FACULTY OF ENGINEERING TECHNOLOGY GROUP T LEUVEN CAMPUS INTRODUCTORY COURSE MATHEMATICS

FACULTY OF ENGINEERING TECHNOLOGY GROUP T LEUVEN CAMPUS INTRODUCTORY COURSE MATHEMATICS

Algebr Content. Rel numbers. The power of rel number with n integer eponent. The n th root of rel number 4. The power of rel number with rtionl eponent 5.4 The power of rel number with n integer eponent 6.5 Rtionlizing the denomintor 7.6 The logrithm of rel number 7.7 Eercise 9. Polynomils with rel coefficients in vrible. Definition. Zeroes of polynomil function. Specil products.4 Polynomil division.5 Eercises 6. Equtions nd inequlities 7. Equtions in IR 7. Inequlities in IR 0. Eercises 6 4. Absolute vlue of rel number 8 4. Definition 8 4. Property 9 4. Opertions 9 4.4 Appliction: grph of rel bsolute vlue function 9 4.5 Eercises 5. Mtrices nd determinnts 5. Mtrices 5. Determinnts 7 5. Eercises 4 6. Systems 4 6. Systems of n equtions in n unknowns 4 6. Systems of liner inequlities in unknown 45 6. Eercises 46 Anlytic Geometry. Vector nd lines. Vectors. Lines 5. Prllel lines 8.4 The Euclidin vectorspce 9.5 Angles.6 Orthogonlity of lines

.7 Distnce from point to line (O.N.B.).8 Eercise 4. Conic sections 9. Introduction 9. The circle 0. The prbol.4 Eercises Clculus. Reltions functions, mppings, - mppings. Epnding IR 4. Continuity of function in IR 5. Continuity t point 5. Right-nd left-continuity 6 4. Limits 7 4. Limits in rel number 7 4. Right-hnd nd left-hnd limits 9 4. Limits t infinity 9 4.4 Infinite limits 9 4.5 Infinite limits t infinity 0 4.6 Elementry rules to clculte limits 0 4.7 Indeterminte cses 4.8 Eercises 4 5. Differentition 7 5. The derivtive in point 7 5. Geometricl interprettion of the derivtive 7 5. Left-hnd nd right-hnd derivtive in point 8 5.4 A verticl tngent 9 5.5 Differentition rules 9 5.6 Eercises 6. Indefinite integrls 4 6. Antiderivtive functions 4 6. The indefinite integrl 4 6. Properties 4 6.4 Bsic integrls 5 6.5 Integrtion by substitution 5 6.6 Eercise 6 6.7 Integrtion by prts 9 6.8 Eercises 9 7. Definite integrls 0 7. The fundmentl theorem of clculus 0 7. Properties of the definite integrl 0 7. The substitution method 7.4 Prtil integrtion 7.5 Eercise

Trigonometry. Angles 4. The trigonometric circle 4. Oriented ngles 4. Conversion between rdins nd degrees 6. The trigonometric numbers 7. Definitions 7. Some specil ngles nd their trigonometric numbers 8. Sign vrition for the trigonometric numbers by qudrnt 9.4 Pythgoren identities 9.5 Specil pirs of ngles 0.6 Eercises. The trigonometric functions 4. Periodic functions 6. Even nd odd functions 6. Sine function 7.4 Cosine function 7.5 Tngent function 8.6 Cotngent function 8.7 The secntfunction 9.8 The cosecnt function 9.9 Eercises 0 4. Right tringles 4. Formuls 4. Eercises 5. Oblique tringles 5 5. The sine rules 5 5. The cosine rules 6 5. Solving oblique tringles 7 5.4 Eercises 8 6. Etr s 9 6. Specil lines in tringle 9 6. Isosceles tringles 6. Equilterl tringles 6.4 Eterior ngles 7. Trigonometric formuls 7. Sum nd difference formuls 7. Double-ngle formuls 4 7. Hlf-ngle formuls 5 7.4 Trigonometric numbers in terms of tn α/ 5 7.5 Conversions sum/difference of ngles into product of ngles nd vice vers 6 7.6 Eercise 7

Algebr Dr. Croline Dnneels

Rel numbers.... The power of rel number with n integer eponent.... The n th rooth of rel number... 4. The power of rel number with rtionl eponent... 5.4 Properties of etrction of roots... 6.5 Rtionlizing the denomintor... 7.6 The logrithm of rel number... 7.7 Eercices... 9 Polynomils with rel coefficients in vrible.... Definition.... Zeroes of polynomil function.... Specil products....4 Polynomil division....5 Eercices... 6 Equtions nd inequlities...7. Equtions in IR... 7. Inequlities in IR... 0. Eercises... 6 4 Absolute vlue of rel number...8 4. Definition... 8 4. Property... 9 4. Opertions... 9 4.4 Appliction: grph of rel bsolute vlue function... 9 4.5 Eercises... 5 Mtrices nd determinnts... 5. Mtrices... 5. Determinnts... 7 5. Eercises... 4 6 Systems...4 6. Systems of n equtions in n unknowns... 4 6. Systems of liner inequlities in unknown... 45 6. Eercises... 46

Rel numbers. The power of rel number with n integer eponent R nd n N : =... (n fctors) R : = 0 0 0 n ( ) ( ) R nd n : = = = n - -n - n 0 N n Properties:,b R nd m,n R m n m+n = m n = m-n ( ) n n n b = b n = b b n n m ( ) n = m n Emple ( ) ( ) - - 4 - b b -5 - = b b 5 Rule of signs: if n is even, then ( ) n n - = if n is odd, then ( ) n n - = - Algebr

. The n th rooth of rel number The n th ( n ) N root of rel number is ech rel number of which the n th power is 0 or, R, n N : is the nth root of = 0 n if n is odd nd R then hs in R one n th root, written s: Emples n 8 = since = 8 8 = since ( ) = 8 If n is even nd : IR 0 + hs n th roots in R which re ech others opposite nd re n n denoted nd. We gree upon 4 so 6 = > 0. n to represent positive rel number; Emple: 4 hs squre roots 4 = nd 4 = = 0 then hs in R one n th root nmely 0 IR 0 then hs in R no n th root. Convention: n n = n n = We will only tret n + with R, since the other forms cn be rewritten to this form since n if < 0 nd n is even then does not eist n if < 0 nd is odd then we write s - n n with R + Algebr 4

. The power of rel number with rtionl eponent Definition: + 0 0 R, m Z, n N : = m n n m Use:. n = n. m - n = n m. np mp mp np m n = = = n m Clcultion rules: + 0, b R en q, q' Q : q q' q+q' = q q' = q-q' ( ) q q q b = b q = b b q ( ) q' = q q q q' Algebr 5

.4 Properties of etrction of roots.4. Multipliction nd division of rdicls + n n n 0, b R, n N : b = b n + + n 0 0 n R, b R ; n N : = b Emples 4 4 4 8 = 6 = b 4 9 = = 4 5.4. Eponentition nd etrction of roots of rdicls n ( ) m n + m 0 R, n N, m N : = + m n m n n m 0 R ; m,n N : = = Emples ( ) 8 = 8 = = 4 8 = 8 =.4. Adding nd subtrction of rdicls ( ) + n n n n 0 R, n R, p, q, r R : p + q r = p + q r Emple: 4 + 5 8 = + 5 = 4 Algebr 6

.5 Rtionlizing the denomintor b = b b b ± b = b b + b = ± b ± b.6 The logrithm of rel number Let be the bse of logrithmic system ( > 0, ), R 0, y R Then y = log = y We cll y the logrithm of to bse. The fct tht log is the inverse of cn be epressed with the following two identities: Logrithmic identy : log = Logrithmic identy : log = Nmes: If = e (number of Euler), the logrithm is clled nturl logrithm, nottion ln. ln = 0 ln e = If = 0, the logrithm is clled Briggs logrithm, nottion log. Properties: log = 0 log 0 =. If the bse is lrger thn ( > ), the function log is incresing everywhere. If the bse is between 0 nd (0 < < ), then the function log is decresing everywhere.. Becuse =, logrithmic identity bove implies log =. Becuse 0 =, logrithmic identity bove implies log = 0 Algebr 7

Opertions:, y R : log y = log + log y + 0 +, y R 0 : log = log log y y + R 0 : log ( ) = log z R ; z Z : log ( ) = z log + 0 R N + n 0; n 0 : log = log n Algebr 8

.7 Eercices + 0.7. Simplify (,b,c R )... 7 4 ( b c ) 0 9 ( bc ) ( b c ) + ( ) + ( ) 4 ( b ) ( 4) 4 4 ( ) ( )( b ) 6 9 6 b c 6 8 4. n n n+ n+ n b b b 5. 6. 4 4 5 b 6 8 8c d 4 ( 5) b, b IR b cd b 4 c + 0.7. Clculte (,b R ). 4 8 b 4 b. 8 7. 4 4 4 6 4 4. 5. 5 0 6 5 60,5 4 8 6. 4 Algebr 9

.7. Rtionlize the denomintor ( R ) +... b + 5 5 + 7.7.4 Clculte, even without knowledge of. log 4 log 6. 5 5 5 log + log log + log 0 6 6 Algebr 0

Polynomils with rel coefficients in vrible. Definition + +... + + with n n- n n- 0 nottion V(). n, n-,...,,0 n N,n 0 is polynomil of degree n in,. Zeroes of polynomil function R is zero of the ssocited polynomil function f : R R : V() if the polynomil epression evluted to, equls 0. In symbols: is zero of V() V() = 0. Emple is zero of V() = 6 7 7 since V = 6 7 7 = 0. Specil products If A, B nd C re polynomils: ( A + B) ( A B) = A B ( A + B) = A + AB + B ( A B) = A AB + B ( A + B+ C) = A + B + C + AB+ BC + AC ( A + B) = A + A B+ AB + B ( A B) = A A B+ AB B A B = ( A B) ( A + AB+ B ) A + B = ( A+ B) ( A AB+ B ) Algebr

.4 Polynomil division For ny polynomils A() nd B() ( B() 0, degree A() degree B() ), there eist unique polynomils Q() nd R() such tht A() = B()Q() + R() nd degree R() < degree B()..4. Polynomil long division A() (dividend) B() (divider) Q()(quotient) R()(reminder) If R() = 0 we sy tht B() evenly divides A(). Emple 4 + 9 + 6 4 6 + 4 + + 6 + 4 + + + + 4 6 9 4 = 6 + Algebr

.4. Division of polynomil by -d with d IR: synthetic division (Horner lgorithm) Emple: ( 4 5 + 6): ( + ) 4 0 5 6 8 6 4 8 6 Q() = 4 8 + R() = 6 4 5 + 6 6 = 4 8 + + +.4. Divisibility by -d with d IR.4.. Reminder theorem If polynomil f() is divided by d, then the reminder is equl to V(d). Consequence: V() is divisible by d R = V( d) = 0 Emples V() = 5 + 7 is not divisible by since V() = 9 0 V() = + 5 + is divisible by + since V( ) = 0.4.. Fctoring polynomils Fctoring polynomils is very importnt issue in lgebr. To fctor polynomil put your polynomil in stndrd form, from highest to lowest power: V() = n n + + 0 After you hve fctored out the gretest common fctor nd specil products, you cn pply the Rtionl Root Test. The Rtionl Root (or Rtionl Zeroes) Test is hndy wy of obtining list of useful first guesses when you re trying to find the zeroes (roots) of polynomil. Given polynomil with integer coefficients, the possible (or potentil) zeroes re found by listing the fctors of the constnt (lst) term over the fctors of the leding coefficient, thus forming list of frctions. This listing gives you list of potentil rtionl (frctionl) roots to test - hence the nme of the Test. Algebr

Cution: Don t mke the Rtionl Root Test out to be more thn it is. It doesn t sy those rtionl numbers re roots, just tht no other rtionl numbers cn be roots. And it doesn t tell you nything bout whether some irrtionl or even comple roots eist. The Rtionl Roots Test is only strting point. Emple Fctor the polynomil: + 6. The fctors of the leding coefficient () re {,}. The fctors of the constnt term (6) re {,,,6}. Therefore the possible rtionl zeroes re ±,, or 6 divided by or : ±/, /, /, 6/, /, /, /, 6/ reduced: ±,,, 6, /, / 4. Check for which vlues V() equls 0: s soon s you find one vlue which stisfies this criterium, pply synthetic division to find the reduced polynomil tht you ll use to find the remining roots: V() 0 V( ) 0 V( ) 0 V() 0 V() = 0 This mens tht is fctor of the given polynomil. The reduced polynomil cn be produced by pplying the synthetic division: 6 d = 6 9 6 0 We find: + 6 = ( )( + ) 5. Agin, serch for the remining cndidte roots in the list ±, /, : As we hve lredy checked ± nd ± (see bove), we check V(/) = 0 Applying synthetic division to + gives: + = ( /)( + 4) 6. All together: + 6 = ( )( /)( + 4) = ( )( )( + ) Algebr 4

.4.. Coefficientrules. A polynomil of degree n is divisible by ( ) if the sum of the coefficients (included the constnt term) equls zero. Emple 5 + is divisible by ( ). If the sum of the coefficients of the terms with odd power of equls the sum of the coefficients of the terms with even power of (included the constnt term), then the polynomil is divisible by ( + ). Emple 5 + 4 + is divisible by ( + ) Algebr 5

.5 Eercices Simplify (be efficient):. ( + )( )( + 4). ( + )( 9 6 + 4). ( 9 4 + + ) : ( + + ) 4. ( 0 7 + ) : ( ) 4 6 7 + 8 Q = 9 0 9, R = 9 + 0 Q = 7, R = (use synthetic division) 5. ( 8 0 + ) : ( ) Q = 4 8, R = 5 (use synthetic division) 6. For which vlue of n R the polynomil ( + n 5n + 0) is divisible by ( + )? Determine for the obtined vlue of n the quotient. 7. Fctor: 4 8 0 8 7 n =, Q = 5 + 0 + + ( )( ) Algebr 6

Equtions nd inequlities. Equtions in IR.. Definition An eqution is n epression tht represents the equlity of two epressions involving constnts nd/or vribles... Solving n eqution To solve n eqution mens to find the vlue or vlues of the unknown quntity (or quntities) tht stisfy the eqution. The solution of n eqution is the set of ll vlues which, when substituted for unknowns, mke n eqution true. Let s cll this set S. Emple S(7 = 5) =, 7 Two equtions re equivlent if their solution set is the sme. Emple 9 4 + = 0 = 4 ( ) since 9 S(4 + = 0) = S = = 4 4 Properties: Following properties mke it possible to simplify nd solve equtions:. ( A = B) ( A + C = B + C). We will pply this property to bring over terms.. ( A = B) ( ma = mb ) with m R 0. ( AiBi C = 0) ( A = 0 B = 0 C = 0) 4. ( AiC = Bi C) ( A = B C = 0) Algebr 7

Avoid following errors: ) If you pply A = B AiC = Bi C, then it s possible tht you import solutions. Emple 4 + = + + = + ( ) 4 ( ) + = 5 6 0 ( )( ) = 0 ( = = ) imported So you should solve this eqution s follows: + = + ( ) 4 ( ) = { } S = Remember: before removing the denomintor which contins the unknown, mention tht the donomintor is different from zero. b) If you pply AiC = Bi C A = B then it is possible tht you loose solutions. Emple ( )( + ) = ( + )( + ) = + + = 0 = : solution is lost So you should solve this eqution s follows: ( )( ) 0 ( 0) ( 0) + + = + = + = = = S =, Algebr 8

.. Discussion of liner eqution tht contins prmeters Consider + b = 0;, b R. 0 + b = 0 = b/ (= the only solution). = 0 0 + b = 0 0 = b if b 0 then S = (flse eqution) if b = 0 then S = R (identicl eqution) Emple p m = 9 p nd m re rel prmeters (p ) = m 9.. p solution: = p = 0 = m 9 m - 9 p - if m = m = then 0 = 0 S = IR if m m then S =..4 Solving second degree eqution in unknown Stndrd form: + b + c = 0 met R0 ;b,c R Determine the discriminnt = b 4c If > 0 then ² + b + c = 0 hs two different roots b b + = nd = = 0 then ² + b + c = 0 hs two equl roots -b = = < 0 then ² + b + c = 0 hs no rel roots If b is even, then one cn use the simplified formuls: If b = b' then = 4b' 4c = 4(b' c) = 4 ' b ' ' b ' + = = ', Algebr 9

. Inequlities in IR.. Definition An inequlity is sttement tht mthemticl epression is less thn or greter thn some other epression A < B (or A B, A > B, A B). A nd B re epressions of which t lest one contins vrible... Solving inequlities To solve n inequlity mens to find the vlue or vlues of the unknown tht stisfy the inequlity. Let s cll this set of solutions S. Emple S(( ) > 0) = ],0[ ], + [ Two inequlities re equivlent if their solution set is the sme. Properties: Following properties mke it possible to simplify nd solve inequlities:. (A < B) (A + C < B + C) ma < mb if m R A < B ma > mb if m R. ( ) 0 In n inequlity we never remove the vrible from the denomintor! + 0 Algebr 0

.. Discussion of the liner inequlity tht contins prmeters Consider + b > 0;, b R + b > 0 > b. b b > 0 > b > S = R >. b b < 0 > b < S = R <. b > 0 S = R = 0 > b 0 > b b 0 S = Emple p m + < p (p, m R) (p ) < m p Discussion:... p > m p S = R < p p < m p S = R > p p = if m 5 > 0 S = R 0 < m 5 if m 5 0 S = Algebr

..4 Solving qudrtic inequlities in unknown Consider b c 0 with R 0;b,c + + R We first emine the evolution of the sign of the left hnd side. Therefore we look t the grph of the function y = + b + c. This function represents prbol with is of symmetry prllel to the y-is. The intersection points of the prbol with the y-is re obtined by solving the set of y = + b + c equtions:. S = {(0, c) } = 0 The intersection points of the prbol with the -is re obtined by solving the set of y = + b + c equtions: : y = 0 b b + if > 0,0 nd,0 re the intersection points with the -is if = 0 b the prbol touches the -is in,0 if < 0 doesn t intersect with the -is Furthermore we know tht if > 0 the concve side of the prbol is directed upwrds nd if < 0 the concve side of the prbol is directed downwrds. Algebr

Summry: D > 0 D = 0 D < 0 > 0 = = < 0 This summry leds us to the following sign chrt of y = + b + c : sign of y sign of 0 opposite sign 0 sign of of. If > 0. If < 0 S( + b + c 0) = ], ] [, + [ S( + b + c³ 0) = [, ] Emple: 0 roots of the left hnd side of this inequlity re, sign chrt: + 0 0 + S = [,] Algebr

..5 Solving rtionl inequlities Solving rtionl inequlities is very similr to solving polynomil inequlities. But becuse rtionl epressions hve denomintors (nd therefore my hve points where they're not defined), you hve to be little more creful in finding your solutions. To solve rtionl inequlity, you first find the zeroes (from the numertor) nd the undefined points (from the denomintor). You use these zeroes nd undefined points to divide the number line into intervls. Then you find the sign of the rtionl on ech intervl. Emple : ( )( + ) f () = > 0 ( ) ( + ) Sign chrt: 0 0 + + + + + + + + + + + + 0 ( + ) 0 + + + + + + + + + ( ) + + + + + + + + + 0 + ( + ) 0 + + + + + + + f( ) 0 + 0 + 0 S( f () > 0) =, ] 0,[ Emple : ( )( + + ) + 6 f () = 0 First find the zeroes of the nomintor nd of the denomintor: zeroes nomintor: ; zeroes denomintor: -, Sign chrt: - + + + 6 f( ) 0 + + + + + + + + + + + 0 0 + + 0 + S(f () 0) = ], [ ], [ Algebr 4

Emple : 4 + + + + f () = < 0 First, fctor the nomintor nd the denomintor: f () = ( )( + )( + ) ( + + ) Sign chrt: 0 0 + + 0 + + + + + 0 + + + + + + f( ) + + + + + + + + 0 + 0 S(f () < 0) = ], 0[ ], + [ Emple 4: + + + + + 5 0 0 + ( + )( ) Sign chrt: 5 5 + 0 + 0 + + + 0 + f () + 0 + 5 S(f () 0) =,, + ] [ Algebr 5

. Eercises.. Solve. ( 5)( + 7) = ( 5)( 5 + ) {,5 }. ( 5) 9 = 0 {,8 }. 6 = 4. + = + {-,4} 5. 0 = 5 ( ) 7, 6 6. + 7 + ( ) + 8 = 0 For which vlue of Z there eists solution? Determine the solution for tht vlue. =, = 5 7. 4 > 5 > 4 5 8. 9. 4 + 5 > 5 8 + < + 7 > 7,, + 0. + + ], [ [, [ ], + [ + >, ], [. ( )( )( ) 0. + < + ],0 [ ], + [ Algebr 6

.. Determine the number of solutions for which vlue(s) of m nd solve. m = 4 m. + = 0 p + m p m. ( ) p p = m p 4. m + m 5. + 4 > m + 8 6. 7. m ( 4) < 4 p 5m + m < m m p Algebr 7

4 Absolute vlue of rel number 4. Definition We define the bsolute vlue (or modulus) = if 0 R : = if 0 of rel number s follows: Remrk: ) for ech rel number different from 0 one of the prts of the functiondefinition is pplicble. Only for 0 the prts re pplicble since they both give the sme functionvlue. ) + R Grph: To obtin the grph of y = you cn reflect tht prt of the grph of y = tht is under the - is round the -is. Algebr 8

4. Property + R, R : 4. Opertions, y R : y = y R 0 : = = R, y R 0 : y y, y R : + y + y 4.4 Appliction: grph of rel bsolute vlue function Given: f () = f ( ) = Asked: mke grph of f Solution: Method : strting from the definition of f without bsolute vlue signs: ( ) ( ) = if 0 f () = = if 0 = if () f () = = if () () represents hlfline through (, 0), (4, ) () represents hlfline through (, 0), (, ) Algebr 9

Method : strting from the grph of y = y = is represented by line through (0, ), (, 0) to get the grph of y =, reflect tht prt of the grph of y = which is under the -is round the -is. Algebr 0

4.5 Eercises Drw the function grph of following functions in n orthogonl coordinte system:. y = +. y = + + 4. y = 4 4. y = + 4 + 5. y = 9 + 9 6. y = + + 7. y = 4 Algebr

5 Mtrices nd determinnts 5. Mtrices 5.. Definition An m n-mtri A is rectngulr rry of m n rel (or comple) numbers, contining m rows nd n columns. In generl:... n ( ij )... n = A mn............... m m m... mn The horizontl nd verticl lines in mtri re clled rows nd columns, respectively. The numbers in the mtri re clled its entries or its elements. Entries re often denoted by vrible with two subscripts, s shown in the generl nottion bove. The first inde denotes the row, while the second one denotes the column. To specify mtri's size, mtri with m rows nd n columns is clled n m-by-n mtri or m n mtri, while m nd n re clled its dimensions. The set of ll m n mtrices is denoted s rel or comple numbers. 5.. Equl mtrices m n R or s Two mtrices A nd B re clled equl if nd only if: They hve the sme dimensions, Elements on the sme position re equl. i =,,,...m A = ( ij ) nd B = ( b ij ) then A = B ij = b ij with j =,,,...m m n C depending on the elements ij re There re number of opertions tht cn be pplied to modify mtrices clled mtri ddition, sclr multipliction nd trnsposition. These form the bsic techniques to del with mtrices. Algebr

5.. Addition of mtrices with the sme dimension The sum A+B of two m-by-n mtrices A nd B is clculted entrywise: ( ij ) ( ij ) ( ij ij ) + b = + b where i m nd j n Emple 4 6 4 0 + 4 + 4 6 + 0 8 6 5 7 + = = 0 + 0 5 + 7 7 4 Properties:.Closure: R + R mn A,B : A B mn. Commuttive property: A + B = B + A. Associtive property: (A + B) + C = A + (B + C) 4. The zero Mtri (ll ij = 0) is the identity element. A + 0 = A 5. Inverse element: for ech mtri A=( ij ) there eists A=( ij ) such tht A + ( A) = 0 m R n, + is commuttive group 5..4 Sclr multipliction of mtri nd rel number r R : r( ) = (ri ) ij ij r is clled sclr. This opertion is clled sclr multipliction. Emple 0 0 6 9 = 0 0 Algebr

Properties:. Closure: R R R m n m n r, A : r A m n. is identity element: A R : A = A. Associtive property: R R = m n r,s, A : (r s)a r(s A) 4. Sclr multipliction is distributive over the ddition in m R n : r(a + B) = r A + r B 5. Sclr multipliction is distributive over the ddition in R : (r + s)a = r A + s A Since m n m n R, + is lso commuttive group we cllrr,,+ is rel vector spce. 5..5 Trnsposed mtri The trnspose of n m-by-n mtri A is the n-by-m mtri A T formed by turning rows into columns nd vice vers: (A T ) ij = A ji... n... m... n T... m A= A =.............................. m m mn n n mn 5..6 Mtri multipliction Multipliction of two mtrices is defined only if the number of columns of the left mtri is the sme s the number of rows of the right mtri. If A=( ij ) is n m-by-n mtri nd B=(b ij ) is n n-by-p mtri, then their mtri product C=A B is the m-by-p mtri of which entries re given by dot-product of the corresponding row of A nd the corresponding column of B: C = ( c ij ) where c = b + b +... + b = b, i =...m, j =...n. ij i j i j ip pj ik kj k= p Algebr 4

Emple c c c b b b c c c = b b b c c c 4 4 c4 c4 c4 4 4 c = b + b with c = b + b Properties:. Associtive property: (A B) C = A (B C). Right distributivity: A (B + C) = A B + A C. Left distributivity: (A + B) C = A C + B C 4. Not-commuttive property: A B B A in generl. Emple 9 while = = 4 6 [ ] [ ] [ ] 5..7 Squre mtrices A squre mtri is mtri which hs the sme number of rows nd columns. An n-by-n mtri is known s squre mtri of order n. The elements ii with i =,,,... n form the min digonl of the mtri. Any two squre mtrices of the sme order cn be dded nd multiplied. Algebr 5

Specil squre mtrices: Identitymtri of order n: The identity mtri or unit mtri of order n is the n-by-n squre mtri with ones on the min digonl nd zeros elsewhere. It is denoted by I n : ii = i, j =,,...,n : ij = 0 if i j The importnt property of mtri multipliction of identity mtri is tht for m-by-n A: I A = A I = A m n Zero mtri of order n: The zero mtri of order n is the n-by-n squre mtri with ll its elements 0. It is denoted by O n. The zero mtri cts s n bsorbing element for mtri multipliction: for m-by-n A: Om A = A On = Om,n O m,n is the zero m-by-n mtri with ll its entries 0. Zero Divisors: There eist mtrices A nd B for which A B = 0 nd A 0 en B 0. Such mtrices re clled zero divisors. Emple: 0 0 = 0 0 Invertible mtri: A squre mtri A is clled invertible or non-singulr if there eists mtri B such tht A B = I n. This is equivlent to B A = I n. Moreover, if B eists, it is unique nd is clled the inverse mtri of A, denoted A. Algebr 6

Digonl mtri: If ll entries outside the min digonl re zero, A is clled digonl mtri. 5. Determinnts 5.. Definitions The determinnt is specil number ssocited with ny squre mtri. The determinnt of mtri A is denoted det(a), or without prentheses det A or A or. An lterntive nottion, used in the cse where the mtri entries re written out in full, is to denote the determinnt of mtri by surrounding the mtri entries by verticl brs insted of the usul brckets or prentheses. The determinnt of squre -by- mtri is given by = To clculte the determinnt of squre -by- mtri A A = continue s follows: Minor of n element: To find the minor of ij. Cross out the entries tht lie in the corresponding row i nd column j.. Rewrite the mtri without the mrked entries.. Obtin the determinnt ij of this new -by- mtri. M ij is clled the minor for element ij. Cofctor of n element: ij = ( ) i+j M ij The cofctor ij is independent of the elements of the i th row nd the elements of the j th column. Algebr 7

Emple α + = ( ) = Definition: The determinnt of -by- mtri is the rel number obtined by tking the elements of chosen row or column nd multiplying ech term by the corresponding cofctor nd then dding these products. So, for chosen fied row vlue i, the determinnt cn be clculted emnting from the i th row: A = i. α i + i. α i + i. α i +... in. α in Emple A = 4 α = = + = 4 + ( ) 4 4 8 α = = = 4 + ( ) (8 6) α = = = + ( ) 4 7 det A = 8 + ( ) + ( 7) = (emnting from the first row). Algebr 8

5.. Properties of determinnts. A squre mtri A nd its trnsposed A T hve the sme determinnt. det A = det A T. If B results from the mtri A by interchnging two rows (or columns) of A, then det B = det A. Consequences: determinnt with equl rows (columns) is zero. the sum of the products of the elements of row (column) nd the corresponding cofctors of nother row (column) is zero. Emple: check the previous emple: α + α + α = 8+ ( ) + ( 7) = 0. If the elements of row (column) cn be written s sum of elements, the determinnt cn be considerd in the sme wy s the sum of determinnts. + b c b c b c ' ' ' ' + = + ' ' + b c b c b c b c b c b c Emple + 0 + = 0 + = + 4 4 4 4 4 4. If we multiply one row (column) with constnt, the determinnt of the new mtri is the determinnt of the old one multiplied by the constnt. kb c b c kb c = k b c kb c b c Consequences: common fctor of ll the elements of row (column) of A cn be tken outside the determinnt. determinnt in which one row (column) is multiple of nother row (column) is zero. Algebr 9

5. To ny row of A we cn dd ny multiple of ny other row without chnging det(a). b c + kb kb c b c = + kb kb c b c + kb kb c 6. det (A.B) = det A. det B 7. The previous properties llow us to clculte determinnt by reducing the order: by pplying the previous properties try to get row or column which contins only one vlue different from zero. Emple 6 9 5 = r r 4 0 5 = r r 4 0 0 = 4 = ( + 4 ) = Algebr 40

5. Eercises. Clculte, y, z if 4 y+5z = y 0 +5 +y+z Solution: =, y =, z =. Fill in the missing numbers:. 5. 6.. 5 + = 4. 4.. If A is p-by-q mtri nd B is r-by-s mtri, under wht condition both mtriproducts eist? Solution: q = r nd s = p 4. Following mtrices re given: 0 0 A = 0 B = C = 4 Clculte A B, (A B) C, B C, A (B C) nd check the ssocitive property. 5. Proof tht (A B) T = B T A T if g h b c A =, B = i j d e f k l 6. Determine ll mtrices B for which A B = B A if A = 4 Solution: z z + z 7. Which reltionship eists between mtrices B (of order ) nd A B if. 0 0 A = 0 k 0 0 0. 0 0 A = 0 0 0 0 Algebr 4

8. Clculte the following determinnts (by reducing the order): 0 0 0 0. 4 4 0 0. 4 0 0 0 0. 0 0 0 0 0 0 0 0 0 9 5 4. 6 5. 8 9 6 4 5 5 4 7 4 b c 6. c b b c Solutions: 49 5 + 5 0 0 + b + c b c 9. Proof tht: b c k b c p q r k = p q r y z k y k z Algebr 4

6 Systems 6. Systems of n equtions in n unknowns 6.. Mtrinottion + +...+ n n = b + +...+ = b... + +...+ = b n n n n nn n n ij R, b R i Set A =... n... R............ n n... nn n n n b b X = R, B = R...... b n n n n Mtrinottion: A X = B Emple - + 5y - z = - 5 - + z = 0 A = 0, X = y, B = 0 -y + z = - 0 - z - Algebr 4

6.. Solving system of equtions by elimintion The gol is to reduce the system until you re left with n equivlent system in which the first eqution contins n unknown, the second eqution contins n - unknown, the third n, the lst but one contins unknown, the lst contins unknown. This is done by combining equtions to cncel out unknown. When you hve reduced the system, you solve the lst eqution, nd bck-substitute this vlue into either of the other equtions nd then solve the remining system for the other unknown. Emple + y z = 9 y z = + 6y + z = 7 + y z = 9 y + z = 8 y + 4z = 0 + y z = 9 y + z = 8 z = + y = 9 + y = 8 z = = 0 y = z = = 8 y = z = 6.. Solving system of equtions by the method of Crmer The system A.X = B hs one unique solution A 0. This solution is given by i = A i A With Ai is the n-by-n mtri defined s: A i =... i- b i+... n... i- b i+... n........................ n n... n i- bn n i+... nn Algebr 44

Emple Tke the emple of 6... A = A = A = A = 6 9 6 9 7 9 6 7 A = 6 A = 48 A = A = 6 A A A = = 8, y = =, z = A A A 6. Systems of liner inequlities in unknown Solve ech inequlity seprtely nd the solution of the system is formed by those vlues which stisfy ll inequlities. Emple 5 4 < < 9 < < 5 5 S = R < < 5 9 - /9 5 Algebr 45

6. Eercises 6.. Solve the following systems of equtions... 4. 5. + y z = + y + z = 4 + 7y + z = 7 y + z = + y + z = 4 + 5y z = 5 y + z = 6 + y + 4z = 7 y + z = 8 + y = y 5z = z = + = y + = y z + = z 9 k,+ k, k 5 5 {(,0,0)} {(5,-,)} 4 4 4,, 5 6.. Solve the following systems of inequlities. 4 0 ( )( + 5) < 0 ]-5,-[ [,[. < < (4 )( + ) 0 0, 4. 7 5 < < ] 4,9[ Algebr 46

Anlytic Geometry Tem Mthemtics Dr. Croline Dnneels

Vectors nd lines.... Vectors.... Lines... 5. Prllel lines... 8.4 The euclidin vectorspce... 9.5 Angles....6 Orthogonlity of lines....7 Distnce from point to line (O.N.B.)....8 Eercises... 4 Conic sections... 9. Introduction... 9. The circle... 0. The prbol....4 Eercises...

Vectors nd lines. Vectors.. The concept of vector π is the set of ll points of the plne. Tke in the plne π fied origin O. A trnsltion t in the plne is now completely determined by the imge point P of O by this trnsltion: t(o) = P. The trnsltion is determined by point P or lso by ny couple of points (A,B) so tht t(a)=b. With trnsltion t, we ssocite free vector. The vector AB is the set of couples of points (A,B) such tht t(a)=b. AB = CD if there is trnsltion t such tht t(a)=b nd t(c)=d. A free vector cn now be represented by mens of n rrow with origin in O nd with terminus point t(o)=p. Now ech point P in the plne cn be seen s the terminus point of P = OP. P is clled bound vector. The vector O is clled the zero vector. The plne in which we consider ech point terminus point of bound vectors is clled πo... Bsis nd coordinte system Let s choose in πo E en For ech P in πo there is only one E, E is clled bsis of πo. { } y Ey which together with O form tringle. (, ) P = E + ye. y R such tht y (,y) re clled the crtesin coordintes of P with respect to the bsis { E, Ey} the notion coordinte is relted to bsis.. We emphsize OE is the -is; OE y is the y-is; together they represent the coordinte es of the ycoordinte system. If (, ) re the coordintes of A we lso write this vector s A(, ) or A = OA = (, ). This mens tht we will use the word vector (with strting point the origin) nd the word point in the y-plne for the sme concept. Annlytic Geometry

.. Opertions with vectors In πo we define the following opertions:. The ddition of vectors A nd B is performed by dding the ssocited vectors OA nd OB using the prllelogrm rule. The result C is determined by its terminus point. From the figure one cn esily see tht, if A = (, ) nd B = ( b, b ), then the sum C = A + B = +, b + b. vector is: ( ) y C A b B O b. The multipliction of vector A of πo nd sclr k R is multiple of A : k A. Furthermore, if k > 0, k A hs the sme sense s A, nd if k < 0, k A hs the opposite sense. A =, k A = k, k. If ( ), then ( )..4 Midpoint of line segment The midpoint of line segment [AB] with A = (, ) nd B = ( b, b ) is determined by M A + B + b + b = =, Annlytic Geometry 4

. Lines.. Vector eqution Tke line e through the origin. Every other point S on tht line is the terminus point of vector S tht determines the direction of e in unique wy. Such vector is clled direction vector of e. If S is direction vector of e, then every ks with k R 0 lso is. e S O Tke line e prllel to e nd two different points P nd P0 on e. P e P 0 e S P P0 If S is direction vector of e, then P P 0 = ks or P = P 0 + ks for certin k R. Otherwise, for every k, the terminus point of the vector P 0 + ks is on the line e. The vector eqution P = P 0 + ks is therefore necessry nd sufficient condition for point to lie on the line e (given by point P 0 nd the direction S ). P = P + ks with k R is clled the vector eqution of the line. 0 Annlytic Geometry 5

.. The prmeter equtions Replcing the vectors by their coordintes in gives Furthermore which is equivlent with 0 0 P = (, y ) R with P = (, y), 0 0 0 (, y) = (, y ) + k(, b) with k R (, y) = ( + k, y + kb) with k R 0 0 = 0 + k with k R y = y0 + kb nd S = (, b) These equtions re clled prmeter equtions of line through given point ( 0, y 0) with direction numbers (, b ), the coordintes of the direction S. It is importnt to see tht the bove equtions re not unique since both P 0 nd S my be chosen... Crtesin eqution(s) So we hve two wys to give the eqution of line: the vector eqution nd the prmeter equtions. A third possibility, lthough less generl, is the Crtesin eqution(s). They my be obtined strting from the prmeter equtions. From the prmeter equtions, k my be eliminted: or 0 y y0 = if b 0 b b y y0 = ( 0 ) if 0 Both equtions re clled the Crtesin eqution of the line. The rtio b (in front of ) is clled the slope m. If = 0, the prmeter equtions re = 0 with k R y = y0 + kb The second of these equtions simply sys tht y my be ny rel number. Since there is no limittion on y, this eqution is superfluous. The other eqution then becomes the Crtesin eqution: = 0 This line is prllel to the y-is. The simplest direction numbers re (0,). The slope m is not defined, since = 0. In the sme wy, if b = 0, the crtesin eqution becomes y = y0. This line is prllel to the -is. The simplest direction numbers re (,0). Its slope is m = 0. Annlytic Geometry 6

Notice tht (, b ) = (0,0) is impossible. The zero vector is ecluded s direction vector of line. If the line is given by two points P0 ( 0, y 0) en P (, y ) then direction vector of the line is given by P P 0, so we obtin: P = P + k( P P ) with k R s vector equtions, s prmeter equtions, 0 0 s crtesin eqution if 0 nd y y 0. = 0 + k( 0 ) met k R y = y0 + k( y y0) 0 y y0 = y y 0 0 If = 0 the eqution is similr to = 0; if y = y0 the eqution is similr to b= 0. So, in generl, the Crtesin eqution of line in R tkes the form with A nd B not both zero t the sme time. A + By + C = Otherwise it cn be proven tht ech eqution of this form in Wht is the slope of this line? Find set of direction numbers. 0 R respresents line. Annlytic Geometry 7

. Prllel lines If is given : line e with direction vector S, direction numbers (,b ), slope m, eqution A +B y+c =0, line e with direction vector S, direction numbers (,b ), slope m, eqution A +B y+c =0 e e then O S S e // e S = ks with k R 0 b = k = kb m = m A B = ka = kb with k R 0 with k R 0 Use: Given line e with eqution A+By+C=0 nd point P 0 ( 0,y 0 ) Determine the eqution of line f // e nd through P 0 f hs eqution: A(- 0 )+B(y- y 0 )=0 Annlytic Geometry 8

.4 The euclidin vectorspce.4. Definitions A(, ) nd B( b, b ) in If π 0 Then these vectors re perpendiculr if Then the norm of the vector is Then A is normlized vector if re given, then the sclr product of these two vectors is A B = b + b R A B A. B = 0 A = A A = + > 0 A = Then the distnce between (, ) B A nd (, ) B b b is ( b ) + ( ) d( A, B) = A B = b O A A - B Annlytic Geometry 9

.4. Normlizing vector If A O then E = A A hs length nd thus is unit vector in the sme sense s A. This process is clled normlizing vector..4. Orthonorml bsis (O.N.B.) { E, E } y is n orthonorml bsis E E y E = E y = Annlytic Geometry 0

.5 Angles First we wnt to mention tht ngles re positive ngles if the terminl ry rottes counterclockwise round the verte from the initil ry..5. The ngle of line with the -is (nlyticl epressions hold in n orthonorml bsis) In the following figure, we see in the right tringle tht m tnα = = m In other words, the slope of the line is equl to the tngent of the ngle between the -is nd the line. y E y e m α 0 E.5. The ngle between two lines If line e hs direction vector S, direction numbers (,b ) nd slope m nd line e hs direction vector S, direction numbers (,b ) nd slope m then the ngle α between these lines is equl to the cute ngle determined by the direction vectors of the lines: S S + bb cosα = = S S + b + b Annlytic Geometry

.6 Orthogonlity of lines If line e hs direction vector S, direction numbers (,b ), slope m, eqution A + B y + C = 0 nd line e hs direction vector S, direction numbers (,b ), slope m, eqution A + B y + C = 0 then (the nlyticl epression with respect to n orthonorml bsis is): e S e O S e e S S m b b + = A A = m B B 0 + = 0 Use: Find the eqution of line f e with eqution A + By + C = 0 nd which psses through point (, ) P y. 0 0 0 f hs eqution: B( ) A( y y ) = 0 0 0 Annlytic Geometry

.7 Distnce from point to line (O.N.B.) Consider in π 0 point P0 ( 0, y 0) nd line e : A + By + C = 0. To find the (orthogonl) distnce from P0 to the line e, one my proceed in the following wy: Construct the line r through P0 perpendiculr to e. Find the intersection S of r nd e. The distnce between P0 nd e is then d( P0, S). So: r my be given by mens of: = + y = y0 + kb 0 ka k R To find the intersection S of r nd e, the prmeter equtions of r is substituted in the eqution of e : A + By + C = 0. Then this eqution is solved for k: ( ) ( ) A ka B y kb C 0 + + 0 + + = 0 gives the vlue of the prmeter k S which corresponds to the intersection S : k S A0 + By0 + C = A + B So the distnce is: d P S P S k A y y k B ( ) ( ) ( ) ( ( )) 0, = 0 = 0 0 + S + 0 0 + S d P S k A B ( 0, ) = S + nd finlly d( P, e) = d( P, S) = 0 0 A + By + C 0 0 A + B Annlytic Geometry

.8 Eercises. Given re A(5,-), B(-,5), C(-7,).. Determine the coordintes of the midpoints of the sides of the tringle ABC. 7,, 4,,, ( ) b. Determine the coordintes of the centroid (or geometricl center or center of re) of the tringle ABC. (, ). Determine the direction numbers nd the slope of the lines determined s follows:. through (-, 7) nd (, -8) (, -5) nd m = -5 b. line (k,0) nd m = 0 c. line y (0,k) nd m does not eist! d. the line with eqution: y + 4 = 0 (, ) nd m = e. the line with eqution: y = (4, ) nd m = 4 4 f. the line with eqution: y = + (, ) nd m =. Mke drwing of the lines with the following equtions:. y = 4 b. + y = 0 c. 4 + y + 5 = 0 4. Determine the ngle which the following lines mke with the -is in n orthonorml coordinte system.. y + 5 = 0 α = 45 ;α = 5 b. y 5 = 0 α = 60 ;α = 0 5. Proof tht the figure formed by A(-,), B(-,4), C(5,6), D(4,) is prllelogrm. 6. Given line e with eqution + y = 4. does A(4,) belong to e? No b. does B(4,0) belong to e? Yes c. determine the bsciss of the point on e with ordinte -5 = 4 d. determine the ordinte of the point C with bsciss y = e. drw line e f. find the intersection points with the -is nd the y-is (4, 0) nd (0, ) Annlytic Geometry 4

7. Given line e with eqution: + y + = 0 with R. Determine, if possible, such tht:. e psses through (, 0) = - b. e psses through (0,0) impossible c. e is prllel to line f with eqution y 5 = 0 = -9 d. e is prllel to the -is = 0 e. e is prllel to the y-is impossible f. the -intercept of e is + = g. the y-intercept of e is +5 impossible h. the y-intercept of e is R 8. Find the eqution of the line :. whose slope is m = - nd psses through the point (, 4) y + 0 = 0 b. which psses through the points (, ) nd (5, ) y + = 0 c. which psses through the origin nd the point (, 6) y = 0 d. which psses through the points (, 5) nd (7, 5) y = 5 e. which psses through the points (-, 4) nd (-, ) = f. which psses through the points (, ) nd (-, -6) y = 0 g. with slope m = nd whose y-intercept is +4 y 4 = 0 h. whose y-intercept is + nd -intercept is + + y 6 = 0 i. whose y-intercept is -6 nd which psses through the point (, 4) y 5 + 6 = 0 j. which psses through the point (-, 6 ) nd is prllel to the line e: + y 5 = 0 y + 6 = 0 k. which psses through the point (0, ) nd is prllel to the line through the points (,0) nd (5, ) y 9 = 0 l. which psses through the point (, ) nd is prllel to the y-is = m. which psses through the point (-, 4) nd is prllel to the -is y = 4 9. Given line e with eqution : y = ( ) + ( + b) with, b R. Determine nd b such tht:. e psses through the points (, ) nd (, 5) = nd b = - b. e psses through the point (, -) nd is prllel to line f: + y + 5 = 0 = nd b = - c. the -intercept of e is - nd e hs (, -0) s direction numbers = - nd b = -7 d. e psses through (0, 0) en (-, -4) = 4 nd b = -4 0. Determine the eqution of line e (in n O.N.B.):. which psses through the point (4, -) nd forms n ngle of 45 with the -is y + 7 = 0; y + = 0 b. whose y-intercept is + nd forms n ngle of 0 with the positive -is y 9 = 0;y + 9 = 0 Annlytic Geometry 5

. Determine the eqution of line e which psses through the point (, 4):. whose positive -intercept is double the positive y-intercept + y 0 = 0 b. whose negtive -intercept nd positive y-intercept re equl y + = 0. Determine such tht e // f: e : ( ) ( + ) y + = 0 f : ( + ) + ( 4 ) y + = 0 = 0 or =. Emine if the following points re colliner:. (8, ), (-6, -), (5, 6) Yes b. (, ), (4, -), (-5, 5) Yes 4. Determine such tht the points (, ), (, ) nd ( +, - ) re colliner. Determine the eqution of the line pssing through these points. = nd y + 5 = 0 = 4nd y + 8 = 0 5. Determine the points of intersection of the following lines: e : + y = 4. f : + y = 7 e : 5 + y = 0 b. f : + 8 = 0 e : 6y = 4 c. f : y + = 0 (, ) (-4, 7) coinciding lines 6. Give the generl eqution of ll lines pssing through (, ). y = m( ) 7. Give the generl eqution of ll lines with slope. y = + q 8. Proof tht the medins of tringle re concurrent. 9. For A(,0), B(,), C(,) given in n O.N.B., determine A B A C C B A A B B C C 4 5 0. Proof tht the points A(4,6), B(,-4), C(-,) form right tringle. Annlytic Geometry 6

. Determine the eqution of the line perpendiculr to. e : 5 + y 6 = 0 nd pssing through (0, 0) 5 5y = 0 nd, b. f : y + 5 = 0 nd pssing through (, -6) + y + 0 = 0 nd ( 4, ) Determine lso the intersection points.. Determine the eqution of the line perpendiculr to the line with eqution + 5y + 4 = 0 in the intersection point with the -is. 5 y + 0 = 0. A line hs slope. Determine the eqution of the line perpendiculr to tht line nd pssing through the point (4,). y + 6 = 0 4. Determine the eqution of the perpendiculr bisector of the line segment [AB] with A(,-)nd B(,). y = 0 5. The equtions of the sides of tringle re: 4y + 4 = 0;4 + y = 0; + 5y 8 = 0. Determine the equtions of the ltitudes nd the coordintes of the orthocenter. y 5 + 5 = 0 y + 4 5 = 0 4y 0 = 0 0 55, 9 9 6. Determine the length of the sides of the tringle formed by A(5, -), B(-, 5), C(-7, ). 0;; 5 7. Determine the distnce from the origin to A(-, 4) nd to the line e : y = 5 5; 0 8. Determine the distnces from A(-, 4), B(, ), C(-, ) to the line e : y + 5 = 0 ; ; 0 9. Determine the distnces from A(5, ) to B(, -), C(-, ) nd to the line BC. 5;5 ;5 Annlytic Geometry 7

0. Determine the distnce from A(-4, 4) to the line e : + y = 0 using. the formul b. the construction method 7 5 5. Determine the distnce from A(, -5) to the line perpendiculr to e : y + = 0 nd through B(, ). 5. Determine the distnce from the origin to the perpendiculr bisector of the line segment [AB] with A(-9, 7), B(5, -).. Determine the distnce between the prllel lines with equtions + y + = 0 nd + y = 0. 5 4. If e : + y + = 0; f : + y = 0, determine p such tht the distnces from A(,p) to the lines e nd f re equl. p = nd p = 5 5. Determine the eqution of the line through A(, ) nd t distnce from the origin. y = nd 5y + 9 = 0 6. Determine the eqution of the line through A(-5, ) such tht the distnce from B(, -) nd C(-, ) to tht line is equl. y + + = 0 nd 0y + 5 = 0 7. Determine the eqution of the line prllel to line e : 5 + y = 0 nd t distnce from A(-, ). 5 + y 5 = 0 5 + y + = 0 Annlytic Geometry 8

Conic sections. Introduction Prbols, ellipses nd hyperbols re conic sections. A conic section is curve which is obtined by the intersection of cone with plne. Circles re lso conic sections, since they re specil cses of ellipses. The type of conic section depends on the ngle between the intersecting plne nd the cone. Figure snijding vn een kegel door een vlk Left: to obtin n ellipse, the ngle between the plne nd the is is lrger thn the ngle between the is of the cone nd line on the cone. Middle: to obtin prbol, the ngle between the plne nd the is of the cone is equl to the ngle between the is nd line on the cone. Right: to obtin n hyperbol the ngle between the plne nd the is of the cone is smller thn the ngle between the plne nd line on the cone. In the following prgrphs we give the Crtesin equtions in stndrd form of circle nd prbol in n orthonorml coordinte system. The equtions represent the condition which points P(,y) must stisfy to be on the grph of the conic section. Annlytic Geometry 9

. The circle Although the circle is only specil cse of n ellipse, yet we will first py ttention to its definition nd its eqution. Definition: A circle C is the set of points P(,y) which re on constnt distnce R from fied point M( 0,y 0 ). This fied distnce R is clled the rdius, the fied point M( 0,y 0 ) the center of the circle. P C d( P, M ) = R ( ) + ( y y ) = R 0 0 If the center M is in the origin, the eqution of the cirlce is + y = R The generl eqution of the circle is: + y + A + By + C = 0 Be wre tht not ll equtions of this form represent circle (see eercise ). Annlytic Geometry 0

. The prbol Definition: A prbol P is the set of points P(,y) for which the distnce to fied line d is equl to the distnce to fied point F which is not on d. The point F is clled the focl point (focus) of the prbol, the line d the directri. Q P d( Q, d) = d( Q, F) The eqution of the prbol P with focl point F(p,0) nd directri d : = p is: This is esily found from the definition: y = 4 p d y Q T F Q P d( Q, d) = d( Q, F) y = Figure prbool ( ) + ( 0) = + p y p ( p) + ( y 0) = ( + p) 4 p So the eqution of the prbol is y = 4 p. The -is is the symmetry is of the prbol. The point F is the focl point, the line d is the directri nd the point hlfwy between the focl point nd the directri is the top of the prbol. It is importnt to see tht the bove eqution is only vlid for this specific choice of the coordinte system nd of the position of the focl point nd the directri. If the prmeter p is negtive, the focl point is t the left of the top nd the directri t the right side. The prbol will open to the negtive -is. The eqution is the sme. If we select horizontl directri, the prbol will be oriented verticlly. Agin, depending on the sign of p, the prbol is open to the positive y-is if p > 0 nd is open to the negtive y-is if p < 0. Annlytic Geometry

So there re four possibilities for prbol with its top in the origin nd one of the coordinte es s symmetry is. y = 4 p with p > 0 y = 4 p with p < 0 = 4 p y with p > 0 = 4 p y with p < 0 If the prbol is shifted to n rbitrry position of its top T( 0,y 0 ) one finds two possible equtions: Horizontl symmetry is: Verticl symmetry is: 0 = 0 with 0 0 ( y y ) 4 p ( ) 0 = 0 with 0 0 ( ) 4 p ( y y ) F( + p, y ) nd d : = 0 p F(, y + p) nd d : y = y0 p The generl eqution of prbol with symmetry is prllel to coordinte is is: Horizontl symmetry is: Verticl symmetry is: = Ay + By + C y = A + B + C Annlytic Geometry

.4 Eercises (we work in n orthonorml coordinte system). Determine the eqution of the circle with center M(, b) nd rdius R.. = 0, b = 0, R = 5 b. = 4, b =, R = + y = 5 ( 4) + ( y ) = 4. Determine the center nd rdius of the following circles:. y 8 6y 0 b. c. + = ( ) + y + y + = 0 6 + 6y 8 5 = 0 d. ( ) 6 + y 48 + 6y 7 = 0 M 4, nd R = 5 M, nd R = 6 M,0 nd R = 4 M, nd R = 7. Emine if following equtions represent circles. If so, determine the center nd the rdius.. + y 6 + 4y + 59 = 0 no circle b. c. 6 + 6y + 8 64y 5 = 0 4 + 4y + 40y + 09 = 0 C,,5 4 C, 5,0 4. Find the eqution of the circle which psses through (, ) nd (5, 7) nd hs its center on the line : y = 5 + 6 6 + 48 = 0 y y 5. Find the eqution of the circle which hs the line segment joining the points A(5,6) nd B(-,0) s dimeter. + y 4 6y 5 = 0 6. Find the eqution of the circle through P(-,4) nd concentric with the circle c : + y + 4y = 0 + + 4 = 0 y y 7. Find the eqution of the circle circumscribing the tringle ABC with A(, ), B(6, -), C(-, -5). + y 5 + y 8 = 0 Annlytic Geometry

8. A circle hs its center in M(, 0) nd psses through P(, ). Determine:. the eqution of the circle. ( ) + y = 5 b. the eqution of the circle with the sme center nd double re. ( ) y + = 0 9. Determine the focl point nd the directri of the following prbols nd mke drwing.. y 8 = 0 (,0); = b. c. d. y + 6 = 0 + y = 0 y = 0,0 ; = 0, ; y = 8 8 0, ; y = 8 8 0. Find the eqution of the prbol with the -is s the symmetry is nd with the top in the origin nd through the point A(-, ). y = 9. Find the eqution of the prbol with focl point 0, nd directri d : y =. + y = 0. Find the eqution of the prbol with focl point (7, -) nd directri d : =. y + 4y 8 + 44 = 0. Find the eqution of the prbol with focl point (7, -) nd s directri d the bisector of the second qudrnt. + y y 8 + 8y + 06 = 0 4. Mke drwing of:. b. 4 y 50 0 + = ( 4) = ( y ) 4y 40y 00 0 + + = ( ) y + 5 = 4 Annlytic Geometry 4

Clculus Dr. Croline Dnneels

Reltions, functions, mppings, - mppings... Epnding IR...4 Continuity of function in IR...5. Continuity t point... 5. Right- nd left-continuity... 6 4 Limits...7 4. Limits in rel number... 7 4. Right-hnd nd left-hnd limits... 9 4. Limits t infinity... 9 4.4 Infinite limits... 9 4.5 Infinite limits t infinity... 0 4.6 Elementry rules to clculte limits... 0 4.7 Indeterminte cses... 4.8 Eercises... 4 5 Differentition...7 5. The derivtive in point... 7 5. Geometricl interprettion of the derivtive... 7 5. Left-hnd nd right-hnd derivtive in point... 8 5.4 A verticl tngent... 9 5.5 Differentition rules... 9 5.6 Eercises... 6 Indefinite integrls...4 6. Antiderivtive functions... 4 6. The indefinite integrl... 4 6. Properties... 4 6.4 Bsic integrls... 5 6.5 Integrtion by substitution... 5 6.6 Eercises... 6 6.7 Integrtion by prts... 9 6.8 Eercises... 9 7 Definite integrls...0 7. The fundmentl theorem of clculus... 0 7. Properties of the definite integrl... 0 7. The substitution method... 7.4 Prtil integrtion... 7.5 Eercices...

Reltions, functions, mppings, - mppings In mthemtics, reltion from A to B is collection of ordered pirs (,y) where A nd y B, in other words it is subset of A B. The domin of the reltion R is the set of ll the first numbers of the ordered pirs. In other = (, y) R. words: dom R { } The rnge of the reltion R is the set of the second numbers in ech pir. In other words: = y (, y) R. rnge R { } If one switches the order of the pirs of reltion, then one gets the inverse reltion from B to A. A function is specil type of reltion where ech -vlue hs one nd only one y-vlue, so no -vlue cn be repeted. All functions re reltions but not ll reltions re functions. The -number is clled the independent vrible, nd the y-number is clled the dependent vrible becuse its vlue depends on the -vlue chosen. A function where for every element of A there is ectly one f() in B is clled mpping from A to B. Then dom f = A. A mpping from A to B is clled - mpping when the inverse is lso mpping from B to A. Clculus

Epnding IR Given the set of rel numbers IR, we dd two elements en +, nd cll the new set the epnded set of rel numbers. Nottie: IR = IR {, + } It is cler tht IR : < < +. Properties:. IR : + ( ) = = ( ) + + ( + ) = + = ( + ) + IR : ( + ) = + = ( + ) + 0 0 ( ) = = ( ) IR : ( + ) = = ( + ) ( ) = + = ( ). ( + ) + ( + ) = + ( ) + ( ) = ( ) ( + ) = = ( + ) ( ) ( + ) ( + ) = + = ( ) ( ) n n + = + = if n is odd = 0 = 0 + So, the following epressions hve no mening, they re indefinite: ( + ) + ( ) ; ( ) + ( + ) ; 0.+ ; +. 0 ; 0. ;.0 ; + + ; + ; + ; Clculus 4

Continuity of function in IR. Continuity t point Definition: function f: IR IR is continuous in point if the grph of the function does not show jump in. If the function is not continuous in, it is defined s discontinuous in. Emples Y f() Y f() b X X f is continuous in f is discontinuous in f is not left-continuous in f is not right-continuous in Y c f() b Y f() b X X f is discontinuous in f is not left-continuous in f is not right-continuous in f is discontinuous in f is right-continuous in Y b f() X f is discontinuous in f is left-continuous in Clculus 5

It is cler tht the point must belong to the domin of the function, otherwise continuity is not possible. The following mthemticl definition cn be understood in terms of the previous intuitive definition of continuous function. f is continuous in domf ε > 0, δ > 0 : < δ f ( ) f ( ) < ε. Right- nd left-continuity f domf is right-continuous in ε > 0, δ > 0 : < + δ f ( ) f ( ) < ε f is left-continuous in domf ε > 0, δ > 0 : δ < f ( ) f ( ) < ε Clculus 6

4 Limits 4. Limits in rel number The concept limit llows us to emine the behvior of function in the close neighbourhood of point. The function vlue in this point is without importnce. Assume tht is rel number which is n dherent point of dom f. This mens tht in ech + intervl round, ] ε, + ε[, ε IR0, there re numbers of dom f which re different from. In other words, f is not necessrily defined in, but it is defined in other points rbitrry close to. We sy tht function hs limit b t n input if the vlues f() come rbitrrily close to b for -vlues which come sufficiently close to. Nottion: lim f ( ) = b Since lim f ( ) = b describes the behvior of the function f in the environment of the point, the concept lim f ( ) is obviously meningless when is n isolted point of dom f. Definition: If f() is defined in n open intervl round, ecept possibly t itself, then lim f ( ) = b IR ε > 0, δ > 0 : 0 < < δ f ( ) b < ε Note tht the limit equls the function vlue in if the function is continuous in : lim f ( ) = f ( ) f is continuous in Clculus 7

Emples Y f() Y f() b X X f is continuous in f is discontinuous in f is not left-continuous in f is not right-continuous in lim f ( ) = f ( ) lim f ( ) = b Y c f() b Y f() b X X f is discontinuous in f is not left-continuous in f is not right-continuous in lim f ( ) = c lim f ( ) = b > < lim f ( ) does not eist f is discontinuous in f is right-continuous in lim f ( ) = f ( ) lim f ( ) = b > < lim f ( ) does not eist Y b f() X f is discontinuous in f is left-continuous in lim f ( ) = b lim f ( ) = f ( ) > < lim f ( ) does not eist Clculus 8

4. Right-hnd nd left-hnd limits We sy the left-hnd (or right-hnd) limit of f() s pproches is b (or the limit of f() s pproches from the left (or the right) is b) if the vlues f() come rbitrrily close to b for - vlues which come sufficiently close to from the left-hnd side (right-hnd side). ] [ lim f ( ) = b ε > 0, δ > 0 :, + δ f ( ) b < ε > ] [ lim f ( ) = b ε > 0, δ > 0 : δ, f ( ) b < ε < The following nottions re lso commonly used: lim f ( ) = b, lim f ( ) = b, f ( ) 4. Limits t infinity Limits t infinity represent the behvior of function vlues f() if ever increses. To define limits t +, we demnd tht dom f contins t lest hlf line of the form [,+ [. To define limits t -, we demnd tht dom f contins t lest hlf line of the form ]-,] We sy the limit of f() s pproches (negtive) infinity is b if the vlues f() come rbitrrily close to b for rbitrrily lrge -vlues. lim f ( ) = b ε > 0, m > 0 : > m f ( ) b < ε + lim f ( ) = b ε > 0, m > 0 : < m f ( ) b < ε 4.4 Infinite limits We sy the limit of f() s pproches is (negtive) infinity, if the vlues f() become rbitrrily lrge for -vlues which come sufficiently close to. lim f ( ) = + n > 0, δ > 0 : 0 < < δ f ( ) > n lim f ( ) = n > 0, δ > 0 : 0 < < δ f ( ) < n Clculus 9

4.5 Infinite limits t infinity We sy the limit of f() s pproches (negtive) infinity is (negtive) infinity, if the vlues f() become rbitrrily lrge for rbitrrily lrge -vlues. lim f ( ) = + n > 0, m > 0 : > m f ( ) > n + lim f ( ) = n > 0, m > 0 : > m f ( ) < n + lim f ( ) = + n > 0, m > 0 : < m f ( ) > n lim f ( ) = n > 0, m > 0 : < m f ( ) < n 4.6 Elementry rules to clculte limits Given two rel functions f nd g for which the limit in IR eists, e.g. lim f ( ) = L nd lim g( ) = L with L, L IR then lim f ( ) + g( ) = lim f ( ) + lim g( ) = L + L. [ ]. lim k f ( ) = k lim f ( ) = k L lim f ( ) g( ) = lim f ( ) lim g( ) = L L. [ ] f ( ) lim f ( ) L 4. lim = = if L 0 g( ) lim g( ) L 5. lim f ( ) = lim f ( ) = L n 6. ( f ) = ( f ) = ( L ) lim ( ) lim ( ) n 7. lim n f ( ) = n lim f ( ) = n L 8. lim k = k n Clculus 0

4.7 Indeterminte cses 4.7. The indeterminte cse 0 0 Rtionl function f () g() : f ( ) 0 lim = mens tht lim ( ) ( ) 0 g( ) 0 f = f = nd lim g ( ) = g ( ) = 0. So this mens tht both f() nd g() re divisible by ( ), so it is possible to write f ( ) = ( ) f( ) nd g( ) = ( ) g ( ). therefore f ( ) 0 ( ) f( ) f( ) lim = = lim = lim g( ) 0 ( ) g ( ) g ( ) Emple 4 ( )( + ) + lim = lim = lim = 4 5 + 6 ( )( ) Irrtionl form: mke nomintor nd/or denomintor rtionl. Emple lim + ( + )( + + ) = lim ( )( + + ) = = = ( + 4) lim ( )( + + ) lim ( + + ) 4 Clculus

4.7. The indeterminte cse Rtionl function: + +... + + lim b b b b b n n n n n 0 n = lim P p p... P + p + + + 0 p = b n p = if n=p if n > p (sign of hs to be determined) = 0 if n < p Irrtionl function: Determine the highest power of in the nomintor nd denomintor nd simplify. Remrk tht = if > 0 = - if < 0 Emple lim lim 5 + + + 5 + = lim 5 + + = + 5 + + lim = 4.7. The indeterminte cse Polynomil: lim + +... + + = lim ( 0 ) n n n n n n Irrtionl function: Multiply nd divide by the conjugte term. Clculus

Emple ( + + ) lim 4 ( ) / lim 4 + + = + + ( ) b / lim 4 + + = lim = lim = lim 4 + ( 4 + + )( 4 + ) 4 + = 4 4 + Clculus

4.8 Eercises... 4. 5. 6. 7. 8. 6 lim 4 lim lim 7 + 4 + lim 4 4 lim + + + + lim + + 6 4 6 4 lim + + 6 4 6 4 lim 0 8 4 0 0 4 m 9. lim 0. lim m n m m n m m m m n n. lim + 4. lim > + 6 0. lim 0 + 4. lim 5 > 4 5 5 5 5. lim 6 4 + 0 Clculus 4

6. + + lim + 7. lim + + 8. lim 4 9. lim > < + 4 4 ± 0. Evlute so tht lim 5 + 8 + 8 = + =. lim + > < + 8 6. lim + 6 + +. lim + + + 4 + 7 5 4 9 4. lim + + + 5. lim 8 + 4 + + ; 6. lim 8 + 7. lim ( ) 4 + ± 8. lim ( ) + + 9. lim ( 4 ) + + + Clculus 5

0. lim ( + ). lim ( ) 4 9 4. + + + + 8 + 4 lim 4 5 + 4. lim ( ) 5 8 5 4 9 ; 4 + + + ± 6 Clculus 6

5 Differentition 5. The derivtive in point y = f ( ) y f ( + ) B f ( ) A f + Consider: f f ( + ) f ( ) lim = lim 0 0 If this limit eists, in other words, if f lim IR, then we cll the limit the derivtive of the 0 function f in the point. We sy tht the function f is differentible or hs derivtive in.. f Nottion: f '( ) = lim 0 or df ( ) Df ( ) or d 5. Geometricl interprettion of the derivtive Consider the curve with eqution y = f(), nd neighboring points A(, f ( )) nd f f ( + ) f ( ) B( +, f ( + )) on tht curve. Then the difference rtio = is the slope of the line AB. When 0 then the line AB rottes round A nd reches the tngent t the function in the point A. Conclusion: f ( + ) f ( ) f '( ) = lim 0 is the slope of the tngent t f in the point A(, f ( )). This mens tht the eqution of the tngent line t (, f ( )) of the function y = f() is given by: y f ( ) = f '( )( ) Clculus 7

Emple y = + 4 f ( + ) f () f '() = lim 0 = lim 0 = lim( ) = 0 ( + ) ( + ) + 4 ( + 4) 5. Left-hnd nd right-hnd derivtive in point f One cn consider in lim 0 eist, we cll: f f ( + ) f ( ) lim = lim = f+ ( ) 0 0 > > f f ( + ) f ( ) lim = lim = f ( ) 0 0 < < seprtely the right-hnd limit nd the left-hnd limit. If they the right-hnd derivtive in, the left-hnd derivtive in. A function is differentible in if nd only if the left-hnd derivtive nd the right-hnd derivtive in re equl. If both one-sided derivtives in eist but they re different, the function is not differentible. This mens tht there eist two tngent lines in. A point like tht is clled corner point. Emple y 4 = is continuous for ech vlue of, buth the right-hnd derivtive in =, is different from the left-hnd derivtive in tht point, nmely lim f '( ) = > lim f '( ) = < Conclusion: If f is continuous in = f is differentible in =. But one cn prove tht, if f is differentible in = f is continuous in =. Clculus 8

5.4 A verticl tngent Emple The function y = is continuous for ech vlue of. We clculte the derivtive of f in = 0: f (0 + ) f (0) lim = lim = lim = + 0 0 0 So the function is not differentible in =0. We cll this n undefined derivtive. Geometriclly this mens tht the function hs verticl tngent in 0. 5.5 Differentition rules To clculte derivtives it is dvisble to use the following differentition rules rther thn using the limit-definition. Let f nd g be two functions tht re differentible in.. the derivtive of constnt function c c ' = 0. the derivtive of the rgument ' =. the derivtive of som of functions ( f + g)'( ) = f '( ) + g '( ) 4. the derivtive of product of functions ( f g)'( ) = f '( ) g( ) + f ( ) g '( ) 5. the derivtive of quotient of functions ' f f '( ) g( ) f ( ) g '( ) ( ) = if g( ) 0 g g ( ) 6. the derivtive of power of function n ( ) ' n f ( ) = n f ( ) f '( ) Clculus 9

7. the derivtive of composite function: the chin rule ( g f )'( ) = g '( f ( )) f '( ) 9. the derivtive of trigonometric functions (sin )' = cos (cos )' = sin (tn )' = cos (cot )' = sin Clculus 0

5.6 Eercises Clculte the derivtive of the following functions:.. = + y y 8 4 = + 4 ' = y ' = + 8 y. ( ) y 4 = + y ' = ( 4 + )( 4) 4. y ( 6) ( ) = + + y ' = ( + 6)( + + 8) 5. y =,5 5 +,5 7,5 y ' = 5 + 6. y = + y ' = 4 4 6 + ( + ) 7. y = 5 + 7 y ' = ( ) 4 + ( + ) 8. y = + + y ' = ( ) ( ) 9. 0. = + + y y = y ' = + y ' = ( ) ( ). y = y ' =. y = + 8 y' = 8 8. 4. y = + y = + 5 5 y ' = + 6 y ' = + 6 + 5 + 5 5 Clculus

5. y = 4 y = ' 4 y = 4 6. ( ) y = + 4 + 7. ( ) y ' 4 y ' = = ( ) 4 ( + )( 9 + 4) 4 + 8. y = 5 4 + y ' = ( ) 5 ( + ) 5 4 9. ( ) ( ) 4 y = 5 6 + 4 + y ' = 5 0 + 5 4 4 + 0.. + 5 y + y = + 0 y ' = + 5 = y ' = + 4 6 9 5 + + + ( ). y = ( + ) 5 y ' = ( ) ( + ) 7. y = ( )( ) + 0 y ' = ( )( 4) ( )( )( ) ( 4) 4. 5. y = y = ( ) 9 4 ( ) ( + ) y ' = + y ' = 4 6 4 9 ( )( 4 + 4 ) 4 ( + ) ( ) 6. y = y' = 4sin 4sin 7. y = cos 5 cos y' = sin + 5sin 8. 9. y = cos y ' = sin cos sin + cos y = sin cos y ' = ( ) sin cos Clculus

0. y = sin y' = cos. y = sin sin y' = cos cos. y = cos sin y ' = cos. y = cos 4 y ' = 8sin 4 cos 4 4. y = 4cos + sin 4 y' = sin + cos 4 5. y = cos + sec y' = sin + tn sec 6. y cos sin sin sin cos cos sin = y ' = 7. y = 4 4 sin cos cos + sin y' = cos sin ( cos sin ) cos sin Clculus

6 Indefinite integrls 6. Antiderivtive functions Definition Let us consider function f: IR IR : f(). An ntidervtive function F of the function f is ny function with the property F () = f(). Property If F is n ntiderivtive of the function f, then F+k with k n rbitrry constnt є IR, lso is. Property If F nd F re both ntiderivtives of f, then they differ only in the constnt. Conclusion If F is n ntiderivtive of f, then ll ntiderivtives of f cn be found by dding to F n rbitrry rel constnt. 6. The indefinite integrl The set of ntidervtives of function f is clled the indefinite integrl of f. Nottion: { } f ( ) d = F( ) + k F '( ) = f ( ) nd k IR or f ( ) d = F( ) + k 6. Properties + = +. k f ( ) d = k f ( ) d. ( ) f ( ) g( ) d f ( ) d g( ) d Clculus 4

6.4 Bsic integrls n+ n d = + k, if n - n + sin d = cos + k cos d = sin + k d tn k cos = + d cot k sin = + 6.5 Integrtion by substitution Let f be continuous function of nd = g(t) is differentible function of t then: f ( ) d = f ( g( t)) g '( t) dt Clculus 5

6.6 Eercises Clculte the indefinite integrl of the following functions:. 4 + k. 4 4 4 7 + k. 5 + k 4. 5. 5 5 4 5 + k + k 6. 9 + k 8 8 7. 4 + k 8. + 7 7 + + k 9. sin + cos cos + sin + k 7 0. cos. cos 7 tn + k cot + k.. 5 8 + 5 cos cos 5 8 + + k 4 4 tn sin + k 4. + cos ± sin + k 5. ( + ) 5 ( + ) 6 6 + k 6. cos sin + k Clculus 6

7. cos + sin + k 4 8. sin 9. ( 5) 4 0. + 6 sin + k 4 ( 5) 5 5 + k ( 6) 6 + + + k. 4 5 8 5 + k. sin cos + k. ( 7) 5 ( 7) 6 + k 4. + 4 5. cos 4 6. ( + 4 ) + 4 + k 9 tn 4 4 ( ) + k + k 7. ( ) ( ) 8. ( + ) + 6 ( ) 5 + k 6 5 + + + k 9. 0. sin cos cos cos + + k sin sin + k. sin cos. 4 cos cos + k + sin + sin 4 + k 8 4 Clculus 7

. 4. 4 sin + cos sin + sin 4 + k 8 4 tn + k 5. 6 5 + + + k + ( ) 6. 5 8 05 + + + k ( )( ) 4 5 8 + + + + k 7. ( + ) + ( )( ) 8. + ( + 0 ) + k 7 Clculus 8

6.7 Integrtion by prts Let u = f() nd v = g() be differentible functions. Since it follows tht nd therefore d( u v) = u dv + v du u dv = d( u v) v du u dv = uv v du 6.8 Eercises Evlute the indefinite integrl of the following functions:. cos sin + cos sin + k. cos sin + cos+ k. sin cos + sin + 6 cos 6sin + k 4. 5. cos cos + sin cos + k cos sin + sin + k 6. sin cos cos + sin + k 4 8 7. cos sin cos 4 4 8 + + + k Clculus 9

7 Definite integrls 7. The fundmentl theorem of clculus Let the function f be continuous on [,b] nd let F be n ntiderivtive of the function f on [,b] then b [ ] f ( ) d = F( b) F( ) = F( ) = F( ) Remrk: The definite integrl is rel number, the indefinite integrl is set of ntiderivtives. b b Emple I π = 0 sin d sin d = sin d( ) = cos + k I = cos π = = 0 Therefore [ ] ( ) 7. Properties of the definite integrl Let f nd g be functions tht re continuous on finite intervl [,b].. f ( ) d = f ( ) d b b b. k f ( ) d = k f ( ) d b with k rel constnt.. ( ) b b b f ( ) + g( ) d = f ( ) d + g( ) d Clculus 0

7. The substitution method b Given the integrl f ( ) d nd f is continuous on [,b]. In this method we pply the substitution rule to the integrnd nd convert the limits s well. Emple I π = 0 sin d let = t then d =dt = 0 t = 0 π = t = π π sin t dt = = cos cos cos 0 = = Therefore I t ( π ) 0 π 0 7.4 Prtil integrtion If u = u() nd v = v() re differentible functions of nd let udv = uv vdu then b b = = [ ] u dv uv v du u v v du b b Emple I = d Set u = du = d dv = d v = d = ( ) 4 4 6 = = = = 5 5 5 5 I ( ) ( ) d ( 0) ( ) ( 0) Clculus

7.5 Eercices Evlute the following definite integrls:. d 4 d. ( + 4). π sin d 0 4. 5. π π d cos d π 6. 0 d π 4 7. 5 + 5d 08 5 8. d 0 + 5 9. π 0 cos sin d 4 5 0. π 0 cos sin d 4 Clculus

Trigonometry Dr. Croline Dnneels

Angles... 4. The trigonometric circle... 4. Oriented ngles... 4. Conversion between rdins nd degrees... 6 The trigonometric numbers... 7. Definitons... 7. Some specil ngles nd their trigonometric numbers... 8. Sign vrition for the trigonometric numbers by qudrnt... 9.4 Pythgoren identities... 9.5 Emples... 0.6 Specil pirs of ngles....7 Eercises... 4 The trigonometric functions... 6. Periodic functions... 6. Even nd odd functions... 6. Sine function... 7.4 Cosine function... 7.5 Tngent function... 8.6 Cotngent function... 8.7 The secntfunction... 9.8 The cosecnt function... 9.9 Eercises... 0 4 Right tringles... 4. Formuls... 4. Eercises... 5 Oblique tringles... 5 5. The sine rules... 5 5. The cosine rules... 6 5. Solving oblique tringles... 7 5.4 Eercises... 8 6 Etr s... 9 6. Specil lines in tringle... 9 6. Isosceles tringles... 6. Equilterl tringles... 6.4 Eterior ngles...

7 Trigonometric formuls... 7. Sum nd difference formuls... 7. Double-ngle formuls... 4 7. Hlf-ngle formuls... 5 7.4 Trigonometric numbers in terms of tn α/... 5 7.5 Conversions sum/difference of ngles into product of ngles nd vice vers... 6 7.6 Eercises... 7

Angles. The trigonometric circle Tke n -is nd n y-is (orthonorml) nd let O be the origin. A circle centered in O nd with rdius =, is clled trigonometric circle or unit circle. Turning counterclockwise is the positive orienttion in trigonometry (fig. ).. Oriented ngles An ngle is the figure formed by two rys tht hve the sme beginning point. Tht point is clled the verte nd the two rys re clled the sides of the ngle (lso legs). If we cll [OA the initil side of the ngle nd [OB the terminl side, then we hve n oriented ngle. This ngle is referred to s AOB nd the orienttion is indicted by n rrow from the initil side to the terminl side. We cn drw the rrow lso in the opposite direction, still strting from the initil side of the ngle [OA. Both ngles represent the sme oriented ngle. The ngle BOA is different oriented ngle which we cll the opposite ngle of AOB (fig. ). Remrk: n oriented ngle is in fct the set of ll ngles which cn be trnsformed to ech other by rottion nd/or trnsltion. y Z + α O B AOB O B BOA A A fig. : the trigonometric circle fig. : ngles AOB nd BOA The introduction of the trigonometric circle mkes it possible to ttch vlue to ech oriented ngle AOB, which we will cll α from now on. Represent the oriented ngle in the trigonometric circle nd let the initil side of this ngle coincide with the -is (see fig. ). Then the terminl side intersects the trigonometric circle in point Z. Then Z is the representtion of the oriented ngle α on the trigonometric circle. Trigonometry 4

If Z I: ngle α belongs to the first qudrnt. If Z II: ngle α belongs to the second qudrnt. If Z III: ngle α belongs to the third qudrnt. If Z IV: ngle α belongs to the fourth qudrnt. fig. : the four qudrnts There re two commonly used units of mesurement for ngles. The more fmilir unit of mesurement is tht of degrees. A circle is divided into 60 equl degrees, so tht right ngle is 90. Ech degree is subdivided into 60 minutes nd ech minute into 60 seconds. The symbols, ' nd " re used for degrees, rcminutes nd rcseconds. In most mthemticl work beyond prcticl geometry, ngles re typiclly mesured in rdins rther thn degrees. An ngle of rdin determines on the circle n rc with length the rdius of the circle. Becuse the length of full circle is πr, circle contins π rdins. Contrriwise, if one drws in the centre of circle with rdius R n ngle of θ rdins, then this ngle determines n rc on the circle with length θ R. Subdivisions of rdins re written in deciml form. Net to Z you cn put n infinite number of vlues which differ from ech other by n integer multiple of 60 or π, becuse you cn mke more turns in one or the other direction strting t the initil side of the ngle nd rriving t the terminl side of the ngle (these ngles re clled coterminl). The set of ll these vlues is clled the mesure of the oriented ngle α. The principl vlue of α is tht vlue which belongs to ]- 80, 80 ], resp. ]- π,π]. Trigonometry 5

. Conversion between rdins nd degrees Becuse π = 60,following conversion formuls cn be pplied: r rd 60ir π πi g g rd 60 Remrk: when n ngle is represented in rdins, one does only mention the vlue, not the term rd. Trigonometry 6

The trigonometric numbers. Definitons Consider the construction of the oriented ngle α s described in the previous prgrph. The terminl side of the ngle α intersects the unit circle in the point Z. The sinα cn be defined s the y-coordinte of this point. The cosα cn be defined s the -coordinte of this point. In this wy, we cn find the sine or cosine of ny rel vlue of α (α ΙR). Conversely, the choice of cosine-vlue nd sine-vlue define in the intervl [0,π[ only one ngle. Overll there re n infinite number of solutions, which one cn find by dding on multiples of π. y sinα α Z cosα fig. 4 : Sine nd cosine in the trigonometric circle Beside sine nd cosine other trigonometric numbers re defined s follows : tngent : sinα tnα = cotngent : cosα cotα = cosα sinα secnt : secα = cosecnt : cosα cscα = sinα Fig. 5 gives grphicl representtion of the bove trigonometric numbers in terms of distnces ssocited with the unit circle. Trigonometry 7

S cscα cotα S tnα sinα α cosα secα fig. 5 : the grphicl representtion of the trigonometric numbers in terms of distnces ssocited with the unit circle Consequently, the trigonometric numbers hve vlues which re in the following res: sin α [ -, ] cos α [ -, ] tn α ] -, + [ cot α ] -, + [ sec α ] -, ] [, + [ csc α ] -, ] [, + [. Some specil ngles nd their trigonometric numbers α 0 0 = π/6 45 = π/4 60 = π/ 90 = π/ sin α 0 / / / cos α / / / 0 tn α 0 / cot α / 0 sec α / csc α / Trigonometric numbers of ngles in the other qudrnts we shell find through the use of the reference ngle (see prgrph.6..) Trigonometry 8

. Sign vrition for the trigonometric numbers by qudrnt Inside qudrnt the trigonometric numbers keep the sme sign (fig. 6). + + - + - + + + + - - - + + - sine cosecnt cosine secnt fig.6 : sign vrition for the trigonometric numbers by qudrnt tngent cotngent.4 Pythgoren identities The bsic reltionship between the sine nd the cosine is the Pythgoren or fundmentl trigonometric identity: cos α + sin α = This cn be viewed s version of the Pythgoren theorem, nd follows from the eqution + y = for the unit circle (see fig. 7): OP + PZ = OZ with OP = cos α ; PZ = sin α ; OZ = Dividing the Pythgoren identity by either cos θ or sin θ yields the following identities: + tn α = sec α + cot α = csc α y Q Z α Ο P fig. 7 : the tringle OPZ Trigonometry 9

.5 Emples.5. Clcultion of the trigonometric numbers Given: sinα = 5 Asked: ll other trigonometric numbers Becuse the sine of this ngle is positive, the ngle is situted in the first or second qudrnt. We determine the other trigonometric numbers s follows: from the Pythgoren trigonometric identity: cos α = sin α = 5 69 = 44 69 we get: cosα = ± 44 69 = ± tnα = sinα cosα = ± 5 cotα = tnα = ± 5 secα = cosα = ± cscα = sinα = 5 The two possible solutions for some of the trigonometric numbers correspond with the vlues of these numbers ccording to the qudrnt in which the ngle is situted. Summry : qudrnt sin cos tn cot sec csc st 5/ / 5/ /5 / /5 nd 5/ -/ -5/ -/5 -/ /5 Trigonometry 0

.5. Proof the following identity sec α + csc α = sec α csc α Proof : sec α + csc α = + cos α sin α sin α + cos α = cos sin α α = cos sin α α = sec α csc α Trigonometry

.6 Specil pirs of ngles The sines, cosines nd tngents, cotngents of some ngles re equl to the sines, cosines nd tngents, cotngents of other ngles..6. Formuls. Supplementry ngles ( = sum is π ) sin(π α) = sin α tn(π α) = - tn α cos(π α) = - cos α cot(π α) = - cot α b. Anti-supplementry ngles ( = difference is π ) sin(π + α) = - sin α tn(π + α) = tn α cos(π + α) = - cos α cot(π + α) = cot α c. Opposite ngles ( = sum is π ) sin(π α) = - sin α tn(π α) = - tn α cos(π α) = cos α cot(π α) = - cot α d. Complementry ngles ( = sum is π / ) sin(π/ α) = cos α tn(π/ α) = cot α cos(π/ α) = sin α cot(π/ α) = tn α y π-α α π+α -α fig. 8 : specil pirs of ngles Trigonometry

.6. Reference ngles The use of reference ngles is wy to simplify the clcultion of the trigonometric numbers t vrious ngles. Associted with every ngle drwn in stndrd position (which mens tht its verte is locted t the origin nd the initil side is on the positive -is) (ecept ngles of which the terminl side lies on the es, clled qudrntl ngles) there is n ngle clled the reference ngle. The reference ngle is the cute ngle formed by the terminl side of the given ngle nd the -is. Angles in qudrnt I re their own reference ngles. For ngles in other qudrnts, reference ngles re clculted this wy: Qudrnt I II III IV β (reference ngle) β = α β = π α β = α π β = - α The reference ngle nd the given ngle form pir of ngles to which you cn pply the properties in the previous prgrph. Due to these properties, the vlue of trigonometric number t given ngle is lwys the sme s the vlue of tht ngle s reference ngle, ecept when there is vrition in sign. Becuse we know the signs of the numbers in different qudrnts, we cn simplify the clcultion of trigonometric number t ny ngle to the vlue of the number t the reference ngle for tht ngle, to be found in the tble in prgrph...6. How to find ll ngles To find the ngle if given certin trigonometric number, usully there re solutions. Clcultors give the most obvious solution, but in prcticl situtions, there cn be second solution, or the second solution cn be the only correct solution. In this cse the user must djust the solution given by the clcultor. The following tble gives for positive nd negtive trigonometric numbers the qudrnt in which the solution given by the clcultor, is situted, nd in the lst column the qudrnt of the second solution: Input Clcultor Second solution --------------------------------------------------------------------------------------------------------- Positive sine or cosecnt Negtive sine or cosecnt 4 Positive cosine or secnt 4 Negtive cosine or secnt Positive tngent or cotngent Negtive tngent or cotngent 4 Trigonometry

.7 Eercises.7. Determine for the given trigonometric numbers the other trigonometric numbers; do not determine the ngle before.. sinα = 6 6. cscα = 4. cotα = 6 4. secα = 5 4.7. Proof the following identities.. csc α + cot β = csc β + cot α ( sinα ) ( + sinα ) ( secα + ) ( secα ) = cos α cot α sec α + tn α sin = secα + tnα = + α secα tnα sinα. ( ) 4. ( cotα )( sec α tnα ) ( + α ) tn + + = tnα.7. Simplify the following epressions by pplying the formuls of pirs of ngles.. π cos + cos ( π ) sin ( π ) cos( π + ) + π π π sin sin ( π ) sin + cos +. csc( π + )sec( π ) sec( π )csc( π ) π π π π csc sec + sec + csc Trigonometry 4

.7.4 Determine the following trigonometric numbers. First find the reference ngle, then pply the properties of specil pirs of ngles.. sin 0. cos ( -5 ). tn 5 4. π cot 4 5. π tn.7.5 Solve in IR. Epress the solution(s) in rdins.. cos 5 =. sin 5 =. sin = sin 4. sin = 5 nd π, π ; sked: sin 5. sin = cos 6. tn ( + ) = Trigonometry 5

The trigonometric functions. Periodic functions Definiton : function f : IR IR is periodic p IRo : dom f : + p dom f f ( + p ) = f() If p stisfies this definition, then ll positive nd negtive numbers which re n integer multiple of p lso stisfy this definition. Therefore we cll the smllest positive number which stisfies this definition the period P of the function. Grphiclly this periodicity mens tht the form of the grph of f() repets itself over subsequent intervls with length P.. Even nd odd functions A function f is clled EVEN if: dom f : - dom f f (- ) = f() Consequently two points with opposite -vlues must hve the sme y-vlue. So the grph must be symmetric bout the y-is. A function f is clled ODD if: dom f : - dom f f (- ) = - f() Consequently two points with opposite -vlues must hve opposite y-vlues. So the grph is symmetric bout the origin. Remrk: we consider the rgument of trigonometric functions lwys in terms of rdins. Trigonometry 6

. Sine function sin : IR [ -, ] : sin The period of this function is π. This function is odd, s opposite ngles hve opposite sines. -π π fig. 9 : the sinusoïde.4 Cosine function cos : IR [ -, ] : cos The period of this function is π. This function is even, s opposite ngles hve the sme cosine. -π π fig. 0 : the cosinusoïde Trigonometry 7

.5 Tngent function tn : IR \ π + kπ, k Z IR : tn The period of this function is π. This function is odd, s opposite ngles hve opposite tngents. π π π π fig. : the tngent function.6 Cotngent function cot : IR \ { kπ, k Z} IR : cot The period of this function is π. This function is odd, s opposite ngles hve lso opposite cotngents. π π π fig. : the cotngent function Trigonometry 8

.7 The secnt function sec : IR \ π + kπ, k Z ], ] [,+ [ : sec The period of this function is π. This function is even, s opposite ngles hve the sme cosines nd so the sme secnts. π π π π fig. : de secnsfunctie.8 The cosecnt function csc : IR \ { kπ, k Z} ], ] [,+ [ : csc The period of this function is π. This function is odd, s opposite ngles hve opposite sines nd so opposite cosecnts. Trigonometry 9

π π π π fig. 4 : the cosecnt function.9 Eercises.9. Determine the period of the following functions nd drw their grph. f() = sin. f() = cos. f() = cos π + Trigonometry 0

4 Right tringles 4. Formuls C C γ b γ b B β c α A B β c α A fig. 5 : orthogonl tringles used to set up the formuls in this prgrph In right tringle with α s the right ngle, the following formuls pply: α = π β + γ = π = b + c If we drw in the tringle bove circle segment with centre in B nd rdius (see the first tringle in fig. 5), then we recognize segment of circle with rdius. The djcent side of the right ngle c nd the opposite side b hve resp. the following lengths: c = cos β nd b = sin β In similr wy, by considering circle segment with centre in C nd rdius (see second tringle in fig. 5), we find: b = cos γ nd c = sin γ In words : The cosine of n cute ngle is the rtio of the length of the djcent rectngle side nd the length of the hypotenuse. The sine of n cute ngle is the rtio of the length of the opposite rectngle side nd the length of the hypothenuse. Trigonometry

By division of the first two formuls we get: b = c tn β or c = b cot β If we do the sme with the lst two formuls, we get: In words : c = b tn γ or b = c cot γ The tngent of n cute ngle is the rtio of the length of the opposite rectngle side nd the length of the djcent rectngle side. The cotngent of n cute ngle is the rtio of the length of the djcent rectngle side nd the length of the opposite rectngle side. Emple Given : α = 90 ; β = ; b = 0 Asked : ll missing ngles nd sides. Solution: γ = 90 - β = 90 - = 77 = c = b 0 sin β = sin = 44.5 b = 4.5 Trigonometry

4. Eercises. Given : ABC with = 45, α = 90, β = 40 0'5" Asked : the remining sides nd ngles.. An etension ldder stnds slntwise to verticl wll on horizontl floor. If the ldder is completely etended, it mkes n ngle of 5 8' with the floor; completely retrcted the ngle is 9 0', while the top t tht moment lens t heigth of 5 meter ginst the wll. If we ssume tht the foot of the ldder does not chnge, clculte the miml height one cn rech the miml length of the ldder. An incident ry is ry of light tht strikes surfce. The ngle between this ry nd the perpendiculr or norml to the surfce is the ngle of incidence (α). The refrcted ry or trnsmitted ry corresponding to given incident ry represents the light tht is trnsmitted through the surfce. The ngle between this ry nd the norml is known s the ngle of refrction (β). The reltionship between the ngles of incidence nd refrction is given by Snell's Lw: sinα sin β = n Emple (see figure below): A ry of light strikes n ir-wter interfce t n ngle of 0 degrees from the norml (α = 0 ). The reltive refrctive inde for the interfce is 4/. At which distnce from P the ry of light hits the bottom if the wter is m deep. Remrk: solve this eercise without clculting the ngle β. fig. 6 : illustrtion for eercise Trigonometry

In mechnics you will del with eercises in which forces must be clculted. In the following eercises such situtions will be sketched. We will confine to the clcultion of ngles between brs.. 4. Clculte: the ngle between FE nd the horizontl plne the ngle between FC nd the verticl plne 5. Clculte the ngle between CD nd DF fig. 7 : illustrtion for eercise 4 6. Clculte the ngle between BC nd CD fig. 8 : illustrtion for eercise 5 fig. 9 : illustrtion for eercise 6 Trigonometry 4

5 Oblique tringles First, remember tht lso for oblique tringles the sum of ngles is 80. An oblique tringle is ny tringle tht is not right tringle. It could be n cute tringle (ll three ngles of the tringle re less thn right ngles) or it could be n obtuse tringle (one of the three ngles is greter thn right ngle). Using the formuls for right tringles, we cn set up formuls for oblique tringles. Let us consider n oblique tringle ABC with sides, b nd c nd ngles α, β nd γ. 5. The sine rules The ltitude from A to the opposite side intersects this side in point S. In this wy the tringle is divided in two right ngles with one common side AS, with length d. Use now the formuls of right tringle in tringles ABS nd ACS to clculte d. A α C b c B d C β γ S γ β B fig. 0 : oblique tringle d = c sin β nd d = b sin γ So we get: sin β sin γ = b c Apply the sme resoning with the ltitude from B to the opposite side b to divide the tringle in two right tringles nd derive similr formuls in which occur nd the opposite ngle α. Then we get: SINE RULES : sinα sin β sinγ = = b c Trigonometry 5

5. The cosine rules This identity cn be derived in different wys. In fig. 0 S divides the side in two prts with length nd. Then nd d cn be written respectively s = b cos γ d = b sin γ In the tringle ABS pply Pythgors theorem: c = d + = d + (- ) = b sin γ + + - = b sin γ + + b cos γ - b cos γ = b + - b cos γ The sme epression cn be derived if S lies outside side. Then, similr epressions cn be derived for the other ngles. Summrized, in this wy we get: COSINE RULES : = b + c - b c cos α b = + c - c cos β c = + b - b cos γ These sttements relte the lengths of the sides of tringle to the cosine of one of its ngles. For emple, the first sttement sttes the reltionship between the sides of lengths, b nd c, where α denotes the ngle contined between sides of lengths b nd c nd opposite to the side with length. These rules look like the Pythgoren theorem ecept for the lst term, nd if you del with right tringle, tht lst term disppers, so these rules re ctully generliztion of the Pythgoren theorem. Trigonometry 6

5. Solving oblique tringles One of the most common pplictions of the trigonometry is solving tringles finding missing sides nd/or ngles, given some informtion bout tringle. The process of solving tringles cn be broken down into number of cses. In these situtions we will use sorts of formuls, pplicble in ll tringles: the sum of ll ngles is 80 the sine rule : reltes two sides to their opposite ngles the cosine rule : reltes the three sides of the tringle to one of the ngles. Nturlly, the given informtion must be such tht the given elements llow tringle: the sum of the given ngles cn not be lrger thn 80, nd the sides must meet the tringle inequlity which sttes tht for ny tringle, the sum of the lengths of ny two sides must be greter thn the length of the remining side.. If you know one ngle nd the two djcent sides. Then, there is solution: you cn determine the opposite side by using the cosine rule, nother ngle by using the sine rule nd the remining ngle s 80 minus the two lredy determined ngles. Attention: the sine rule gives two solutions for the second ngle (supplementry ngles). Test the solutions by verifying the properties of tringle (see eercises). b. If you know one side nd the two djcent ngles. Then there is solution: the third ngle is immeditely known s 80 minus the two given ngles; the two remining sides cn be determined by using the sine rule. c. If you know ll three sides of tringle. Then there is solution: determine one ngle by using cosine rule, the second ngle cn be determined by using nother cosine rule or by using the sine rule. The lst ngle cn be determined by the property of tringles tht the sum of ll ngles must be 80. d. If you know sides nd b nd β (one of the djcent ngles). In this cse, there cn be 0, or solutions. Determine the ngle α by using the sine rule. You will get 0 (if sin α > ) or solutions (supplementry ngles hve the sme sine). For ech solution determine the missing ngle γ, nd then the length of side c by using the sine rule. Finlly you test if ech solution which you find is cceptble: you cn not hve negtive ngles or sides (see eercises). Trigonometry 7

5.4 Eercises. A tower is seen from the ground under n ngle of. If one pproches the tower by 4 meter, then this ngle becomes 5. Determine the height of the tower.. Two plnes deprt from the sme point, ech in different direction. The directions form n ngle of. The velocity of the first plne is 600 km/hour, the velocity of the second is 900 km/hour. Determine their mutul distnce fter one hour nd hlf.. The pole of flg reches up from fcde with n ngle of 45 (see fig. ). Five meter bove the bse point of the pole in the wll, there is cble fied to the wll with length of,60 meter. At which distnce of the bse point, mesured long the pole, the other end of the cble cn be fied. fig. : illustrtion for eercise 4. Solve the previous eercise for cble with length of m, respectively with cble with length of 8 m. 5. Three observers re t mutul distnces of, nd 4 meters. Determine for ech observer the ngle under which he sees the other observers. 6. A bot sils north nd sees lighthouse 40 estwrds. After hving siled 0 km, this ngle hs incresed to 80. Determine t both positions the distnce from the bot to the lighthouse. 7. The following figure demonstrtes sitution in mechnics. Determine the ngle between the ropes AC nd AD (d = m). fig. : illustrtion of eercise 7 Trigonometry 8