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1 Highr Mthmtics UNIT Mthmtics HSN000 This documnt ws producd spcilly for th HSN.uk.nt wbsit, nd w rquir tht ny copis or drivtiv works ttribut th work to Highr Still Nots. For mor dtils bout th copyright on ths nots, pls s

2 Highr Mthmtics Unit Mthmtics Contnts Vctors 09 Vctors nd Sclrs 09 Componnts 09 Mgnitud 0 Equl Vctors Addition nd Subtrction of Vctors Multipliction by Sclr 7 Position Vctors 8 Bsis Vctors 9 Collinrity 0 Dividing Lins in Rtio 7 Th Sclr Product 9 Th Angl Btwn Vctors Prpndiculr Vctors Proprtis of th Sclr Product Furthr Clculus Diffrntiting sin nd cos Intgrting sin nd cos Th Chin Rul 7 Spcil Css of th Chin Rul 8 A Spcil Intgrl 9 Intgrting sin( + b) nd cos( + b) Eponntils nd Logrithms Eponntils Logrithms Lws of Logrithms Eponntils nd Logrithms to th Bs 9 Eponntil nd Logrithmic Equtions 0 Formul from Eprimntl Dt 7 Grph Trnsformtions Th Wv Eqution Eprssing pcos + qsin in th form kcos( ) Eprssing pcos + qsin in othr forms 7 Multipl Angls 8 Mimum nd Minimum Vlus 9 Solving Equtions 0 Sktching Grphs of y pcos + qsin - ii - HSN000

3 Highr Mthmtics Unit Mthmtics OUTCOME Vctors Vctors nd Sclrs A sclr is quntity with mgnitud (siz) only for mpl, n mount of mony or lngth of tim. Somtims siz lon is not nough to dscrib quntity for mpl, dirctions to th nrst shop. For this w nd to know mgnitud (i.. how fr), nd dirction. Quntitis with mgnitud nd dirction r clld vctors. A vctor is nmd ithr by using th lttrs t th nd of dirctd lin sgmnt (.g. AB ) or by using bold lttr (.g. u). You will s bold lttrs usd in printd tt, howvr w show this s u whn w writ it (i.. w undrlin th lttr). B Throughout ths nots, w will show vctors in bold s wll us undrlining thm (.g. u). Componnts A vctor my b rprsntd by its componnts. Ths dfin how to gt from on nd of th vctor to th othr. For mpl: is vctor in two dimnsions is vctor in thr dimnsions Th componnts r writtn in column, to void confusing vctors with coordints. Zro Vctors Any vctor with ll its componnts zro is clld zro vctor nd cn b writtn s 0,.g A u Pg 09 HSN000

4 Highr Mthmtics Unit Mthmtics Mgnitud Th mgnitud (lngth) of vctor is writtn s u or AB. It cn b clcultd s follows: EXAMPLES. Givn If PQ b thn PQ If PQ b thn PQ c u, find u. u ( ) + ( ) 9 units. Find th lngth of units Unit Vctors. + b + b + c Any vctor with mgnitud of on is clld unit vctor. For mpl: If 0 u thn So u is unit vctor. u unit Pg 0 HSN000

5 Highr Mthmtics Unit Mthmtics Equl Vctors Vctors with th sm mgnitud nd dirction r qul. For mpl, ll th vctors shown to th right r qul. If vctors r qul to ch othr, thn ll of thir componnts r qul, i.. b c d d If b thn d, b nd c f c f Th convrs is lso tru. Addition nd Subtrction of Vctors Considr th following vctors: b c Addition W cn construct + b s follows: b + b + b mns followd by b Similrly, w cn construct + b + c s follows: b + b + c c + b + c mns followd by b followd by c To dd vctors, w position thm nos-to-til. Thn th sum of th vctors is th vctor btwn th first til nd th lst nos. Pg HSN000

6 Highr Mthmtics Unit Mthmtics Subtrction Now considr w cn us vctor ddition to obtin b. This cn b writtn s + ( ) b. b, so if w first find b, b b is just b but in th opposit dirction. b b nd b hv th sm mgnitud, i.. b b. Thrfor w cn construct b b s follows: b b mns followd by b Using Componnts If w hv th componnts of vctors, thn things bcom much simplr. Th following ruls cn b usd for ddition nd subtrction. d + d b + b + c f c + f dd th componnts d d b b c f c f subtrct th componnts EXAMPLES. Givn u nd v, clcult u + v nd u v. 0 u + v + u v Pg HSN000

7 Highr Mthmtics Unit Mthmtics. Givn p nd p q Multipliction by Sclr q, clcult p q nd q + p. + + q p 9 9 A vctor u which is multiplid by sclr k will giv th rsult ku. This vctor will b k tims s long, i.. th mgnitud will b k u. For mpl: u u u u k If u b thn ku kb c kc Ech componnt is multiplid by th sclr. EXAMPLES. Givn v, find v. v 9 Pg HSN000

8 Highr Mthmtics Unit Mthmtics. Givn r, find r. r Ngtiv Vctors Th ngtiv of vctor is th vctor multiplid by. If w writ vctor s dirctd lin sgmnt AB, thn AB BA : 7 Position Vctors OA is clld position vctor of point A rltiv to th origin O, nd is writtn s. OB is clld th position vctor of point B, writtn b. z Givn P (, y, z ), th position vctor y P OP or p hs componnts y. z y O A b AB A B B To mov from point A to point B w cn mov bck long th vctor to th origin, nd long vctor b to point B. AB AO + OB OA + OB + b b B AB BA For th vctor joining ny two points P nd Q, PQ q p. A O p Pg HSN000

9 Highr Mthmtics Unit Mthmtics EXAMPLE R is th point (,, ) nd S is th point (,, ) From th coordints, RS s r 8 r nd s.. Find RS. Not Thr is no nd to writ this down 8 Bsis Vctors A vctor my lso b dfind in trms of th bsis vctors i, j nd k. Ths r thr mutully prpndiculr unit vctors (i.. thy r prpndiculr to ch othr). k i j Ths bsis vctors cn b writtn in componnt form s i 0, 0 0 j nd 0 0 k 0. Any vctor cn b writtn in bsis form using i, j nd k. For mpl: 0 0 v i j + k 0 0 Thr is no nd for th working bov if th following is usd: i + bj + ck b c Pg HSN000

10 Highr Mthmtics Unit Mthmtics 9 Collinrity In Stright Lins (Unit Outcom ), w lrnd tht points r collinr if thy li on th sm stright lin. not collinr not collinr collinr Th points A, B nd C in D spc r collinr if AB is prlll to BC, with B common point. Not tht w cnnot find grdints in thr dimnsions instd w us th fct tht vctors r prlll if thy r sclr multipls of th sm vctor. For mpl: u, v u So u nd v r prlll. 0 p 9, q 8 So p nd q r prlll. EXAMPLE A is th point (,, ), B( 8,, 9) nd C(,,7 ). Show tht A, B nd C r collinr. AB b BC c b BC AB, so AB nd BC r prlll nd sinc B is common point, A, B nd C r collinr. Pg HSN000

11 Highr Mthmtics Unit Mthmtics 0 Dividing Lins in Rtio Thr is simpl procss for finding th coordints of point which divids lin sgmnt in givn rtio. EXAMPLE. P is th point (,, ) nd R is th point ( 8,,9 ). Th point T divids PR in th rtio :. Find th coordints of T. Stp Mk sktch of th lin, showing th rtio in which th point divids th lin. Stp Using th sktch, qut th rtio of th two lins with th givn rtio. Stp Cross multiply, thn chng dirctd lin sgmnts to position vctors. Stp Rrrng to giv th position vctor of th unknown point. Stp From th position vctor, stt th coordints 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 Pg 7 HSN000

12 Highr Mthmtics Unit Mthmtics Using th Sction Formul Th prvious mthod cn b condnsd into formul s shown blow. If th point P divids th lin AB in th rtio m : n, thn: n + mb p n + m whr is, b nd p r th position vctors of A, B nd P. This is rfrrd to s th sction formul. It is not ncssry to know this, sinc th pproch plind bov will lwys work. EXAMPLE. P is th point (,, ) nd R is th point ( 8,,9 ). Th point T divids PR in th rtio :. Find th coordints of T. Th rtio is :, so lt m nd n, thn: n p + mr t n + m p + r [ ( ) + ( 8) ] ( ) + ( ) [ ( ) + ( 9) ] 7 So T is th point (,, 7 ). Not If you r confidnt with rithmtic, this stp cn b don mntlly Pg 8 HSN000

13 Highr Mthmtics Unit Mthmtics Th Sclr Product So fr w hv ddd nd subtrctd vctors nd multiplid vctor by sclr. Now w will considr th sclr product, which is form of vctor multipliction. Th sclr product is dfind s. b (somtims it is clld th dot product) nd cn b clcultd s follows:. b b cosθ whr θ is th ngl btwn th two vctors nd b. This is givn on th formul list. Th dfinition bov ssums tht th vctors nd b r positiond so tht thy both point wy from th ngl, or both point into th ngl. θ b θ b Howvr, if on vctor is pointing wy from th ngl, whil th othr points into th ngl θ b θ b w find tht. b b cosθ. EXAMPLE. Two vctors, nd b hv mgnituds 7 nd units rspctivly nd r t n ngl of 0 to ch othr s shown blow. b 0 Wht is th vlu of. b?. b b cosθ 7 cos 0 Pg 9 HSN000

14 Highr Mthmtics Unit Mthmtics Th Componnt Form of th Sclr Product Th sclr product cn lso b clcultd s follows:. b b + b + b whr This is givn on th formul list. nd b b b b EXAMPLES. Find p. q, givn tht p. q p q + p q + p q p nd ( ) + ( ) + ( ) + 9 q.. If A is th point (,, 9 ), B(,, ) nd C(,, ), clcult AB.AC. C(,, ) B(,, ) A (,, 9) W nd to us th position vctors of th points: AB b AC c 9 AB.AC (( ) ( ) ) + ( 0) + ( ) ( ) Pg 0 HSN000

15 Highr Mthmtics Unit Mthmtics Th Angl Btwn Vctors Th sclr product cn b rrrngd to giv th following qutions, both of which cn b usd to clcult θ, th ngl btwn th two vctors.. cosθ b b b + b + b b or cosθ Look bck to th formul for finding th sclr product, givn on th prvious two pgs. Notic tht th first qution is simply rrrngd form of th on which cn b usd to find th sclr product. Also notic tht th scond simply substituts. b for on of its quivlncs. Ths formul r not givn in th m but cn both b sily drivd from th formul on th prvious pgs (which r givn). EXAMPLES. Clcult th ngl θ btwn vctors p i + j k nd q i + j + k. p i + j k nd p q + p q + p q cosθ p q ( ) + ( ) + (( ) ) + + ( ) q i + j + k 0 θ cos 9 8. (to d. p.) (or. 98 rdins (to d. p.)) Pg HSN000

16 Highr Mthmtics Unit Mthmtics. K is th point (, 7, ), L(,, ) nd M(,, ). Find Strt with sktch: cosθ L(,, ) θ K (, 7, ) LK.LM LK LM LK k l 7 0 ( ) + ( 0 ) + ( ( ) ) + ( 0) + ( ) + + ( ) 0 8 M(,,) θ cos (to d. p.) (or. 8 rdins (to d. p.)) ɵ KLM. LM m l Pg HSN000

17 Highr Mthmtics Unit Mthmtics Prpndiculr Vctors If nd b r prpndiculr thn. b 0. This is bcus. b b cosθ b cos90 ( θ 90 sinc prpndiculr) 0 (sinc cos90 0) Convrsly, if. b 0 thn nd b r prpndiculr. EXAMPLES. Two vctors r dfind s i + j k nd b i + j + k. Show tht nd b r prpndiculr.. b b + b + b ( ) ( ) (( ) ) Sinc. b 0, nd b r prpndiculr.. PQ 7 nd RS nd RS whr is constnt. r prpndiculr, find th vlu of. Givn tht PQ PQ.RS 0 Sinc PQ nd RS r prpndiculr + ( ) Pg HSN000

18 Highr Mthmtics Unit Mthmtics Proprtis of th Sclr Product Som proprtis of th sclr product r s follows: EXAMPLE AMPLES. Clcult.( + ). b b.. b + c. b +. c (Epnding brckts). p q r whn p, r nd q nd th vctors r rrngd s shown in th digrm blow. p. q + r p. q + p. r p q cosθ + cos 0 + cos + + p r cosθ. In th digrm blow c nd b. p q r c 0 b 0 b c. Clcult.( + + ) ( + + ). b c. +. b +. c + b cosθ c cosθ + cos0 cos Rmmbr. c c cosθ sinc points into θ nd c points wy Pg HSN000

19 Highr Mthmtics Unit Mthmtics OUTCOME Furthr Clculus Diffrntiting sin nd cos Thr r two nw ruls which r usd to diffrntit prssions involving trigonomtric functions: d d d d ( sin ) cos ( cos ) sin EXAMPLES. Diffrntit 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 + + π π. Find th qution of th tngnt to th curv y sin whn At π : y sin π π So th point is, W lso nd th grdint t th point whr π : dy cos d At π, mtngnt cos π ( ) π. Pg HSN000

20 Highr Mthmtics Unit Mthmtics Now w hv th point π, nd th grdint m tngnt y b m ( π ) y y π y π + 0, so: Intgrting sin nd cos Sinc w know th drivtivs of sin nd cos, it follows tht th intgrls r: EXAMPLES cos d sin + c sin d cos + c. Find sin + cos d. sin + cos d cos + sin + c π. Find cos + sin d. 0 π 0 Not [ ] cos + sin d sin cos π π π [ ] ( ) ( ) [ ] 0 sin cos sin0 cos Not It is good prctic to rtionlis th dnomintor With trigonomtric drivtivs nd intgrls, w should lwys work in rdins. Pg HSN000

21 Highr Mthmtics Unit Mthmtics Th Chin Rul W hv not yt covrd how to diffrntit composit functions,.g. f g ( ). If th functions f nd g r dfind on suitbl domins, thn: EXAMPLE. If y cos( π ) d d +, find dy d. ( π ) ( π ) ( π ) y cos + dy sin + d sin + f ( g ( )) f g ( ) g [ ] n For prssions of th form f ( ), whr n is constnt, w cn us simplr vrsion of th chin rul: d d [ ] n n ( f ( )) n f ( ) f Sttd simply: th powr ( n ) multiplis to th front, th brckt stys th sm, th powr lowrs by on ( n ) nd vrything is multiplid by th drivtiv of th brckt ( f ( )). EXAMPLES. A function f is dfind on suitbl domin by f ( ) +. Find f ( ). + ( + ) f f + + ( ) ( ). Diffrntit ( )( ) y sin ( sin ) dy ( sin ) cos d 8sin cos y sin with rspct to. Not Th coms from d ( + d π ) Pg 7 HSN000

22 Highr Mthmtics Unit Mthmtics Spcil Css of th Chin Rul Th rul for diffrntiting n prssion of th form ( + b, whr, b nd n r constnts, is s follows: EXAMPLES d d n n b n b ( + ) ( + ). Diffrntit y ( + ) with rspct to.. If y ( + ) dy ( + ) d ( + ) y ( + ), find dy d. y ( + ) ( + ) dy ( + ) d ( + ) ( + ). A function f is dfind by f ( ) ( ) for R. Find f ( ). f ( ) ( ) ( ) f ( ) ( ) ( ) Th following ruls cn b usd to diffrntit trigonomtric functions. d d sin( + b) cos( + b) d cos( + b) sin( + b) Ths r givn in th m. EXAMPLE. Diffrntit y sin( 9 + π). dy 9cos( 9 + π) d d ) n Pg 8 HSN000

23 Highr Mthmtics Unit Mthmtics A Spcil Intgrl Th mthod for intgrting n prssion of th from ( + b is: ( + ) n+ n b ( + b) d + c whr 0 nd n ( n + ) Sttd simply: ris th powr ( n ) by on, divid by th nw powr multiplid by th drivtiv of th brckt ( ( n + )), dd c. EXAMPLES d.. Find ( + ) 7 ( ) 7 ( + ) + d + c 8 ( + ) c 8. Find ( + ) d.. Find ( + ) ( + ) d + c ( + ) + c 9 d + whr 9. ) n d + 9 ( + 9) ( + 9 ) ( + 9) + c c 0 d d c Pg 9 HSN000

24 Highr Mthmtics Unit Mthmtics. Evlut Not + d whr. 0 + d + d 0 0 ( + ) 9 0 ( + ) 0 ( ) + ( ( 0) + ) ( 8 ) (or 8. 8 to d. p.) For n prssion of th form ( + b : n n b n b d ( + ) ( + ) d ( + ) n+ n b + b d + c ( n + ) EXAMPLES. () Diffrntit y ( ) (b) Hnc, or othrwis, find () y ( ) ( ) dy ( )( ) d 0 ( ) ) n with rspct to. ( ) d. Pg 0 HSN000

25 Highr Mthmtics Unit Mthmtics (b) From prt () w know 0 d c ( ) ( ) +. So: 0 d + c ( ) ( ). () Diffrntit (b) Hnc, find () d + ( ) 0 c ( ) + c ( ) ( ) y ( ) ( ) ( ) y ( ) dy ( ) d d. whr c with rspct to. Not W could lso hv usd th spcil intgrl to obtin this nswr is som constnt (b) From prt () w know d + c. So: ( ) ( ) d + c ( ) ( ) d + c + c ( ) ( ) whr c is som constnt Not In this cs, th spcil intgrl cnnot b usd Pg HSN000

26 Highr Mthmtics Unit Mthmtics Intgrting sin( + b) nd cos( + b) Sinc w know th drivtivs of sin( + b) nd cos( b) tht th intgrls r: Ths r givn on th formul list. EXAMPLES. Find ( + ) sin d. cos + b d sin + b + c sin + b d cos + b + c sin + d cos + + c. Find cos ( + π ) d. ( π ) π cos + d sin + + c. Find th r nclosd b th grph of y sin( π ) th lins 0 nd π. y +, it follows +, th -is, nd y sin + π O π π ( + π ) d ( + π ) sin cos 0 cos( π π ( ) ) ( cos 0 π ) So th r is π 0 squr units Pg HSN000

27 Highr Mthmtics Unit Mthmtics d.. Find cos( ) ( ) ( ) sin( ) cos d sin + c + c. Find cos( ) + sin( ) d. cos( ) sin( ) sin( ) cos( ) + d + c Pg HSN000

28 Highr Mthmtics Unit Mthmtics OUTCOME Eponntils nd Logrithms Eponntils W hv lrdy mt ponntil functions in Unit Outcom. Rcll tht function in th form f ( ) is clld n ponntil function to th bs, whr > 0. If > thn th grph looks lik this: y y, > O (, ) This is somtims clld growth function If 0 < < thn th grph looks lik this: y y, 0 < < O (, ) This is somtims clld dcy function Rmmbr tht th grph of n ponntil function f ( ) lwys psss through ( 0, ) nd (, ) sinc: 0 f ( 0) f Pg HSN000

29 Highr Mthmtics Unit Mthmtics EXAMPLES. Th ottr popultion on n islnd incrss by % pr yr. How mny full yrs will it tk th popultion to doubl? Lt u 0 b th initil popultion. u u (% s dciml) 0 u u u u 0 0 u u u u u 0 0 n n u 0 For th popultion to doubl ftr n yrs, w rquir u u0. So th cofficint of u 0, which is n, must b t lst, i.. n. Try vlus of n until this is stisfid (strting with n sinc w know < ): If n, < If n, < If n, 8 < If n, 0 > Thrfor ftr yrs th popultion will doubl. On clcultor: n ANS. Th fficincy of mchin dcrss by % ch yr. Whn th fficincy drops blow 7%, th mchin nds to b srvicd. Aftr how mny yrs will th mchin nd srvicd? Lt u 0 b th initil fficincy. u 0 9 u (9% s dciml) 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 initil vlu) th mchin must b srvicd. So th mchin nds srvicd ftr n yrs if 0 9 n 0 7. Pg HSN000

30 Highr Mthmtics Unit Mthmtics Try vlus of n until this is stisfid: If n, > 0 7 If n, > 0 7 If n, > 0 7 If n, > 0 7 If n, < 0 7 Thrfor ftr yrs, th mchin will hv to b srvicd. Logrithms Hving prviously dfind wht logrithm is (s Unit Outcom ) w wnt to look in mor dtil t th proprtis of ths importnt functions. Th rltionship btwn logrithms nd ponntils is prssd s: y log whr, > 0. Hr, y is th powr of which givs. EXAMPLES. Writ in logrithmic form. log. Evlut log. Th powr of which givs is, so log Lws of Logrithms Thr r thr lws of logrithms which you must know. Rul y log + log y log y whr,, y > 0 If two logrithmic trms with th sm bs numbr ( bov) r bing ddd togthr, thn th trms cn b combind by multiplying th rgumnts ( nd y bov). EXAMPLE. Simplify log + log. log + log log log 8 Pg HSN000

31 Highr Mthmtics Unit Mthmtics Rul ( y ) log log y log whr,, y > 0 If logrithmic trm is bing subtrctd from nothr logrithmic trm with th sm bs numbr ( bov), thn th trms cn b combind by dividing th rgumnts ( nd y in this cs). Not tht th rgumnt which is bing tkn wy (y bov) pprs on th bottom of th frction whn th two trms r combind. EXAMPLE. Evlut log log. Rul log log log log (Sinc ) log n nlog whr, > 0. Th powr of th rgumnt (n bov) cn com to th front of th trm s multiplir, nd vic-vrs. EXAMPLE. Eprss log7 in th form log7. log 7 log7 log 9 7 Sqush, Split nd Fly Hr is simplr wy to rmmbr th lws of logrithms: If logrithmic trms, with th sm bs numbr, r bing ddd, th rgumnts r squshd togthr by multiplying thm If logrithmic trm is bing subtrctd from nothr, with th sm bs numbr, th rgumnts r split into frction Th powr of n rgumnt cn fly to th front of th log trm nd vicvrs. Pg 7 HSN000

32 Highr Mthmtics Unit Mthmtics Not Whn working with logrithms, th following fcts r usful: EXAMPLE log 0 sinc. Evlut log7 7 + log. log 7 + log + 7 Combining svrl log trms 0 log sinc Whn dding nd subtrcting svrl log trms in th form log b, thr is simpl wy to combin ll th trms in on stp. Multiply th rgumnts of th positiv log trms in th numrtor, Multiply th rgumnts of th ngtiv log trms in th dnomintor. EXAMPLES. Evlut log0 + log log log 0 + log log 0 log log. Evlut log + log. log + log log + log log + log 9 log 9 log log (Sinc ) rgumnts of positiv log trms rgumnts of ngtiv log trms log OR log + log log + log log + log ( ) ( ) log + log log log + log 0 + log log (Sinc log ) Pg 8 HSN000

33 Highr Mthmtics Unit Mthmtics Eponntils nd Logrithms to th Bs Th constnt is n importnt numbr in Mthmtics, prticulrly whn dling with rl-lif situtions. Its vlu is roughly (to 9 d.p.), nd is dfind s: n + s n n If you try vry lrg vlus of n on your clcultor, you will gt clos to th vlu of. Lik π, is n irrtionl numbr. Throughout this sction, w will us in prssions of th form:, which is clld n ponntil to th bs log, which is clld logrithm to th bs. This is lso known s th nturl logrithm of, nd is oftn writtn s ln (i.. ln log ). EXAMPLES. Writ down th vlu of log 8 log (to d.p.). Solv log 9. log 9 so (to d.p.).. Simplify log ( ) log ( ) prssing your nswr in th form + log b log c whr, b nd c r whol numbrs. log ( ) log ( ) log + log log log log + log + log log + log log + log log 7 On clcultor: ln 8 On clcultor: 9 OR log ( ) log ( ) log ( ) log ( ) ( ) log ( ) Rmmbr log 7 n n n b b log 7 log + log log 7 + log log 7 Pg 9 HSN000

34 Highr Mthmtics Unit Mthmtics Eponntil nd Logrithmic Equtions Mny mthmticl modls of rl-lif situtions us ponntils nd logrithms. It is importnt to bcom fmilir with using th lws of logrithms to hlp solv qutions. EXAMPLES. Solv log + log log 7 for > 0. log + log log 7 log log 7 7 (Sinc log log y y). Solv ( ) ( ) log + log for >. ( ) ( ) log + log log + + (Sinc log y y ) + ( ) + 8 log p + + log p 0 log p for p >.. Solv log ( p + ) + log ( p 0) log ( p) log ( p + )( p 0) log ( p) ( p + )( p 0) p p 0 p + p 0 p 0 p + 0 p 8 p 0 0 ( p )( p ) + 0 or p 0 p p Sinc w rquir p >, p is th solution. Pg 0 HSN000

35 Highr Mthmtics Unit Mthmtics Dling with Constnts Somtims it my b ncssry to writ constnts s logs, in ordr to solv qutions. EXAMPLE. Solv log 7 log + for > 0. Writ in logrithmic form: log (Sinc log ) log log 8 Us this in th qution: log 7 log + log 8 log 7 log OR log 7 log + log 7 log 7 log Convrting from log to ponntil form: Solving Equtions with unknown Eponnts If n unknown vlu (.g. ) is th powr of trm (.g. or 0 ), nd its vlu is to b clcultd, thn w must tk logs on both sids of th qution to llow it to b solvd. Th sm solution will b rchd using ny bs, but clcultors cn b usd for vluting logs to th bs nd 0. EXAMPLES. Solv 7. Tking log of both sids log log 7 log log 7 ( log ) log 7 9 (to d.p.) OR Tking log0 of both sids log log log log log0 7 log 0 9 (to d.p.) Pg HSN000

36 Highr Mthmtics Unit Mthmtics. Solv log log 0 ( + ) log log 0 log 0 + log (to d.p.) OR log log 0 ( + ) log log log0 0 + log (to d.p.) t 7. For th formul s ( t ) 00 : () Evlut s ( 0). 0 (b) Clcult th vlu of t whr s ( t ) s. () s ( 0) ( 0) (Sinc ) (b) Clcult th vlu of t whr s ( t ) s ( 0) 0 : log 0 00 t t 0 00 t log t log log ( ) t log (Sinc log ) t 0 (to d.p.) Formul from Eprimntl Dt Rsults from primnts oftn rlt to ponntil functions, ithr of th form b y or y b. It is oftn mor usful to hv this informtion s stright lin grph, bcus it is thn sir to find its qution. To obtin stright lin, th qutions must b writtn in th form y m + c. Th vlus of nd b cn thn b dtrmind from this qution. Pg HSN000

37 Highr Mthmtics Unit Mthmtics Equtions in th form y b EXAMPLE. Th rsults from n primnt wr notd s follows: Th rltionship btwn ths dt cn b writtn in th form y. Find th vlus of nd b, nd stt th formul for y in trms of. W nd to prss th qution in th form y m + c : y b log y log (tking logs of both sids) 0 0 log log log y log + log log y log + b log Compr to y m + c : b i.. log y b log + log b W cn find th grdint b, using two points on th lin: y y Using ( 70, ) nd ( 8, 0 ), b (to d.p.) log y 0 9 log + log. So y Now w cn work out by substituting point into this qution: Using ( 70, ), log0 log (to d.p.) b Thrfor y 0 9. Not Dpnding on th points usd, slightly diffrnt nswrs my b obtind. Pg HSN000

38 Highr Mthmtics Unit Mthmtics Equtions in th form y b EXAMPLE. Th rsults from n primnt wr notd s follows: Th rltionship btwn ths dt cn b writtn in th form y b. Find th vlus of nd b, nd stt th formul for y in trms of. W nd to prss th qution in th form y m + c : log y log b (tking logs of both sids) y b log y log + log b log y log + log b Compr to y m + c : i.. log y log b + log W cn find th grdint log b (nd hnc b), using two points on th lin: y y Using ( 0, 0 ) nd ( 80, ), log b So log y log. So b (to d.p.) (to d.p.) Now w cn work out log (nd hnc ) by substituting point into this qution: Using ( 0, 0 ), Thrfor y log y 0 78 log y 0 nd 0 so log log so 09 (to d.p.) (to d.p.) Pg HSN000

39 Highr Mthmtics Unit Mthmtics 7 Grph Trnsformtions Grph trnsformtions wr covrd in Unit Outcom Functions nd Grphs, but w cn now look in mor dtil t pplying trnsformtions to grphs of ponntil nd logrithmic functions. EXAMPLES. Shown blow is th grph of y f ( ) whr f ( ) log. y y f ( ) O ( 9,) () Stt th vlu of. (b) Sktch th grph of y f ( + ) +. () log 9 (sinc 9) (b) Th grph shifts two units to th lft, nd on unit upwrds: y ( 7,) O y f ( + ) +. Shown blow is prt of th grph of y log. y y log O (,) Sktch th grph of y log ( ) y y log ( ). log log O So rflct in th -is y log (, ) ( ) Pg HSN000

40 Highr Mthmtics Unit Mthmtics OUTCOME Th Wv Eqution Eprssing pcos + qsin in th form kcos( ) An prssion of th form p cos + q sin cn b writtn in th form k cos( ) whr: k sin k p + q nd tn k cos Th following mpl shows how to chiv this. EXAMPLES. Writ cos + sin in th form k cos( ) whr 0 0. Stp Epnd k cos( ) using th cos + sin compound ngl formul. k cos( ) Stp Rrrng to compr with p cos + q sin. Stp Compr th cofficints of cos nd sin with p cos + qsin. Stp Mrk th qudrnts on CAST digrm, ccording to th signs of k cos nd k sin. Stp Find k nd using th formul bov ( lis in th qudrnt mrkd twic in Stp ). Stp Stt p cos + q sin in th form k cos( ) using ths vlus. k cos cos + k sin sin k cos cos + k sin sin k cos k sin 80 S A T C k + 9 k sin tn k cos tn 7. (to d. p.) cos + sin cos 7. Pg HSN000

41 Highr Mthmtics Unit Mthmtics. Writ cos sin in th form k cos( ) whr 0 π. cos sin k cos( ) k cos k sin π S A T C π + π Hnc is in th fourth qudrnt. k cos cos + k sin sin k cos cos + k sin sin k + ( ) Hnc cos sin cos( 7).. Eprssing pcos + qsin in othr forms k sin tn k cos First qudrnt nswr is: tn 0. 0 (to d. p.) So π (to d. p.) An prssion in th form p cos + q sin cn lso b writtn in ny of th following forms using similr mthod. EXAMPLES k cos( + ) k sin( ) k sin( + ). Writ cos + sin in th form k sin( + ) whr 0 0. cos + sin k sin( + ) k cos k sin 80 S A T C Hnc is in th first qudrnt. k sin cos + k cos sin k cos sin + k sin cos k + Hnc cos + sin sin( +. ). k sin tn k cos So: tn. (to d. p.) Pg 7 HSN000

42 Highr Mthmtics Unit Mthmtics. Writ cos sin in th form k cos( + ) whr 0 π. cos sin k cos( + ) k cos k sin π S A T C π + π Hnc is in th first qudrnt. k cos cos k sin sin k cos cos k sin sin k + + Hnc cos sin cos( π ) Multipl Angls +. k sin tn k cos So: tn π Th sm mthod is usd with prssions of th form pcos( n) + qsin( n), whr n is constnt. EXAMPLE Writ cos + sin in th form k sin( + ) whr 0 0. cos + sin k sin( + ) k cos k sin 80 S A T C Hnc is in th first qudrnt. k sin cos + k cos sin k cos sin + k sin cos k + 9 Hnc cos + sin sin( +. ). k sin tn k cos So: tn. (to d. p.) Pg 8 HSN000

43 Highr Mthmtics Unit Mthmtics Mimum nd Minimum Vlus To work out th mimum or minimum vlus of p cos + qsin, w cn rwrit it s singl trigonomtric function,.g. k cos( ). Rcll tht th mimum vlu of th sin nd cosin functions is, nd thir minimum is. y y sin y y cos m. m. O EXAMPLE Writ sin + cos in th form k cos( ) whr 0 π nd stt: (i) th mimum vlu nd th vlu of 0 < π t which it occurs (ii) th minimum vlu nd th vlu of 0 < π t which it occurs. sin + cos k cos( ) k cos k sin π π S A T C π + π Hnc is in th first qudrnt. k cos cos + k sin sin k cos cos + k sin sin k ( ) + 7 O Hnc sin + cos 7 cos(. ). min. π min. k sin tn k cos So: tn ( ). (to d. p.) Th mimum vlu occurs whn: cos(. ). cos. 0. (to d. p.) Th minimum vlu occurs whn: cos(. ). cos ( ). π. 8 (to d. p.) Pg 9 HSN000

44 Highr Mthmtics Unit Mthmtics Solving Equtions Th wv qution cn b usd to hlp solv trigonomtric qutions involving both sin( n ) nd cos( n ) trm. EXAMPLES. Solv cos + sin whr 0 0. First, w writ cos + sin in th form k cos( ) : cos + sin k cos( ) k cos k sin 80 S A T C Hnc is in th first qudrnt. k cos cos + k sin sin k cos cos + k sin sin k + Hnc cos + sin cos(. ). Now w us this to hlp solv th qution: cos + sin cos(. ) cos(. ).. 9 or or or k sin tn k cos So: tn. (to d. p.) S A T C. cos. 9 (to d. p.) Pg 0 HSN000

45 Highr Mthmtics Unit Mthmtics. Solv cos + sin whr 0 π. First, w writ cos + sin in th form k cos( ) : cos + sin k cos( ) k cos k sin π S A T C π + π Hnc is in th first qudrnt. k cos cos + k sin sin k cos cos + k sin sin k + ( ) + 9 Hnc cos + sin cos( 0. 98). Now w us this to hlp solv th qution: k sin tn k cos So: tn (to d. p.) cos + sin π S A 0 < < π cos( 0. 98) cos( 0. 98) π + T C 0 < < π π cos ( ). 90 (to d. p.) or π. 90 or π or π + π. 90 or π + π or. 99 or 7. 7 or or. 97 or 8. or or. 988 or. 78 or. 0 Pg HSN000

46 Highr Mthmtics Unit Mthmtics Sktching Grphs of y pcos + qsin Eprssing pcos + q sin in th form k cos( ) nbls us to sktch th grph of y pcos + q sin. EXAMPLES. () Writ 7cos + sin in th form k cos( ), 0 0. (b) Hnc sktch th grph of y 7cos + sin for 0 0. () First, w writ 7 cos + sin in th form k cos( ) : 7 cos + sin k cos( ) k cos 7 k sin 80 S A T C Hnc is in th first qudrnt. k cos cos + k sin sin k cos cos + k sin sin k Hnc 7 cos + sin 8 cos( 0 ) k sin tn k cos So: (b) Now w cn sktch th grph of y 7cos + sin : y y 7 cos + sin 8 7 tn 7 0. (to d. p.) O Pg HSN000

47 Highr Mthmtics Unit Mthmtics. Sktch th grph of y sin + cos for 0 0. First, w writ sin + cos in th form k cos( ) : sin + cos k cos( ) k cos k sin 80 S A T C Hnc is in th first qudrnt. k cos cos + k sin sin k cos cos + k sin sin k + + Hnc sin + cos cos( 0 ). k sin tn k cos So: tn 0 Now w cn sktch th grph of y sin + cos : y y sin + cos O 0 0 Pg HSN000

48 Highr Mthmtics Unit Mthmtics. () Writ sin cos in th form k sin( ), 0 0. (b) Hnc sktch th grph of y sin cos +, 0 0. () sin cos k sin( ) k cos k sin 80 S A T C Hnc is in th first qudrnt. k sin cos + k cos sin k cos sin + k sin cos k + + Hnc sin cos sin(. ). (b) Now sktch th grph of y sin cos + sin(. ) + : y 8 y sin cos + k sin tn k cos So: ( ) tn. (to d. p.) O. 0 Pg HSN000

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