Mathematics. Mathematics 3. hsn.uk.net. Higher HSN23000


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1 hsn uknt Highr Mathmatics UNIT Mathmatics HSN000 This documnt was producd spcially for th HSNuknt wbsit, and w rquir that any copis or drivativ works attribut th work to Highr Still Nots For mor dtails about th copyright on ths nots, plas s
2 Unit Mathmatics Contnts Vctors 8 Vctors and Scalars 8 Componnts 8 Magnitud 0 Equal Vctors Addition and Subtraction of Vctors Multiplication by a Scalar 7 Position Vctors 8 Basis Vctors 9 Collinarity 7 0 Dividing Lins in a Ratio 8 Th Scalar Product Th Angl Btwn Vctors Prpndicular Vctors 7 Proprtis of th Scalar Product 8 Furthr Calculus 9 Diffrntiating sin and cos 9 Intgrating sin and cos 0 Th Chain Rul Spcial Cass of th Chain Rul A Spcial Intgral Intgrating sin(a + b) and cos(a + b) 7 Eponntials and Logarithms 0 Eponntials 0 Logarithms Laws of Logarithms Eponntials and Logarithms to th Bas Eponntial and Logarithmic Equations Graphing with Logarithmic As 9 7 Graph Transformations 7 Wav Functions 7 Eprssing pcos + qsin in th form kcos( a) 7 Eprssing pcos + qsin in othr forms 7 Multipl Angls 77 Maimum and Minimum Valus 78 Solving Equations 79 Sktching Graphs of y pcos + qsin 8  ii  HSN000
3 Unit Vctors OUTCOME Vctors Vctors and Scalars A scalar is a quantity with magnitud (siz) only for ampl, an amount of mony or a lngth of tim Somtims siz alon is not nough to dscrib a quantity for ampl, dirctions to th narst shop For this w nd to know a magnitud (i how far), and a dirction Quantitis with both magnitud and dirction ar calld vctors A vctor is namd ithr by using th points at th nd of a dirctd lin sgmnt (g AB rprsnts th vctor starting at point A and nding at point B) or by using a bold lttr (g u) You will s bold lttrs usd in printd tt, but in handwriting you should just undrlin th lttr (g ) B Throughout ths nots, w will show vctors bold and undrlind (g u ) Componnts A A vctor may b rprsntd by its componnts, which w writ in a column For ampl, is a vctor in two dimnsions In this cas th first componnt is and this tlls us to mov units in th dirction Th scond componnt tlls us to mov units in th y dirction So if th vctor starts at th origin, it will look lik: y u O Pag 8 HSN000
4 Unit Vctors Not that w writ th componnts in a column to avoid confusing thm with coordinats Th following diagram also shows th vctor, but in this cas it dos not start at th origin y (, ) (, ) O Vctors in Thr Dimnsions In a vctor with thr componnts, th first two tll us how many units to mov in th  and ydirctions, as bfor Th third componnt spcifis how far to mov in th zdirction Whn looking at a pair of (, y ) as, th zais points out of th pag from th origin z A st of D as can b drawn on a pag as shown to th right z O y For ampl, is a vctor in thr dimnsions This vctor is shown in th diagram, starting from th origin z O y Zro Vctors Any vctor with all its componnts zro is calld a zro vctor and can b 0 writtn as 0, g Pag 9 HSN000
5 Unit Vctors Magnitud Th magnitud (or lngth) of a vctor u is writtn as u It can b calculatd as follows EXAMPLES Givn a If PQ b thn PQ a + b a If PQ b thn PQ a + b + c c u, find u u + ( ) 9 units Find th lngth of a a units Unit Vctors Any vctor with a magnitud of on is calld a unit vctor For ampl: if 0 u thn So u is a unit vctor u unit Pag 0 HSN000
6 Unit Vctors Distanc in Thr Dimnsions Th distanc btwn th points A and B is d AB AB units So givn AB, w find d AB In fact, thr is a thr dimnsional vrsion of th distanc formula Th distanc d btwn th points (, y, z ) and (,, ) EXAMPLE y z is units d + y y + z z Find th distanc btwn th points (,,) and ( 0,, 7) Th distanc is Equal Vctors ( ) + ( y y ) + ( z z ) ( 0 ( ) ) + ( ) + ( 7 ) + + ( 8) + + units Vctors with th sam magnitud and dirction ar said to b qual For ampl, all th vctors shown to th right ar qual If vctors ar qual to ach othr, thn all of thir componnts ar qual, i p r q s t a d if b thn a d, b and c f c f Convrsly, two vctors ar only qual if all of thir componnts ar qual Pag HSN000
7 Unit Vctors Addition and Subtraction of Vctors Considr th following vctors: a b c Addition W can construct a + b as follows: b a a + b a + b mans a followd by b Similarly, w can construct a + b + c as follows: a a + b + c mans a followd by b followd by c To add vctors, w position thm nostotail Thn th sum of th vctors is th vctor btwn th first tail and th last nos Subtraction b a + b + c c Now considr a b This can b writtn as a + ( b ), so if w first find b w can us vctor addition to obtain a b b b is just b but in th opposit dirction b b and b hav th sam magnitud, i b b Thrfor w can construct b a b as follows: a b a a b mans a followd by b Pag HSN000
8 Unit Vctors Using Componnts If w hav th componnts of vctors, thn things bcom much simplr Th following ruls can b usd for addition and subtraction a d a + d b + b + c f c + f add th componnts a d a d b b c f c f subtract th componnts EXAMPLES Givn u and u + v + 0 Givn 0 7 p and p q v, calculat u + v and 0 u v 0 u v q, calculat p q and q + p q + p Pag HSN000
9 Unit Vctors Multiplication by a Scalar A vctor u which is multiplid by a scalar k > 0 will giv th rsult ku This vctor will b k tims as long, i its magnitud will bk u Not that if k < 0 this mans that th vctor ku will b in th opposit dirction to u For ampl: u u u u a ka If u b thn ku kb c kc Each componnt is multiplid by th scalar EXAMPLES Givn v, find v v 9 Givn r, find r r Pag HSN000
10 Unit Vctors Ngativ Vctors Th ngativ of a vctor is th vctor multiplid by If w writ a vctor as a dirctd lin sgmnt AB, thn AB BA : AB A 7 Position Vctors B OA is calld th position vctor of point A rlativ to th origin O, and is writtn as a A B AB BA OB is calld th position vctor of point B, writtn b Givn P (, y, z ), th position vctor OP or p has componnts y z z O y P y O a A b B To mov from point A to point B w can mov back along th vctor a to th origin, and along vctor b to point B, i AB AO + OB OA + OB a + b b a For th vctor joining any two points P and Q, PQ q p Pag HSN000
11 Unit Vctors EXAMPLE R is th point (,, ) From th coordinats, RS s r 8 and S is th point (,, ) r and s Find RS Not You don t nd to writ this lin down in th am 8 Basis Vctors A vctor may also b dfind in trms of th basis vctors i, j and k Ths ar thr mutually prpndicular unit vctors (i thy ar prpndicular to ach othr) Ths basis vctors can b writtn in componnt form as i 0, 0 0 j and 0 0 k 0 Any vctor can b writtn in basis form using i, j and k For ampl: k i j 0 0 v i j + k 0 0 Thr is no nd for th working abov if th following is usd: a ai + bj + ck b c Pag HSN000
12 Unit Vctors 9 Collinarity In Straight Lins (Unit Outcom ), w larnd that points ar collinar if thy li on th sam straight lin Th points A, B and C in D spac ar collinar if AB is paralll to BC, with B a common point Not that w cannot find gradints in thr dimnsions instad w us th following Nonzro vctors ar paralll if thy ar scalar multipls of th sam vctor For ampl: u, v u So u and v ar paralll 0 p 9, q 8 So p and q ar paralll EXAMPLE A is th point (,, ), B( 8,, 9) and C(,,7 ) Show that A, B and C ar collinar AB b a BC c b BC AB, so AB and BC ar paralll, and sinc B is a common point, A, B and C ar collinar Pag 7 HSN000
13 Unit Vctors 0 Dividing Lins in a Ratio Thr is a simpl procss for finding th coordinats of a point which divids a lin sgmnt in a givn ratio EXAMPLE P is th point (,, ) and R is th point ( 8,, 9) Th point T divids PR in th ratio : Find th coordinats of T Stp Mak a sktch of th lin, showing th ratio in which th point divids th lin sgmnt Stp Using th sktch, quat th ratio of th two lin sgmnts with th givn ratio Stp Cross multiply, thn chang dirctd lin sgmnts to position vctors Stp Rarrang to giv th position vctor of th unknown point Stp From th position vctor, stat th coordinats of th unknown point R PT TR PT TR ( t p) ( r t ) T t p r t t + t r + p 8 t + 9 t t 0 t 7 So T is th point (,, 7 ) P Pag 8 HSN000
14 Unit Vctors Using th Sction Formula Th prvious mthod can b condnsd into a formula as shown blow If th point P divids th lin AB in th ratio m : n, thn na + mb p, n + m whr a, b and p ar th position vctors of A, B and P rspctivly This is rfrrd to as th sction formula It is not ncssary to know this, sinc th approach plaind abov will always work EXAMPLE P is th point (,, ) and R is th point ( 8,, 9) Th point T divids PR in th ratio : Find th coordinats of T Th ratio is :, so m and n Hnc: np + mr t n + m p + r ( ( ) + ( 8) ) ( ( ) + ( ) ) ( ( ) + ( 9) ) 7 So T is th point (,, 7 ) Not If you ar confidnt with arithmtic, this stp can b don mntally Pag 9 HSN000
15 Unit Vctors Furthr Eampls EXAMPLES Th cuboid OABCDEFG is shown in th diagram blow E F H D A G B O C Th point A has coordinats ( 0,0, ), C ( 8,0,0 ) and G( 8,,0 ) Th point H divids BF in th ratio : Find th coordinats of H From th diagram: OH OA + AB + BF OA + OC + CG ( g c) h a + c + a + c + g c a + c + g So H has coordinats 8 8,, Th points P(,, ), Q ( 8,,) and R ( 9,,) ar collinar Find th ratio in which Q divids PR Sinc th points ar collinar PQ kqr for som k Working with th first componnts: 8 k ( 9 8) k Thrfor PQ QR so Q divids PR in th ratio : Not BH, so BH BF BF Not Th ratio is : sinc PQ QR Pag 0 HSN000
16 Unit Vctors Th points A ( 7,, ), B(,, 7) and C ar collinar Givn that B divids AC in th ratio :, find th coordinats of C AB AC b a ( c a) b a c a c b a A c b a So C has coordinats ( 7,, 9) Not A sktch may hlp you to s this: B C Th Scalar Product So far w hav addd and subtractd vctors and multiplid a vctor by a scalar Now w will considr th scalar product, which is a form of vctor multiplication Th scalar product is dnotd by ab (somtims it is calld th dot product) and can b calculatd using th formula: ab a b cos θ, whr θ is th angl btwn th two vctors a and b This is givn in th am Pag HSN000
17 Unit Vctors Th dfinition abov assums that th vctors a and b ar positiond so that thy both point away from th angl, or both point into th angl a θ b a θ b Howvr, if on vctor is pointing away from th angl, whil th othr points into th angl, a θ b a θ b w find that ab a b cosθ EXAMPLES Two vctors, a and b hav magnituds 7 and units rspctivly and ar at an angl of 0 to ach othr as shown blow What is th valu of ab? ab a b cosθ 7 cos 0 b 0 Th vctor u has magnitud k and v is twic as long as u Th angl btwn u and v is 0, as shown blow v a 0 u Find an prssion for uv in trms of k uv u v cosθ k k cos0 k k Rmmbr Whn on vctor points in and on points out, uv u v cosθ Pag HSN000
18 Unit Vctors Th Componnt Form of th Scalar Product Th scalar product can also b calculatd as follows: ab a b + a b + a b whr This is givn in th am a a and a a b b b b EXAMPLES Find pq, givn that pq p q + p q + p q p and ( ) + ( ) + ( ) + 9 If A is th point ABAC C(,, ) B(,, ) A (,, 9) q,, 9, B(,, ) and C(,, ), calculat W nd to us th position vctors of th points: AB b a AC c a 9 ABAC (( ) ( ) ) + ( 0) + ( ) ( ) Pag HSN000
19 Unit Vctors Th Angl Btwn Vctors Th formula for th scalar product can b rarrangd to giv th following quations, both of which can b usd to calculat θ, th angl btwn two vctors cosθ ab a b or a b + a b + a b a b cos θ Look back to th formula for finding th scalar product, givn on th prvious pags Notic that th first quation is simply a rarrangd form of th on which can b usd to find th scalar product Also notic that th scond simply substituts ab for th componnt form of th scalar product Ths formula ar not givn in th am but can both b asily drivd from th formula on th prvious pags (which ar givn in th am) EXAMPLES Calculat th angl θ btwn vctors p i + j k and q i + j + k p and q p q + p q + p q cosθ p q ( ) + ( ) + (( ) ) + + ( ) θ cos 9 8 (to d p ) (or 98 radians (to d p )) Pag HSN000
20 Unit Vctors K is th point (, 7, ) Start with a sktch: L(,, ) θ, L(,, ) and M(,,) M,, Find ɵ KLM K (, 7, ) Now find th vctors pointing away from th angl: LK k l 7 0, LM m l Us th scalar product to find th angl: ɵ LKLM cosklm LK LM ( ) + ( 0 ) + ( ( ) ) + ( 0) + ( ) + + ( ) 0 8 ɵ KLM cos (to d p ) (or 8 radians (to d p )) Pag HSN000
21 Unit Vctors Th diagram blow shows th cub OPQRSTUV Th point R has coordinats (,0,0 ) (a) Writ down th coordinats of T and U (b) Find th componnts of RT and RU O (c) Calculat th siz of angl TRU (a) From th diagram, T( 0,, ) and U(,, ) z S P T V R U Q y (b) (c) 0 RT t r 0 0, 0 RU u r 0 0 RTRU cos TRU RT RU ( 0) + ( ) + ( ) ( ) TRU cos (to d p ) (or 0 radians (to d p )) Pag HSN000
22 Unit Vctors Prpndicular Vctors If a and b ar prpndicular thn ab 0 This is bcaus ab a b cosθ a b cos90 ( θ 90 sinc prpndicular) 0 (sinc cos90 0) Convrsly, if ab 0 thn a and b ar prpndicular EXAMPLES Two vctors ar dfind as a i + j k and b i + j + k Show that a and b ar prpndicular ab a b + a b + a b ( ) ( ) (( ) ) Sinc ab 0, a and b ar prpndicular PQ a 7 and RS a and RS whr a is a constant Givn that PQ Sinc PQ and RS ar prpndicular, PQRS 0 + ( a) + 7a 0 8 a + 7a a 0 a ar prpndicular, find th valu of a Pag 7 HSN000
23 Unit Vctors Proprtis of th Scalar Product Som usful proprtis of th scalar product ar as follows: ab ba a b + c ab + ac (Epanding brackts) aa a Not that ths ar not givn in th am, so you nd to rmmbr thm EXAMPLES In th diagram, p, r and q Calculat p ( q + r ) p q r p q + r pq + pr p q cosθ p r cosθ cos 0 + cos In th diagram blow, a c and b a c 0 b 0 a a b c Calculat ( + + ) ( + + ) a a b c aa + ab + ac a + a b cosθ a c cosθ cos0 cos0 Rmmbr a c a c cosθ sinc a points into θ and c points away Pag 8 HSN000
24 Unit Furthr Calculus OUTCOME Furthr Calculus Diffrntiating sin and cos In ordr to diffrntiat prssions involving trigonomtric functions, w us th following ruls: d d d d ( sin ) cos, ( cos ) sin Ths ruls only work whn is an angl masurd in radians A form of ths ruls is givn in th am EXAMPLES Diffrntiat y sin with rspct to dy cos d A function f is dfind by f ( ) sin cos for R Find f π f ( ) cos ( sin ) cos + sin f π cosπ + sinπ + + Find th quation of th tangnt to th curv y sin whn Whn, y sin π So th point is ( ) π W also nd th gradint at th point whr dy cos d π m π tangnt Whn, cos π, π : Rmmbr Th act valu triangl: π π π Pag 9 HSN000
25 Now w hav th point π, and th gradint y b m a ( π ) y y π y π + 0 Unit Furthr Calculus tangnt m, so: Intgrating sin and cos W know th drivativs of sin and cos, so it follows that th intgrals ar: cos d sin + c, sin d cos + c Again, ths rsults only hold if is masurd in radians EXAMPLES Find ( sin + cos ) d ( sin + cos ) d cos + sin + c d Find ( cos + sin ) π 0 0 π ( cos + sin ) [ sin cos ] d π ( sin π cos π ) ( sin0 cos0) (( ) ( )) Not It is good practic to rationalis th dnominator Find th valu of 0 0 sin d sin d cos 0 cos + cos 0 ( 0 + ) 0 87 (to dp) Rmmbr W must us radians whn intgrating or diffrntiating trig functions Pag 0 HSN000
26 Unit Furthr Calculus Th Chain Rul W will now look at how to diffrntiat composit functions, such as f g ( ) If th functions f and g ar dfind on suitabl domains, thn d d f ( g ( )) f g ( ) g Statd simply: diffrntiat th outr function, th brackt stays th sam, thn multiply by th drivativ of th brackt This is calld th chain rul You will nd to rmmbr it for th am EXAMPLE If y cos( π ) +, find dy ( π ) ( π ) ( π ) y cos + dy sin + d sin + d Not Th coms from d d( + π ) Spcial Cass of th Chain Rul W will now look at how th chain rul can b applid to particular typs of prssion Powrs of a Function [ ] n For prssions of th form f ( ), whr n is a constant, w can us a simplr vrsion of th chain rul: d d [ ] n n ( f ( )) n f ( ) f Statd simply: th powr ( n ) multiplis to th front, th brackt stays th sam, th powr lowrs by on (giving n ) and vrything is multiplid by th drivativ of th brackt ( f ( )) Pag HSN000
27 Unit Furthr Calculus EXAMPLES A function f is dfind on a suitabl domain by f + Find f ( ) + ( + ) f f + + ( ) ( ) ( )( ) Diffrntiat y sin with rspct to y sin ( sin ) dy ( sin ) cos d 8sin cos Powrs of a Linar Function Th rul for diffrntiating an prssion of th form ( a + b, whr a, b and n ar constants, is as follows: EXAMPLES d d n n a b an a b ( + ) ( + ) Diffrntiat y ( + ) with rspct to y ( + ) dy ( + ) d ( + ) ) n Pag HSN000
28 Unit Furthr Calculus If y ( + ), find dy d y ( + ) ( + ) dy ( + ) d ( + ) ( + ) A function f is dfind by f ( ) ( ) for R Find f ( ) f ( ) ( ) ( ) f ( ) ( ) Trigonomtric Functions Th following ruls can b usd to diffrntiat trigonomtric functions d d sin( a + b) a cos( a + b) d cos( a + b) a sin( a + b) Ths ar givn in th am EXAMPLE Diffrntiat y sin( 9 + π ) with rspct to dy 9cos( 9 + π) d d Pag HSN000
29 Unit Furthr Calculus A Spcial Intgral Th mthod for intgrating an prssion of th form ( a + b is: ( + ) n+ n a b ( a + b) d + c whr a 0 and n a ( n + ) Statd simply: rais th powr ( n ) by on, divid by th nw powr and also divid by th drivativ of th brackt ( a ( n + )), add c EXAMPLES d Find ( + ) 7 ( ) 7 ( + ) + d + c 8 8 ( + ) Find ( + ) d Find + c 8 ( + ) ( + ) d + c ( + ) + c 9 d + 8 whr 9 ) n d + 9 ( + 9) ( + 9 ) ( + 9) + c c 0 d d c Pag HSN000
30 Unit Furthr Calculus Evaluat + d whr d + d ( + ) 0 ( + ) ( ) ( 0 ) (or 8 8 to d p ) Not Changing powrs back into roots hr maks it asir to valuat th two brackts Rmmbr To valuat, it is asir to work out first Warning Mak sur you don t confus diffrntiation and intgration this could los you a lot of marks in th am Rmmbr th following ruls for diffrntiating and intgrating prssions of th form ( a + b) n : d d n ( a b) + an( a + b) ( + ) n+ n, n a b a + b d + c a( n + ) Ths ruls will not b givn in th am Pag HSN000
31 Unit Furthr Calculus Using Diffrntiation to Intgrat Rcall that intgration is just th procss of undoing diffrntiation So if w diffrntiat f ( ) to gt g ( ) thn w know that g ( ) d f ( ) + c EXAMPLES (a) Diffrntiat y ( ) (b) Hnc, or othrwis, find (a) y ( ) ( ) dy ( )( ) d 0 ( ) with rspct to ( ) d (b) From part (a) w know 0 d c ( ) ( ) + So: 0 d + c ( ) ( ) (a) Diffrntiat (b) Hnc, find (a) d ( ) 0 + c ( ) + c ( ) ( ) y ( ) ( ) ( ) y ( ) dy ( ) d d whr c with rspct to Not W could also hav usd th spcial intgral to obtain this answr is som constant Pag HSN000
32 Unit Furthr Calculus (b) From part (a) w know d + c So: ( ) ( ) d + c ( ) ( ) d + c + c ( ) ( ) ( ) whr c Not In this cas, th spcial intgral cannot b usd is som constant Intgrating sin(a + b) and cos(a + b) Sinc w know th drivativs of sin( a + b) and cos( a b) that thir intgrals ar: Ths ar givn in th am EXAMPLES Find sin( + ) d cos a + b d a sin a + b + c, sin a + b d a cos a + b + c sin + d cos + + c Find cos ( + π ) d π π cos + d sin + + c +, it follows Find th valu of cos ( ) d 0 0 cos( ) d sin( ) 0 (to dp) sin( ) sin( ) ( ) 0 Rmmbr W must us radians whn intgrating or diffrntiating trig functions Pag 7 HSN000
33 Unit Furthr Calculus Find th ara nclosd by th graph of y sin( π ) th lins 0 and π y +, th ais and y sin + π O π π 0 π ( + π ) d ( + π ) 0 ( cos( π ( ) + π )) ( cos( ( 0) + π )) (( ) ( )) + ( ) sin cos So th ara is squar units d Find cos( ) ( ) ( ) ( ) cos d sin + c sin + c ( ) d ( ) Find cos( ) + sin( ) cos + sin d sin cos + c Pag 8 HSN000
34 Unit Furthr Calculus 7 (a) Diffrntiat (b) Hnc find cos tan d cos with rspct to (a) ( cos ), and d ( cos ) ( cos ) sin cos d sin cos sin tan cos sin (b) cos cos cos sin From part (a) w know d + c cos cos Thrfor tan d + c cos cos Pag 9 HSN000
35 Unit Eponntials and Logarithms OUTCOME Eponntials and Logarithms Eponntials W hav alrady mt ponntial functions in Unit Outcom A function of th form f ( ) a whr a, R and a > 0 is known as an ponntial function to th bas a If a > thn th graph looks lik this: y y a, a > O (, a) This is somtims calld a growth function If 0 < a < thn th graph looks lik this: y y a, 0 < a < O (, a) This is somtims calld a dcay function Rmmbr that th graph of an ponntial function f ( ) a always passs through ( 0, ) and (, a ) sinc 0 f ( 0) a, f ( ) a a Pag 0 HSN000
36 Unit Eponntials and Logarithms EXAMPLES Th ottr population on an island incrass by % pr yar How many full yars will it tak for th population to doubl? Lt u 0 b th initial population u u (% as a dcimal) 0 u u u u 0 0 u u u u u 0 0 n n u 0 For th population to doubl aftr n yars, w rquir u u0 W want to know th smallst n which givs n a valu of or mor, sinc this will mak u n at last twic as big as u 0 Try valus of n until this is satisfid If n, < If n, < If n, 8 < If n, 0 > Thrfor aftr yars th population will doubl On a calculator: n ANS Th fficincy of a machin dcrass by % ach yar Whn th fficincy drops blow 7%, th machin nds to b srvicd Aftr how many yars will th machin nd srvicd? Lt u 0 b th initial fficincy u 0 9 u (9% as a dcimal) u u u u n 0 9 n 0 u u u u u u Whn th fficincy drops blow 0 7u 0 (7% of th initial valu) th machin must b srvicd So th machin nds srvicd aftr n yars if 0 9 n 0 7 Pag HSN000
37 Unit Eponntials and Logarithms Try valus of n until this is satisfid If n, > 0 7 If n, > 0 7 If n, > 0 7 If n, > 0 7 If n, < 0 7 Thrfor aftr yars, th machin will hav to b srvicd Logarithms Having prviously dfind what a logarithm is (s Unit Outcom ) w want to look in mor dtail at th proprtis of ths important functions Th rlationship btwn logarithms and ponntials is prssd as: y log a whr a, > 0 Hr, y is th powr of a which givs EXAMPLES Writ in logarithmic form log Evaluat log a Th powr of which givs is, so log Laws of Logarithms Thr ar thr laws of logarithms which you must know Rul y log + log y log y whr a,, y > 0 a a a If two logarithmic trms with th sam bas numbr (a abov) ar bing addd togthr, thn th trms can b combind by multiplying th argumnts ( and y abov) EXAMPLE Simplify log + log log + log log log 8 Pag HSN000
38 Unit Eponntials and Logarithms Rul ( y ) log log y log whr a,, y > 0 a a a If a logarithmic trm is bing subtractd from anothr logarithmic trm with th sam bas numbr (a abov), thn th trms can b combind by dividing th argumnts ( and y in this cas) Not that th argumnt which is bing takn away (y abov) appars on th bottom of th fraction whn th two trms ar combind EXAMPLE Evaluat log log Rul log log log log (sinc ) log a n nlog whr a, > 0 a Th powr of th argumnt (n abov) can com to th front of th trm as a multiplir, and vicvrsa EXAMPLE Eprss log7 in th form log7 a log 7 log7 log 9 7 Squash, Split and Fly You may find th following nams ar a simplr way to rmmbr th laws of logarithms log log y log ( y) + th argumnts ar squashd togthr by multiplying a a a loga loga y log a ( y ) log a th argumnts ar split into a fraction n nloga th powr of an argumnt can fly to th front of th log trm and vicvrsa Pag HSN000
39 Unit Eponntials and Logarithms Not Whn working with logarithms, you should rmmbr: EXAMPLE log 0 a sinc Evaluat log7 7 + log log 7 + log + 7 Combining svral log trms 0 a, log a a sinc a a Whn adding and subtracting svral log trms in th form log a b, thr is a simpl way to combin all th trms in on stp Multiply th argumnts of th positiv log trms in th numrator Multiply th argumnts of th ngativ log trms in th dnominator EXAMPLES Evaluat log0 + log log log 0 + log log 0 log log Evaluat log + log log + log log + log log + log 9 log 9 log log a (sinc ) argumnts of positiv log trms argumnts of ngativ log trms log OR log + log log + log log + log ( ) ( ) log + log log log + log0 + log log (sinc log ) Pag HSN000
40 Unit Eponntials and Logarithms Eponntials and Logarithms to th Bas Th constant is an important numbr in Mathmatics, and occurs frquntly in modls of rallif situations Its valu is roughly (to 9 dp), and is dfind as: ( n ) n + as n If you try larg valus of n on your calculator, you will gt clos to th valu of Lik π, is an irrational numbr Throughout this sction, w will us in prssions of th form:, which is calld an ponntial to th bas, log, which is calld a logarithm to th bas This is also known as th natural logarithm of, and is oftn writtn as ln (i ln log ) EXAMPLES Calculat th valu of log 8 log 8 08 (to dp) Solv log 9 log 9 so (to dp) Simplify log ( ) log ( ) prssing your answr in th form a + log b log c whr a, b and c ar whol numbrs log ( ) log ( ) log + log log log log + log + log log + log log + log log 7 On a calculator: ln 8 On a calculator: 9 OR log ( ) log ( ) log ( ) log ( ) ( ) log ( ) Rmmbr log 7 n n n ( ab) a b log 7 log + log log 7 + log log 7 Pag HSN000
41 Unit Eponntials and Logarithms Eponntial and Logarithmic Equations Many mathmatical modls of rallif situations us ponntials and logarithms It is important to bcom familiar with using th laws of logarithms to hlp solv quations EXAMPLES Solv log + log log 7 for > 0 a a a log + log log 7 a a a log log 7 a 7 (sinc log log y y) Solv ( ) ( ) a log + log for > ( ) ( ) log + log a log + + (sinc log y a y a ) + ( ) for p > Solv loga ( p ) loga ( p 0) loga ( p) loga ( p + ) + loga ( p 0) loga ( p) loga (( p + )( p 0) ) loga ( p) ( p + )( p 0) p p p + p p p + 0 p p ( p )( p ) p Sinc w rquir + 0 or p 0 p p >, p is th solution a Pag HSN000
42 Unit Eponntials and Logarithms Daling with Constants Somtims it may b ncssary to writ constants as logs, in ordr to solv quations EXAMPLE Solv log 7 log + for > 0 Writ in logarithmic form: log (sinc log ) log log 8 Us this in th quation: log 7 log + log 8 log 7 log OR log 7 log + log 7 log 7 log Convrting from log to ponntial form: Solving Equations with Unknown Eponnts If an unknown valu (g ) is th powr of a trm (g or 0 ), and its valu is to b calculatd, thn w must tak logs on both sids of th quation to allow it to b solvd Th sam solution will b rachd using any bas, but calculators can b usd for valuating logs ithr in bas or bas 0 EXAMPLES Solv 7 Taking log of both sids OR Taking log 0 of both sids log log 7 log log 7 ( log ) log 7 9 (to dp) 0 0 log log 7 log log log0 7 log 0 9 (to dp) Pag 7 HSN000
43 Unit Eponntials and Logarithms Solv log log 0 ( + ) log log 0 log 0 + log (to dp) Not log 0 could hav bn usd instad of log Eponntial Growth and Dcay Rcall from Sction that ponntial functions ar somtims known as growth or dcay functions Ths oftn occur in modls of rallif situations For instanc, radioactiv dcay can b modlld using an ponntial function An important masurmnt is th halflif of a radioactiv substanc, which is th tim takn for th mass of th radioactiv substanc to halv 7 Th mass G grams of a radioactiv sampl aftr tim t yars is givn by t th formula G 00 (a) What was th initial mass of radioactiv substanc in th sampl? (b) Find th halflif of th radioactiv substanc (a) Th initial mass was prsnt whn t 0 : G So th initial mass was 00 grams (b) Th halflif is th tim t at which G 0, so 0 00 t t 0 00 t log ( ) (convrting to log form) t 0 (to dp) So th halflif is 0 yars, roughly 0 8 days Pag 8 HSN000
44 Unit Eponntials and Logarithms 8 Th world population, in billions, t yars aftr 90 is givn by 0 078t P (a) What was th world population in 90? (b) Find, to th narst yar, th tim takn for th world population to doubl (a) For 90, t 0 : P So th world population in 90 was billion (b) For th population to doubl: t 0 078t 0 078t log (convrting to log form) t 8 9 (to dp) So th population doubld aftr 9 yars (to th narst yar) Graphing with Logarithmic As It is common in applications to find an ponntial rlationship btwn variabls; for instanc, th rlationship btwn th world population and tim in th prvious ampl Givn som data (g from an primnt) w would lik to find an plicit quation for th rlationship Rlationships of th form y ab Suppos w hav an ponntial graph y y ab, whr a, b > 0 y ab a O Pag 9 HSN000
45 Unit Eponntials and Logarithms Taking logarithms w find that log y log ( ab ) log a + log log a + log b W can scal th yais so that Y b ais Now our rlationship is of th form straight lin in th (, Y )plan Y log y ; th Yais is calld a logarithmic Y log b + log a, which is a ( log ) Y b + log a log a gradint is log b Sinc this is just a straight lin, w can us known points to find th gradint log b and th Yais intrcpt log a From ths w can asily find th valus of a and b, and hnc spcify th quation EXAMPLES y ab Th rlationship btwn two variabls, and y, is of th form y ab An primnt to tst this rlationship producd th data shown in th graph, whr log y is plottd against Find th valus of a and b W nd to obtain a straight lin quation: log y log ab (taking logs of both sids) y ab log y log a + log b log y log a + log b i Y log b + log a O log y From th graph, th Yais intrcpt is log a ; so O ( 7,) a Pag 70 HSN000
46 Unit Eponntials and Logarithms Using th gradint formula: log b b 7 Th rsults from an primnt wr notd as follows: Th rlationship btwn ths data can b writtn in th form y ab Find th valus of a and b, and stat th formula for y in trms of W nd to obtain a straight lin quation: log y log ab (taking logs of both sids) y ab log y log a + log b log y log a + log b i Y log b + log a W can find th gradint log b (and hnc b), using two points on th lin: using ( 0, 0 ) and ( 80, ), So log y log a log b So b (to dp) (to dp) Now w can work out log a (and hnc a) by substituting a point into this quation: using ( 0, 0 ), Thrfor y log y 0 78 log y 0 and 0 so log log a so a 09 (to dp) (to dp) a Not Dpnding on th points usd, slightly diffrnt valus for a and b may b obtaind Pag 7 HSN000
47 Unit Eponntials and Logarithms Equations in th form y a b Anothr common rlationship is y a, whr a, > 0 In this cas, th rlationship can b rprsntd by a straight lin if w chang both as to logarithmic ons EXAMPLE Th rsults from an primnt wr notd as follows: Th rlationship btwn ths data can b writtn in th form y a Find th valus of a and b, and stat th formula for y in trms of W nd to obtain a straight lin quation: y a log y log a (taking logs of both sids) 0 0 log y log a + log log y log a + b log i Y bx + log a b log0 log b 0 b W can find th gradint b using two points on th lin: using ( 70, ) and ( 8, 0 ), So log0 0 9 log0 log0 0 y + a b y b (to dp) Now w can work out a by substituting a point into this quation: using ( 70, ), log0 a log a a 0 0 (to dp) b Thrfor y 0 9 Pag 7 HSN000
48 Unit Eponntials and Logarithms 7 Graph Transformations Graph transformations wr covrd in Unit Outcom Functions and Graphs, but w will now look in mor dtail at applying transformations to graphs of ponntial and logarithmic functions EXAMPLES Shown blow is th graph of y f ( ) whr f ( ) log y y f ( ) ( 9,a) O (a) Stat th valu of a (b) Sktch th graph of y f ( + ) + (a) a log 9 (sinc 9) (b) Th graph shifts two units to th lft, and on unit upwards: y y f ( + ) + y ( 7,) O (,) Shown blow is part of th graph of y log y y f ( ) Sktch th graph of y log ( ) y y log ( ) y log ( ) log O log So rflct in th ais O (,) (, ) Pag 7 HSN000
49 Unit Eponntials and Logarithms Th diagram shows th graph of y y y (, ) O On sparat diagrams, sktch th graphs of: (a) y ; (b) y (a) Rflct in th yais: y (b) y (, ) So scal th graph from (a) by in th ydirction: (,8 ) y O y y O Pag 7 HSN000
50 Unit Wav Functions OUTCOME Wav Functions Eprssing pcos + qsin in th form kcos( a) An prssion of th form pcos + q sin can b writtn in th form k cos( a) whr k sina k p + q and tan a k cosa Th following ampl shows how to achiv this EXAMPLES Writ cos + sin in th form k cos( a ) whr 0 a 0 Stp Epand k cos( a) using th cos + sin compound angl formula k cos( a ) Stp Rarrang to compar with pcos + q sin Stp Compar th cofficints of cos and sin with pcos + q sin Stp Mark th quadrants on a CAST diagram, according to th signs of k cos a and k sin a Stp Find k and a using th formula abov (a lis in th quadrant markd twic in Stp ) Stp Stat pcos + q sin in th form k cos( a) using ths valus k cos cosa + k sin sin a k cos a cos + k sina sin k cosa k sin a 80 a a S A T C 80 + a 0 a k + 9 k sin a tana k cos a a tan 7 (to d p ) cos + sin cos 7 Pag 7 HSN000
51 Unit Wav Functions Writ cos sin in th form k cos( a) whr 0 a π cos sin k cos( a) k cosa k sina π a a S A T C π + a π a Hnc a is in th fourth quadrant k cos cosa + k sin sina k cosa cos + k sina sin k + ( ) Hnc cos sin cos( 7) Eprssing pcos + qsin in othr forms k sin a tan a k cos a First quadrant answr is: tan Not 0 0 (to d p ) Mak sur your calculator is in radian So a π 0 0 mod 7 (to d p ) An prssion in th form pcos + q sin can also b writtn in any of th following forms using a similar mthod: EXAMPLES k cos ( + a), k sin ( a), k sin ( + a) Writ cos + sin in th form k sin( + a ) whr 0 a 0 cos + sin k sin( + a ) k cosa k sina 80 a a S A T C 80 + a 0 a Hnc a is in th first quadrant k sin cos a + k cos sina k cos a sin + k sina cos k + Hnc cos + sin sin( + ) k sina tana k cosa So: a tan (to d p ) Pag 7 HSN000
52 Unit Wav Functions Writ cos sin in th form k cos( + a) whr 0 a π cos sin k cos( + a) k cosa k sin a π a a S A T C π + a π a Hnc a is in th first quadrant k cos cos a k sin sin a k cos a cos k sin a sin k + + Hnc cos sin cos( π ) Multipl Angls + k sin a tan a k cos a So: a tan π W can us th sam mthod with prssions involving th sam multipl angl, i pcos( n) + q sin( n), whr n is a constant EXAMPLE Writ cos + sin in th form k sin( + a ) whr 0 a 0 cos + sin k sin( + a ) k cosa k sina 80 a a S A T C 80 + a 0 a Hnc a is in th first quadrant k sin cos a + k cos sina k cosa sin + k sina cos k + 9 Hnc cos + sin sin( + ) k sina tana k cosa So: a tan (to d p ) Pag 77 HSN000
53 Unit Wav Functions Maimum and Minimum Valus To work out th maimum or minimum valus of pcos + q sin, w can rwrit it as a singl trigonomtric function, g k cos( a) Rcall that th maimum valu of th sin and cosin functions is, and thir minimum is y y sin y y cos ma ma O EXAMPLE Writ sin + cos in th form k cos( a) whr 0 a π and stat: (i) th maimum valu and th valu of 0 < π at which it occurs (ii) th minimum valu and th valu of 0 < π at which it occurs sin + cos k cos( a) k cosa k sin a π a π a S A T C π + a π a Hnc a is in th first quadrant k cos cosa + k sin sina k cosa cos + k sina sin k + 7 O Hnc sin + cos 7 cos( ) min π k sin a tan a k cosa So: a tan ( ) (to d p ) min Th maimum valu of 7 occurs whn: cos( ) cos 0 (to d p ) Th minimum valu of 7 occurs whn: cos( ) cos ( ) π 8 (to d p ) Pag 78 HSN000
54 Unit Wav Functions Solving Equations Th mthod of writing two trigonomtric trms as on can b usd to hlp solv quations involving both a sin( n ) and a cos( n ) trm EXAMPLES Solv cos + sin whr 0 0 First, w writ cos + sin in th form k cos( a ) : cos + sin k cos( a ) k cos a k sina 80 a a S A T C 80 + a 0 a Hnc a is in th first quadrant k cos cosa + k sin sina k cos a cos + k sina sin k + Hnc cos + sin cos( ) Now w us this to hlp solv th quation: cos + sin cos( ) cos( ) 9 or or 9 78 or 0 80 k sina tana k cosa So: a tan (to d p ) S A T C cos 9 (to d p ) Pag 79 HSN000
55 Unit Wav Functions Solv cos + sin whr 0 π First, w writ cos + sin in th form k cos( a ) : cos + sin k cos( a) k cosa k sina π a a S A T C π + a π a Hnc a is in th first quadrant k cos cos a + k sin sina k cosa cos + k sina sin k + ( ) + 9 Hnc cos + sin cos( 0 98) Now w us this to hlp solv th quation: k sina tana k cosa So: a tan 0 98 (to d p ) cos + sin π S A 0 < < π cos( 0 98) cos( 0 98) π + T C 0 < < π π 0 98 cos ( ) 90 (to d p ) or π 90 or π + 90 or π + π 90 or π + π or 99 or 7 7 or 7 7 or 97 or 8 or 9 7 or 988 or 78 or 0 Pag 80 HSN000
56 Unit Wav Functions Sktching Graphs of y pcos + qsin Eprssing pcos + q sin in th form k cos( a) nabls us to sktch th graph of y pcos + q sin EXAMPLES (a) Writ 7cos + sin in th form k cos( a ), 0 a 0 (b) Hnc sktch th graph of y 7cos + sin for 0 0 (a) First, w writ 7 cos + sin in th form k cos( a ) : 7 cos + sin k cos( a ) k cosa 7 k sina 80 a a S A T C 80 + a 0 a Hnc a is in th first quadrant k cos cosa + k sin sina k cosa cos + k sina sin k Hnc 7 cos + sin 8 cos( 0 ) k sina tana k cosa So: (b) Now w can sktch th graph of y 7cos + sin : y y 7 cos + sin 8 7 a tan 7 0 (to d p ) O Pag 8 HSN000
57 Unit Wav Functions Sktch th graph of y sin + cos for 0 0 First, w writ sin + cos in th form k cos( a ) : sin + cos k cos( a ) k cos a k sina 80 a a S A T C 80 + a 0 a Hnc a is in th first quadrant k cos cosa + k sin sin a k cosa cos + k sin a sin k + + Hnc sin + cos cos( 0 ) k sina tan a k cosa So: a tan 0 Now w can sktch th graph of y sin + cos : y y sin + cos O 0 0 Pag 8 HSN000
58 Unit Wav Functions (a) Writ sin cos in th form k sin( a ), 0 a 0 (b) Hnc sktch th graph of y sin cos +, 0 0 (a) sin cos k sin( a ) k cosa k sin a 80 a a S A T C 80 + a 0 a Hnc a is in th first quadrant k sin cosa + k cos sin a k cosa sin + k sin a cos k + + Hnc sin cos sin( ) (b) Now sktch th graph of y sin cos + sin( ) + : y 8 O y sin cos + 0 k sina tana k cosa So: ( ) a tan (to d p ) Pag 8 HSN000
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