ENGINEERING MATHEMATICS-II APPLED MATHEMATICS
ENGINEERING MATHEMATICS-IIAPPLED MATHEMATICSDIPLOMA COURSE IN ENGINEERINGSECOND SEMESTERA Publication underGovernment of TamilnaduDistribution of Free Textbook Programme(NOT FOR SALE)Untouchability is a sinUntouchability is a crimeUntouchability is a inhumanDIRECTORATE OF TECHNICAL EDUCATIONGOVERNMENT OF TAMILNADU
Government of TamilnaduFirst Edition – 2015Thiru. PRAVEEN KUMAR I.A.SPrincipal Secretary / Commissioner of Technical EducationDirectorate of Technical EducationGuindy, Chennai- 600025Dr. K SUNDARAMOORTHY M.E., Phd.,Additional Director of Technical Education (Polytechnics)Directorate of Technical EducationGuindy, Chennai- 600025ConvenerThiru P.L.SANKARLecturer (S.G) / MathematicsRajagopal Polytechnic CollegeGudiyathamCo-ordinatorEr. R.SORNAKUMAR M.E.,PrincipalDr. Dharmambal GovernmentPolytechnic College for WomenTharamani, Chennai—113ReviewerProf. Dr. E. THANDAPANIUGC EMERITUS FELLOWRamanujan Institute for Advanced Study in MathematicsUniversity of Madras, Chennai 600 005AuthorsThiru M.RAMALINGAMLecturer (Sr.Gr) / MathematicsBharathiyar Centenary MemorialPolytechnic CollegeEttayapuramTHIRU.V. SRINIVASANLecturer (S.G) / MathematicsArasan Ganesan Polytechnic CollegeSivakasiThirumathi R.S.SUGANTHILecturer/MathematicsGovernment Polytechnic CollegeKrishnagiriThiru K.NITHYANANDAMHOD/MathematicsAdhiparasakthi Polytechnic CollegeMelmaruvathurThiru B.R. NARASIMHANLecturer (S.G)/MathematicsArulmigu Palaniandavar PolytechnicCollege, Palani.Thirumathi N.KANCHANDEVEILecturer / MathematicsRajagopal Polytechnic CollegeGudiyathamThis book has been prepared by the Directorate of Technical EducationThis book has been printed on 60 G.S.M PaperThrough the Tamil Nadu Text book and Educational Services Corporationii
FOREWORDWe take great pleasure in presenting this book of mathematics to thestudents of polytechnic colleges. This book is prepared in accordance with thenew syllabus under ‘M” scheme framed by the Directorate of TechnicalEducation, Chennai.This book has been prepared keeping in mind, the aptitude and attitude ofthe students and modern method of education. The lucid manner in which theconcepts are explained, make the teaching and learning process more easyand effective. Each chapter in this book is prepared with strenuous efforts topresent the principles of the subject in the most easy to understand and themost easy to workout manner.Each chapter is presented with an introduction, definitions, theorems,explanation, solved examples and exercises given are for better understandingof concepts and in the exercises, problems have been given in view of enoughpractice for mastering the concept.We hope that this book serve the purpose keeping in mind the changingneeds of the society to make it lively and vibrating. The language used isvery clear and simple which is up to the level of comprehension of students.We extend our deep sense of gratitude to Thiru. R. SornakumarCoordinator and Principal, Dr. Dharmambal Government Polytechnic Collegefor women, Chennai and to Thiru P.L. Sankar, Convener and Lecturer / SG,Rajagopal Polytechnic College, Gudiyattam who took sincere efforts inpreparing and reviewing this book.Valuable suggestions and constructive criticisms for improvement of thisbook will be thankfully acknowledged.AUTHORSiii
30022 ENGINEERING MATHEMATICS – IIDETAILED SYLLABUSUNIT—I: ANALYTICAL GEOMETRYChapter - 1.1 EQUATION OF CIRCLE5 Hrs.Equation of circle – given centre and radius. General equation of circle – finding centreand radius. Equation of circle on the line joining the pointsas diameter. Simple Problems.(x ,y )(x ,y )1 1 and 2 2Chapter - 1.2 FAMILY OF CIRCLES4 Hrs.Concentric circles, contact of two circles(Internal and External) -Simple problems.Orthogonal circles (results only). Problems verifying the condition .Chapter - 1.3 INTRODUCTION TO CONIC SECTION5 Hrs.Definition of a Conic, Focus, Directrix and Eccentricity. General equation of a conica x2 2 h x y b y2 2 g x 2 f y c 0conic (i) for circle:ahghbfgfca bandh 0(statement only). Condition for(ii) for pair of straight line: 0(iii) for parabola:h2 a b 0h2 a b 0h2 a b 0(iv) for ellipse:and (v) for hyperbola:. SimpleProblems.UNIT II: VECTOR ALGEBRA – IChapter - 2.1 VECTOR - INTRODUCTION5 Hrs.Definition of vector - types, addition, and subtraction of Vectors, Properties of additionand subtraction. Position vector. Resolution of vector in two and three dimensions.Directions cosines, Direction ratios. Simple problems.Chapter - 2.2 SCALAR PRODUCT OF VECTORS5 Hrs.Definition of Scalar product of two vectors – Properties – Angle between two vectors.Simple Problems.Chapter - 2.3 APPLICATION OF SCALAR PRODUCT4 Hrs.Geometrical meaning of scalar product. Work done by Force. Simple Problems.UNIT III: VECTOR ALGEBRA – IIChapter - 3.1 VECTOR PRODUCT OF TWO VECTORS5 Hrs.Definition of vector product of two vectors. Geometrical meaning. Properties – Anglebetween two vectors – unit vector perpendicular to two vectors. Simple Problems.Chapter - 3.2APPLICATION OF VECTOR PRODUCT OF TWO VECTORS & SCALARTRIPLE PRODUCT5 Hrs.Definition of moment of a force. Definition of scalar product of three vectors –Geometrical meaning – Coplanar vectors. Simple Problems.iv
Chapter - 3.3 VECTOR TRIPLE PRODUCT & PRODUCT OF MORE VECTORS4 Hrs.Definition of Vector Triple product, Scalar and Vector product of four vectors Simple Problems.UNIT IV: INTEGRAL CALCULUS – IChapter - 4.1 INTEGRATION – DECOMPOSITION METHOD5 Hrs.Introduction - Definition of integration – Integral values using reverse process of differentiation – Integration using decomposition method. Simple Problems.Chapter - 4. 2 INTEGRATION BY SUBSTITUTION5 Hrs.Integrals of the formn [ f ( x)] f ' ( x)dx, n 1 ,f ' ( x)dxf ( x)and F[ f ( x)] f ' ( x)dx . Simple Problems.Chapter - 4.3 STANDARD INTEGRALSdx4 Hrs.dx 2 22a xx a 2 andIntegrals of the form, dxa2 x2UNIT V: INTEGRAL CALCULUS – IIChapter - 5.1 INTEGRATION BY PARTSIntegrals of the form xsin nx dx , x cos nx dx , x eChapter - 5.2 BERNOULLI’S FORMULAEvaluation of the integrals. Simple Problems5 Hrs.nx dx,n x log x dx and4 Hrs.mmm nx x sin nx dx , x cos nx dx and x e dx wherem 2using Bernoulli’s formula. Simple Problems.Chapter - 5.3 DEFINITE INTEGRALS5 Hrs.Definition of definite Integral. Properties of definite Integrals - Simple Problems.v
30023 APPLIED MATHEMATICSDETAILED SYLLABUSUNIT—I: PROBABILITY DISTRIBUTION – IChapter - 1.1 RANDOM VARIABLE5Hrs.Definition of Random variable – Types – Probability mass function –Probability densityfunction. Simple Problems.Chapter - 1.2 MATHEMATICAL EXPECTATION4Hrs.Mathematical Expectation of discrete random variable, mean and variance. SimpleProblems.Chapter - 1.3 BINOMIAL DISTRIBUTION5Hrs.Definition of Binomial distributionx 0 ,1, 2 , .P( X x) nC p x q n xxwhereStatement only. Expression for mean and variance. Simple Prob-lems.UNIT—II: PROBABILITY DISTRIBUTION – IIChapter - 2.1 POISSION DISTRIBUTIONP ( X x) 5 Hrs.e . xx!x 0 ,1, 2 , .Definition of Poission distributionwhere(statement only). Expressions of mean and variance. Simple Problems.Chapter - 2.2 NORMAL DISTRIBUTION5 Hrs.Definition of normal and standard normal distribution – statement only. Constants ofnormal distribution (Results only). Properties of normal distribution - Simple problemsusing the table of standard normal distribution.Chapter - 2.3 CURVE FITTING4 Hrs.Fitting of straight line using least square method (Results only). Simple problems.UNIT III : APPLICATION OF DIFFERENTIATIONChapter – 3.1 VELOCITY AND ACCELERATION5 Hrs.Velocity and Acceleration – Simple Problems.Chapter - 3.2 TANGENT AND NORMAL4 Hrs.Tangent and Normal – Simple Problems.Chapter - 3.3 MAXIMA AND MINIMA5 Hrs.Definition of increasing and decreasing functions and turning points. Maxima andMinima of single variable only – Simple Problems.UNIT IV: APPLICATION OF INTEGRATION – IChapter - 4.1 AREA AND VOLUME5 Hrs.Area and Volume – Area of Circle. Volume of Sphere and Cone – Simple Problems.Chapter - 4.2 FIRST ORDER DIFFERENTIAL EQUATION5 Hrs.Solution of first order variable separable type differential equation .Simple Problems.Chapter - 4.3 LINEAR TYPE DIFFERENTIAL EQUATION4 Hrs.Solution of linear differential equation. Simple problems.vi
UNIT V : APPLICATION OF INTEGRATION – IIChapter – 5.1 SECOND ORDER DIFFERENTIAL EQUATION – I4 Hrs.Solution of second order differential equation with constant co-efficients in the formd2ydya b cy 02dxdxa, bcwhereandare constants. Simple Problems.Chapter - 5.2 SECOND ORDER DIFFERENTIAL EQUATION – II5 Hrs.Solution of second order differential equations with constant co- efficients in the formad2ydy b cy f ( x)dxdx 2wherea, bandcare constantsandf ( x) k e mx. Simple Problems.Chapter - 5.3 SECOND ORDER DIFFERENTIAL EQUATION – III5 Hrs.Solution of second order differential equation with constant co-efficients in the formd2ydya b cy f ( x)dxdx 2f ( x) k sin mxor kcos mxwherea, bandcare constantsand. Simple Problems.vii
NOTESviii
UNIT – IANALYTICAL GEOMETRY1.1 Equation of CricleEquation of circle given centre and radius. General equation of circle finding centre and radius.Equation of circle on the line joining the points (x1, y1) and (x2, y2) as diameter. Simple problems.1.2 Family of Circles:Concentric circles, contact of two circles (internal and external). Simple problems, Orthogonal circles (results only), problems verifying the condition.1.3 Introduction to conic section:Definition of a conic, Focus, Directrix and Eccentricity, General equation of a conic ax2 by2 2hxy 2gx 2fy c 0 (Statement only). Condition for conic (i) for circle : a b and h 0.a h g(ii) for pair of straight line h b f 0g f c(iii) for parabola : h2 –ab 0 (iv) for ellipse : h2 –ab 0(v) for hyperbola : h2 – ab 0 – Simple problems.1.1 EQUATION OF CIRCLEDefinition:A circle is the locus of a point which moves in a plane such that its distance from a fixed point iscontant. The fixed point is the centre of the circle and the constant distance is the radius of the circle.1.1.1 Equation of the circle with centre (h, k) and radius 'r' units:Given: The centre and radius of the circle are (h, k) and 'r' units. Let P (x, y) be any point on the circle.From Fig.(1.1)CP rr(x – h) 2 (y – k) 2 r(x – h) (y – k) rNote:222C(h, k)(Using distance formula)–––––– (1)Fig.(1.1)The equation of the circle with centre (0, 0) and radius 'r' units is x2 y2 r2.P (x, y)
2 Engineering Mathematics-II1.1.2 General Equation of the circle:The General Equation of the circle isx2 y2 2gx 2fy c 0 –––––– (2)Add g2, f2 on both sidesx2 2gx g2 y2 2fy f2 g2 f2 – c(x g)2 (y f)2 g2 f2 – c[x – (–g)]2 [y – (–f )]2 g 2 f 2 – c 2.(3)Equation (3) is of the form equation (1)The equation (2) represents a circle with centre (– g, – f) and radiusNote:g2 f 2 – c .1. In the equation of the circle co-efficient of x2 co-efficient of y2.112. Centre of the circle – co-efficient of x, – co-efficient of y 2 2 3. radius of the circleg2 f 2 – c .1.1.3 Equation of the circle on the line joining the points (x1, y1) and (x2, y2) as diameter.Let A(x1, y1) and B (x2, y2) be the given end points of a diameter. Let P (x, y) be any point on thecircle.P (x, y) Angle in a semi-circle is 90o APB 90oA (x1, y1) AP BP(Slope of AP) (Slope of BP) – 1. y – y1 y – y 2 x – x x – x –112(y – y1) (y – y2) – (x – x1) (x – x2)(x –x1) ( x – x2) (y – y1) (y – y2) 0This is the Diameter form of the equation of a circle.PART – A1. Find the equation of the circle whose centre is (– 1, 2) and radius 5 units.Solution:Equation of the circle is(x – h) 2 (y – k) 2 r 222(x 1) (y – 2) 52(–1, 2), r 5(h, K)x2 2x 1 y2 – 4y 4 25x2 y2 2x – 4y – 20 0CB (x2, y2)
Analytical Geometry 2. Find the centre and radius of the circlex2 y2 –4x 8y – 7 0Solution:Centre C (–g, –f) 2g –4, 2f 8, c –7 C(2, – 4) g –2 f 4r g2 f 2 – c (–2) 2 (4) 2 – (–7) 4 16 7 273. Obtain the equation of the circle on the line joining (– 1, 2) and (– 3, 5) as diameter.Solution:Equation of the circle is(x –x1) ( x – x2) (y – y1) (y – y2) 0(x 1) (x 3) (y – 2) (y – 5) 0x2 4x 3 y2 – 7y 10 0x2 y2 4x – 7y 13 0PART – B1. Find the equation of the circle passing through (2, 1) and having its centre at (– 3, – 4).Solution:r (–3 – 2) 2 (–4 – 1) 2 25 25 50Equation of the circle (x – h)2 (y – k)2 r2(x 3)2 (y 4)2 ( 50 )2x2 6x 9 y2 8y 16 50x2 y2 6x 8y – 25 02. Show that the line 4x – y 17 is the diameter of the circle x2 y2 – 8x 2y 3 0.Solution:x2 y2 – 8x 2y 3 0Centre C (–g, –f) 2g –8, 2f 2 C(4, –1)g –4, f 1Put x 4, y – 1 in4x – y 174 (4) – (–1) 1716 1 1717 17 (4, – 1) lies on the line 4x – y 17. 4x –y 17 is the diameter of the circle.C4yx–7 13
4 Engineering Mathematics-II3. Find the centre and radius of the circle 4x2 4y2 – 8x 16y 19 0.Solution:4x2 4y2 – 8x 16y 19 0Divide by 419x 2 y 2 – 2x 4y 04Centre C (–g, –f) 2g –2, 2f 4, c C (1, –2)g –1, f 2r g 2 f 2 – c (–1) 2 (2) 2 –1941919 1 4 – 4420 – 19 41 1 4 2PART – C1. Find the equation of the circle, two of whose diameters are x y 6 and x 2y 4 and whose radiusis 10 units.Solution:x y 6 ––––– (1)x 2y 4 ––––– (2)(1) – (2) – y 2 y –2Substitute y – 2 in (1)x (– 2) 6x 6 2x 8Centre C (8, –2)Equation of the circleh, k 8, –2222(x – h) (y – k) rr 10222(x –8) (y 2) 10x2 – 16x 64 y2 4y 4 100x2 y2 – 16x 4y – 32 06y 10 P(x, y)x C2y 4x 2. Show that the point (8, 9) lies on the circle x2 y2 – 10x – 12y 43 0. Find the co-ordinates of theother end of the diameter of the circle through this point.Solution:Put x 8, y 9 inx2 y2 –10x – 12y 43 0(8)2 (9)2 – 10(8) – 12(9) 43 064 81 – 80 – 108 43 0188 –188 00 0(8, 9) lies on the circle x2 y2 –10x – 12y 43 0.(8, 9)A(5, 6)B (x, y)
Analytical Geometry 2g –10Centre C (–g,–f) g –5 C (5, 6)2f –12f –6Centre C is the mid point of diameter AB. 8 x 9 y , (5,6) 2 28 x9 yi.e. 5 , 6 2210 8 x,12 9 yx 2,y 3 The other end of the diameter is (2, 3).Note:(i) If (x1, y1) lies on the circle x2 y2 2gx 2fy c 0 then x12 y12 2gx1 2fy1 c 0.(ii) If (x1, y1) lies outside the circle x12 y12 2gx1 2fy1 c 0.(iii) If (x1, y1) lies inside the circle, then x12 y12 2gx1 2fy1 c 01.2 FAMILY OF CIRCLES1.2.1 Concentric Circles:Two or more circles having the same centre but differ in radii are called concentric circles.For examples:x2 y2 2gx 2fy c 0x2 y2 2gx 2fy p 0x2 y2 2gx 2fy q 0are concentric circles which have same centre (– g, – f).Note: Equation of concentric circles differ only by the constant term.r2r1C rFig.1.2.11.2.2 Contact of circles:Case (i)Two circles touch externally if the distance between their centres is equal to sum of their radii.i.e c1c2 r1 r2.r1c1r2c2Fig.1.2.25
6 Engineering Mathematics-IICase (ii)dii.Two circles touch internally if the distance between their centres is equal to difference of their rai.e c1c2 r1 – r2 or r2 – r1r1c2 r2c1Fig.1.2.31.2.3 Orthogonal Circles:Two circles are said to be orthogonal if the tangents at their point of intersection are perpendicularto each other.Condition for two circles to cut orthogonally. (Results only).Let the equation of the circles bex2 y2 2g1x 2f1y c1 0x2 y2 2g2x 2f2y c2 0Pr1r2A(–g1, –f1)B(–g2, –f2)Fig.1.2.4Let the circles cut each other at the point P.The centres and radii of the circles are A1(–g1, –f1), B (–g2, –f2)r1 AP g12 f12 – c1 , r2 PB g 22 f 22 – c2 .From fig.(1 2 4) APB is a right angled triangle.AB2 AP2 PB2(–g1 g2)2 (–f1 f2)2 g12 f12 – c1 g22 f22 c2. g12 f12 – c1 g22 f22 – c2g12 g22 – 2g1g2 f12 f22 – 2f1f2– 2g1g2 – 2f1f2 – c1 – c22g1g2 2f1f2 c1 c2This is the required condition for two circles to cut orthogonally.Note:When the centre of any one circle is at the origin then condition for orthogonal circles is c1 c2 0.
Analytical Geometry 7PART – A1. Find whether the circles x2 y2 – 4x 6y 8 0 and x2 y2 – 4x 6y – 8 0 are concentric.Solution:From the two given equations of the circles we observe that the constant term alone differs i.e withthe same centre. (2, – 3). The given circles are concentric circles.2. Find the distance between the centres of the circles x2 y2 – 4x – 6y 9 0 andx2 y2 2x 2y – 7 0.Solution:x2 y2 – 4x – 6y 9 0 and x2 y2 2x 2y – 7 0.2g1 – 4, 2f1 – 6 2g 2 2, 2f 2 2g1 – 2, f1 – 3g 2 1,f2 1Centres: c1(2, 3)c2 (–1, – 1)Distance c1c 2 (2 1) 2 (3 1) 2 9 16 53. Find the equation of the circle concentric with the circle x2 y2 5 0 and passing through (1, 0).Solution:Equation of the circle concentric withx2 y2 5 0 isx2 y2 k 0Put x 1 and y 0 in x2 y2 k 0(1)2 0 k 0k –122 x y –1 0which is the required equation of the circle.4. Verify whether the circles x2 y2 10 0 and x2 y2 – 10 0 cut orthogonally.Solution:When the centre of any one of the circle is at the origin thenCondition for orthogonallityc1 c2 0i.e 10 – 10 00 0 The circles cut orthogonally.PART – B1. Find the equation of the circle which is concentric with the circle x2 y2 – 8x 12y 15 0 andpasses through (5, 4).Solution:Equation of the concentric circle bex2 y2 – 8x 12y k 0–––––––– (1)
8 Engineering Mathematics-IIPut x 5, y – 4 in (1)x2 y2 – 8x 12y – 29 0(5)2 (4)2 – 8 (5) 12 (4) k 025 16 – 40 48 k 029 k 0k – 49 The required equation of the circle is x2 y2 – 8x 12y – 49 0.2. Find the equation of the circle concentric with the cicle x2 y2 3x – 7y 1 0 and having radius5 units.Solution:–3 7Centre of the circle x2 y2 3x – 7y 1 0 is , 2 2 –3 7 Centre of the concentric circle is , amd radius 5 units. 2 2 Equation of the circle(x – h) 2 (y – k) 2 r 2223 7 2 x y – 522 –3 7 , , r 52 2(4, k)949 3x y 2 – 7y 25444x 2 4y 2 12x – 28y – 42 0x2 3. Show that the circles x2 y2 – 8x 6y – 23 0 and x2 y2 – 2x – 5y 16 0 are orthogonal.Solution:x2 y2 – 8x 6y – 23 0 and x2 y2 – 2x – 5y 16 0g1 – 4, f1 3, c1 – 23, g2 – 1, f2 –5/2, c2 16Condition for orthogonallity2g1g2 2f1f2 c1 c2( 2 ) –23 162(– 4)(–1) 2(3) –58 – 15 –7–7 –7 The circles cut orthogonally.PART – C1. Show that the circles x2 y2 – 4x 6y 8 0 and x2 y2 – 10x – 6y 14 0 touch each other.Solution:x2 y2 – 4x 6y 8 0 and x2 y2 – 10x – 6y 14 0c1 (2, – 3)c2 (5, 3)r2 (–5) 2 (–3) 2 – 14r1 (–2) 2 (3) 2 – 8 5 20 2 5
Analytical Geometry 9c1c 2 (2 – 5) 2 (–3 – 3) 2 9 36 45 3 5 c1c 2 r1 r2 the circles touch each other externally.2. Show that the circles x2 y2 – 2x 6y 6 0 and x2 y2 – 5x 6y 15 0 touch each other.Solution:2x 2 y 2 – 2x 6y 6 0 x 2 y 2 – 5x 6y 15 0 5 2 cc–11 2 (–3 3)c1 (1, –3) 5 2c 2 , –3 2222r1 (–1) (–3) – 6 3 25 1 9 – 62r2 9 – 1543 4c1c 2 252–6 r1 241r1 – r2 2 –12r2 24 –1 3 22 c1c 2 the circles touch each other internally.1.3 INTRODUCTION TO CONIC SECTION1.3.1 DefinitionsConicA conic is defined as the locus of a point which moves such that its distance from a fixed point isalways 'e' times its distance from a fixed straight line.Focus:The fixed point is called the focus of the conic.Directrix:The fixed straight line is called the directirx of the conic.Eccentricity:The constant ratio is called the eccentricity of the conic.From Fig.(1.3)S is the focus line XM is the Directrix 'e' eccentricity of the conic thenXSP e.PMPMSFig.(1.3)
10 En
ENGINEERING MATHEMATICS-II APPLED MATHEMATICS DIPLOMA COURSE IN ENGINEERING SECOND SEMESTER Untouchability is a sin Untouchability is a crime Untouchability is a inhuman DIRECTORATE OF TECHNICAL EDUCATION GOVERNMENT OF TAMILNADU A Publication under Government of Tamilnadu Distribution of Free Textbook Programme (NOT FOR SALE)
Materials Science and Engineering, Mechanical Engineering, Production Engineering, Chemical Engineering, Textile Engineering, Nuclear Engineering, Electrical Engineering, Civil Engineering, other related Engineering discipline Energy Resources Engineering (ERE) The students’ academic background should be: Mechanical Power Engineering, Energy .
Careers in Engineering Guide the brighter choice. Contents ABOUT LSBU 4–5 BUILDING SERVICES ENGINEERING 6–7 CHEMICAL AND PETROLEUM ENGINEERING 8–9 CIVIL ENGINEERING 10–11 ELECTRICAL AND ELECTRONIC ENGINEERING 12–13 MECHANICAL ENGINEERING 14–15 MECHATRONICS ENGINEERING 16–17 PRODUCT DESIGN ENGINEERING 18–19 An engineering degree is a big challenge to take on. There is no denying .
OLE MISS ENGINEERING RECOMMENDED COURSE SCHEDULES Biomedical engineering Chemical engineering Civil engineering Computer engineering Computer science Electrical engineering General engineering Geological engineering Geology Mechanical engineering Visit engineering.olemiss.edu/advising for full course information.
IBDP MATHEMATICS: ANALYSIS AND APPROACHES SYLLABUS SL 1.1 11 General SL 1.2 11 Mathematics SL 1.3 11 Mathematics SL 1.4 11 General 11 Mathematics 12 General SL 1.5 11 Mathematics SL 1.6 11 Mathematic12 Specialist SL 1.7 11 Mathematic* Not change of base SL 1.8 11 Mathematics SL 1.9 11 Mathematics AHL 1.10 11 Mathematic* only partially AHL 1.11 Not covered AHL 1.12 11 Mathematics AHL 1.13 12 .
as HSC Year courses: (in increasing order of difficulty) Mathematics General 1 (CEC), Mathematics General 2, Mathematics (‘2 Unit’), Mathematics Extension 1, and Mathematics Extension 2. Students of the two Mathematics General pathways study the preliminary course, Preliminary Mathematics General, followed by either the HSC Mathematics .
2. 3-4 Philosophy of Mathematics 1. Ontology of mathematics 2. Epistemology of mathematics 3. Axiology of mathematics 3. 5-6 The Foundation of Mathematics 1. Ontological foundation of mathematics 2. Epistemological foundation of mathematics 4. 7-8 Ideology of Mathematics Education 1. Industrial Trainer 2. Technological Pragmatics 3.
Advanced Engineering Mathematics Dr. Elisabeth Brown c 2019 1. Mathematics 2of37 Fundamentals of Engineering (FE) Other Disciplines Computer-Based Test (CBT) Exam Specifications. Mathematics 3of37 1. What is the value of x in the equation given by log 3 2x 4 log 3 x2 1? (a) 10 (b) 1(c)3(d)5 E. Brown . Mathematics 4of37 2. Consider the sets X and Y given by X {5, 7,9} and Y { ,} and the .
for use in animal nutrition. Regulation (EC) No 178/2002 laying down the general principles and requirements of food law. Directive 2002/32/EC on undesirable substances in animal feed. Directive 82/475/EEC laying down the categories of feed materials which may be used for the purposes of labelling feedingstuffs for pet animals The Animal Feed (Hygiene, Sampling etc and Enforcement) (England .
