-1 0 notations used to described a very large small numbers.

sexagesimal number system a number system .using a place value of 60. This was used by the Babylonians or Mesopotamians and is considered be the oldest number system which dates back to 2 millenium B.C.

sine curve (syn. sine wave) a curve with equation y =sin x. sinking fund method a method of depreciation where a fixed sum of money is regularly deposited at compound interest in a real or imaginary fund in order to accumulate an amount equal to the total depreciation of an asset at the end of the asset's estimated life. skew lines two lines that are not coplanar. This is also known as "crossing lineS" slant height (syn. element) the length of a generator of a circular cone.

1

" 620 · 1001 Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas

slope ratio of the vertical distance to horizontal distance. It also refers to the tangent function of the angle of inclination. Also refers to rise over run. In Analytic Geometry, slope of line is denoted as m. m = Y2 -Y1 x2 -x1

Smith number a composite number, the sum of whose digits equal to the sum of the digits of its prime factors. snowball prime a prime number whose digits can be chopped off, one by one, from the righthand side, yet still leave a prime number.

spherical excess the sum of the angles of a polygon over (n-2) 1so· with n as the number. of sides of the polygon spherical polygon the portion of a spherical surface bounded by three or more great circle arcs spherical pyramid the portion of a sphere bounded by a spherical polygon and the plane of the sides spherical sector a solid generated by rotating a sector of a circle about an axis which passes through the center of the circle but which contains no point inside the sector

spherical segment asolid bounded by a zone and Soddy's formula a formula used for four circles of the planes of the zone's bases radii, r1, r2, rJ and r4 drawn not overlapping but each touches the other three. The following is the Soddy's formula: spherical triangle a triangle extracted from the 2 (b1 +b2 +b3 +b4) = 2(b/ +b/+bl+b/) surface of a sphere. where b1 = 11r1 and so on ... solid is any limited portion of space, bounded by surfaces

spherical wedge the portion bounded by a lune and the planes of two great circles spheroid another term for ellipsoid.


statically indeterminate the condition exists in structures where the reactive forces or the internal resisting forces over the estimated life of the asset in terms of the periods or units of output stationary point a point on the graph of a function where the tangent to the graph is parallel to the x-axis or, equivalently, where the derivative of the function is zero. statistics the study of ways that lots of data can be represented using a few numbers and the study how such numbers can be chosen and used to draw reasonable conclusion about the data stellation the process of constructing new polyhedron by extending the face planes of a given polyhedron past their edges steradian ( sr ) a unit of measure of solid angle. Then maximum value for a solid angle is a full sphere which is equal to 4n steradians. strain the change of relative positions of points due to stress. Unit strain is equal to the ratio of the deformation to the total length.

solid angle an angle formed by three or more planes intersecting at a common point. Solid angles are measured in steradians

spherometer an instrument for measuring the curvature of a surface.

stress a force per unit area

solidus the slant line in a fraction such as alb dividing the numerator from the denominator

square a rectangular all of whose sides are congruent

strobogrammatic prime a prime number that remain the same when rotated through 180 degrees. Example is 619,' when rotated 180° remain the same- 619

space a set of all points

square free an integer that is not divisible by a perfect square, n2, for n > 1

specific gravity the ratio of the density of the substance to the density of water. Specific gravity of water at densed condition (4.C) = 1.0

square matrix (syn. determinant) a matrix with the same number of rows columns

sphere a solid bounded by a closed surface every point of which is equidistant from a fixed point called the center spherical angle the opening between two great circle arcs. A spherical angle is measured by the plane angle formed by the tangents to the arcs at their point of intersection

square numbers (see figurate numbers)

subset a set that contains some of the elements of a given set subtrahend the number to be subtracted. Example: 7 - 5 =2, 5 is the subtrahend. sum the result of addition.

square pyramidal numbers (see figurate numbers) standard deviation a quantitative measure defining the extent to which scores are dispersed throughout in relation to the arithmetic mean. This is also equal to the square root of the variance.

supplementary chords two chords which join a point on a circle to the end points of a diameter.

sum-of-years' digits method (syn. SYD method ) a method of computing depreciation in which the amount for any year is based on !he ratio: (years of remaining life/(1 + 2 + 3 + .. + n), with n being the total anticipated life of the equipment

Angle 0 is the supplemental angle

supplementary angle two angle whose sum is equal to 180° surd an irrational number which is a root of a positive integers or fraction or it is a radical expressing an irrational number. Types of surds: Example Quadratic

J2

Cubic

~ (2

Quartic The type of surd is named after the index of the radical.

Pure surd is a surd that contains no rational number (i.e. all its factors or terms are surds). Example:

J2 , J3 + J2

.•Mixed surd is a surd that contains at least one rational term. Example: 2 + ..J 3 symmedian reflection of a median of a triangle about the corresponding angle bisector

T table c!Wlpilation of values such as trigonometric table, logarithmic table, etc. tangent a line (in the same plane) which intersect the curve in one and only one point. In trigonometry, it is the ratio of the side opposite to side adjacent in a right triangle. tangent plane of a sphere a plane which intersects the sphere in exactly one point

622· 1001 Solved Problems in En'gineering Mathematics (2nd Edition) by Tiong & Rojas tension force in longitudinal direction.

~

terminal speed the final speed Vt attained by the ·falling body. The principle is that when a. body first. start to move, v =0, the resisting force is zero and the initial acceleration is a = g. As it speed increases, the resisting force also increase until finally it equals the weight in magnitude. AI this time, the acceleration becomes zero and there is no more increase in its speed. ternary a number system using a place value noiation with 3 as the base.

transverse axis the axis of the hyperbola which passes through the foci, vertices and center

truss a framework composed of members joined at their ends to form a rigid structure.

trapezium (syn. trapezoid) commonly used term in United Kingdom rather than trapezoid. In United States of America, the tem1 trapezoid is used.

Tschirnhaus's cubic a curve that has a rectangular

theorem a statement of truth of which must be established by proof

trapezoid a quadrilateral in which one and only one pair of opposite sides are parallel. The parallel sides of the trapezoids are called bases.

time value of money the cumulative effect of elapsed time on the money value of an event, based on the earning power of equivalent invested funds capital should or will earn

tree a graph which the property that there is a unique path from any vertex to any other vertex traveling along the edges trefoil curve a plane curve that has a rectangular

torque (syn. moment of force) a force times a moment ami. torr a unit of pressure which is equivalent to mm of mercury (Hg). 1 torr = 1 mm of Hg. 1 torr is equivalent to 133 Pascals. torus (syn. anchor ring or doughnut) a solid formed by revolving a circle about a line not intersecting it trace the sum of the terms along the main diagonal of a matrix trajectory the path followed by a projectile. It is a graph of a parabola. translation a parallel displacement of the original system along one or more of its axes. transpose to transfer to the other side of the equation. When a term is transpose, the sign must be changed. transversal the intersecting line of two parallel or non parallel lines

Appendix A- Glossary 623 truncated value the value of number when written with the further digits have been suppressed and replaced with three dots. For example, the number 1t = 3.141592653... can also be written by truncation as 3.14159 ... The truncated value is not a rounded off value and therefore always smaller than the exact value.

tetrahedral numbers (see figurate numbers)

ton a mass of 1000 kilograms.

~

~

equation of x4 + x2 y2 + y4 = x( x2

-

y2 )

triangular numbers (see figurate numbers) triangular region is the union of a triangle and its interior trident of Newton 'a curve that has a rectangular equation of xy = cx 3 + dx 2 +ex+ f trigonometry branch of mathematics which deals with triangles and trigonometric functions.

equatior. of 3ay 2 = x (x- a)

2

twin primes prime numbers that appear in pair and differ by 2. Examples are 3 and 5, 11 and

13, ...

u undecagon polygon of eleven sides. undulating number an integer whose digits are alternate. Example 343,434 unimodal sequence a sequence that first increases and then decreases unimodular matrix a square matrix whose determinant is 1 union a set consisting of all elements that appear at least once in the original set. union of two sets a set of all objects that belong to one or both sets

trillion a million million or 1012

universal set the set that contains all elements untouchable number a number that is not the sum of the aliquot part of any other number.

v variable an expression than is assigned a certain set of values. variance a measure of the dispersion of scort:ti tn a distribution away from the arithmetic mean. The mean of the squared deviations about the mean. Varignon's Theorem The moment of the resultant or two concurrent forces with respect to the center in their planes is equal to the algebraic sum of the moments of the components with respect to the same center. Named after the French mathematician, Pierre Varignon (1654-

1722). vector quantity a physical quantity the described a magnitude ('row much' or "how big") and the direction in space vector product of A and B is denoted as A x B. Because of this notation, vector product is also called the cross product. velocity rate of change of displacement. Venn diagram a pictorial description of the probability concepts of independent and dependent events. This was named after English logician, John Venn (1834 -1923).

unit circle a circle of radius one unit and is used to determine the sign of all trigonometric functions in all quadrants

vertex point of intersection of two sides of a polygon.

trivial considering the values of all the variables as zero.

unit fraction a fraction whose numerator is 1

vertex figure the polygon. that appears if a polyhedron is truncated at a vertex

Truncatable prime a prime number, n that remains a prime when digits are deleted from it one at a time.

unit vector a vector having a magnitude of unity, with no units and is used only to described a direction in space

vertical angles angles that are opposite to each other and formed by two intersecting straight lines. Vertical angles are equal.

truncated prism ihe portion of a prism included between the base and a plane not parallel to the base cutting all the edges or elements

unity referring to one.

trinomial having three terms.

universe (statistics, see population)

:Jl

,f

~

l:\

624· 1001 Solved Problems in Engineering Mathematics (2"d Edition) by Tiong & Rojas Weight =mass x gravitational acceleration

AppendixB:

whole number another term for natural number

Vertical angles (a = 8)

.vigesimal pertaining to the number 20. vigesimal number system a number system using the base 20 vinculum the bar that is placed over repeatinp decimal .fractions to indicate the portion of the pattern that repeats. Vlnogradov's theorem 'Every sufficiently large odd number can be expressed as the sum of three prime numbers." Viviani's curve a space curve the marks the intersection of the cylinder (x - a)2 + yz =a2 and the sphere xz + y2 + z2 = a2. The parametric equations of Viviani's curve are: x=a(1+cost) y =asint

Wilson's theorem 'Any number p is a prime number if, and only if, (p - 1)! + 1 is divisible byp.' word problems (worded problems) real problems that are usually given orally or written in words work the force time a distance.

X x-axis the horizontal · axis of the ~ rectangular coordinate system. x-intercept the value of the abscissa of the point where the curve crosses the x-axis

y yard a unit of distance equivalent to 3 feet

z=2asin(4J volume space occupied by a solid. Volume is expressed in cubic units.

w Waring's conjecture 'For every number k, there is another number s such that every natural number can be represented as the sum of s kth powers.' · watt the Sl unit of power, This is equivalent to joules per second. weak inequality an inequality that permits the equalitt:Case. weight the force · of the earth's gravitational attraction for the body. It is .. 1ownward force acting at the centroid or center of gravity of the Wdy.

y-axis the vertical axis of the rectangular ·coordinates system.

y-intercept the value of the ordinate of the point where the curve crosses the y-axis Young's modulus (see modulus of elasticity)

z zenith a point in the celestial sphere directly above the observer. · zero (syn. cipher) void, emptiness or nothing. Zero is derived from Hindu word ·sunya"

Appendix 13 - Conv~sion_ .§25 ·

+

of~eigh~l

The English System and Measures

Units

A. !.!near Measure,.t!:J!nm.!:!l 1 inch = 1,000 mils .

Definitions:

1 foot= 12 inches 1 yard "' 3 feet

Inch - the length of three barley grains placed end to end

1 rod= 5.5_yards 1 chain = 4 rods

Digit- the breadth of a finger, about· 0.75 inch

1 furlong "' 10 chains 1 furlong :::: 40 rods

Palm ... the breadth of a hand, about 4 inches

1 mile = 8 furlongs 1 mile

Hand - the length from the wrist to the end of the middle finger, about 8 inches Cubit- the length of the forearm from the point of the elbow to the end of the middle finger, about 18 inches

= 5280 feet

1 mile= 1,760 yards

1 league = 3 miles B.Naut~

1 fathom :::: 6 feet

Foot- the length of man's foot, about 12 inches_

1 cable = 120 fathoms

Fathom -the length of rope when pulled between a man's two outstretch~d arms, about 6 feet

1 nautical mile per hour = 1 knot

Furlong - the length of a short race, about 1/8 of a mile Mile - the distance of a thousand paces,· as marked off by a length 5 feet between lifts of the same foot. The mile is 5280 feet.

league -the distance a person can see across a flat field, about 3 miles

1 nautical mile

C.

Surve~~!l

Measure

1 link= 7.92 inches

= 100 links 1 chain = 66 feet 1 furlong = 10 chains 1 chain

1 mile

==

80 chains

p.' Square Measure (Area} 1 square foot= 144 square inches 1 square yard

zone a portion of the surface of a sphere included between two parallel planes

= 6,080 feet

==

9 square feet

1 square rod = 30.25 square yards 1 acre = ·t60 square rods

Appendix B - Conversion 62!

626 · IOO,f Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas 1 acre = 4,840 square yards ·1 square mile = 640 acres

E. Cubic Measure (Volume)

I, Weifiht !Avoirdupois) The avoirdupois system is used for general weighing

M. Angular or Circular Measure 1 minute = 60 seconds 1 degree = 60 minutes ·

1 dram= 27.3438 grains

1 zodiac sign = 30 degrees

1 cubic foot= 1,728 cubic inches

1 ounce = 16 drams

1 radian= 57.2958 degrees

1 cubic yarc.t= 27 cubic feet

1 pound = 16 ounces

1 quadrant I right angle = 90 degrees

1 gallon (US)= 231 cubic inches

1 stone= 14 pounds

1 circle :::: 360 degrees

1 gallon (UK)::: 277.27 cubic inches

1 long hundredweight (UK)

1 bushel (US) = 2,150.42 cubic inches 1 bushel (UK)= 2,219.36 cubic inches

= 4 quarters 1 short hundredweight (US) = 100 pounds

F. Liquid Measure (Capacity) 1 tablespoon = 3 teaspoon

1 long ton (UK) = 2,240 pounds

Diamond= 10 Corundum= 9

J. Weight (Troy)

Topaz= 8 Quartz= 7

1 cup = 8 fluid ounces

The troy system is used for weighing precious metals or gems

1 pint = 2 cups

Labradorite = 6 Smithsonite = 5

1 pint= 4 gills

1 carat= 3.086 grains

1 quart = 2 pints

1 pennyweight = 24 grains

1 gallon

The scale for hardness of precious metals or gems runs from 10 to 1, with 10 being the hardest.

1 short ton (US) = 2,000 pounds

1 fluid ounce = 2 tablespoons 1 gill = 4 fluid ounces

N. Hardness

= 4 quarts

1 ounce = 20 pennyweights

Fluorite= 4 Calcite= 3 Alabaster = 2

1 pound= 12 ounces G. Dry Measure (Capacity) 1 quart

=2 pints

K. Weight (Apothecaries)

1 peck = 8 quarts 1 bushel

The apothecaries system was formerly used by pharmacists

= 4 pecks

1 UK dry quart = 1.032

us dry quarts

1 scruple

=20 grains

1 dram = 3 scruples H. Dry Measure (Cooking) 1 pinch = 1/8 teaspoon

1 ounce = 8 drams

Definitions: The following are the basic units: Meter - the distance traveled by light in vacuum in 1/299.792.458 second

1 pound= 12 ounces

1 dash= 1/16 teaspoon 1 sprinkle= 1/32 teaspoon

The Metric. System of Weights and Measures

L. Wood Measure 1 board foot = 144 cubic feet 1 cord foot= 16 cubic feet 1 cord = 8 cord feet

Kilogram ..;. the unit of mass equal to the mass of the platinum-iridium cylinder kept by the International Bureau of Weights and Measures in France.

Second.,.. the duration of 9,192,631,770 periods of radiation corresponding to the specifiC transition of the cesium-133 atom Ampere - the constant current that, flowing in two parallel conductors 1 meter apart in a vacuum, will produce a force\ of· 2 x 10"7 newtons per meter of length. Kelvin - base on the triple point of water, which is the point at which water can exist in three states: liquid, vapor and ice. It is defined as 1/273.16 of the temperatl)re of the triple point of water. Candela - intensity of the black-body radiation from a surface of 1/600,000 square meters at the temperature of freezing platinum and at a pressure of 101,325 pascals. Mole - an amount of substance in a system that contains as many elementary entities as there are atoms in 0.012 kilogram of carbon 12. The following are supplementary units: Radian - a unit of angular measure equal to the central angle whose sides are two radii of a circle that cut off an arc whose length is equal to the radius of the circle. Steradlans - a unit of measure equal to the solid angle whos.e vertex is in the center of a sphere, which cuts off an area equal to the rad.ius squared on the surface of the sphere.

Appendix B - Conversion 629 628 1001. Solved Problems in Engineering Mathematics (2"ct Edition) by Tiong & Rojas The following are derived units: Newton - the unit of force equal to the force needed to accelerate 1 kilogram by 1 meter per second squared. Joule- the unit of energy and work equal to the work done when the point of application 'of a force on 1 newton moves 1 meter in the direction of the force.

Poundal - the force needed to accelerate a 1-pound mass 1 foot per second squared. 1 poundaf = 13,825.5 dynes 1 poundal = 0. 138255 newtons Newton - the force needed to accelerate 1-kilogram mass 1 meter per second squared 5

1 newton = 10 dynes Hertz -·the unit of frequency in the field of electromagnetism defined as 1 cycle per second Watt- the unit of power defined as the power of 1 joule per second Volt - the u!'lit of voltage defined as the difference of electrical potential between two points of a conductor carrying a constant current of 1 ampere when the ·power used between them equals 1 watt, Ohm - the unit of electrical .resistance equal to a resistance that passes a current of 1 ampere when there is an electrical potential difference of 1 volt across it.

[ nne~$~-;e-oTFo~cea~dP~essu;e] Force and pressure are closely related. Force is anything that changes the motion or state of rest in a body while pressure is a force acting on a surface per unit area .. Dyne - the force needed to accelerate a 1-gram mass 1 centimeter per second squared. 1 dyne=: 0.0000723 poundal 1 dyne = 1o- newtons 5

1 newton= 7.23300 pouridals Pascal -.unit of pressure which is equal to 1 newton per square meter. 1 pascal= 0.020855 pounditf Atmosphere- unit of pressure at sea level. 1 atm = 14.6952 pounds/in

Daylight Saving Time -In the US, standard time plus one hour. Daylight Saving Time seems tc:> make the day last longer by ari hour. Equivalent values: 1 nanosecond

= 1 billionth of a second

1 microsecond = 1 millionth of a second 1 millisecond = 1 thousandth of a second 1 minute = 60 seconds

2

1 atm = 2,166.102 poiJnds/W

1 hour= 60 minutes

1 atm = 1.0332 kg/cm2

1 day = 24 hours

1 atm = 101,325 newtons/m

2

c--~-----J

~

Standard time- Introduced in 1883 by international agreement, Standard Time divided the Earth into 24 time zones. Calculated on solar time, the base of the system is the zero meridian that passes through the Royal Greenwich Observatory at Greenwich, England. Time is measured east or west of this Prime Meridian according to the time zones.

Measure of Time _

Solar time ·- A measurement of time in which noon occurs when the sun is at is highest point over a given location. Atomic time -A measurement of time using atomic clocks. Atomic clocks calculate time with extreme accuracy based on !he natural resonance frequency of !he cesium 133 atom

oc

oF

-273.15 -11 -5 0 4

-459.67 12 23 32 39.2

7 10 20

45 50 68

35 37

95 98.6

38 40 45

101 104 113

58

137

60 66 80 100 190 327

140 151 176 212 374 621

I

[Y~rthq~~k~-l~t~~sit!J

1 week = 7 days 1 month = 4 weeks

The Richter Scale, developed by seismologist Charles Richter, is used to express the amount of energy released at the focus of an earthquake. The scale is logarithmic and based on a numerical system of exponents. For example, the difference between 6.0 and 7.0 on the Richter scale is not one but a factor of ten .. Thus, an earthquake that measures 6.0 on the Richter scale is a hundred times more powerful than an earthquake that measures 4.0. An earthquake of 8.0 is 10 million times greater than a 1.0 earthquake.

1 calendar month = 30 days 1 year = 52 weeks

Sidereal time -Also known as astronomical time. This method of measurement uses the. movement of stars to calculate time. An average sidereal day is 23 hours, 56 minutes and 4.09 .seconds long.

Reference point Absolute zero Frozen yogurt A snowy day Water freezes Water at densed condition Cold water A cool fall day Room temperature on winter day A hot day Normal body temperature A warm bath A high fever A hot bath Highest air temperature recorded on Earth Broiled steak Hot faucet water Hot soup Water boils Hot oven Lead melts

1 year = 12 months 1 common year = 365 days 1 leap year = 366 days 1 decade

= 10 years

1 century = 100 years 1 millennium = 1000 years

L

Temp-erature

1

The two most common scales used for measunng temperature are the Celsius and Fahrenheit scales.

The great San Frar1cisco earthquake in 1906 has a magnitude of 8 ..3 which destroyed the city. The earthquake that causes the sinking of the island of Krakatoa in .1883 has a magnitude of 9.9.


~

I

Comment Richter Number . Near total devastation 9 A disaster, few buildings left 8 standina Many buildinas.destroyed 7 Buildings shakes; roads and walls 6 crack Strong rumbli11g; china and dishes 5 break . Weak; much like a passing truck 4 Very weak; less than 3.5 3 Detectable only bv seismograph 2

1·

M;~~ry

Computer

Onionskin paper = 9 pounds Standard cc;>pier paper = 20 pounds· Typewriting paper = 20 pounds Letterhead paper = 24 pounds For printinv on both sides = 60 pounds Typical business card = 65 pounds Poster .paper = 120 pounds

I

Metric prefix yotta zetta exa peta tera giga mega kilo hecto deka deci centi milli micro nano pico femto · alto zepto

8 bits = 1 character = 1 byte

= 1,024J>ytes = 210

1 megabyte

= 1,048,576 bytes or2 20

1 gigabyte= 1,073,741,832 bytes or 230

I

Paper Weight~-n-]

Standard Amounts: 1 quire

= 24 sheets

1 printer's quire = 25 sheets 1 ream = 20 quires 1 printer's ream= 21.5 quires 1 bundle= 2 reams. 1 case = 4 bundles 1 printer's bundle

= 4 reams

1 bale = 10 reams 1 short ream = 480 sheets 1 long ream

= 500 sheets

Metric Prefixes ··--] .

I

· Computer memory is based on bits and bytes:

1 kilobyte

The following is a table for conversion of units from English System to Metric System and from Metric System to English System.

Guidelines for Commercial Paper Weights:

1

_

yoct~_

Symbol y

z E p T G M k h dk

d c m ll

n p

f a z y

Value· 1024 102 . 10'" 1015 10

Appendix B - Conversion . 631

12

10~

1()6 10J 102 10 10' 10'2 10'~

10'6 10'" 10·12

to_,,

10·1ll

1o:z· 10·24

-

632 1001 Solved Problems in Eng_ineering Mathematics (2nd Edition) by 'I'iong & Rojas

Appendix C- Physical Constants 633

Appendix C:

Physical Constants +!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Appel\dix F- Greek Alphabets 635

634 IOO 1 So~ved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas

Appendix D:

+

Numeration

Appendix F:- · !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Appendix E:

+

n!! ! !

!!!!!!!!!!!!!!!M!!!!!!.!!!!!!at!!!!!!h!!!!!!N!!!!!!o!!!!!!t!!!!!!a!!!!!!ti!!!!!!o

•i

Greek Alphabets

+

. .

636 · 100 1 Solved Problems in Engineering Mathematics (2nd Edition) by Tiong & Rojas

Appendix G: !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

+Divisibility Rules

ei

!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

"

PHIUPI?lNE'

& MARKETrNG ~ ft EXCELLENCE('!; ~~

AWARDS

~

~~-~

An integer is evenly divisible by 2

- if it is even or it ends with an even number

3

- if the sum of its digits is divisible by 3

4

- if the number formed by the last two digits is divisible by 4

5

- if it ends with either 0 or 5

6

- if it is divisible by 2 and 3

7

- if the number formed after cancellation of the units digit and subtraction of twice the value of the units digit is divisible by 7

8

- if the number formed by the last three digits is divisible by 8

9

- if the sum of its digit is divisible by 9

10

- if it ends with zero

11

- if the difference between the cross sums of alternate digits is divisible by 11

12

- if it divisible by 3 and 4

~~

~-~-~-~.~(~SIQ

THE PHILIPPINE MARKETlNG EXCELLENCE AWARD is given to

EXCEL Review ·center Most Outstanding Engineering Review Center In recognition of its outstanding marketing performance leading towards market dominance, goodwill, high-level customer confidence, and market acceptability. Awarded by the Philippine Marketing Excellence Awards Institute, Sales and Marketing Magazine and the Asian Institute of Marketing and Entrep}:eneurship this 7th day of July, 2005 at the Westin

· Philippine Plaza, City of Pasay, Philippines.

~

FELIX ~LAO, JR., Ph.D. Chairman, Awards Committee National President, Philippine Marketing Association (1987)

Engineering Mathematics: Loolsolveo Problems

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