Introduction to calculus

Transcription

1 Chpter6 Introdution to lulus Sllus referene: 7., 7.5 Contents: A B C D Limits Finding smptotes using limits Rtes of hnge Clultion of res under urves

2 6 INTRODUCTION TO CALCULUS (Chpter 6) Clulus is mjor rnh of mthemtis whih uilds on lger, trigonometr, nd nlti geometr. It hs widespred pplitions in siene, engineering, nd finnil mthemtis. The stud of lulus is divided into two fields, differentil lulus nd integrl lulus, oth of whih we will stud in this ourse. These fields re linked the fundmentl theorem of lulus whih we will stud in Chpter. HISTORICAL NOTE Clulus is Ltin word mening pele. Anient Romns used stones for ounting. The histor of lulus egins with the Egptin Mosow pprus from out 85 BC. Its stud ontinued in Egpt efore eing tken up the Greek mthemtiin Arhimedes of Sruse. It ws further developed through the enturies mthemtiins of mn ntions. Two of the most importnt ontriutors were Gottfried Wilhelm Leiniz nd Sir Is Newton who independentl developed the fundmentl theorem of lulus. A LIMITS The ide of limit is essentil to differentil lulus. We will see tht it is neessr for finding the grdient of tngent to urve t n point on the urve. Consider the following tle of vlues for f() = in the viinit of =. :9 :99 :999 :9999 : : : : f() :6 :96 :996 :999 6 : : : : 9 Notie tht s pprohes from the left, then f() pprohes from elow. Likewise, s pprohes from the right, then f() pprohes from ove. We s tht s pprohes from either diretion, f() pprohes limit of, nd write lim! =.

3 INFORMAL DEFINITION OF A LIMIT INTRODUCTION TO CALCULUS (Chpter 6) 6 The following definition of limit is informl ut dequte for the purposes of this ourse: If f() n e mde s lose s we like to some rel numer A mking suffiientl lose to, we s tht f() pprohes limit of A s pprohes, nd we write lim f() =A.! We lso s tht s pprohes, f() onverges to A. Notie tht we hve not used the tul vlue of f() when =, or in other words f(). This is ver importnt to the onept of limits. 5 + For emple, if f() = nd we wish to find the limit s!, it is tempting for us to simpl sustitute = into f(). Not onl do we get the meningless vlue of Oserve tht if f() = 5 + then f() f() hs the grph shown longside. It is the stright line = +5 with the point (, 5) missing, lled point of disontinuit of the funtion. However, even though this point is missing, the limit of f() s pprohes does eist. In prtiulr, s! from either diretion, f()! 5. ) lim! 5 + =5, ut lso we destro the si limit method. ½ =5+ if 6= is undefined if =. 5 ( ) missing point In prtie we do not need to grph funtions eh time to determine limits, nd most n e found lgerill. Emple Evlute: lim 5 + lim!! Self Tutor n e mde s lose s we like to mking suffiientl lose to. ) lim =.! ½ 5 + =5+ if 6= is undefined if =. ) lim! 5 + = lim! (5 + ) = lim! 5+ sine 6= =5

4 6 INTRODUCTION TO CALCULUS (Chpter 6) RULES FOR LIMITS If f() nd g() re funtions nd is onstnt: ² lim! = ² lim! f() = lim! f() ² lim! [f() g()] = lim! f() lim! g() ² lim! [f()g()] = lim! f() lim! g() f() ² lim! g() = lim f() lim g() provided lim g() 6= :!!! Emple Self Tutor Use the rules for limits to evlute: lim! ( + )( ) lim! + Desrie the results in terms of onvergene. As!, +! 5 nd! ) lim! ( + )( ) = 5 = As!, ( + )( ) onverges to. As!, +! nd! ) lim! + = = As!, + onverges to. LIMITS AT INFINITY We n use the ide of limits to disuss the ehviour of funtions for etreme vlues of. We write! to men when gets s lrge s we like nd positive, nd! to men when gets s lrge s we like nd negtive. We red! s tends to plus infinit nd! s tends to minus infinit. Notie tht s!, << < < :::::: ) lim! =, lim! =, nd so on.

5 INTRODUCTION TO CALCULUS (Chpter 6) 65 Emple Self Tutor Evlute the following limits: lim! + lim! + lim! = lim! = = lim! = lim! = = fdividing eh term in oth numertor nd denomintor g fs!,! nd! g fdividing eh term g fs!,!,!, nd! g or lim! = lim! = lim! + ( )+8+ µ + = fs lim! or lim! = lim! = lim! µ = lim! =g + + ( ) + + = fs lim! =g EXERCISE 6A Evlute: lim ( +) lim (5 ) lim ( )!!! d g lim 5 + e lim h ( h) f lim! h!! lim! ( +5) h lim! + Evlute the following limits looking for ommon ftor in the numertor nd denomintor: d lim! lim! e h +6h lim h! h lim! f h 8h lim h! h 6 lim! 5 +6

6 66 INTRODUCTION TO CALCULUS (Chpter 6) Evlute: d lim! lim! e lim! lim! + + f lim! lim! B FINDING ASYMPTOTES USING LIMITS Rtionl funtions re funtions of the form f() g() where f() nd g() re polnomils. Rtionl funtions re hrterised the eistene of smptotes whih m e vertil, horizontl, or olique. An olique smptote is neither horizontl nor vertil, ut we will not disuss these in this ourse. We n investigte the smptotes of funtion using limits. + Consider the funtion f : 7! whih hs domin f R, 6= g. There is vertil smptote (VA) t =. To disuss the ehviour ner the VA, we find wht hppens to f() s! from the left nd right. ² First we drw sign digrm of f() ² Hene s! (left), f()!! (right), f()! +: Is there nother tpe of smptote? + ( ) + Now f() = = =+. ² As! +, f()! (ove) fs! from oveg ² As!, f()! (elow) fs! from elowg Hene, there is horizontl smptote (HA) t =. This horizontl smptote orresponds to the nswer in Emple prt. Emple Self Tutor Find n smptotes of the funtion f : 7! + nd disuss the ehviour of f() ner these smptotes. -\Ew_ We notie tht f() = ( )( ) ( ) = ( )( + ) + provided 6=.

7 INTRODUCTION TO CALCULUS (Chpter 6) 67 So, there is point of disontinuit t =. Also, when =, f() is undefined. This indites tht = is vertil smptote. The sign digrm for f() is: As! (left), f()!. As! (right), f()! +. ) = is VA. For 6=, f() = As!, As!, ( ) + = + + fdividing zerog = ( +)+ + = + +! nd is + > ) f()! (ove)! nd is + < ) f()! (elow) ) HA is = (see Emple prt ) The open irle indites disontinuit. EXERCISE 6B For eh of the following, determine ll smptotes nd disuss the ehviour of the grph ner its smptotes: f() = + f() = + = + d f() = + INVESTIGATION LIMITS IN NUMBER SEQUENCES The sequene :, :, :,... n e defined the generl term: n =:::: where there re n s. Wht to do: Cop nd omplete the tle longside: n n n n Consider whih ontins s. In the numer ( ), how mn s do we need etween the deiml point nd the? In the limit s n tends to infinit, n ontins n infinite numer of s. In the numer ( n ), 5 how mn s do we need to write efore the? Using our nswer to, stte the limit of n s n!. 5 Hene stte lim n! n, whih is the et vlue of :.

8 68 INTRODUCTION TO CALCULUS (Chpter 6) C RATES OF CHANGE A rte is omprison etween two quntities with different units. We often judge performnes using rtes. For emple: ² Sir Donld Brdmn s tting rte t Test riket level ws 99:9 runs per innings. ² Mihel Jordn s sketll soring rte ws : points per gme. ² Rngi s tping rte is 6 words per minute with n error rte of : errors per pge. Speed is ommonl used rte. It is the rte of hnge in distne per unit of time. We re fmilir with the formul verge speed = distne trvelled. time tken However, if r hs n verge speed of 6 km h for journe, it does not men tht the r trvels t etl 6 km h for the whole time. In ft, the speed will prol vr ontinuousl throughout the journe. So, how n we lulte the r s speed t n prtiulr time? Suppose we re given grph of the r s distne trvelled ginst time tken. If this grph is stright line, then we know the speed is onstnt nd is given the grdient of the line. distne trvelled h 6 km time If the grph is urve, then the r s instntneous speed is given the grdient of the tngent to the urve t tht time. distne trvelled time distne time INVESTIGATION INSTANTANEOUS SPEED When ll ering is dropped from the top of tll uilding, the distne it hs fllen fter t seonds is reorded, nd the following grph of distne ginst time otined. In this investigtion we will tr to mesure the speed of the ll t the instnt when t =seonds.

9 INTRODUCTION TO CALCULUS (Chpter 6) Wht to do: D hord M (, ) F (, ) The verge speed in the time intervl 6 t 6 is distne trvelled = time tken (8 ) m = ( ) s = 6 ms =ms Clik on the ion to strt the demonstrtion. F is the point where t =seonds, nd M is nother point on the urve. To strt with M is t t =seonds. The numer in the o mrked grdient is the grdient of the hord [FM]. This the verge speed of the ll ering in the intervl from F to M. For M t t = seonds, ou should see the verge speed is ms. Clik on M nd drg it slowl towrds F. Cop nd omplete the tle longside with the grdient of the hord when M is t vrious times t. Oserve wht hppens s M rehes F. Eplin wh this is so. When t =seonds, wht do ou suspet the instntneous speed of the ll ering is? 5 Move M to the origin, nd then slide it towrds F from the left. Cop nd omplete the tle longside with the grdient of the hord when M is t vrious times t. 6 Do our results gree with those in? From the investigtion ou should hve disovered tht: urve 6 t t :5 : : t :5 :9 :99 DEMO grdient of [FM] grdient of [FM] The instntneous rte of hnge of vrile t prtiulr instnt is given the grdient of the tngent to the grph t tht point. THE TANGENT TO A CURVE A hord or sent of urve is stright line segment whih joins n two points on the urve. The grdient of the hord [AB] mesures the verge rte of hnge of the funtion for the given hnge in -vlues. A tngent is stright line whih touhes urve t point. The grdient of the tngent t point A mesures the instntneous rte of hnge of the funtion t point A. In the limit s B pprohes A, the grdient of the hord [AB] will e the grdient of the tngent t A. urve hord (sent) A B tngent

10 7 INTRODUCTION TO CALCULUS (Chpter 6) The grdient of the tngent t = is defined s the grdient of the urve t the point where =, nd is the instntneous rte of hnge in f() with respet to t tht point. INVESTIGATION Wht to do: ƒ( ) X A (, ) THE GRADIENT OF A TANGENT Given urve f(), we wish to find the grdient of the tngent t the point (, f()). For emple, the point A(, ) lies on the urve f() =. Wht is the grdient of the tngent t A? DEMO ƒ( ) X Suppose B lies on f() = nd B hs oordintes (, ). B (, X) A (, ) Show tht the hord [AB] hs grdient f() f() or. Cop nd omplete the tle longside: Comment on the grdient of [AB] s gets loser to. Repet the proess s gets loser to, ut from the left of A. Clik on the ion to view demonstrtion of the proess. 5 Wht do ou suspet is the grdient of the tngent t A? Point B grdient of [AB] 5 (5, 5) 6 :5 : : : Fortuntel we do not hve to use grph nd tle of vlues eh time we wish to find the grdient of tngent. Insted we n use n lgeri nd geometri pproh whih involves limits. LIMIT ARGUMENT From the investigtion, the grdient of [AB] = ( + )( ) ) grdient of [AB] = = + provided tht 6=

11 A = B B B B tngent t A INTRODUCTION TO CALCULUS (Chpter 6) 7 In the limit s B pprohes A,! nd the grdient of [AB]! the grdient of the tngent t A. So, the grdient of the tngent t the point A is m T = lim! = lim! +, 6= = As B pprohes A, the grdient of [ AB] pprohes or onverges to. Limit rguments like tht ove form the foundtion of differentil lulus. D CALCULATION OF AREAS UNDER CURVES Consider the funtion f() = +. We wish to estimte the re A enlosed = f(), the -is, nd the vertil lines = nd =. Suppose we divide the -intervl into three strips of width unit s shown. 5 5 f ( ) A The digrm longside shows upper retngles, whih re retngles with top edges t the mimum vlue of the urve on tht intervl. 5 f ( ) The re of the upper retngles, A U = f() + f() + f() =5++7 =units The net digrm shows lower retngles, whih re retngles with top edges t the minimum vlue of the urve on tht intervl. The re of the lower retngles, A L = f() + f() + f() =+5+ =7units 5 5 f ( ) 7 5 Now lerl A L <A<A U, so the required re lies etween 7 units nd units. If the intervl 6 6 ws divided into 6 equl intervls, eh of length, then

12 7 INTRODUCTION TO CALCULUS (Chpter 6) A U = f( )+ f() + f( )+ f() + f( )+ f() = ( =7:875 units ) nd A L = f() + f( )+ f() + f( )+ f() + f( ) = 9 ( ) =:75 units From this refinement we onlude tht the required re lies etween :75 nd 7:875 units. As we rete more sudivisions, the estimtes A L nd A U will eome more nd more urte. In ft, s the sudivision width is redued further nd further, oth A L nd A U will onverge to A. EXERCISE 6D. Consider the re etween = nd the -is from = to =. Divide the intervl into 5 strips of equl width, then estimte the re using: i upper retngles ii lower retngles. Clulte the tul re nd ompre it with our nswers in. Consider the re etween = nd the -is from = to =. Divide the intervl into 6 strips of equl width, then estimte the re using: upper retngles lower retngles. USING TECHNOLOGY B sudividing the horizontl is into smll enough intervls, we n in theor find estimtes for the re under urve whih re s lose s we like to the tul vlue. We illustrte this proess estimting the re A etween the grph of = nd the -is for 6 6. This emple is of historil interest. Arhimedes (87 - BC) found the et re. In n rtile tht ontins propositions he developed the essentil theor of wht is now known s integrl lulus. Consider f() = nd divide the intervl 6 6 into equl sudivisions. () X () X (, ) (, ) A L = () + ( ) + ( ) + ( ) nd A U = ( ) + ( ) + ( ) + () ¼ :9 ¼ :69

13 INTRODUCTION TO CALCULUS (Chpter 6) 7 Now suppose there re n sudivisions, eh of width n. We n use tehnolog to help lulte A L nd A U for lrge vlues of n. Clik on the pproprite ion to ess our re finder softwre or instrutions for the proedure on grphis lultor. AREA FINDER TI-nspire TI-8 Csio The tle longside summrises the results ou should otin for n =,, 5 nd 5. The et vlue of A is in ft. Notie how oth A L nd A U re onverging to this vlue s n inreses. n A L A U Averge :8 75 :68 75 : 75 :85 :85 :5 5 : 6 :5 6 : 6 5 : : : EXERCISE 6D. Use retngles to find lower nd upper sums for the re etween the grph of = nd the -is for 6 6. Use n =, 5, 5, nd 5. Give our nswers to deiml ples. As n gets lrger, oth A L nd A U onverge to the sme numer whih is simple frtion. Wht is it? Use retngles to find lower nd upper sums for the res etween the grphs of eh of the following funtions nd the -is for 6 6. Use vlues of n =5,, 5,, 5, nd. Give our nswer to 5 deiml ples in eh se. i = ii = iii = iv = For eh se in, A L nd A U onverge to the sme numer whih is simple frtion. Wht frtions re the? On the sis of our nswer to, onjeture wht the re etween the grph of = nd the -is for 6 6 might e for n numer >. Consider the qurter irle of entre (, ) nd rdius units illustrted. Its re is (full irle of rdius ) = ¼ = ¼ units Estimte the re using lower nd upper retngles for n =, 5,,, nd. Hene, find rtionl ounds for ¼. Arhimedes found the fmous pproimtion 7 <¼< 7. For wht vlue of n is our estimte for ¼ etter thn tht of Arhimedes?

14 7 INTRODUCTION TO CALCULUS (Chpter 6) THE DEFINITE INTEGRAL Consider the lower nd upper retngle sums for funtion whih is positive nd inresing on the intervl 6 6. We divide the intervl into n sudivisions of width w = n. =ƒ( ) =ƒ( ) n n n n n n Sine the funtion is inresing, A L = wf( )+wf( )+wf( )+:::: + wf( n )+wf( n ) n X = w i= f( i ) nd A U = wf( )+wf( )+wf( )+:::: + wf( n )+wf( n ) nx = w f( i ) i= Notie tht A U A L = w (f( n ) f( )) = ( )(f() f()) n ) lim (A U A L )= fsine lim n! n! n =g ) lim A L = lim A U n! n! ) sine A L <A<A U, lim n! A L = A = lim n! A U This ft is true for ll positive ontinuous funtions on the intervl 6 6. The vlue A is known s the definite integrl of f() from to, written A = Z f() d. If f() > for ll 6 6, then Z f() d is the shded re. ƒ( )

15 INTRODUCTION TO CALCULUS (Chpter 6) 75 HISTORICAL NOTE The word integrtion mens to put together into whole. An integrl is the whole produed from integrtion, sine the res f( i ) w of the thin retngulr strips re put together into one whole re. Z The smol is lled n integrl sign. In the time of Newton nd Leiniz it ws the strethed out letter s, ut it is no longer prt of the lphet. Emple 5 Sketh the grph of = for 6 6. Shde the re desried Z d. Use tehnolog to lulte the lower nd upper retngle sums for n equl sudivisions where n = 5,, 5, nd 5. Use the informtion in to find Z d to signifint figures. Self Tutor AREA FINDER n A L A U : :5 :9 : :5 :. A d :95 :5. 5 :99 : When n = 5, A L ¼ A U ¼ :, to signifint figures. ) sine A L < Z d < A U, Z d ¼ : Emple 6 Use grphil evidene nd known re fts to find: Z ( +)d Z p d Self Tutor 5 (, ) Z ( +)d = shded re = +5 =6

16 76 INTRODUCTION TO CALCULUS (Chpter 6) If = p then = nd so + = whih is the eqution of the unit irle. = p is the upper hlf. =~``-`!`X Z p d = shded re = (¼r ) where r = = ¼ EXERCISE 6D. Sketh the grph of = p for 6 6. Z p Shde the re desried d. Find the lower nd upper retngle sums for n =5,, 5, nd 5. Use the informtion in to find Z p d to signifint figures. Sketh the grph of = p + nd the -is for 6 6. Find the lower nd upper retngle sums for n =5,, 5. Z p Wht is our est estimte for + d? Use grphil evidene nd known re fts to find: Z ( + ) d Z ( ) d Z p d INVESTIGATION ESTIMATING Z e d Wht to do: The integrl Z e d is of onsiderle interest to sttistiins. In this investigtion we shll estimte the vlue of this integrl using upper nd lower retngulr sums for n = 5. This vlue of n is too lrge for most lultors to hndle in single list, so we will perform it in setions. Sketh the grph of = e for 6 6 : TI-nspire Clulte the upper nd lower retngulr sums for the three intervls 6 6, 6 6 nd 6 6 using n = 75 for eh. TI-8 Csio

17 INTRODUCTION TO CALCULUS (Chpter 6) 77 Comine the upper retngulr sums nd the lower retngulr sums ou found in to otin n upper nd lower retngulr sum for 6 6 for n = 5. Use the ft tht the funtion = e retngulr sums for 6 6 for n = 5. 5 Use our results of nd to find n estimte for How urte is our estimte? 6 Compre our estimte in 5 with p ¼. is smmetri to find upper nd lower Z e d. AREA FINDER REVIEW SET 6 Evlute the limits: lim! 6 lim! d lim! lim! p Find n smptotes of the following funtions nd disuss the ehviour of the grph ner them: = e =ln( +) f() =e ln d = +ln( ) Sketh the region etween the urve = nd the -is for Divide the intervl into 5 equl prts nd displ the 5 upper nd lower retngles. Find the lower nd upper retngle sums for n =5, 5, nd 5. Z Give our est estimte for d nd ompre this nswer with ¼. + Consider the grph of f() =e. Sketh the grph of = f() for 6 6. Find upper nd lower ounds for Z e d using upper nd lower retngles with 8 sudivisions. Use tehnolog to find, orret to signifint figures, upper nd lower ounds for Z e d when n =. 5 The grph of = f() is illustrted: Evlute the following using re interprettion: Z f() d Z 6 f() d semiirle 6

18 78 INTRODUCTION TO CALCULUS (Chpter 6) 6 Use four upper nd lower retngles to find rtionls A nd B suh tht: () X A< Z ( ) d < B. Hene, find good estimte for Z ( ) d.

19 Chpter7 Differentil lulus Sllus referene: 7., 7., 7.7 Contents: A B C D E F G H The derivtive funtion Derivtives t given -vlue Simple rules of differentition The hin rule The produt rule The quotient rule Tngents nd normls The seond derivtive

20 8 DIFFERENTIAL CALCULUS (Chpter 7) In the previous hpter we disussed how the topi of lulus is divided into two fields: differentil lulus nd integrl lulus. In this hpter we egin to emine differentil lulus nd how it reltes to rte prolems nd the grdient of urves. HISTORICAL NOTE The topi of differentil lulus originted in the 7th entur with the work of Sir Is Newton nd Gottfried Wilhelm Leiniz. These mthemtiins developed the neessr theor while ttempting to find lgeri methods for solving prolems deling with: ² the grdients of tngents to urves t n point on the urve, nd ² finding the rte of hnge in one vrile with respet to nother. Is Newton 6 77 Gottfried Leiniz A THE DERIVATIVE FUNCTION For non-liner funtion with eqution = f(), the grdients of the tngents t vrious points re different. Our tsk is to determine grdient funtion so tht when we reple some vlue then we will e le to find the grdient of the tngent to = f() t =. ƒ( ) Consider generl funtion = f() where A is (, f()) nd B is ( + h, f( + h)). ƒ( ) B f( + h) f() ƒ( h) The hord [AB] hs grdient = + h ƒ( ) A f( + h) f() = : h If we now let B pproh A, then the grdient of [AB] pprohes the grdient of the tngent t A. So, the grdient of the tngent t the vrile point (, f()) is the limiting vlue of f( + h) f() h h h s h pprohes, or f( + h) f() lim. h! h

21 DIFFERENTIAL CALCULUS (Chpter 7) 8 This formul gives the grdient of the tngent for n vlue of the vrile. Sine there is onl one vlue of the grdient for eh vlue of, the formul is tull funtion. The grdient funtion, lso known s the derived funtion or derivtive funtion or simpl the derivtive is defined s f f( + h) f() () = lim. h! h We red the derivtive funtion s eff dshed. Emple Consider the grph longside. Find f() nd f (). ( ) (, ) Self Tutor The grph shows the tngent to the urve = f() t the point where =. The grdient of this tngent is f (). The tngent psses through (, ) nd (6, ), so f () = 6 =. The eqution of the tngent is = ) = When =, =, so the point of ontt is (, ) ) f()= So, f() = nd f () =. EXERCISE 7A. Using the grph longside, find: f() f () ( ) Using the grph longside, find: f() f () ( )

22 8 DIFFERENTIAL CALCULUS (Chpter 7) Consider the grph longside. Find f() nd f (). (, ) INVESTIGATION FINDING GRADIENTS OF FUNCTIONS The softwre on the CD n e used to find the grdient of the funtion f() t n point. B sliding the point long the grph we n oserve the hnging grdient of the tngent. We n hene generte the grdient funtion f (). Wht to do: Consider the funtions f() =, f() = nd f() =. For eh of these funtions, wht is the grdient? Is the grdient onstnt for ll vlues of? Consider the funtion f() = m +. Stte the grdient of the funtion. Is the grdient of the tngent onstnt for ll vlues of? Use the CD softwre to grph the following funtions nd oserve the grdient funtion f (). Hene verif tht our nswer in is orret. i f() = ii f() = + iii f() = + Oserve the funtion f() = using the CD softwre. Wht tpe of funtion is the grdient funtion f ()? Oserve the following qudrti funtions using the CD softwre: i f() = + ii f() = iii f() = + iv f() = +6 Wht tpes of funtions re the grdient funtions f () for the funtions in? 5 Oserve the funtion f() =ln using the CD softwre. Wht tpe of funtion is the grdient funtion f ()? Wht is the domin of the grdient funtion f ()? 6 Oserve the funtion f() =e using the CD softwre. Wht is the grdient funtion f ()? GRADIENT FUNCTIONS To find the grdient funtion f () for generl funtion f(), we need to evlute the f( + h) f() limit lim. We ll this the method of first priniples. h! h

23 DIFFERENTIAL CALCULUS (Chpter 7) 8 Emple Use the definition of f () to find the grdient funtion of f() =. f () = lim h! f( + h) f() h = lim h! ( + h) h = lim h! +h + h h = lim h! h( + h) h = lim h! ( + h) fs h 6= g = Self Tutor EXERCISE 7A. Find, from first priniples, the grdient funtion of f() where f() is: 5 d Reminder: ( + ) = ( + ) = Use our nswer to to predit formul for f () where f() = n, n N. Find, from first priniples, f () given tht f() is: +5 + d + e +5 f + B DERIVATIVES AT A GIVEN -VALUE Suppose we re given funtion f() nd sked to find its derivtive t the point where =. This is tull the grdient of the tngent to the urve t =, whih we write s f (). There re two methods for finding f () using first priniples: The first method is to strt with the definition of the grdient funtion. Sine f f( + h) f() () = lim, h! h we n simpl sustitute = to give f f( + h) f() () = lim. h! h ƒ( ) tngent t point of ontt

24 8 DIFFERENTIAL CALCULUS (Chpter 7) The seond method is to onsider two points on the grph of = f(): fied point A(, f()) nd vrile point B(, f()). ƒ( ) ƒ( ) A B ƒ( ) (, ƒ( )) tngent t A with grdient ƒ'( ) The grdient of hord [AB] = f() f(). In the limit s B pprohes A,! nd the grdient of hord [AB]! the grdient of the tngent t A. ) f f() f() () = lim.! Thus f () = lim! f() f() is n lterntive definition for the grdient of the tngent t =. You n lso find the grdient of the tngent t given point on funtion using our grphis lultor. Instrutions for doing this n e found in the grphis lultor hpter t the strt of the ook. Emple Self Tutor Find, from first priniples, the grdient of the tngent to = + t the point where =. Let f() = + ) f() = () += nd f f() f() () = lim! ) f + () = lim! 8 = lim! ( + )( ) = lim! = =8 fs 6= g Emple Self Tutor Use the first priniples formul f f( + h) f() () = lim h! h to find the instntneous rte of hnge in f() = + t the point where =5. So, f(5)=5 + (5) = 5 f (5) = lim h! f(5 + h) f(5) h = lim h! (5 + h) + (5 + h) 5 h

25 DIFFERENTIAL CALCULUS (Chpter 7) 85 = lim h! 5 + h + h ++h 5 h = lim h! h +h h h(h + ) = lim h! h = fs h 6= g ) the instntneous rte of hnge in f() t =5 is. EXERCISE 7B Use the formul f () = lim! f() f() to find the grdient of the tngent to: f() = +5 t = f() =5 t = f() = + t = d f() =5 t = Use the first priniples formul f () = lim h! f( + h) f() h to find: f () for f() = f () for f() =. C SIMPLE RULES OF DIFFERENTIATION Differentition is the proess of finding derivtive or grdient funtion. There re numer of rules ssoited with differentition. These rules n e used to differentite more omplited funtions without hving to resort to the tedious method of first priniples. INVESTIGATION SIMPLE RULES OF DIFFERENTIATION In this investigtion we ttempt to differentite funtions of the form n, n where is onstnt, nd funtions whih re sum or differene of terms of the form n. Wht to do: Differentite from first priniples: Consider the inomil epnsion: ( + h) n = n n + n n h + n n h + :::: + n n h n = n + n n h + n n h + :::: + h n

26 86 DIFFERENTIAL CALCULUS (Chpter 7) Use the first priniples formul f () = f( + h) f() lim h! h to find the derivtive of f() = n. Find, from first priniples, the derivtives of: d Use to op nd omplete: If f() = n, then f () =:::::: 5 Use first priniples to find f () for: f() = + f() = 6 Use nd 5 to op nd omplete: If f() =u()+v() then f () =:::::: f() f () Nme of rule ( onstnt) differentiting onstnt n n n differentiting n u() u () onstnt times funtion u()+v() u ()+v () ddition rule Eh of these rules n e proved using the first limit definition of f (). For emple: ² If f() =u() where is onstnt then f () =u (). Proof: f () = lim h! f( + h) f() h u( + h) u() = lim h! h u( + h) u() = lim h! h = lim h! u( + h) u() h = u () ² If f() =u()+v() then f () =u ()+v () Proof: f f( + h) f() () = lim h! h µ u( + h)+v( + h) [u()+v()] = lim h! h µ u( + h) u()+v( + h) v() = lim h! h u( + h) u() v( + h) v() = lim + lim h! h h! h = u ()+v ()

27 DIFFERENTIAL CALCULUS (Chpter 7) 87 Using the rules we hve now developed we n differentite sums of powers of. For emple, if f() = then f () = ( ) + ( ) 5() + 7() + = Emple 5 Self Tutor Find f () for f() equl to: f() = ) f () = 5( ) + 6() () =5 + f() =7 + =7 + ) f () = 7() ( )+( ) =7+ 9 Rememer tht n = n. =7+ 9 Emple 6 Find the grdient funtion of f() = to the funtion t the point where =. Self Tutor nd hene find the grdient of the tngent f() = = ) f () = ( ) = + = + Now f ()=+=5. So, the tngent hs grdient =5. Emple 7 Find the grdient funtion for eh of the following: f() = p + g() = p Self Tutor f() = p + = + ) f () =( )+( ) = = p g() = p = ) g () = ( ) = + = + p

28 88 DIFFERENTIAL CALCULUS (Chpter 7) ALTERNATIVE NOTATION If we re given funtion f() then f () represents the derivtive funtion. If we re given in terms of then or derivtive. d d d d reds dee dee or the derivtive of with respet to. d d is not frtion. However, the nottion d d is result of tking the limit of frtion. If we reple h ± nd f( + h) f() ±, then f f( + h) f() () = lim h! h f () = lim ±! = d d. ± ± eomes re ommonl used to represent the ( ) u u Emple 8 If =, find d d nd interpret its mening. Self Tutor As = d, =6. d d d is: ² the grdient funtion or derivtive of = from whih the grdient t n point n e found ² the instntneous rte of hnge in s hnges. EXERCISE 7C Find f () given tht f() is: 7 d 6 p e p f + g h + 5 i 6 j 6 k +5 l m + n p o ( ) p ( +) Find d d for: =:5 : : = ¼ = 5 d = e = ( +) f =¼

29 Differentite with respet to : DIFFERENTIAL CALCULUS (Chpter 7) p (5 ) d 6 9 e ( + )( ) f Find the grdient of the tngent to: +6p g = t = = 8 t the point = +7 t = d = 5 e = t the point, Chek our nswers using tehnolog. h ( + )( 5) t the point 9, 8 8 f = 8 t = 5 Suppose f() = +( +) +, f() =, nd f ( ) =. Find the onstnts nd. 6 Find the grdient funtion of f() where f() is:, p + p p d p e p 5 f p g 7 If = d, find d 5 p nd interpret its mening. h p The position of r moving long stright rod is given S =t +t metres where t is the time in seonds. Find ds nd interpret its mening. dt The ost of produing tosters eh week is given C = : dollrs. Find dc nd interpret its mening. d D THE CHAIN RULE In Chpter we defined the omposite of two funtions g nd f s (g ± f)() or g(f()). We n often write omplited funtions s the omposite of two or more simpler funtions. For emple =( +) ould e rewritten s = u where u = +, or s =(g ± f)() =g(f()) where g() = nd f() = +. Emple 9 Self Tutor Find: g(f()) if g() = p nd f() = g() nd f() suh tht g(f()) =.

30 9 DIFFERENTIAL CALCULUS (Chpter 7) g(f()) = g( ) = p g(f()) = = f() ) g() = nd f() = EXERCISE 7D. Find g(f()) if: g() = nd f() = +7 g() = +7 nd f() = g() = p nd f() = d g() = nd f() = p e g() = nd f() = + f g() = + nd f() = Find g() nd f() suh tht g(f()) is: ( + ) + p d ( ) DERIVATIVES OF COMPOSITE FUNCTIONS The reson we re interested in writing omplited funtions s omposite funtions is to mke finding derivtives esier. INVESTIGATION DIFFERENTIATING COMPOSITES The purpose of this investigtion is to gin insight into how we n differentite omposite funtions. Bsed on our previous rule if = n then d d = nn, we might suspet tht if =( +) then d d = ( +) = ( +). But is this so? Wht to do: Consider =(+). Epnd the rkets nd then find d d.isd d = (+)? Consider =(+). Epnd the rkets nd then find d d.is d d = (+)? Consider =( +). Epnd the rkets nd find d d. Is d If = u where u is funtion of, wht do ou suspet 5 Consider =( +). Epnd it nd find d d. Does our nswer gree with the rule ou suggested in? d d d =( +)? will e equl to?

31 In the previous investigtion ou prol found tht if = u DIFFERENTIAL CALCULUS (Chpter 7) 9 d du =u d d = d du du d. then Now onsider =( +) whih hs the form = u where u = +. Epnding we hve =( +) ) =() + () + () + = d d = + +6 = 6( + +) = 6( +) = ( +) =u du d du whih is gin d du d : From the ove emples we formulte the hin rule: finomil epnsiong If = g(u) where u = f() then d d = d du du d. This rule is etremel importnt nd enles us to differentite omplited funtions muh fster. For emple, we n redil see tht for n funtion f(): If =[f()] n then d d = n[f()]n f (). A non-eminle proof of the hin rule is inluded for ompleteness. Proof: Consider = g(u) where u = f(). For smll hnge of ± in u, there is smll hnge u( ) of f( +±) f() =±u uu in u nd smll hnge of u ± in. u g() u u u uu u Now ± ± = ± ±u ±u ± ffrtion multiplitiong As ±!, ±u! lso. ) lim ±! ) ± ± = lim ± ±u! d d = d du du d ±u lim ±! ±u ± flimit ruleg

32 9 DIFFERENTIAL CALCULUS (Chpter 7) Emple Find d d if: =( ) = p Self Tutor =( ) ) = u where u = Now d d = d du fhin ruleg du d = =u ( ) =( ) ( ) p The rkets round re essentil. Now ) =u where u = d d = d du du d fhin ruleg = ( u ) ( ) =u = ( ) EXERCISE 7D. Write in the form u n, lerl stting wht u is: ( ) p d p e Find the grdient funtion d d for: =( 5) = 5 ( ) f p = p d =( ) e = 6(5 ) f = p µ 6 g = (5 ) h = i = Find the grdient of the tngent to: = p t = =( +) 6 t = = e = ( ) t = d =6 p t = + p t = f = Chek our nswers using tehnolog. µ + t =

33 DIFFERENTIAL CALCULUS (Chpter 7) 9 If = then =. Find d d d nd d d Eplin wh d d d = funtion = f(). nd hene show tht d d d d =. whenever these derivtives eist for n generl E THE PRODUCT RULE If f() =u()+v() then f () =u ()+v (): So, the derivtive of sum of two funtions is the sum of the derivtives. We now onsider the se f() =u()v(). Is f () =u ()v ()? In other words, does the derivtive of produt of two funtions equl the produt of the derivtives of the two funtions? The following emple shows tht this nnot e true: If f() = p we ould s f() =u()v() where u() = nd v() = p. Now f() = so f () =. But u ()v () = = 6= f (). THE PRODUCT RULE If u() nd v() re two funtions of nd = uv then d d = du d v + u dv or = u ()v() +u()v (). d Consider gin the emple f() = p. This is produt u()v() where u() = nd v() = ) u () = nd v () = Aording to the produt rule f () =u v + uv = + For ompleteness we prove the produt rule: = + = whih is orret X Proof: Let = u()v(). Suppose there is smll hnge of ± in whih uses orresponding hnges of ±u in u, ±v in v, nd ± in. As = uv, + ± =(u + ±u)(v + ±v) ) + ± = uv +(±u)v + u(±v)+±u±v ) ± =(±u)v + u(±v)+±u±v

34 9 DIFFERENTIAL CALCULUS (Chpter 7) ) ± ± = ± ) lim ±! ± = ) µ ±u ± µ lim ±! v + u d d = du d v + u dv d µ ±v + ± µ ±u v + u ± µ ±u ±v fdividing eh term ±g ± lim ±! ±v ± + fs ±!, ±v! lsog Emple Find d d if: = p ( +) = ( ) Self Tutor = p ( +) is the produt of u = nd v =( +) ) u = nd v = ( +) fhin ruleg Now d d = u v + uv = 6( +) fprodut ruleg = ( +) + 6( +) = ( +) +6 ( +) = ( ) is the produt of u = nd v =( ) ) u = nd v =( ) ( ) fhin ruleg Now d d = u v + uv fprodut ruleg =( ) + ( ) ( ) =( ) + ( ) ( ) EXERCISE 7E Find d d using the produt rule: = ( ) =( +) = p d = p ( ) e =5 ( ) f = p ( ) Find the grdient of the tngent to: = ( ) t = = p ( +) t = = p t = d = p 5 t = If = p ( ) show tht d d ( )( 5) = p. Find the -oordintes of ll points on = p ( ) where the tngent is horizontl.

35 F Epressions like + 5, Quotient funtions hve the form Q() = u() v(). Notie tht nd the produt rule So, if DIFFERENTIAL CALCULUS (Chpter 7) 95 p nd ( ) re lled quotients. u() =Q() v() u () =Q () v()+q() v () ) u () Q() v () =Q () v() or if ) Q () v() =u () u() v() v () ) Q () v() = u () v() u() v () v() Q() = u() v() = u v ) Q () = u () v() u() v () [v()] then nd this formul is lled the quotient rule. Q () = u ()v() u()v () [v()] where u nd v re funtions of then THE QUOTIENT RULE d d = u v uv. v Emple Use the quotient rule to find d d if: = + + = Self Tutor p ( ) = + + is quotient with u =+ nd v = + ) u = nd v = Now d d = u v uv v fquotient ruleg = ( +) ( + ) ( +) = + 6 ( +) = ( +) p = ( ) is quotient where u = nd v =( ) ) u = nd v = ( ) ( ) fhin ruleg = ( )

36 96 DIFFERENTIAL CALCULUS (Chpter 7) Now d d = u v uv v = = = = = fquotient ruleg ( ) ( ( )) ( ) ( ) + ( ) ( ) µ p ( ) p +p p flook for ommon ftorsg ( ) +8 p ( ) 6 + p ( ) Note: d Simplifition of s in the ove emple is often unneessr, espeill if d ou wnt to find the grdient of tngent t given point. In suh ses ou n sustitute vlue for without simplifing. EXERCISE 7F Use the quotient rule to find d d = + d = p Find the grdient of the tngent to: = = if: = + e = f = t = = + p + t = d = p +5 If = p, show tht d d = + p ( ) : For wht vlues of is d d If = +, show tht + d For wht vlues of is d i zero ii undefined? d d = + 7 ( +) : i zero ii undefined? Wht is the grphil signifine of our nswers in? = p t = t =

37 G DIFFERENTIAL CALCULUS (Chpter 7) 97 TANGENTS AND NORMALS ƒ( ) tngent grdient m T Consider urve = f(). If A is the point with -oordinte, then the grdient of the tngent t this point is f () =m T. point of ontt A (,ƒ ( )) The eqution of the tngent is norml or f() = f () fequting grdientsg f() =f ()( ): Alterntivel, the eqution of the tngent t the point A(, ) is = f () or = f ()( ). A norml to urve is line whih is perpendiulr to the tngent t the point of ontt. The grdients of perpendiulr lines re negtive reiprols of eh other. Thus, the grdient of the norml t = is m N = f (). For emple, if f() = then f () =. At =, m T = f () = nd m N f () =. So, t = the tngent hs grdient nd the norml hs grdient. Sine f() =, the tngent hs eqution =( ) or = nd the norml hs eqution = ( ) or = + 9. Reminder: If line hs grdient 5 s, nd psses through (, ) s, nother quik w to write down its eqution is 5 = () 5( ), or 5 =. If the grdient ws 5, we would hve: +5 = () + 5( ), or +5 = 7. You n lso find the equtions of tngents t given point using our grphis lultor. Instrutions for doing this n e found t the strt of the ook.

38 98 DIFFERENTIAL CALCULUS (Chpter 7) Emple Self Tutor Find the eqution of the tngent to f() = + t the point where =. ƒ( ) X Sine f()=+=, the point of ontt is (, ). Now f () =, so m T = f ()= (, ) ) the tngent hs eqution = whih is = or =. Emple Self Tutor Find the eqution of the norml to = 8 p t the point where =. When =, = 8 p = 8 =. So, the point of ontt is (, ). tngent grdient m T norml grdient m N Now s =8, d d = ) when =, m T = = (, ) ) the norml t (, ) hs grdient m N =. ) the eqution of the norml is = () () or =: Emple 5 Self Tutor Find the equtions of n horizontl tngents to = +. Sine = +, d d = Horizontl tngents hve grdient, so = ) ( ) = ) ( + )( ) = ) = or When =, =8 + = When =, = = 8 ) the points of ontt re (, ) nd (, 8) ) the tngents re = nd =8.

39 DIFFERENTIAL CALCULUS (Chpter 7) 99 Emple 6 Self Tutor Find the eqution of the tngent to = p t the point where =. Let f() = ( ) ) f () = ( ) ( ) ) m T = f () = () ( ) = So, the tngent hs eqution = When =, = p 9= ) the point of ontt is (, ). whih is = +9 or + = EXERCISE 7G Find the eqution of the tngent to: = + t = = p + t = = 5 t = d = p t (, ) e = t (, ) f = t =. Chek our nswers using tehnolog. Find the eqution of the norml to: = t the point (, 9) = 5 + t = = 5 p p t the point (, ) d =8 p t =: Find the equtions of the horizontl tngents to = + +. Find the points of ontt where horizontl tngents meet the urve = p + p. Find k if the tngent to = +k t the point where =hs grdient. d Find the eqution of the tngent to = + 8 whih is prllel to the tngent t (, ): The tngent to the urve = ++ where nd re onstnts, is + =6 t the point where =. Find the vlues of nd. The norml to the urve = p + p where nd re onstnts, hs eqution + = t the point where =. Find the vlues of nd. Show tht the eqution of the tngent to = t the point where =, is = +. 5 Find the eqution of the tngent to: = p + t = = f() = t (, ) d f() = t = t (, ).

40 5 DIFFERENTIAL CALCULUS (Chpter 7) 6 Find the eqution of the norml to: = ( +) t (, ) = p t = f() = p ( ) t = d f() = + t =. 7 = p where nd re onstnts, hs tngent with eqution + =5 t the point where =. Find nd. Emple 7 Self Tutor Find the oordintes of the point(s) where the tngent to = + + t (, ) meets the urve gin. Let f() = + + ) f () = + ) m T = f ()=+= ) the tngent t (, ) hs grdient nd its eqution is = () or =: Now = meets = + + where + += ) += Sine the tngent touhes the urve when =, = must e repeted zero of +. ) ( ) ( +)= f inspetiong (, ) = ( ) = ) =or When =, =( ) +( ) + = 8 ) the tngent meets the urve gin t (, 8). (, ) Emple 8 Self Tutor Find the equtions of the tngents to = from the eternl point (, ). X (, X) (, ) Let (, ) lie on f() =. Now f () =, so f () = ) t (, ) the grdient of the tngent is ) its eqution is = or = or =

41 DIFFERENTIAL CALCULUS (Chpter 7) 5 But this tngent psses through (, ). ) =() ) += ) ( )( ) = ) =or If =, the tngent hs eqution = with point of ontt (, ). If =, the tngent hs eqution =6 9 with point of ontt (, 9). 8 Find where the tngent to the urve = t the point where =, meets the urve gin. Find where the tngent to the urve = + + t the point where =, meets the urve gin. Find where the tngent to the urve = + the urve gin. t the point where =, meets 9 Find the eqution of the tngent to = +9 t the point where =. Hene, find the equtions of the two tngents from (, ) to the urve. Stte the oordintes of the points of ontt. Find the equtions of the tngents to = from the eternl point (, ). Find the eqution(s) of the norml(s) to = p from the eternl point (, ). Consider f() = 8. Sketh the grph of the funtion. Find the eqution of the tngent t the point where =. If the tngent in uts the -is t A nd the -is t B, find the oordintes of A nd B. d Find the re of tringle OAB nd disuss the re of the tringle s!. H THE SECOND DERIVATIVE Given funtion f(), the derivtive f () is known s the first derivtive. The seond derivtive of f() is the derivtive of f (), or the derivtive of the first derivtive. We use f () or or d d to represent the seond derivtive. f () reds f doule dshed of. d d = d µ d reds dee two dee squred. d d

42 5 DIFFERENTIAL CALCULUS (Chpter 7) THE SECOND DERIVATIVE IN CONTEXT Mihel rides up hill nd down the other side to his friend s house. The dots on the grph show Mihel s position t vrious times t. t Mihel s ple t t t t t friend s house DEMO The distne Mihel hs trvelled t vrious times is given in the following tle: Time (t min) :5 5 7:5 : Distne trvelled (s m) The model s ¼ :8t :7t + 8:5t 6:8 metres fits this dt well, lthough the model gives s() ¼ 6:8 m wheres the tul dt gives s() =. This sort of prolem often ours when modelling from dt. A grph of the dt points nd the model is given elow: Now ds dt ¼ :5t 6:9t + 8:5 metres per minute is the instntneous rte of hnge in displement per unit of time, or instntneous veloit. The instntneous rte of hnge in veloit t n point in time is Mihel s elertion, so d dt µ ds = d s dt dt d s In this se ¼ 7:8t 6:9 dt We see tht when t =, s ¼ 5 m, ds dt is the instntneous elertion. ¼ 6 metres per minute, metres per minute per minute. nd d s ¼ metres per minute per minute. dt We will emine displement, veloit nd elertion in greter detil in the net hpter.

43 DIFFERENTIAL CALCULUS (Chpter 7) 5 Emple 9 Find f () given tht f() = : Self Tutor Now f() = ) f () = + ) f () =6 6 =6 6 EXERCISE 7H Find f () given tht: f() = 6 + f() = +5 f() = p d f() = e f() =( ) f f() = + Find d d given tht: = = 5 = p d = Find when f () = for: e =( ) f = + f() = f() = + Consider the funtion f() =. Complete the following tle inditing whether f(), f () nd f () re positive (+), negtive ( ), or zero () t the given vlues of. f() f () f () REVIEW SET 7A NON-CALCULATOR If f() =7+, find: f() f () f (). Find the eqution of the tngent to = t the point where =. Find d d for: = = Find, from first priniples, the derivtive of f() = +. 5 Find the eqution of the norml to = t the point where =.

44 5 DIFFERENTIAL CALCULUS (Chpter 7) 6 The tngent to = + p t = is =. Find nd. 7 Determine the derivtive with respet to of: p +5 f() =( +) g() = 8 Find f () for: f() = f() = p 9 The urve f() = + + hs tngent with grdient t the point (, ). Find the vlues of nd. Find if f () = nd f() = The line through A(, ) nd B(, 8) is tngent to = Find the oordintes of P nd Q.. Find. ( +) P (, ) 5 Q REVIEW SET 7B Differentite with respet to : CALCULATOR 5 ( + ) ( + )( ) Determine the eqution of n horizontl tngents to the urve with eqution = 9 +. The tngent to = p t = uts the es t A nd B. Determine the re of tringle OAB. Find where the tngent to = + t (, 5) uts the urve gin. 5 Find given tht the tngent to = ( +) t = psses through (, ). 6 Find ll points on the urve = + + where the grdient of the tngent is. 7 If = p 5, find: d d d d 8 Find the eqution of the norml to = + t the point where =. 9 Show tht = 7 hs no horizontl tngents. +

45 DIFFERENTIAL CALCULUS (Chpter 7) 55 Find the eqution of the qudrti g() in the form g() = + +. (, ) g() Sketh the grph of 7! for >. Find the eqution of the tngent to the funtion t the point where = k, k>. If the tngent in uts the -is t A nd the -is t B, find the oordintes of A nd B. d Wht n e dedued out the re of tringle OAB? e Find k if the norml to the urve t = k psses through the point (, ). REVIEW SET 7C Differentite with respet to : = p = p + Find the eqution of the norml to = p t the point where =. Use the rules of differentition to find d d = p = for: µ = p Use the grph longside to find f() nd f (). ( ) Differentite with respet to : f() = ( +) p f() = p + 6 = is tngent to the urve = + + t =. Find nd. 7 The tngent to = + + t = is prllel to the line =. Find the vlue of nd the eqution of the tngent t =. Where does the tngent ut the urve gin?

46 56 DIFFERENTIAL CALCULUS (Chpter 7) 8 If the norml to f() = + length of [BC]. 9 Find d d for: = t (, ) uts the es t B nd C, determine the = + p The urve f() = + + hs tngent with grdient t the point (, ). Find nd nd hene f ( ). Show tht the urves whose equtions re = p + nd = p 5 hve the sme grdient t their point of intersetion. Find the eqution of the ommon tngent t this point.

47 Chpter8 Applitions of differentil lulus Sllus referene: 7., 7.6, 7.7 Contents: A B C D E F G Time rte of hnge Generl rtes of hnge Motion in stright line Some urve properties Rtionl funtions Infletions nd shpe Optimistion

48 58 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chpter 8) We sw in the previous hpter tht one pplition of differentil lulus is in finding the equtions of tngents nd normls to urves. There re mn other uses, however, inluding the following whih we will onsider in this ourse: ² funtions of time ² rtes of hnge ² motion in stright line (displement, veloit nd elertion) ² urve properties (monotoniit nd onvit) ² optimistion (mim nd minim) ² pplitions in eonomis A TIME RATE OF CHANGE There re ountless quntities in the rel world tht vr with time. For emple: ² temperture vries ontinuousl ² the height of tree vries s it grows ² the pries of stoks nd shres vr with eh d s trding. Vring quntities n e modelled using funtions of time. For emple, we ould use: ² s(t) to model the distne trvelled runner ² H(t) to model the height of person riding in Ferris wheel ² C(t) to model the pit of person s lungs, whih hnges when the person rethes. We sw in the previous hpter tht if = f() then f () or d is the grdient of the tngent t n vlue d of, nd lso the rte of hnge in with respet to. ƒ( ) We n likewise find the derivtive of funtion of time to tell us the rte t whih something is hppening. For the emples ove: ² ds or s (t) is the instntneous speed of the runner. dt It might hve units metres per seond or m s. ² dh or H (t) is the instntneous rte of sent of the person in the Ferris wheel. dt It might lso hve units metres per seond or m s. ² dc or C (t) is the person s instntneous rte of hnge in lung pit. dt It might hve units litres per seond or L s.

49 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chpter 8) 59 EXERCISE 8A The estimted future profits of smll usiness re given P (t) =t t + 8 thousnd dollrs, where t is the time in ers from now. Wht is the urrent nnul profit? Find dp nd stte its units. dt dp Wht is the signifine of dt? d For wht vlues of t will the profit: i derese ii inrese on the previous er? e Wht is the minimum profit nd when does it our? f Find dp when t =, nd 5. Wht do these figures represent? dt Wter is drining from swimming pool suh tht the remining volume of wter fter t minutes is V = (5 t) m. Find: the verge rte t whih the wter leves the pool in the first 5 minutes the instntneous rte t whih the wter is leving t t =5minutes. When ll is thrown, its height ove the ground is given s(t) =:+8:t :9t metres where t is the time in seonds. From wht distne ove the ground ws the ll relesed? Find s (t) nd stte wht it represents. Find t when s (t) =. Wht is the signifine of this result? d Wht is the mimum height rehed the ll? e Find the ll s speed: i when relesed ii t t =s iii t t =5s. Stte the signifine of the sign of the derivtive. f How long will it tke for the ll to hit the ground? g Wht is the signifine of d s dt? A shell is identll fired vertill from mortr t ground level nd rehes the ground gin fter : seonds. Its height ove the ground t n time t seonds is given s(t) =t :9t metres where is onstnt. Show tht the initil veloit of the shell is ms upwrds. Find the initil veloit of the shell. B GENERAL RATES OF CHANGE We hve seen previousl tht if s(t) is displement funtion then s ds (t) or is the dt instntneous rte of hnge in displement with respet to time, whih is lled veloit. In generl, d d gives the rte of hnge in with respet to.

50 5 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chpter 8) We n see tht: If inreses s inreses, then Emple If dereses s inreses, then d d d d will e positive. will e negtive. Aording to pshologist, the ilit of person to understnd sptil onepts is given A = p t where t is the ge in ers, 5 6 t 6 8. Find the rte of improvement in ilit to understnd sptil onepts when person is: i 9 ers old ii 6 ers old. Eplin wh da >, 5 6 t 6 8. Comment on the signifine of this result. dt Eplin wh d A dt <, 5 6 t 6 8. Comment on the signifine of this result. A = p t = t ) da dt = 6 t = 6 p t da i When t =9, dt = 8 ) the rte of improvement is 8 units per er for 9 er old. da ii When t =6, dt = ) the rte of improvement is units per er for 6 er old. As p t is never negtive, 6 p is never negtive t ) da > for ll 5 6 t 6 8. dt This mens tht the ilit to understnd sptil onepts inreses with ge. da dt = 6 t so d A dt = t = t p t ) d A < for ll 5 6 t 6 8. dt This mens tht while the ilit to understnd sptil onepts inreses with time, the rte of inrese slows down with ge. Emple The ost of produing items in ftor eh d is given Self Tutor Self Tutor C() =: {z +: } ost of lour rw mteril osts fied or overhed osts suh s heting, ooling, mintenne, rent Find C (), whih is lled the mrginl ost funtion. Find the mrginl ost when 5 items re produed. Interpret this result. Find C(5) C(5). Compre this with the nswer in.

51 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chpter 8) 5 The mrginl ost funtion is C () =: 9 +: +5 C (5) = $:8 This is the rte t whih the osts re inresing with respet to the prodution level when 5 items re mde per d. It gives n estimte of the ost for mking the 5st item. C(5) C(5) ¼ $8:9 $:75 ¼ $: tngent ( nswer) hord ( nswer) 5 5 C( ) C( ) This is the tul ost of mking the 5st item eh week, so the nswer in gives good estimte. EXERCISE 8B You re enourged to use tehnolog to grph the funtion for eh question. This is often useful in interpreting results. The quntit of hemil in humn skin whih is responsile for its elstiit is given Q = p t where t is the ge of person in ers. Find Q t: i t = ii t =5 iii t = ers. At wht rte is the quntit of the hemil hnging t the ge of: i 5 ers ii 5 ers? Show tht the rte t whih the skin loses the hemil is deresing for ll t>. The height of pinus rdit, grown in idel onditions, is given H = 97:5 t +5 metres, where t is the numer of ers fter the tree ws plnted from n estlished seedling. How high ws the tree t the time of its plnting? Find the height of the tree t t =, t =8 nd t =ers. Find the rte t whih the tree is growing t t =, 5 nd ers. d Show tht dh dt > for ll t >. istokphoto/niol Strtford Wht is the signifine of this result? The totl ost of running trin from Pris to Mrseille is given C(v) = 5 v + euros where v is the verge speed of the trin in km h. v Find the totl ost of the journe if the verge speed is: i 5 km h ii km h. Find the rte of hnge in the ost of running the trin t speeds of: i km h ii 9 km h. At wht speed will the ost e minimum?

52 5 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chpter 8) Alongside is lnd nd se profile where se hill lke the -is is se level. The funtion = ( )( ) km gives the height of the lnd or se ed reltive to se level. Find where the lke is loted reltive to the shore line of the se. Find d d nd interpret its vlue when = nd when = km. Find the deepest point of the lke nd the depth t this point. 5 A tnk ontins 5 litres of wter. The tp is left full on nd ll the wter drins from the tnk in 8 minutes. The volume of wter remining in the tnk fter t minutes µ is given V = 5 t where 6 t Find dv dv nd drw the grph of dt dt ginst t. At wht time ws the outflow fstest? Show tht d V dt is lws onstnt nd positive. Interpret this result. 6 A fish frm grows nd hrvests rrmundi in lrge dm. The popultion of fish fter t ers is given the funtion P (t). The rte of hnge in the popultion dp is modelled dp µ dt dt = P P P where, nd re known onstnts. is the irth rte of the rrmundi, is the mimum rring pit of the dm nd is the hrvest rte eh er. Eplin wh the fish popultion is stle when dp dt =. If the irth rte is 6%, the mimum rring pit is, nd 5% is hrvested eh er, find the stle popultion. If the hrvest rte hnges to %, wht will the stle popultion inrese to? 7 Selue mke denim jens. The ost model for mking pirs per d is C() =: +: dollrs. Find the mrginl ost funtion C (): Find C (). Wht does it estimte? Find C() C(). Wht does this represent? d Find C () nd the vlue of when C () =. Wht is the signifine of this point?

53 C APPLICATIONS OF DIFFERENTIAL CALCULUS (Chpter 8) MOTION IN A STRAIGHT LINE 5 DISPLACEMENT Suppose n ojet P moves long stright line so tht its position s from n origin O is given s some funtion of time t. We write s = s(t) where t >. s(t) is displement funtion nd for n vlue of t it gives the displement from O. s(t) is vetor quntit. Its mgnitude is the distne from O, nd its sign indites the diretion from O. It is ler tht: if s(t) >, P is loted to the right of O if s(t) =, P is loted t O if s(t) <, P is loted to the left of O. MOTION GRAPHS O origin st () P Consider s(t) =t +t m. s() = m, s() = m, s() = 5 m, s() = m, s()=m. To ppreite the motion of P we drw motion grph t t t t t DEMO Clik on the demo ion to get etter ide of the motion. Full nimted, we not onl get good ide of the position of P, ut lso of wht is hppening to its veloit nd elertion. VELOCITY AND ACCELERATION AVERAGE VELOCITY The verge veloit of n ojet moving in stright line in the time intervl from t = t to t = t is the rtio of the hnge in displement to the time tken. If s(t) is the displement funtion then verge veloit = s(t ) s(t ) t t. On grph of s(t), the verge veloit is the grdient of hord. INSTANTANEOUS VELOCITY ds In Chpter 7 we estlished tht dt = s (t) is the instntneous rte of hnge in displement per unit of time, or instntneous veloit.

54 5 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chpter 8) If s(t) is the displement funtion of n ojet moving in stright line, then v(t) =s s(t + h) s(t) (t) = lim is the instntneous veloit or h! h veloit funtion of the ojet t time t. On grph of s(t), the instntneous veloit is the grdient of tngent. AVERAGE ACCELERATION If n ojet moves in stright line with veloit funtion v(t) then its verge elertion on the time intervl from t = t to t = t is the rtio of the hnge in veloit to the time tken. verge elertion = v(t ) v(t ). t t INSTANTANEOUS ACCELERATION If prtile moves in stright line with veloit funtion v(t), instntneous elertion t time t is Emple then the (t) =v v(t + h) v(t) (t) = lim. h! h A prtile moves in stright line with displement from O given s(t) =t t metres t time t seonds. Find: the verge veloit in the time intervl from t = to t =5 seonds the verge veloit in the time intervl from t = to t =+h seonds s( + h) s() lim h! h nd omment on its signifine. Self Tutor verge veloit s(5) s() = 5 (5 5) (6 ) = = = ms verge veloit s( + h) s() = +h = ( + h) ( + h) h = 6+h h h h h h = h = h ms provided h 6= s( + h) s() lim = lim ( h) h! h h! = ms This is the instntneous veloit of the prtile t time t = seonds.

55 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chpter 8) 55 EXERCISE 8C. A prtile P moves in stright line with displement funtion of s(t) =t +t metres, where t >, t in seonds. Find the verge veloit from t = to t = seonds. Find the verge veloit from t = to t =+h seonds. Find the vlue of s( + h) s() lim h! h nd omment on its signifine. d Find the verge veloit from time t to time t + h seonds nd interpret s(t + h) s(t). h lim h! A prtile P moves in stright line with displement funtion of s(t) =5 t m, where t >, t in seonds. Find the verge veloit from t = to t =5 seonds. Find the verge veloit from t = to t =+h seonds. Find the vlue of s( + h) s() lim h! h nd stte the mening of this vlue. d s(t + h) s(t) Interpret lim. h! h A prtile moves in stright line with veloit funtion v(t) = p t+ m s, t >. Find the verge elertion from t = to t = seonds. Find the verge elertion from t = to t =+h seonds. Find the vlue of v( + h) v() lim. h! h Interpret this vlue. d Interpret v(t + h) v(t) lim. h! h An ojet moves in stright line with displement funtion s(t) nd veloit funtion v(t), t >. Stte the mening of: s( + h) s() lim h! h v( + h) v() lim h! h VELOCITY AND ACCELERATION FUNCTIONS If prtile P moves in stright line nd its position is given the displement funtion s(t), t >, then: ² the veloit of P t time t is given v(t) =s (t) fthe derivtive of the displement funtiong ² the elertion of P t time t is given (t) =v (t) =s (t) fthe derivtive of the veloit funtiong ² s(), v() nd () give us the position, veloit nd elertion of the prtile t time t =, nd these re lled the initil onditions.

56 56 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chpter 8) SIGN INTERPRETATION Suppose prtile P moves in stright line with displement funtion s(t) reltive to n origin O. Its veloit funtion is v(t) nd its elertion funtion is (t). We n use sign digrms to interpret: ² where the prtile is loted reltive to O ² the diretion of motion nd where hnge of diretion ours ² when the prtile s veloit is inresing or deresing. SIGNS OF s(t): s(t) Interprettion = PistO > P is loted to the right of O < P is loted to the left of O SIGNS OF v(t): v(t) Interprettion = P is instntneousl t rest > P is moving to the right < P is moving to the left v(t) = lim h! s(t + h) s(t) h If v(t) > then s(t + h) s(t) > ) s(t + h) >s(t) For h> the prtile is moving from s(t) to s(t + h). st () st ( + h) ) P is moving to the right. SIGNS OF (t): (t) Interprettion > veloit is inresing < veloit is deresing = veloit m e mimum or minimum A useful tle: Phrse used in question t s v initil onditions t the origin sttionr reverses mimum height onstnt veloit m. or min. veloit When prtile reverses diretion, its veloit must hnge sign. We need sign digrm of to determine if the point is mimum or minimum.

57 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chpter 8) 57 SPEED As we hve seen, veloities hve size (mgnitude) nd sign (diretion). In ontrst, the speed of n ojet is mesure of how fst it is trvelling, regrdless of the diretion of trvel. Speed is therefore slr quntit whih hs size ut no sign. Speed nnot e negtive. The speed t n instnt is the mgnitude of the ojet s veloit. If S(t) represents speed then we write S = jvj. To determine when the speed of n ojet is inresing or deresing, we need to emplo sign test. ² If the signs of v(t) nd (t) re the sme (oth positive or oth negtive), then the speed of P is inresing. ² If the signs of v(t) nd (t) re opposite, then the speed of P is deresing. We prove the first of these s follows: ½ v if v > Proof: Let S = jvj e the speed of P t n instnt, so S = v if v<. Cse : If v>, S = v nd ) ds dt = dv dt = (t) If (t) > then ds > whih implies tht S is inresing. dt Cse : If v<, S = v nd ) ds dt = dv dt = (t) If (t) < then ds > whih lso implies tht S is inresing. dt Thus if v(t) nd (t) hve the sme sign then the speed of P is inresing. INVESTIGATION DISPLACEMENT, VELOCITY AND ACCELERATION GRAPHS In this investigtion we emine the motion of projetile whih is fired in vertil diretion. The projetile is ffeted grvit, whih is responsile for the projetile s onstnt elertion. MOTION DEMO We then etend the investigtion to onsider other ses of motion in stright line. Wht to do: Clik on the ion to emine vertil projetile motion in stright line. Oserve first the displement long the line, then look t the veloit or rte of hnge in displement.

58 58 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chpter 8) Emine the three grphs: ² displement v time ² veloit v time ² elertion v time Comment on the shpes of these grphs. Pik from the menu or onstrut funtions of our own hoosing to investigte the reltionship etween displement, veloit nd elertion. You re enourged to use the motion demo ove to nswer questions in the following eerise. Emple Self Tutor A prtile moves in stright line with position reltive to some origin O given s(t) =t t +m, where t is the time in seonds, t >. Find epressions for the prtile s veloit nd elertion, nd drw sign digrms for eh of them. Find the initil onditions nd hene desrie the motion t this instnt. Desrie the motion of the prtile t t = seonds. d Find the position of the prtile when hnges in diretion our. e Drw motion digrm for the prtile. f For wht time intervl(s) is the prtile s speed inresing? g Wht is the totl distne trvelled in the time from t = to t =seonds? s(t) =t t +m ) v(t) =t fs v(t) =s (t)g =(t ) =(t + )(t ) m s whih hs sign digrm nd (t) =6t m s whih hs sign digrm fs (t) =v (t)g vt () t t When t =, s() = m v() = m s () = m s ) the prtile is m to the right of O, moving to the left t speed of m s. When t =, s()=8 6+=m v()= =9m s ()=m s ) the prtile is m to the right of O, moving to the right t speed of 9 m s. Sine nd v hve the sme sign, the speed is inresing. d Sine v(t) hnges sign when t =, hnge of diretion ours t this instnt. s() = +=, so the prtile hnges diretion when it is m to the left of O. t () t > ) the ritil vlue t = is not required.

59 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chpter 8) 59 e f g As t!, s(t)! nd v(t)!: Speed is inresing when v(t) nd (t) hve the sme sign. This is for t >. Totl distne trvelled =+=6m. The motion is tull on the line, not ove it s shown. Note: In lter hpters on integrl lulus we will see nother tehnique for finding the distnes trvelled nd displement over time. EXERCISE 8C. An ojet moves in stright line with position given s(t) =t t +m from n origin O, where t is in seonds, t >. Find epressions for the ojet s veloit nd elertion t n instnt nd drw sign digrms for eh funtion. Find the initil onditions nd eplin wht is hppening to the ojet t tht instnt. Desrie the motion of the ojet t time t = seonds. d At wht time(s) does the ojet reverse diretion? Find the position of the ojet t these instnts. e Drw motion digrm for the ojet. f For wht time intervls is the speed of the ojet deresing? A stone is projeted vertill so tht its position ove ground level fter t seonds is given s(t) =98t :9t metres, t >. Find the veloit nd elertion funtions for the stone nd drw sign digrms for eh funtion. Find the initil position nd veloit of the stone. Desrie the stone s motion t times t = 5 nd t = seonds. d Find the mimum height rehed the stone. e Find the time tken for the stone to hit the ground. A prtile moves in stright line with displement funtion s(t) =t t entimetres where t is in seonds, t >. Find veloit nd elertion funtions for the prtile s motion. Find the initil onditions nd interpret their mening. Find the times nd positions when the prtile reverses diretion. d At wht times is the prtile s: i speed inresing ii veloit inresing? The position of prtile moving long the -is is given (t) =t 9t +t metres where t is in seonds, t >. Drw sign digrms for the prtile s veloit nd elertion funtions. s() t m

60 5 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chpter 8) Find the position of the prtile t the times when it reverses diretion, nd hene drw motion digrm for the prtile. At wht times is the prtile s: i speed deresing ii veloit deresing? d Find the totl distne trvelled the prtile in the first 5 seonds of motion. 5 In n eperiment, n ojet ws fired vertill from the erth s surfe. From the results, two-dimensionl grph of the position s(t) metres ove the erth s surfe ws plotted, where t ws the time in seonds. It ws noted tht the grph ws proli. Assuming onstnt grvittionl elertion g nd n initil veloit of v(), show tht: v(t) =v() + gt s(t) =v() t + gt. Hint: Assume tht s(t) =t + t +. st () When finding the totl distne trvelled, lws look for diretion reversls first. t D SOME CURVE PROPERTIES In this setion we onsider some properties of urves whih n e estlished using derivtives. These inlude intervls in whih urves re inresing nd deresing, nd the sttionr points of funtions. INCREASING AND DECREASING INTERVALS The onepts of inresing nd deresing re losel linked to intervls of funtion s domin. Some emples of intervls nd their grphil representtions re: Algeri form Geometri form > > 6 < <

61 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chpter 8) 5 Suppose S is n intervl in the domin of f(), sof() is defined for ll in S. ² f() is inresing on S, f() <f() for ll, S suh tht <. ² f() is deresing on S, f() >f() for ll, S suh tht <. For = is deresing for 6 nd inresing for >. Importnt: People often get onfused out the point =. The wonder how the urve n e oth inresing nd deresing t the sme point when it is ler tht the tngent is horizontl. The nswer is tht inresing nd deresing re ssoited with intervls, not prtiulr vlues for. We must lerl stte tht = is deresing on the intervl 6 nd inresing on the intervl >. We n dedue when urve is inresing or deresing onsidering f () on the intervl in question. For most funtions tht we del with in this ourse: ² f() is inresing on S, f () > for ll in S ² f() is deresing on S, f () 6 for ll in S. MONOTONICITY Mn funtions re either inresing or deresing for ll R. re monotone inresing or monotone deresing. For emple: We s these funtions = is inresing for ll. = is deresing for ll. Notie tht: ² for n inresing funtion, n inrese in produes n inrese in ² for deresing funtion, n inrese in produes derese in. inrese in derese in inrese in inrese in

62 5 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chpter 8) Emple 5 Find intervls where f() is: inresing deresing. (, ) ƒ( ) Self Tutor (, ) f() is inresing for 6 nd for > sine f () > on these intervls. f() is deresing for 6 6. (, ) ƒ( ) (, ) Sign digrms for the derivtive re etremel useful for determining intervls where funtion is inresing or deresing. Consider the following emples: ² f() = f () = whih hs sign digrm deresing inresing ) f() = is deresing for 6 nd inresing for >. ² f() = f () = whih hs sign digrm ) f() = is inresing for 6 nd deresing for >. ² f() = f () = whih hs sign digrm ) f() is monotone inresing. ² f() = + f () = =( ) =( + )( ) DEMO whih hs sign digrm DEMO DEMO DEMO ) f() is inresing for 6 nd for >, nd deresing for 6 6. inresing deresing inresing for ll (never negtive) inresing deresing inresing

63 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chpter 8) 5 Emple 6 Find the intervls where the following funtions re inresing or deresing: f() = + +5 f() = 8 + f() = + +5 ) f () = +6 ) f () = ( ) whih hs sign digrm So, f() is deresing for 6 nd for >, nd inresing for 6 6. f() = 8 + ) f () = = ( ) whih hs sign digrm So, f() is deresing for 6, nd inresing for >. Self Tutor Rememer tht f() must e defined for ll on n intervl efore we n lssif the intervl s inresing or deresing. We must elude points where funtion is undefined, nd need to tke re with vertil smptotes. Emple 7 Consider f() = +. Show tht f ( ) () = ( +) ( ) nd drw its sign digrm. Hene, find intervls where = f() is inresing or deresing. Self Tutor f() = + f () = ( + ) ( )( +) ( + ) fquotient ruleg = + 6 [ 6] (( )( + )) = = +6 ( ) ( +) ( ) ( ) ( +) whih hs sign digrm

64 5 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chpter 8) f() is inresing for 6 < nd for <6. f() is deresing for < nd for <6 nd for >. EXERCISE 8D. Write down the intervls where the grphs re: i inresing ii deresing. ( ) d e f ( ) ( ) ( ) = Find intervls where f() is inresing or deresing: f() = f() = f() = + d f() = p e f() = p f f() = 6 g f() = + h f() = i f() = 6 + j f() = k f() = 6 + l f() = p m f() = n f() = Consider f() = + : Show tht f () = ( + )( ) ( +) nd drw its sign digrm. Hene, find intervls where = f() is inresing or deresing.

65 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chpter 8) 55 Consider f() = ( ). Show tht f () = ( +) ( ) nd drw its sign digrm. Hene, find intervls where = f() is inresing or deresing. 5 Consider f() = + 7. Show tht f ( + )( ) () = ( ) nd drw its sign digrm. Hene, find intervls where = f() is inresing or deresing. 6 Find intervls where f() is inresing or deresing if: f() = f() = + STATIONARY POINTS A sttionr point of funtion is point suh tht f () =. It ould e lol mimum, lol minimum, or horizontl infletion. TURNING POINTS (MAXIMA AND MINIMA) Consider the following grph whih hs restrited domin of B (, ) D (, ) ƒ( ) C (, ) A (, \Qw_\) Aisglol minimum s it is the minimum vlue of on the entire domin. Bislol mimum s it is turning point where the urve hs shpe nd f () = t tht point. Cislol minimum s it is turning point where the urve hs shpe nd f () = t tht point. Disglol mimum s it is the mimum vlue of on the entire domin. For mn funtions, lol mimum or minimum is lso the glol mimum or minimum. For emple, for = the point (, ) is lol minimum nd is lso the glol minimum. HORIZONTAL OR STATIONARY POINTS OF INFLECTION It is not lws true tht whenever we find vlue of where f () = we hve lol mimum or minimum.

66 56 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chpter 8) For emple, f() = hs f () = nd f () = when =: The -is is tngent to the urve whih tull rosses over the urve t O(, ). This tngent is horizontl ut O(, ) is neither lol mimum nor lol minimum. It is lled horizontl infletion (or infleion) s the urve hnges its urvture or shpe. SIGN DIAGRAMS Consider the grph longside. The sign digrm of its grdient funtion is shown diretl eneth it. We n use the sign digrm to desrie the sttionr points of the funtion. lol mimum horizontl infletion lol minimum lol lol mimum minimum horizontl infletion Sttionr point Sign digrm of f () ner = Shpe of urve ner = lol mimum = lol minimum = horizontl infletion or sttionr infletion or = or = Emple 8 Self Tutor Find nd lssif ll sttionr points of f() = 9 +5: f() = 9 +5 ) f () = 6 9 =( ) =( )( +) whih hs sign digrm: So, we hve lol mimum t = nd lol minimum t =.

67 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chpter 8) 57 f( ) = ( ) ( ) 9( ) + 5 = f()= 9 +5 = There is lol mimum t (, ). There is lol minimum t (, ). If we re sked to find the gretest or lest vlue on n intervl, then we should lws hek the endpoints. We seek the glol mimum or minimum on the given domin. Emple 9 Self Tutor Find the gretest nd lest vlue of 6 +5 on the intervl First we grph = 6 +5 on In this se the gretest vlue is lerl t the lol mimum when d d =. Now d d = =( ) d nd = when = or : d So, the gretest vlue is f()=5 when =. The lest vlue is either f( ) or f(), whihever is smller. Now f( ) = 7 nd f() = 7 ) lest vlue is 7 when = nd when =. EXERCISE 8D. The tngents t points A, B nd C re horizontl. A (, ) ƒ() Clssif points A, B nd C. Drw sign digrm for the grdient funtion f () for ll. B Stte intervls where = f() is: i inresing ii deresing. d Drw sign digrm for f() for ll. C (, ) e Comment on the differenes etween the sign digrms found ove.

68 58 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chpter 8) For eh of the following funtions, find nd lssif the sttionr points, nd hene sketh the funtion showing ll importnt fetures. f() = f() = + f() = + d f() = e f() = f f() = p + g f() = p h f() = 6 +8 i f() = p j f() = 8 At wht vlue of does the qudrti funtion f() = + +, 6=, hve sttionr point? Under wht onditions is the sttionr point lol mimum or lol minimum? f() = + + hs lol mimum t =. Find : 5 f() = + + hs sttionr point t (, ). Find the vlues of nd. Find the position nd nture of ll sttionr points. 6 The ui polnomil P () = d touhes the line with eqution =9 + t the point (, ), nd hs sttionr point t (, 7). Find P (). 7 Find the gretest nd lest vlue of: for for A mnufturing ompn mkes door hinges. The hve stnding order filled produing 5 eh hour, ut prodution of more thn 5 per hour is useless s the will not sell. The ost funtion for mking hinges per hour is: C() =:7 :796 +: dollrs where Find the minimum nd mimum hourl osts, nd the prodution levels when eh ours. E RATIONAL FUNCTIONS Rtionl funtions hve the form f() = g() h() where g() nd h() re polnomils. For emple, f() = + nd f() = + re rtionl funtions. In this ourse we onsider rtionl funtions for whih g() nd h() re either liner or qudrti. We hve lred onsidered some rtionl funtions of the form = liner liner erlier in the tet. We sw tht one feture of these funtions is the presene of smptotes. These re lines (or urves) tht the grph of the funtion pprohes when or tkes lrge vlues.

69 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chpter 8) 59 Vertil smptotes n e found solving h() =. Horizontl smptotes n e found finding wht vlue f() pprohes s!. Olique smptotes re neither horizontl nor vertil. The re not overed in this ourse. FUNCTIONS OF THE FORM = liner liner Emple Consider f() = +. Write f() in the form + nd hene find the equtions of its smptotes. + Find f () nd drw its sign digrm. Find the es interepts. d Sketh its grph. Self Tutor f() = + ( +) 7 = + = 7 + Vertil smptote is += or =. As!, f()!. ) horizontl smptote is =. It uts the -is when = ) = ) = = It uts the -is when = ) = = f () = d = = ( +) ( ) ( +) + + ( +) 7 ( +) whih hs sign digrm: As f () is never zero, = f() hs no turning points. -\Qw_ \Qw_ EXERCISE 8E. Write the following funtions in the form f() = + + d the equtions of their smptotes: f() = + f() = nd hene determine f() =

70 5 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chpter 8) For eh of the following funtions: i ii iii iv stte the equtions of the smptotes, giving resons for our nswers find f () nd drw its sign digrm find the es interepts sketh the grph of = f(). f() = + + f() = FUNCTIONS OF THE FORM = liner qudrti f() = + d f() = + For funtions of this form, we notie tht s!, f()!. The will therefore ll hve the horizontl smptote =. Emple Consider f() = 9. Determine the equtions of n smptotes. Find f () nd determine the position nd nture of n sttionr points. Find the es interepts. d Sketh the grph of the funtion. f() = 9 = 9 ( )( +) Vertil smptotes re = nd = Horizontl smptote is = f () = ( ) ( 9)( ) ( ) ( +) fquotient ruleg = 6 [6 +9] ( ) ( +) = +8 5 ( ) ( +) = ( 6 +5) ( ) ( +) = ( 5)( ) ( ) ( +) fwhen the denomintor is g fs!, f()! g f () hs sign digrm: There is lol mimum when =5 nd lol minimum when = The lol mimum is (5, ). The lol minimum is (, ). Self Tutor Sign digrms must show vertil smptotes.

71 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chpter 8) 5 Cuts the -is when = ) 9= or = So, the -interept is. Cuts the -is when = ) = 9 = So, the -interept is. d 9 \Qw_ min (' ) m (5' Qe_) EXERCISE 8E. Determine the equtions of the smptotes of: = For eh of the following funtions: i ii iii iv = + ( +) = + determine the eqution(s) of the smptotes find f () nd hene determine the position nd nture of n sttionr points find the es interepts sketh the grph of the funtion, showing ll informtion in i, ii nd iii. f() = + f() = f() = 5 ( ) d f() = ( +) FUNCTIONS OF THE FORM = qudrti qudrti Funtions suh s = + hve horizontl smptote whih n e found + dividing ever term. Notie tht = + + so s!,! =. Emple Consider f() = Determine the equtions of its smptotes. d Find f () nd determine the position nd nture of n turning points. Find the es interepts. Sketh the grph of the funtion. Self Tutor

72 5 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chpter 8) f() = = so s!,! ) horizontl smptote is =. f() = + ( + )( +) ) vertil smptotes re = nd = : f () = ( )( + +) ( + )( +) ( +) ( +) = 6 ( +) ( +) fon simplifingg = 6( + p )( p ) ( +) ( +) nd hs sign digrm So, we hve lol mimum t = p nd lol minimum t = p. The lol mimum is ( p, :). The lol minimum is ( p, : 9). Cuts the -is when = ) += ) ( )( ) = ) =or So, the -interepts re nd. Cuts the -is when = ) = = So, the -interept is. d lol m (-~`' -") -~` ~` lol min (~`' -"9) EXERCISE 8E. Determine the equtions of the smptotes of: = + For eh of the following funtions: i ii iii iv = = + ( +) determine the eqution(s) of the smptotes determine the position nd nture of n sttionr points find the es interepts sketh the funtion, showing ll informtion otined in i, ii nd iii. = 6 = = + d = 6 +5 ( +)

73 F When urve, or prt of urve, hs shpe: APPLICATIONS OF DIFFERENTIAL CALCULUS (Chpter 8) 5 INFLECTIONS AND SHAPE we s tht the shpe is onve downwrds we s tht the shpe is onve upwrds. TEST FOR SHAPE Consider the onve downwrds urve: m = m = m = m = m = X Likewise, if the urve is onve upwrds: X m = m = m = m = m = Wherever we re on the urve, s is inresed, the grdient of the tngent dereses. ) f () is deresing ) its derivtive is negtive, whih mens f () <. Wherever we re on the urve, s is inresed, the grdient of the tngent inreses. ) f () is inresing ) its derivtive is positive, whih mens f () >. POINTS OF INFLECTION (INFLEXION) A point of infletion is point on urve t whih there is hnge of urvture or shpe. or DEMO point of infletion point of infletion If the tngent t point of infletion is horizontl then this point is lso sttionr point. We s tht we hve horizontl or sttionr infletion. sttionr infletion tngent grdient = ƒ( ) SD '() SD ''()

74 5 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chpter 8) If the tngent t point of infletion is not horizontl we s tht we hve non-horizontl or non-sttionr infletion. non-sttionr infletion tngent grdient ƒ( ) SD '() SD ''() The tngent t the point of infletion, lso lled the infleting tngent, rosses the urve t tht point. There is point of infletion t = if f () = nd the sign of f () hnges on either side of =. The point of infletion orresponds to hnge in urvture. In the viinit of, f () hs sign digrm either or Oserve tht if f() = then f () = nd f () = nd f () hs sign digrm Although f () = we do not hve point of infletion t (, ) euse the sign of f () does not hnge on either side of =. In ft the grph of f() = is: SUMMARY lol minimum (, ) V ( ) ( ) turning points horizontl infletions non-horizontl infletions sttionr points'( ) infletions''( ) A urve is onve downwrds on n intervl S if f () 6 for ll in S. A urve is onve upwrds on n intervl S if f () > for ll in S. If f () hnges sign t =, nd f () =, then we hve ² horizontl infletion if f () = ² non-horizontl infletion if f () 6=.

75 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chpter 8) 55 Clik on the demo ion to emine some stndrd funtions for turning points, points of infletion, nd intervls where the funtion is inresing, deresing, nd onve up or down. DEMO Emple Self Tutor Find nd lssif ll points of infletion of f() = +5. f() = +5 ) f () = = ( ) ) f () = =( ) ) f () = when =or onve up onve down onve up '( ) ''( ) Sine the signs of f () hnge out =nd =, these two points re points of infletion. Also f () =, f () = 8 6= nd f() = 5, f() = = Thus (, 5) is horizontl infletion, nd (, ) is non-horizontl infletion. Emple Self Tutor Consider f() = Find nd lssif ll points where f () =: Find nd lssif ll points of infletion. Find intervls where the funtion is inresing or deresing. d Find intervls where the funtion is onve up or down. e Sketh the funtion showing ll importnt fetures. f() = ) f () = 8 +8 =( +) Now f() = 9 nd f() = 7 =( ) ) (, 9) is lol minimum nd (, 7) is horizontl infletion. f () = = ( 8 +) We_ = ( )( ) ) (, 7) is horizontl infletion nd, f( ) or (, :8) is non-horizontl infletion.

76 56 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chpter 8) f() is deresing for 6 f() is inresing for >. d f() is onve up for 6 nd > f() is onve down for 6 6. e ( ) ('-9) lol min sttionr infletion (' 7) (We_\'-"8) non-sttionr infletion EXERCISE 8F. Find nd lssif ll points of infletion of: f() = + f() = f() = d f() = e f() = 8 + f f() = p For eh of the following funtions: i find nd lssif ll points where f () = ii find nd lssif ll points of infletion iii find intervls where the funtion is inresing or deresing iv find intervls where the funtion is onve up or down v sketh the funtion showing ll importnt fetures. f() = f() = f() = p d f() = + e f() = + f f() =( ) g f() = + h f() = p ZEROS OF f () AND f () f () = orresponds to the sttionr points of = f (), so f () will hnge sign t lol mimum or minimum of = f (). Suh points orrespond to the points of infletion of = f(). If lol mimum or minimum of = f () touhes the -is, then it orresponds to sttionr point of infletion of = f(). Otherwise it orresponds to non-sttionr point of infletion. Emple 5 Using the grph of = f() longside, sketh the grphs of = f () nd = f (). Self Tutor ( )

77 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chpter 8) 57 The lol minimum orresponds to f () = nd f () 6=. The non-sttionr point of infletion orresponds to f () 6= nd f () =. The sttionr point of infletion orresponds to f () = nd f () =. '( ) non-sttionr point of infletion ''( ) ( ) sttionr point of infletion Emple 6 Self Tutor The grph longside shows grdient funtion = f (). Sketh grph whih ould e = f(), showing lerl the -vlues orresponding to ll sttionr points nd points of infletion. '( ) Sign digrm of f () is: f () is mimum when = nd minimum when ¼. lol m '( ) At these points f () = ut f () 6=, so the orrespond to non-sttionr points of infletion. lol min ( ) non-sttionr point of infletion lol min EXERCISE 8F. Using the grphs of = f() elow, sketh the grphs of = f () nd = f (). Show lerl the es interepts nd turning points. ( ) ( ) ( )

78 58 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chpter 8) For the grphs of = f () elow, sketh grph whih ould e = f(). Show lerl the lotion of n sttionr points nd points of infletion. '( ) '( ) G OPTIMISATION There re mn prolems for whih we need to find the mimum or minimum vlue of funtion. We n solve suh prolems using differentil lulus tehniques. The solution is often referred to s the optimum solution nd the proess is lled optimistion. Consider the following prolem: An industril shed is to hve totl floor spe of 6 m nd is to e divided into retngulr rooms of equl size. The wlls, internl nd eternl, will ost $6 per metre to uild. Wht dimensions should the shed hve to minimise the ost of the wlls? We let eh room e m m s shown. m Clerl > nd >. The totl length of wll mteril is L =6 + m. We know tht the totl re is 6 m, so = 6 nd hene =. Knowing this reltionship enles us to write L in terms of one vrile, in this se. µ µ L =6 + = m µ The ost is $6 per metre, so the totl ost is C() =6 Now C() = ) C () = 6 8 ) C 8 () = when 6 = ) 8 = 6 ¼ : ) ¼ :57 m dollrs.

79 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chpter 8) 59 Now when ¼ :57, ¼ ¼ 7: nd C(:57) ¼ 8:8 dollrs. :57 So, the minimum ost is out $8 when the shed is :6 m 7: m. WARNING The mimum or minimum vlue does not lws our when the first derivtive is zero. It is essentil to lso emine the vlues of the funtion t the endpoint(s) of the domin for glol mim nd minim. For emple: d d = d d = ƒ( ) The mimum vlue of ours t the endpoint =. The minimum vlue of ours t the lol minimum = p. p TESTING OPTIMAL SOLUTIONS If one is tring to optimise funtion f() nd we find vlues of suh tht f () =, there re severl tests we n use to see whether we hve mimum or minimum solution: SIGN DIAGRAM TEST If ner to = where f () = the sign digrm is: ² we hve lol mimum ² we hve lol minimum. SECOND DERIVATIVE TEST At = where f () =: d ² If < we hve shpe, d whih is lol mimum d ² If > we hve shpe, d whih is lol minimum. GRAPHICAL TEST If the grph of = f() shows: ² we hve lol mimum ² we hve lol minimum.

80 5 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chpter 8) OPTIMISATION PROBLEM SOLVING METHOD Step : Step : Step : Drw lrge, ler digrm of the sitution. Construt formul with the vrile to e optimised (mimised or minimised) s the sujet. It should e written in terms of one onvenient vrile, s. You should write down wht restritions there re on. Find the first derivtive nd find the vlues of when it is zero. Step : If there is restrited domin suh s 6 6, the mimum or minimum m our either when the derivtive is zero or else t n endpoint. Show using the sign digrm test, the seond derivtive test or the grphil test, tht ou hve mimum or minimum sitution. Emple 7 A retngulr ke dish is mde utting out squres from the orners of 5 m m retngle of tin-plte, nd then folding the metl to form the ontiner. Wht size squres must e ut out to produe the ke dish of mimum volume? Self Tutor DEMO Step : Step : Step : Let m e the side lengths of the squres tht re ut out. ) dv d = 6 + = ( ) = ( 5)( 5) dv d ( ) m ( ) m m = when = 5 =6 Step : Volume = length width depth = ( )(5 ) = ( ) = + m Notie tht > nd 5 > ) <<:5 or =5 Sign digrm test dv hs sign digrm: d or Seond derivtive test d V d = 6 so when =5, d V = whih is < d ) the shpe is nd we hve lol mimum. So, the mimum volume is otined when =5, whih is when 5 m squres re ut from the orners.

81 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chpter 8) 5 Emple 8 A litre ontiner must hve squre se, vertil sides, nd n open top. Find the most eonomil shpe whih minimises the surfe re of mteril needed. open Self Tutor Step : m m m Let the se lengths e m nd the depth e m. The volume V = length width depth ) V = ) =... () fs litre m g Step : The totl surfe re A = re of se +(re of one side) = + µ = + fusing ()g ) A() = + 6 where > Step : A () = 6 ) A 6 () = when = ) = 6 ) = p 8 = Step : Sign digrm test or Seond derivtive test A () = + =+ whih is lws positive If =, If =, s > for ll >. A 6 () = = 6 = A 6 () = 6 9 ¼ 6 7:8 ¼ : Both tests estlish tht the minimum mteril is used to mke the ontiner when = nd = =. So, m m m is the most eonomil shpe.

82 5 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chpter 8) Sometimes the vrile to e optimised is in the form of single squre root funtion. In these situtions it is onvenient to squre the funtion nd use the ft tht if A>, the optimum vlue of A() ours t the sme vlue of s the optimum vlue of [A()]. Emple 9 Self Tutor An niml enlosure is right ngled tringle with one leg eing drin. The frmer hs m of fening ville for the other two sides, [AB] nd [BC]. A drin C If AB = m, show tht AC = p 9 6. Find the mimum re of the tringulr enlosure. B (AC) + = ( ) fpthgorsg ) (AC) = = 9 6 ) AC = p 9 6 The re of tringle ABC is A() = (se ltitude) = (AC ) = p 9 6 m A B << C ( ) m ) ) [A()] = (9 6) = 5 5 d d [A()] = 5 5 = 5( ) with sign digrm: A() is mimised when = so A m = ()p 9 6 ¼ 866 m m m Emple A squre sheet of metl hs smller squres ut from its orners s shown. Wht sized squre should e ut out so tht when the sheet is ent into n open o it will hold the mimum mount of liquid? Self Tutor m m

83 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chpter 8) 5 ( ) m Let m m squres e ut out. Volume = length width depth =( ) ( ) ) V () =( ) Now V () =( ) + ( ) ( ) fprodut ruleg =( )[ ] =( )( 6) ) V () = when = or 6 However, must e > nd so < ) = 6 is the onl vlue in << with V () =. Seond derivtive test: Now V () = ( 6)+( )( 6) fprodut ruleg = ) ³ = 8 V = 8 = 6 whih is < ) the volume is mimised when = 6. 6 onve downwrds Conlusion: The resulting ontiner hs mimum pit when = 6. Use lulus tehniques to nswer the following prolems. In ses where finding the zeros of the derivtives is diffiult ou m use grphis lultor or grphing pkge to help ou. GRAPHING PACKAGE EXERCISE 8G A mnufturer n produe fittings per d where 6 6. The prodution osts re: ² E per d for the workers ² E per d per fitting ² E 5 per d for running osts nd mintenne. How mn fittings should e produed dil to minimise osts? For the ost funtion C() = 7 + +: dollrs nd prie funtion p() =5 : dollrs, find the prodution level tht will mimise profits.

84 5 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chpter 8) The totl ost of produing lnkets per d is +8 + dollrs, nd for this prodution level eh lnket m e sold for ( ) dollrs. How mn lnkets should e produed per d to mimise the totl profit? The ost of running ot is $ v per hour where v km h is the speed of the ot. All other osts mount to $6:5 per hour. Find the speed whih will minimise the totl ost per kilometre. 5 A duk frmer wishes to uild retngulr enlosure of re m. The frmer must purhse wire netting for three of the sides s the fourth side is n eisting fene. Nturll, the frmer wishes to minimise the length (nd therefore ost) of fening required to omplete the jo. If the shorter sides hve length m, show tht the required length of wire netting to e purhsed is L = + : Use tehnolog to help ou sketh the grph of = + : Find the minimum vlue of L nd the orresponding vlue of when this ours. d Sketh the optimum sitution showing ll dimensions. 6 Rdiotive wste is to e disposed of in full enlosed led oes of inner volume m. The se of the o hs dimensions in the rtio :. Wht is the inner length of the o? Eplin wh h =. Eplin wh the inner surfe re of the o is given A() = + 6 m. d Use tehnolog to help sketh the grph of = + 6 : e Find the minimum inner surfe re of the o nd the orresponding vlue of. f Sketh the optimum o shpe showing ll dimensions. 7 Consider the mnufture of lindril tin ns of L pit where the ost of the metl used is to e minimised. This mens tht the surfe re must e s smll s possile. Eplin wh the height h is given h = ¼r m. r m Show tht the totl surfe re A is given A =¼r + m h m. r Use tehnolog to help ou sketh the grph of A ginst r. d Find the vlue of r whih mkes A s smll s possile. e Sketh the n of smllest surfe re. h m m

85 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chpter 8) 55 8 Sm hs sheets of metl whih re 6 m 6 m squre. He wnts to ut out identil squres whih re m m from the orners of eh sheet. He will then end the sheets long the dshed lines to form n open ontiner. Show tht the pit of the ontiner is given V () =(6 ) m. Wht sized squres should e ut out to produe the ontiner of gretest pit? 6 m 6 m 9 An thletis trk hs two strights of length l m nd two semiirulr ends of rdius m. The perimeter of the trk is m. Show tht l = ¼ nd hene write down the possile vlues tht m hve. Show tht the re inside the trk is A = ¼. Wht vlues of l nd produe the lrgest re inside the trk? m l m A setor of rdius m nd ngle µ o is ent to form onil up s shown. setor eomes m join when edges [AB] nd [CB] re joined with tpe B A m C Suppose the resulting one hs se rdius r m nd height h m. Show using the setor tht r AC = µ¼ 8 : Eplin wh r = µ q 6 : Show tht h = µ : 6 d Find the one s pit V in terms of µ onl. e Use tehnolog to sketh the grph of V (µ): f Find µ when V (µ) is mimum. B is row ot 5 km out t se from A. [AC] is stright snd eh 6 km long. Peter n row the ot t 8 km h nd run long the eh t 7 km h. Suppose Peter rows diretl from B to point X on [AC] suh tht AX = km. Eplin wh If T () is the totl time Peter tkes to row to X nd then run long the eh p +5 to C, show tht T () = dt Find suh tht d sttement. 5km B hours. A km X 6km =. Wht is the signifine of this vlue of? Prove our C

86 56 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chpter 8) A pumphouse is to e pled t some point X long river. Two pipelines will then onnet the pumphouse to homesteds A nd B. How fr from M should point X e so tht the totl length of pipeline is minimised? A km M river X 5km B N km Open lindril ins re to ontin litres. Find the rdius nd height of the in shpe whih minimises the surfe re nd thus the mteril needed. p 5 p Two lmps with intensities nd 5 ndle-power re pled 6 m prt. If the intensit of illumintion I t n point is diretl proportionl to the power of the 6m soure, nd inversel proportionl to the squre of the distne from the soure, find the drkest point on the line joining the two lmps. 5 A right ngled tringulr pen is mde from mof fening, ll used for sides [AB] nd [BC]. Side [AC] is n eisting rik wll. C If AB = m, find D() in terms of. D ( ) metres Find d[d()] nd hene drw sign digrm d for it. Find the smllest nd the gretest possile vlue of D() nd the design of the pen in eh se. B A wll 6 At : pm ship A leves port P. It sils in the diretion o t km h. At the sme time, ship B is km due est of P, nd is siling t 8 km h towrds P. Show tht the distne D(t) etween the two ships is given D(t) = p t 8t + km, where t is the numer of hours fter : pm. Find the minimum vlue of [D(t)] for ll t >. At wht time, to the nerest minute, re the ships losest? 7 [AB] is m high fene whih is m from vertil wll [RQ]. An etension ldder [PQ] is pled on the fene so tht it touhes the ground t P nd the wll t Q. If AP = m, find QR in terms of. If the ldder hs length L m, show µ tht [L()] =( +) +. Q wll m B R m A P d[l()] Show tht = onl when = p. d d Find, orret to the nerest entimetre, the shortest length of the etension ldder. You must prove tht this length is the shortest.

87 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chpter 8) 57 REVIEW SET 8A NON-CALCULATOR A prtile P moves in stright line with position reltive to the origin O given s(t) =t 9t +t 5 m, where t is the time in seonds, t >. Find epressions for the prtile s veloit nd elertion nd drw sign digrms for eh of them. Find the initil onditions. Desrie the motion of the prtile t time t = seonds. d Find the times nd positions where the prtile hnges diretion. e Drw digrm to illustrte the motion of P. f Determine the time intervls when the prtile s speed is inresing. Retngle ABCD is insried within the prol = k nd the -is, s shown. If OD =, show tht the retngle ABCD hs re funtion A() =k. If the re of ABCD is mimum when AD = p, find k. k B A D C A retngulr gutter is formed ending m wide sheet of metl s shown in the illustrtion. end view m Where must the ends e mde in order to mimise the pit of the gutter? Consider the funtion f() = +. Stte the eqution of the vertil smptote. Find the es interepts. Find f () nd drw its sign digrm. d Find the position nd nture of n sttionr points. 5 Given the grph of = f () drwn longside, sketh possile urve for = f(). Show lerl '( ) n turning points nd points of infletion. 6 Consider f() = +. Determine the equtions of n smptotes. Find the position nd nture of its turning points. Find its es interepts. d Sketh the grph of the funtion showing the importnt fetures of, nd. e For wht vlues of p does = p hve two rel distint solutions? +

88 58 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chpter 8) 7 A prtile moves in stright line with position reltive to O given s(t) =t t m, where t> is the time in seonds. d e Find veloit nd elertion funtions for the prtile s motion nd drw sign digrms for eh of them. Desrie the motion of the prtile t t =seond. Does the prtile ever hnge diretion? If so, where nd when does it do this? Drw digrm to illustrte the motion of the prtile. Find the time intervls when the: i veloit is inresing ii speed is inresing. 8 A retngulr sheet of tin-plte is k m k m. Four squres, eh with sides m, re ut from its orners. The reminder is ent into the shpe of n open retngulr ontiner. Find the vlue of whih will mimise the pit of the ontiner. k m m k m REVIEW SET 8B A tringulr pen is enlosed two fenes [AB] nd [BC], eh of length 5 m, with the river eing the third side. If AC = m, show tht the re of tringle ABC is A() = p 5 m. Find d[a()] nd hene find suh tht the d re is mimum. A CALCULATOR B 5 m river C Suppose f() = +, < hs turning point when = p. Find. Find the position nd nture of ll sttionr points of = f(). Sketh the grph of = f(). A prtile moves long the -is with position reltive to origin O given (t) =t p t m, where t is the time in seonds, t >. Find epressions for the prtile s veloit nd elertion t n time t, nd drw sign digrms for eh funtion. Find the initil onditions nd hene desrie the motion t tht instnt. Desrie the motion of the prtile t t =9 seonds. d Find the time nd position when the prtile reverses diretion. e Determine the time intervl when the prtile s speed is deresing.

89 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chpter 8) 59 The ost per hour of running freight trin is given C(v) = v + 9 v where v is the verge speed of the trin in km h. Find the ost of running the trin for: i two hours t 5 km h ii 5 hours t 6 km h. Find the rte of hnge in the hourl ost of running the trin t speeds of: i 5 km h ii 66 km h. At wht speed will the ost per hour e minimum? 5 Consider the funtion f() = +. Find the es interepts. Eplin wh f() hs no vertil smptotes. Find the position nd nture of n sttionr points. q d Show tht = f() hs non-sttionr infletion points t = : dollrs e Sketh the grph of = f() showing ll fetures found in,, nd d ove. 6 The grph of = f() is given. On the sme es sketh the grph of = f (). ( ) 7 A m fene is pled round lwn whih hs the shpe of retngle with semi-irle on one of its sides. Using the dimensions shown on the figure, show tht = ¼. Find the re of the lwn A in terms of onl. Find the dimensions of the lwn if it hs the mimum possile re. m m REVIEW SET 8C Consider the funtion f() = Find nd lssif ll sttionr points nd points of infletion. Find intervls where the funtion is inresing nd deresing. Find intervls where the funtion is onve up or down. d Sketh the grph of = f() showing ll importnt fetures. Consider the funtion f() = +. Find ll es interepts.

90 55 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chpter 8) Find nd lssif ll sttionr points nd points of infletion. Sketh the grph of = f() showing fetures from nd. A mnufturer of open steel oes hs to mke one with squre se nd pit of m. The steel osts $ per squre metre. If the se mesures m m nd the height is m, find in terms of. open m m Hene, show tht the totl ost of the steel is C() = + 8 dollrs. Find the dimensions of the o whih would ost the lest in steel to mke. A prtile P moves in stright line with position from O given s(t) =5t 6 m, where t is the time in seonds, t >. (t ) Find veloit nd elertion funtions for P s motion. Desrie the motion of P t t = seonds. For wht vlues of t is the prtile s speed inresing? 5 The height of tree t ers fter it is plnted is given µ H(t) =6 metres, t >. t + How high ws the tree when it ws plnted? Determine the height of the tree fter t =, 6 nd 9 ers. Find the rte t whih the tree is growing t t =,, 6 nd 9 ers. d Show tht H (t) > nd eplin the signifine of this result. e Sketh the grph of H(t) ginst t. 6 A mhinist hs spheril ll of rss with dimeter m. The ll is pled in lthe nd mhined into linder. If the linder hs rdius m, show tht the linder s volume is given V () =¼ p m. Hene, find the dimensions of the linder of lrgest volume whih n e mde. m h m 7 The grph of = f () is drwn. On the sme es lerl drw possile '( ) grph of = f(). Show ll turning points nd points of infletion.

91 Chpter9 Derivtives of eponentil nd logrithmi funtions Sllus referene: 7., 7., 7. Contents: A B C D Eponentil e Nturl logrithms Derivtives of logrithmi funtions Applitions

92 55 DERIVATIVES OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS (Chpter 9) In Chpter we sw tht the simplest eponentil funtions hve the form where is n positive onstnt, 6=. f() = The grphs of ll memers of the eponentil fmil f() = hve the following properties: ² pss through the point (, ) ² smptoti to the -is t one end ² lie ove the -is for ll ² onve up for ll ² monotone inresing for > ² monotone deresing for <<. For emple, (. ) (. 5) 5 (. ) A EXPONENTIAL e INVESTIGATION THE DERIVATIVE OF = This investigtion ould e done using grphis lultor or liking on the ion. The purpose of this investigtion is to oserve the nture of the derivtives of f() = for vrious vlues of. Wht to do: For = find the grdient of the tngent t =, :5,, :5, nd :5. Use modelling tehniques from our grphis lultor or d the softwre provided to show tht d ¼ :69. CALCULUS DEMO Repet for the funtions = =5 =(:5) : Use our oservtions from nd to write sttement out the derivtive of the generl eponentil = for >, 6=. From the investigtion ou should hve disovered tht: If f() = then f () =k where k is onstnt equl to f ().

93 DERIVATIVES OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS (Chpter 9) 55 Proof: then But If f() =, f () = lim h! f( + h) f() h = lim h! +h h ( h ) = lim h! h µ = h lim h! h f () = lim h! f( + h) f() h = lim h! h h ) f () = f () ffirst priniples definition of derivtiveg fs is independent of hg grdient is ƒ'() Given this result, if we n find vlue of suh tht f () =, then we hve found funtion whih is its own derivtive. INVESTIGATION FINDING WHEN = d AND d = Clik on the ion to grph f() = nd its derivtive funtion = f (). DEMO Eperiment with different vlues of until the grphs of f() = nd = f () pper the sme. Estimte the orresponding vlue of to deiml ples. From Investigtion ou should hve disovered tht f() =f () = when ¼ :7: To find this vlue of more urtel we return to the lgeri pproh: µ We showed tht if f() = then f () = h lim : h! h So if f () = ) h h h we require lim =: h! h ¼ for vlues of h whih re lose to ) h ¼ +h for h lose to :

94 55 DERIVATIVES OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS (Chpter 9) Letting h = n, n ¼ + n for lrge vlues of n fh =! s n!g µ n ) ¼ + n n for lrge vlues of n: We now emine µ + n n s n!. n µ + n n n µ + n n In ft s n!, : : : : : : : : : : : : µ + n n! : :::: nd this irrtionl numer is denoted the smol e. e =: :::: nd is lled eponentil e. Alterntive nottion: If f() =e then f () =e. e is sometimes written s ep(). For emple, ep( ) =e. e is n importnt numer with similrities to the numer ¼. Both numers re irrtionl (not surds) with non-reurring, non-terminting deiml epnsions, nd oth re disovered nturll. We lso sw e in n erlier hpter when looking t ontinuous ompound interest. PROPERTIES OF = e d Notie tht d = e =. As!,! ver rpidl, d nd so d!. This mens tht the grdient of the urve is ver lrge for lrge vlues of. The urve inreses in steepness s gets lrger. 5 5 e

95 DERIVATIVES OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS (Chpter 9) 555 As!,! nd so d d!. This mens for lrge negtive, the grph eomes fltter nd pprohes the smptote =. e > for ll, so the rnge of f : 7! e is R +. THE DERIVATIVE OF e f () The funtions e, e + nd e re ll of the form e f(). Suh funtions re often used in prolem solving. In generl, e f() > for ll, no mtter wht the funtion f(). Consider = e f() : If we let u = f() then = e u : But d d = d du du d fhin ruleg ) d du = eu d d = ef() f () Summr: Funtion Derivtive e e e f() e f() f () Emple Find the grdient funtion for equl to: e + e e e Self Tutor If =e + e then d d =e + e ( ) =e e If = e then d d =e + e ( ) =e e fprodut ruleg If = e then d d = e () e () fquotient ruleg = e ( )

96 556 DERIVATIVES OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS (Chpter 9) Emple Find the grdient funtion for equl to: (e ) p e + Self Tutor =(e ) = u where u = e d d = d du du d fhin ruleg =u du d =(e ) e =e (e ) =(e +) = u where u =e + d d = d du du d = u du d fhin ruleg = (e +) e ( ) = e (e +) Emple Self Tutor Find the et position nd nture of the sttionr points of =( )e. So, d d = ()e +( )e ( ) fprodut ruleg = e ( ( )) = e where e is positive for ll. d = when =: d The sign digrm of d d is: ) t = we hve lol mimum. But when =, = ()e = e ) the lol mimum is (, e ). EXERCISE 9A Find the grdient funtion for f() equl to: e e + ep ( ) d e e e f e g e e h e + e i e j e k ( + e ) l ( e ) m e + n e o e p e :

97 DERIVATIVES OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS (Chpter 9) 557 Find the derivtive of: e e e d e e f e p g p e h e e + e + Find the grdient funtion for f() equl to: (e +) d ( e ) e Suppose = Ae k where A nd k re onstnts. d Show tht: d = k d d = k. 5 If =e +5e, show tht e p e + p e f p e d d 7 d + =. d 6 Given f() =e k + nd f () = 8, find k. 7 Find the position nd nture of the turning point(s) of: = e = e = e d = e ( +) B NATURAL LOGARITHMS In Chpter we found tht: ² If e = then =ln nd vie vers. So, e =, =ln. ² The grph of = ln is the refletion of the grph of = e e in the mirror line =. ln ² = e nd =ln re inverse funtions. ² From the definition e =, =ln we oserve tht e ln =. This mens tht n positive rel numer n e written s power of e, or lterntivel, the nturl logrithm of n positive numer is its power of e, i.e., ln e n = n.

98 558 DERIVATIVES OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS (Chpter 9) You should rememer tht the lws of logrithms in se e re identil to those for se nd indeed for n se. For >, > : ² ln() =ln +ln ³ ² ln =ln ln ² ln ( n )=n ln Notie lso tht: ² ln = nd ln e = µ ² ln = ln ² log = ln ln, 6= Emple Self Tutor Find lgerill the et points of intersetion of = e nd = e. Chek our solution using tehnolog. The funtions meet where e = e ) e +e = ) e e +== fmultipling eh term e g ) (e )(e ) = ) e =or ) =lnor ln ) =or ln GRAPHING PACKAGE When =, = e = When =ln, e = ) = = ) the funtions meet t (, ) nd t (ln, ). Emple 5 Self Tutor Consider the funtion = e. Find the -interept. Find the -interept. Show lgerill tht the funtion is inresing for ll. d Show lgerill tht the funtion is onve down for ll. e Use tehnolog to help grph = e : f Eplin wh = is horizontl smptote.

99 DERIVATIVES OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS (Chpter 9) 559 The -interept ours when =, ) e = ) =ln ) = ln ) the -interept is ln ¼ :69 The -interept ours when = ) = e = = d d = e ( ) = e = e Now e > for ll, so d > for ll d ) the funtion is inresing for ll. d d d = e ( ) = e whih is < for ll ) the funtion is onve down for ll. e f As!, e! nd e! )! (elow) Hene, the horizontl smptote is =: The first three questions of the following eerise revise logrithmi nd eponentil funtions. If ou need help with these ou should onsult Chpters nd. EXERCISE 9B Without using lultor, evlute: ln e ln p e ln µ e µ p d ln e e e ln f e ln g e ln 5 h e ln Write s power of e: d Solve for : e = e = e = d e =e e e = e f e 5e +6= g e +=e h +e = e i e + e = Find lgerill the point(s) of intersetion of: = e nd = e 6 =e + nd =7 e = e nd =5e Chek our nswers using tehnolog. 5 The funtion f() =e uts the -is t A nd the -is t B. Find the oordintes of A nd B. Show lgerill tht the funtion is inresing for ll.

100 56 DERIVATIVES OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS (Chpter 9) d e Find f () nd hene eplin wh f() is onve up for ll. Use tehnolog to help grph = e. Eplin wh = is horizontl smptote. 6 Suppose f() =e nd g() = 5e. Find the nd -interepts of oth funtions. Disuss f() nd g() s! nd s!. Find lgerill the point(s) of intersetion of the funtions. d Sketh the grph of oth funtions on the sme set of es for 6 6. Show ll importnt fetures on our grph. 7 The funtion = e e uts the -is t P nd the -is t Q. Determine the oordintes of P nd Q. Prove tht the funtion is inresing for ll. Show tht d d =. Wht n e dedued out the onvit of the funtion ove nd elow the -is? C d Use tehnolog to help grph = e e. Show the fetures of, nd on the grph. DERIVATIVES OF LOGARITHMIC FUNCTIONS INVESTIGATION THE DERIVATIVE OF ln If =ln, wht is the grdient funtion? CALCULUS DEMO Wht to do: Clik on the ion to see the grph of =ln. A tngent is drwn to point on the grph nd the grdient of this tngent is given. Wth s the point moves from left to right nd grph of the grdient of the tngent t eh point is displed. Wht do ou think the eqution of the grdient funtion is? Find the grdient t =:5, =:5, =, =, =, =, =5. Do our results onfirm our suggestion in? From the investigtion ou should hve oserved tht The proof of this result is eond the sope of this ourse. if =ln then d d =. B use of the hin rule, we n show tht if =lnf() then d d = f () f().

101 DERIVATIVES OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS (Chpter 9) 56 Proof: Suppose =lnf() If we let u = f(), then =lnu. Now d d = d du du d = u f () = f () f() fhin ruleg Summr: Funtion ln ln f() Derivtive f () f() Emple 6 Find the grdient funtion of: =ln(k) where k is onstnt If =ln(k) then d d = k k = If =ln( ) then then d d = = If = ln d d = ln + = ln + = ( ln +) =ln( ) = ln µ fprodut ruleg ln(k) =lnk +ln =ln + onstnt so ln(k) nd ln oth hve derivtive. Self Tutor The lws of logrithms n help us to differentite some logrithmi funtions more esil. Emple 7 Differentite with respet to : =ln(e ) =ln ( + )( ) Self Tutor If =ln(e ) then =ln +lne flog of produt lwg ) =ln f ln e = g d Differentiting with respet to, we get d = If =ln then =ln ln[( + )( )] ( + )( ) =ln [ln( + ) + ln( )] ) =ln ln( +) ln( ) d d = +

102 56 DERIVATIVES OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS (Chpter 9) EXERCISE 9C Find the grdient funtion of: =ln(7) =ln( +) =ln( ) d = ln e = ln f = ln g = e ln h =(ln) i = p ln j = e ln k = p ln() l = p ln m = ln( ) n = ln( +) Find d d for: = ln 5 =ln( ) =ln( + ) d = ln( 5) e = [ln( +)] f = ln() µ g =ln h =ln(ln) i = ln Use the logrithm lws to help differentite with respet to : =ln p µ =ln =ln(e p ) + d =ln( p µ µ + ) e =ln f =ln µ g f() = ln (( ) ) h f() = ln (( + + )) i f() =ln 5 B sustituting e ln for in = find d d. Show tht if = then d d = ln. 5 Consider f() = ln( ). Find the -interept. Cn f() e found? Wht is the signifine of this result? Find the grdient of the tngent to the urve t =. d e f For wht vlues of does f() hve mening? Find f () nd hene eplin wh f() is onve down whenever f() hs mening. Grph the funtion. 6 Consider f() = ln. For wht vlues of is f() defined? Show tht the minimum vlue of f() is e.

103 DERIVATIVES OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS (Chpter 9) 56 ln 7 Prove tht 6 for ll >. e Hint: Let f() = ln nd find its gretest vlue. 8 Consider the funtion f() = ln : D Show tht the grph of = f() hs lol minimum nd tht this is the onl turning point. Hene prove tht ln 6 for ll >. The pplitions we onsider here re: ² tngents nd normls ² rtes of hnge ² urve properties ² displement, veloit nd elertion ² optimistion (mim nd minim) Emple 8 APPLICATIONS Self Tutor Show tht the eqution of the tngent to =ln t the point where = is = e. When =, ln = ) = e = e µ ) the point of ontt is e,. e ln Now f() =ln hs derivtive f () = ) the tngent t µ e, ) the tngent hs eqution hs grdient e e = e = e ) +=e( e ) ) = e EXERCISE 9D Find the eqution of the tngent to the funtion f : 7! e t the point where =. Find the eqution of the tngent to =ln( ) t the point where =. The tngent to = e t = uts the nd -es t A nd B respetivel. Find the oordintes of A nd B. Find the eqution of the norml to =ln p t the point where =.

104 56 DERIVATIVES OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS (Chpter 9) 5 Find the eqution of the tngent to = e t the point where =. Hene, find the eqution of the tngent to = e whih psses through the origin. 6 Consider f() =ln. For wht vlues of is f() defined? Find the signs of f () nd f () nd omment on the geometril signifine of eh. Sketh the grph of f() =ln nd find the eqution of the norml t the point where =. 7 Find, orret to deiml ples, the ngle etween the tngents to =e nd =+e t their point of intersetion. 8 A rdiotive sustne des ording to the formul W =e kt grms where t is the time in hours. Find k given tht fter 5 hours the weight is grms. Find the weight of rdiotive sustne present t: i t =hours ii t =hours iii t =week. How long will it tke for the weight to reh grm? d Find the rte of rdiotive de t: i t = hours ii t = hours. dw e Show tht is proportionl to the weight of sustne remining. dt 9 The temperture of liquid fter eing pled in refrigertor is given T =5+95e kt o C where k is positive onstnt nd t is the time in minutes. Find k if the temperture of the liquid is o C fter 5 minutes. Wht ws the temperture of the liquid when it ws first pled in the refrigertor? Show tht dt = (T 5) for some onstnt. dt d At wht rte is the temperture hnging t: i t =mins ii t =mins iii t =mins? The height of ertin speies of shru t ers fter it is plnted is given H(t) = ln(t +)+m, t >. How high ws the shru when it ws plnted? How long will it tke for the shru to reh height of m? At wht rte is the shru s height hnging: i ers fter eing plnted ii ers fter eing plnted? In the onversion of sugr solution to lohol, the hemil retion oes the lw A = s( e kt ), t > where t is the numer of hours fter the retion ommened, s is the originl sugr onentrtion (%), nd A is the lohol produed, in litres. Find A when t =. If s = nd A =5 fter hours, find k. If s =, find the speed of the retion t time 5 hours. d Show tht the speed of the retion is proportionl to A s.

105 DERIVATIVES OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS (Chpter 9) 565 Consider the funtion f() = e. Does the grph of = f() hve n or -interepts? Disuss f() s! nd s! : Find nd lssif n sttionr points of = f(). d Sketh the grph of = f() showing ll importnt fetures. e Find the eqution of the tngent to f() = e t the point where = : A prtile P moves in stright line. Its displement from the origin O is given s(t) = t + e t 5 m where t is the time in seonds, t >. Find the veloit nd elertion funtions. Find the initil position, veloit nd elertion of P. Disuss the veloit of P s t!. d Sketh the grph of the veloit funtion. e Find when the veloit of P is 8 m per seond. A pshologist lims tht the ilit A to memorise simple fts during infn ers n e lulted using the formul A(t) =t ln t + where <t6 5, t eing the ge of the hild in ers. At wht ge is the hild s memorising ilit minimum? Sketh the grph of A(t). 5 One of the most ommon funtions used in sttistis is the norml distriution funtion f() = p ¼ e. Find the sttionr points of the funtion nd find intervls where the funtion is inresing nd deresing. Find ll points of infletion. Disuss f() s! nd s!. d Sketh the grph of = f() showing ll importnt fetures. 6 A mnufturer of eletri kettles performs ost ontrol stud. The disover tht to produe kettles per d, the ost per kettle C() is given µ C() =ln + hundred dollrs with minimum prodution pit of kettles per d. How mn kettles should e mnuftured to keep the ost per kettle to minimum? 7 Infinitel mn retngles whih sit on the -is n e insried under the urve = e. Determine the oordintes of C suh tht B C e retngle ABCD hs mimum re. A D

106 566 DERIVATIVES OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS (Chpter 9) 8 The revenue generted when mnufturer sells torhes per d is given ³ R() ¼ ln dollrs. Eh torh osts the mnufturer $:5 to produe plus fied osts of $ per d. How mn torhes should e produed dil to mimise the profits mde? 9 A qudrti of the form =, >, touhes the logrithmi funtion =ln. If the -oordinte of the point of ontt is, eplin wh =ln nd =. Dedue tht the point of ontt is ( p e, ). Wht is the vlue of? d Wht is the eqution of the ommon tngent? ln A smll popultion of wsps is oserved. After t weeks the popultion is modelled 5 P (t) = + e :5t wsps, where 6 t 6 5. Find when the wsp popultion is growing fstest. Consider the funtion = Ate t, t > where A nd re positive onstnts. Find the t nd -interepts of the funtion. Find nd desrie the sttionr point in terms of A nd. Find the point of infletion in terms of A nd. d e Grph the funtion showing the fetures found ove. When new pin killing injetion is dministered, the effet is modelled E(t) = 75te :5t units, where t > is the time in hours fter the injetion. At wht time is the drug most effetive? REVIEW SET 9A NON-CALCULATOR Find d µ d if: = + + e =ln Find the eqution of the norml to = e t the point where =. Find the et oordintes of the point of intersetion P of the grphs of f() =e nd g() = e +6. Show tht the eqution of the tngent to f() t P is given =8 + 8ln. Consider the funtion f() = e. Find the nd -interepts. For wht vlues of is f() defined? Find the signs of f () nd f () nd omment on the geometril signifine of eh.

107 DERIVATIVES OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS (Chpter 9) 567 d Sketh the grph of = f(): e Find the eqution of the tngent t the point where =. 5 Find where the tngent to =ln( +) t = uts the -is. 6 At the point where =, the tngent to f() =e + p + q hs eqution =5 7. Find p nd q. 7 Find the et roots of the following equtions: e 5= e ln ln µ = REVIEW SET 9B The height of tree t ers fter it ws plnted is given H(t) = 6 + ln(t +)m, t >. How high ws the tree when it ws plnted? CALCULATOR How long does it tke for the tree to reh: i 5 m ii m? At wht rte is the tree s height inresing fter: i ers ii ers? A prtile P moves in stright line with position given s(t) =8e t t m where t is the time in seonds, t >. Find the veloit nd elertion funtions. Find the initil position, veloit, nd elertion of P. Disuss the veloit of P s t!. d e Sketh the grph of the veloit funtion. Find when the veloit is metres per seond. A shirt mker sells shirts per d with revenue funtion ³ R() = ln + + dollrs. The mnufturing osts re determined the ost funtion C() =( ) + dollrs. How mn shirts should e sold dil to mimise profits? Wht is the mimum dil profit? The vlue of r t ers fter its purhse is given V = e :t dollrs. Clulte: the purhse prie of the r the rte of derese of the vlue of the r ers fter it ws purhsed. 5 A mnufturer determines tht the totl weekl ost C of produing ³ loks per d is given C() = ln + dollrs. How mn loks per d should e produed to minimise the osts given tht t lest 5 loks per d must e mde to fill fied dil orders?

108 568 DERIVATIVES OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS (Chpter 9) 6 A prtile P moves in stright line with position given s(t) =5t ln t m, t >, where t is the time in minutes. Find the veloit nd elertion funtions. Find the position, veloit, nd elertion when t = e minutes. Disuss the veloit s t!. d Sketh the grph of the veloit funtion. e Find when the veloit of P is m per minute. REVIEW SET 9C Find d d if: =ln( ) = e Find where the tngent to =ln( +) t = uts the -is. Solve etl for : e =e e 7e += Consider the funtion f() =e. Find nd lssif n sttionr points of = f(). Disuss f() s!. Find f () nd drw its sign digrm. Give geometril interprettion for the sign of f (). d Sketh the grph of = f(). e Dedue tht e > + for ll. 5 Differentite with respet to : f() = ln(e +) ( +) f() =ln 6 Consider the funtion f() = +ln. Find the vlues of for whih f() is defined. Find the signs of f () nd f () nd omment on the geometril signifine of eh. Sketh the grph of = f(): d Find the eqution of the norml t the point where =. 7 Infinitel mn retngles n e insried under the urve = e s shown. Determine the oordintes of A suh tht the retngle OBAC hs mimum re. C A B e

109 Chpter Derivtives of trigonometri funtions Sllus referene: 7., 7., 7. Contents: A B Derivtives of trigonometri funtions Optimistion with trigonometr

110 57 DERIVATIVES OF TRIGONOMETRIC FUNCTIONS (Chpter ) INTRODUCTION In Chpter we sw tht sine nd osine urves rise nturll from motion in irle. Clik on the ion to oserve the motion of point P round the unit irle. Oserve the grphs of P s height reltive to the -is, nd then P s horizontl displement from the -is. The resulting grphs re those of = sin t nd = os t. Suppose P moves ntilokwise round the unit irle with onstnt liner speed of unit per seond. After ¼ seonds, P will trvel ¼ units whih is one full revolution. So, fter t seonds P will trvel through t rdins, nd t time t, P is t (os t, sin t). The ngulr veloit of P is the time rte of hnge in AOP. Angulr veloit is onl meningful in motion long irulr or elliptil r. For the emple ove, the ngulr veloit of P is dµ dt nd dµ =rdin per seond. dt If we let l e the r length AP, the liner speed of P dl is the time rte of hnge in l, or dt. For the emple ove, l = µr = µ = µ nd dl dt = dµ =unit per seond. dt DEMO P( os, sin) P l A (, ) A DERIVATIVES OF TRIGONOMETRIC FUNCTIONS INVESTIGATION DERIVATIVES OF sint AND ost Our im is to use omputer demonstrtion to investigte the derivtives of sin t nd os t. Wht to do: Clik on the ion to oserve the grph of = sin t. A tngent with t-step of length unit moves ross the urve, nd its -step is trnslted onto the grdient grph. Suggest the derivtive of the funtion = sin t. DERIVATIVES DEMO

111 DERIVATIVES OF TRIGONOMETRIC FUNCTIONS (Chpter ) 57 Repet the proess in for the grph of = os t. Hene suggest the derivtive of the funtion = ost. From the investigtion ou m hve dedued tht d dt (sin t) = os t nd d (os t) = sin t. dt We will now demonstrte these derivtives using lger. THE DERIVATIVE OF sin A& os ( h), sin( h)* C( os, sin) B h A' C' Consider f(µ) = sin µ. Now f(µ + h) f(µ) h sin(µ + h) sin µ = h = AA CC h = AA BA h = AB h = AB... () r AC Now ngle Á = OAC OAA But OAC = ¼ h ) Á = ¼ h = µ + h fse ngle of isoseles tringle OACg ³ ¼ µ h on simplifing As h!, Á! µ nd os Á! os µ... () Also, s h!, ) ) sin(µ + h) sin µ h sin(µ + h) sin µ h sin(µ + h) sin µ h! AB AC ) f (µ) = os µ fr AC! ACg! os Á fusing ABCg! os µ fusing ()g For in rdins, if f() = sin then f () = os.

112 57 DERIVATIVES OF TRIGONOMETRIC FUNCTIONS (Chpter ) THE DERIVATIVE OF os Consider ) = os ) = sin( ¼ ) f sin( ¼ ) = os g ) = sin u where u = ¼ d d = d du du d = os u ( ) = os u = os( ¼ ) = sin fhin ruleg For in rdins, if f() = os then f () = sin. THE DERIVATIVE OF tn Consider = tn = sin os We let u = sin, v = os so ) du dv = os, d d = sin d d = u v uv v fquotient ruleg os os sin ( sin ) = [os ] = os + sin os = os f sin + os =g DERIVATIVE DEMO is often lled os sent or se, so d d (tn ) = se. Summr: Funtion Derivtive sin os os sin tn os THE DERIVATIVES OF sin[f()], os[f()] AND tn[f()] Given = sin[f()] we n write = sin u where u = f(). Now d d = d du du d fhin ruleg = os u f () = os[f()] f ()

113 DERIVATIVES OF TRIGONOMETRIC FUNCTIONS (Chpter ) 57 We n perform the sme proedure for os[f()] nd tn[f()], giving the results in the following tle. Funtion sin[f()] os[f()] tn[f()] Derivtive os[f()] f () sin[f()] f () f () os [f()] Emple Self Tutor Differentite with respet to : sin tn () If = sin then the produt rule d = () sin +() os d = sin + os If = tn () = [tn()] then the hin rule d d = 8[tn()] d d [tn()] = 8 tn() = sin() os () os () Emple Find the eqution of the tngent to = tn t the point where = ¼. Self Tutor Let f() = tn so f( ¼ ) = tn ¼ = f () = os so f ( ¼ )= os ( ¼ ) = ( p ) = At ( ¼, ), the tngent hs grdient ) the eqution is ¼ = whih is = ¼ or = +( ¼ ) Emple Find the rte of hnge in the re of tringle ABC s µ hnges, t the time when µ =6 o. A m B Self Tutor m C

114 57 DERIVATIVES OF TRIGONOMETRIC FUNCTIONS (Chpter ) Are A = sin µ fare = sin C or sin Ag ) A = 6 sin µ da µ must e in ) = 6 os µ dµ When µ = ¼, os µ = da ) dµ =m per rdin rdins so the dimensions re orret. EXERCISE A Find d d for: = sin() = sin + os = os() sin d = sin( +) e = os( ) f = tn(5) g = sin os h = tn(¼) i = sin os() Differentite with respet to : + os tn sin e os d e sin ³ e ln(sin ) f e tn g sin() h os i tn() j os k sin Differentite with respet to : l tn sin( ) os( p ) p os d sin e os f os sin() g os(os ) h os () i sin j os() Suppose f() = sin sin. k sin () Show tht f () = os os. Find f (). l 8 tn ( ) d 5 If = sin( +), show tht + =. d If = sin + os, show tht + = where represents d d. Show tht the urve with eqution = os nnot hve horizontl tngents. + sin 6 Find the eqution of: the tngent to = sin t the origin the tngent to = tn t the origin the norml to = os t the point where = ¼ 6 d the norml to = sin() t the point where = ¼.

115 DERIVATIVES OF TRIGONOMETRIC FUNCTIONS (Chpter ) On the Indonesin ost, the depth of wter t time t hours fter midnight is given d = 9: + 6:8 os(:57t) metres. Is the tide rising or flling t 8: m? Wht is the rte of hnge in the depth of wter t 8: m? 8 The voltge in iruit is given V (t) = sin(¼t) where t is the time in seonds. At wht rte is the voltge hnging: when t =: when V (t) is mimum? 9 A piston is operted rod [AP] tthed to flwheel of rdius m. AP =m. P hs oordintes (os t, sin t) nd point Ais(, ): Show tht = p sin t os t. Find the rte t whih is hnging t the instnt when: i t = ii t = ¼ iii t = ¼ piston For eh of the following funtions, determine the position nd nture of the sttionr points on the intervl 6 6 ¼, then show them on grph of the funtion. f() = sin f() = os() f() = sin d f() =e sin e f() = sin() + os A prtile P moves long the -is with position given (t) = os t m where t is the time in seonds. Stte the initil position, veloit nd elertion of P. Desrie the motion when t = ¼ seonds. Find the times when the prtile reverses diretion on <t<¼ position of the prtile t these instnts. nd find the d When is the prtile s speed inresing on 6 t 6 ¼? Show tht =e sin hs mimum when = ¼. A m m P( os t, sint) t (, ) B OPTIMISATION WITH TRIGONOMETRY Emple Self Tutor Two orridors meet t right ngles nd re m nd m wide respetivel. µ is the ngle mrked on the given figure. [ AB] is thin metl tue whih must e kept horizontl nd nnot e ent s it moves round the orner from one orridor to the other. B m Show tht the length AB is given L = os µ + sin µ. A m

116 576 DERIVATIVES OF TRIGONOMETRIC FUNCTIONS (Chpter ) dl ³ Show tht dµ = when µ = tn ¼ : o. ³ q Find L when µ = tn nd omment on the signifine of this vlue. q DEMO os µ = nd sin µ = so = os µ nd = sin µ ) L = + = os µ + sin µ L = [os µ] + [sin µ] ) Thus dl dµ = [os µ] ( sin µ) [sin µ] os µ = sin µ os µ os µ sin µ = sin µ os µ os µ sin µ dl dµ =, sin µ = os µ ) tn µ = q ) tn µ = µ q nd so µ = tn ¼ : o A B Sign digrm of dl dµ : 6. 9 dl dµ ¼ :9 <, dl dµ ¼ 9:6 > Thus, AB is minimised when µ ¼ : o. At this time L ¼ 7: metres, so if we ignore the width of the rod then the gretest length of rod le to e horizontll rried round the orner is 7: m. EXERCISE B A irulr piee of tinplte of rdius m hs segments removed s illustrted. If µ is the mesure of ngle COB, show tht the remining re is given A = 5(µ + sin µ). Hene, find µ to the nerest of degree when the re A is mimum. m B C

117 DERIVATIVES OF TRIGONOMETRIC FUNCTIONS (Chpter ) 577 A smmetril gutter is mde from sheet of metl m wide ending it twie s shown. Dedue tht the ross-setionl re is given A = os µ( + sin µ): da Show tht dµ = when sin µ = or : For wht vlue of µ does the gutter hve mimum rring pit? m end view Hieu n row ot ross irulr lke of rdius km t km h. He n wlk round the edge of the lke t 5 km h. Wht is the longest possile time Hieu ould tke to get from P to R rowing from P to Q nd then wlking from Q to R? P km Q km R Fene [AB] is m high nd is m from house. [XY] is ldder whih touhes the ground t X, the house t Y, nd the fene t B. If L is the length of [XY], show tht L = os µ + sin µ dl Show tht dµ = sin µ os µ sin. µ os µ X Wht is the length of the shortest ldder [XY] whih touhes t X, B nd Y? B A m m Y house 5 In Emple, suppose the orridors re those in hospitl nd re m wide nd m wide respetivel. Wht is the mimum length of thin metl tue tht n e moved round the orner? Rememer it must e kept horizontl nd must not e ent. m m REVIEW SET Differentite with respet to : sin(5)ln() sin() os() e tn µ d sin() e ln f sin(5)ln() os

118 578 DERIVATIVES OF TRIGONOMETRIC FUNCTIONS (Chpter ) Show tht the eqution of the tngent to = tn t = ¼ is ( + ¼) = ¼. Find f () nd f () for: f() = sin os() f() = p os() A prtile moves in stright line long the -is with position given (t) = + sin(t) m fter t seonds. Find the initil position, veloit nd elertion of the prtile. Find the times when the prtile hnges diretion during 6 t 6 ¼ seonds. Find the totl distne trvelled the prtile in the first ¼ seonds. 5 Consider f() = p os for 6 6 ¼. For wht vlues of is f() meningful? Find f () nd hene find intervls where f() is inresing nd deresing. Sketh the grph of = f() on 6 6 ¼. 6 A ork os up nd down in uket of wter suh tht the distne from the entre of the ork to the ottom of the uket is given s(t) = + os(¼t) m, t > seonds. Find the ork s veloit t times t =,,,, s. Find the time intervls when the ork is flling. st () 7 A light ul hngs from the eiling t height h metres ove the floor, diretl ove point N. At n point A on the floor whih is metres from the light ul, the illumintion I is given p 8 os µ I = units. If NA =metre, show tht t A, I = p 8 os µ sin µ. The light ul m e lifted or lowered to hnge the intensit t A. Assuming NA = metre, find the height the ul should e ove the floor for gretest illumintion t A. eiling L light ul h floor N A 8 Find the eqution of: the tngent to = sin t the point where = ¼ the norml to = os( ) t the point where = ¼. 9 The funtion f is defined f : 7! sin os, 6 6 ¼. Write down n epression for f() in the form k sin. Solve f () =, giving et nswers.

119 Chpter Integrtion Sllus referene: 7., 7.5 Contents: A B C D E Antidifferentition The fundmentl theorem of lulus Integrtion Integrting f ( + ) Definite integrls

120 58 INTEGRATION (Chpter ) In the previous hpters we used differentil lulus to find the derivtives of mn tpes of funtions. We lso used it in prolem solving, in prtiulr to find the grdients of grphs nd rtes of hnges, nd to solve optimistion prolems. In this hpter we onsider integrl lulus. This involves ntidifferentition whih is the reverse proess of differentition. Integrl lulus lso hs mn useful pplitions, inluding: ² finding res where urved oundries re involved ² finding volumes of revolution ² finding distnes trvelled from veloit funtions ² finding hdrostti pressure ² finding work done fore ² finding entres of mss nd moments of inerti ² solving prolems in eonomis nd iolog ² solving prolems in sttistis ² solving differentil equtions. A ANTIDIFFERENTIATION In mn prolems in lulus we know the rte of hnge of one vrile with respet to nother, ut we do not hve formul whih reltes the vriles. In other words, we know d, ut we need to know in terms of. d Emples of prolems we need to solve inlude: ² The grdient funtion f () of urve is + nd the urve psses through the origin. Wht is the funtion = f()? dt ² The rte of hnge in temperture is dt =e t o C per minute where t >. Wht is the temperture funtion given tht initill the temperture ws o C? The proess of finding from d d or f() from f () is the reverse proess of differentition. We ll it ntidifferentition. differentition or f() ntidifferentition or integrtion d d or f ()

121 Consider the following prolem: If INTEGRATION (Chpter ) 58 d d =, wht is in terms of? From our work on differentition, we know tht when we differentite power funtions the inde redues. We hene know tht must involve. Now if = then d d =, so if we strt with = then d d =. However, in ll of the ses = +, = + nd = 7 we find d d =. In ft, there re infinitel mn funtions of the form = + where is n ritrr onstnt whih will give d d =. Ignoring the ritrr onstnt, we s tht is the ntiderivtive of. It is the simplest funtion whih when differentited gives. If F () is funtion where F () =f() we s tht: ² the derivtive of F () is f() nd ² the ntiderivtive of f() is F (): Emple Find the ntiderivtive of: e p Self Tutor We know tht the derivtive of involves. d Sine = d, d d = ) the ntiderivtive of is. Sine d e = e, d ) the ntiderivtive of e is e : d d e = e =e p = Now d d ( )= ) ) the ntiderivtive of p is p : d d ( )=( ) = EXERCISE A Find the ntiderivtive of: i ii iii 5 iv v vi vii From our nswers in, predit generl rule for the ntiderivtive of n.

122 58 INTEGRATION (Chpter ) Find the ntiderivtive of: i e ii e 5 iii e iv e : v e ¼ vi e From our nswers in, predit generl rule for the ntiderivtive of e k where k is onstnt. Find the ntiderivtive of: 6 + differentiting + e + differentiting e + p differentiting p d ( +) differentiting ( +) : B THE FUNDAMENTAL THEOREM OF CALCULUS Sir Is Newton nd Gottfried Wilhelm Leiniz showed the link etween differentil lulus nd the definite integrl or limit of n re sum we sw in Chpter 6. This link is lled the fundmentl theorem of lulus. The eut of this theorem is tht it enles us to evlute omplited summtions. We hve lred oserved in Chpter 6 tht: If f() is ontinuous positive funtion on n intervl 6 6 then the re under the urve etween = nd = is Z f() d. ƒ( ) INVESTIGATION Consider the onstnt funtion f() =5. The orresponding re funtion is A(t) = Z t 5 d = shded re in grph =(t )5 =5t 5 THE AREA FUNCTION t t ) we n write A(t) in the form F (t) F () where F (t) =5t or equivlentl, F () =5. Wht to do: Wht is the derivtive F () of the funtion F () =5? How does this relte to the funtion f()?

123 INTEGRATION (Chpter ) 58 Consider the simplest liner funtion f() =. The orresponding re funtion is A(t) = Z t d = shded re in grph µ t + = (t ) Cn ou write A(t) in the form F (t) F ()? If so, wht is the derivtive F ()? How does it relte to the funtion f()? Consider f() = +. The orresponding re funtion is A(t) = Z t ( +)d = shded re in grph µ t ++ + = (t ) Cn ou write A(t) in the form F (t) F ()? If so, wht is the derivtive F ()? How does it relte to the funtion f()? Repet the proedure in nd for finding the re funtions of f() = + f() =5 Do our results fit with our erlier oservtions? 5 If f() = + +5, predit wht F () would e without performing the lgeri proedure. t t t t t t From the investigtion ou should hve disovered tht, for f() >, Z t f() d = F (t) F () where F () =f(). F () is the ntiderivtive of f(). The following rgument shows wh this is true for ll funtions f() >. Consider funtion = f() whih hs ntiderivtive F () nd n re funtion A(t) whih is the re from = to = t, Z t ƒ( ) So, A(t) = f() d. A(t) is lerl n inresing funtion nd A() =... () At () t

124 58 INTEGRATION (Chpter ) Now onsider nrrow strip of the region etween = t nd = t + h. The re of this strip is A(t + h) A(t). ƒ( ) Sine the nrrow strip is ontined within two retngles then re of smller re of lrger 6 A(t + h) A(t) 6 retngle retngle t th ) hf(t) 6 A(t + h) A(t) 6 hf(t + h) ) f(t) 6 A(t + h) A(t) h 6 f(t + h) Tking the limit s h! gives h ƒ( ) f(t) 6 A (t) 6 f(t) ) A (t) =f(t) The re funtion A(t) is n ntiderivtive of f(t), so A(t) nd F (t) differ onstnt. ƒ( t) ƒ( th) ) A(t) =F (t)+ Letting t = ; A() =F ()+ But A() = ffrom ()g so = F () ) A(t) =F (t) F () t th enlrged strip ) letting t =, Z f() d = F () F () This result is in ft true for ll ontinuous funtions f(), nd n e stted s the fundmentl theorem of lulus: For ontinuous funtion f() with ntiderivtive F (), Z f() d = F () F (). The fundmentl theorem of lulus hs mn pplitions eond the lultion of res. For emple, given veloit funtion v(t) we know tht ds dt = v. So, s(t) is the ntiderivtive of v(t) nd the fundmentl theorem of lulus, Z t t v(t) dt = s(t ) s(t ) gives the displement over the time intervl t 6 t 6 t.

125 INTEGRATION (Chpter ) 585 PROPERTIES OF DEFINITE INTEGRALS The following properties of definite integrls n ll e dedued from the fundmentl theorem of lulus: ² ² ² ² Z Z Z Z f() d = f() d = f() d + Z Z [f() g()]d = f() d f() d = Z Z f() d f() d ² ² Z Z Z g() d d= ( ) f() d = Z f is onstntg f() d Emple proof: Z Z f() d + f() d = F () F ()+F() F () = F () F () = Z f() d Z f() d + ƒ( ) A A Z f() d = A + A = Z f() d Emple Use the fundmentl theorem of lulus to find the re: etween the -is nd = from = to = etween the -is nd = p from = to =9: Self Tutor f() = hs ntiderivtive F () = ) the re = Z d = F () F () = = units

126 586 INTEGRATION (Chpter ) f() = p = hs ntiderivtive 9 F () = ) the re = = p Z 9 d = F (9) F () = 7 =7 units EXERCISE B Use the fundmentl theorem of lulus to show tht: d e Z Z Z Z Z f() d = nd eplin the result grphill d= ( ) where is onstnt Z f() d = f() d = Z [f()+g()] d = f() d f() d where is onstnt Z Z f() d + g() d Use the fundmentl theorem of lulus to find the re etween the -is nd: = from = to = = from = to = = + + from = to = d = p from = to = e = e from = to =:5 f = p from = to = g = from = to =:5: Using tehnolog, find orret to signifint figures, the re etween the -is nd: = e from = to =:5 =(ln) from = to = = p 9 from = to =: Chek eh nswer using tehnolog.

127 INTEGRATION (Chpter ) 587 Use the fundmentl theorem of lulus to show tht Z ( f()) d = Z Use the result in to show tht if f() 6 for ll on 6 6 then the shded re = Z f() d f() d. ( ) d Clulte the following integrls, nd give grphil interprettions of our nswers: i Z ( ) d ii Z ( ) d iii Use grphil evidene nd known re fts to find Z Z d ³ p d: C INTEGRATION Erlier we showed tht the ntiderivtive of ws. We showed tht n funtion of the form + where is n onstnt, hs derivtive. We s tht the integrl of is + nd write Z d = +. We red this s the integrl of with respet to. In generl, if F () =f() then Z f() d = F ()+. This proess is known s indefinite integrtion. It is indefinite euse it is not eing pplied over prtiulr intervl. DISCOVERING INTEGRALS Sine integrtion is the reverse proess of differentition we n sometimes disover integrls differentition. For emple: ² if F () = then F () = Z ) d = + ² if F () = p = then F () = = p Z ) p d = p +

128 588 INTEGRATION (Chpter ) The following rules m prove useful: ² An onstnt m e written in front of the integrl sign. Z Z kf() d = k f() d, k is onstnt Proof: Consider differentiting kf() where F () =f(). d d (kf()) = kf () =kf() Z ) kf() d = kf() Z = k f() d ² The integrl of sum is the sum of the seprte integrls. This rule enles us to integrte term term. Z Z Z [f()+g()] d = f() d + g() d Emple If = +, find d Z d : Hene find ( + ) d: If = +, d then d = +6 Z ) +6 d = + + Z ) ( + ) d = + + Z ) ( + ) d = + + Z ) ( + ) d = + + Self Tutor EXERCISE C. If = 7, find d Z : Hene find 6 d. d If = +, find d Z : Hene find ( +) d. d If = e +, find d Z : Hene find e + d. d If =( +) find d Z. Hene find ( +) d: d 5 If = p, find d Z pd. : Hene find d We n lws hek tht n integrl is orret differentiting the nswer. It should give us the integrnd, the funtion we originll integrted.

129 6 If = p, find d Z : Hene find d 7 If = os, find d. Hene find d 8 If = sin( 5), find 9 Prove the rule Find d d Z d. Hene find d Z [f() +g()] d = if = p : Hene find B onsidering d d ln(5 + ), find p d. Z sin d. Z os( 5) d. Z f() d + Z Z p d. INTEGRATION (Chpter ) 589 g() d d. RULES FOR DIFFERENTIATION In erlier hpters we developed rules to help us differentite funtions more effiientl. Following is summr of these rules: Funtion Derivtive Nme, onstnt m +, m nd re onstnts m n n n power rule u() u () u()+v() u ()+v () ddition rule u()v() u ()v()+u()v () produt rule u() v() = f(u) where u = u() e e f() ln ln f() [f()] n sin os tn u ()v() u()v () [v()] d d = d du du d e e f() f () f () f() n[f()] n f () os sin os quotient rule hin rule

130 59 INTEGRATION (Chpter ) These rules or omintions of them n e used to differentite lmost ll funtions. However, the tsk of finding ntiderivtives is not so es nd nnot e written s simple list of rules s we did ove. In ft, huge ooks of different tpes of funtions nd their integrls hve een written. Fortuntel our ourse is restrited to few speil ses. RULES FOR INTEGRATION For k onstnt, d (k + )=k ) d Z kd= k + If n 6=, µ d n+ d n + + (n +)n = n + = n d d (e + )=e ) ) Z Z n d = n+ +, n 6= n + e d = e + If >, d d (ln + )= d (sin + ) = os ) d d ( os + ) = sin + ) d ) Z d =ln +, > Z os d = sin + Z sin d = os + Funtion Integrl k, onstnt k + n n+ +, n 6= n + e e + ln +, > os sin + sin os + In eh se, is n ritrr onstnt lled the onstnt of integrtion or integrting onstnt. You should rememer tht ou n lws hek our integrtion differentiting the resulting funtion.

131 INTEGRATION (Chpter ) 59 Emple Find: Z ( +5)d Z µ p d Self Tutor Z ( +5)d = +5 + Z µ p d Z = ( ) d = + = + Emple 5 Integrte with respet to : Self Tutor sin os +e Z [ sin os ]d Z +e d =( os ) sin + = os sin + = ln +e + provided > There is no produt or quotient rule for integrtion. Consequentl we often hve to rr out multiplition or division efore we integrte. Emple 6 Find: Z µ + d Z µ + d Z = µ9 ++ d Z = (9 ++ ) d = = + + Z µ p d = = Z µ p d Z µ p p d Z ( ) d = = 5 p p + Self Tutor We epnd the rkets nd simplif to form tht n e integrted.

132 59 INTEGRATION (Chpter ) We n find the onstnt of integrtion if we re given point on the urve. Emple 7 Self Tutor Find f() given tht f () = + nd f() =. Sine f () = +, Z f() = ( +)d ) f() = + + But f() =, so ++ = nd so = Thus f() = + + Emple 8 Self Tutor Find f() given tht f () = sin p nd f() =. f() = Z h i sin d ) f() = ( os ) ) f() = os + But f() = os + ) = + nd so =6 Thus f() = os +6: + If we re given the seond derivtive we need to integrte twie to find the funtion. This retes two integrting onstnts, so we need two other fts out the urve in order to determine these onstnts. Emple 9 Self Tutor Find f() given tht f () =, f () = nd f() =. If f () = then f () = + fintegrting with respet to g ) f () = + But f () = so + = nd so = Thus f () =

133 INTEGRATION (Chpter ) 59 ) f() = + d fintegrting ging ) f() = + d But f()= so +d = nd so d =6 Thus f() = +6 EXERCISE C. Find: Z ( +)d d g Z µ p d Z µ + d Integrte with respet to : Z ( p + e ) d Z µ e p + d Z µ h + e d f i Z µ e d Z µ + Z µ 5e + sin os sin os + e d p sin e Find: Z ( + ) d ( ) + os f sin + p Z µ p p d Z Z d p d e ( +) d f Z Z + g p d h p d i Find: Z µ p + os d 5 Find if: d 6 Find if: Z e t sin t dt d d =6 d d = d d = e d d =e 5 d d =( ) d d = p p 7 Find f() if: f Z µe d Z + d Z ( +) d Z µ os t dt t d d =5p d d = + d d = + 5 f () = 5 p + f () = p ( ) f () =e d d

134 59 INTEGRATION (Chpter ) 8 Find f() given tht: f () = nd f() = f () = + nd f() = 5 f () =e + p nd f()= d f () = p nd f()=. 9 Find f() given tht: f () = os nd f() = f () = os sin nd f ¼ = p Find f() given tht: f () = +, f ()= nd f() = 7 f () =5 p + p, f ()= nd f()=5 d f () = os, f ( ¼ )= nd f()= f () = nd the points (, ) nd (, 5) lie on the urve. D INTEGRATING f( + ) In this setion we del with integrls of funtions whih re omposite with the liner funtion +. µ d Notie tht d e+ = e+ = e + ) Z e + d = e + + Likewise if n 6=, µ d ( + )n+ = d (n +) (n +) (n + )( + )n, =( + ) n ) Z ( + ) n d = ( + ) n+ n + +, n 6= Also, µ d ln( + ) = µ = d + + for + > ) Z + d = ln( + ) +, + >

135 We n perform the sme proess for the irulr funtions: d (sin( + )) = os( + ) d Z ) os( + ) d = sin( + )+ Z ) os( + ) d = sin( + )+ INTEGRATION (Chpter ) 595 So, Likewise we n show Z Z os( + ) d = sin( + )+ sin( + ) d = os( + )+ SUMMARY OF INTEGRALS FOR FUNCTIONS OF THE FORM f( + ) Funtion Integrl e + e+ + ( + ) n ( + ) n+ +, n 6= n + ln( + )+, + > + os( + ) sin( + )+ sin( + ) os( + )+ Emple Find: Z ( +) d Z p d Self Tutor Z ( +) d Z p d = ( +)5 5 + Z = ( ) d = ( +)5 + = ( ) + = p +

136 596 INTEGRATION (Chpter ) Emple Z Find: (e e ) d Z (e e ) d =( )e ( )e + = e + e + Z d Z d Z = d ³ = Self Tutor ln( )+, > = ln( )+, < Emple Integrte with respet to : sin() + os( + ¼) Self Tutor Z ( sin() + os( + ¼)) d = ( os()) + sin( + ¼)+ = os()+ sin( + ¼)+ INTEGRALS OF POWERS OF CIRCULAR FUNCTIONS Integrls involving sin ( + ) nd os ( + ) n e found first using sin µ = os(µ) or os µ = + os(µ). These formule re simpl rerrngements of os(µ) formule. For emple, ² sin ( ¼ ) eomes os(6 ¼) ³ ³ ² os eomes + os = + os : Emple Integrte ( sin ). Z Z = ( sin ) d ( sin + sin ) d Self Tutor = Z sin + os() d = Z 9 sin os() d = 9 + os sin()+ = 9 + os sin()+

137 INTEGRATION (Chpter ) 597 EXERCISE D Find: Z Z d Z g ( +5) d ( ) 7 d e ( ) d h Z Z ( ) d Z Z p d f Z p d ( ) d p 5 d Integrte with respet to : sin() os( )+ os d sin() e e sin + ¼ 6 f os ¼ g os() + sin() h sin() + 5 os() i os(8) sin Find = f() given d d = p 7 nd tht = when =8. Funtion f() hs grdient funtion p (, ). Find the point on the grph of = f() with -oordinte 8. Integrte with respet to : nd psses through the point os sin + os () d sin () e os () f ( + os ) 5 Find: Z Z d ( ) d ( ) d e Z Z ( ) d p 5 d f Z Z ( ) d ( +) d 6 Find: Z e +5e d d g Z Z d e (e + e ) d h Z e 5 d Z Z 5 d f (e +) d i Z e 7 d Z µ e Z µ 5 d + d 7 Find given tht: d d =( e ) d = + d + d d = e +

138 598 INTEGRATION (Chpter ) Z Z 8 To find d, Tr s nswer ws d = ln()+, > Z Z nd Ndine s nswer ws d = d = ln()+, >. Whih of them hs found the orret nswer? Prove our sttement. 9 Suppose f () =p sin, f()= nd f(¼) =. Find p nd f(). Consider funtion g suh tht g () = sin. Show tht the grdients of the tngents to = g() when = ¼ nd = ¼ re equl. If f () =e nd f()=, find f(). If f () = nd f( ) =, find f(). If urve hs grdient funtion p + e nd psses through (, ), find the eqution of the funtion. Z Show tht (sin + os ) = + sin, nd hene determine (sin + os ) d. Show tht (os +) = os + os +, nd hene determine Z (os +) d. E DEFINITE INTEGRALS We hve seen from the fundmentl theorem of lulus: If F () is the ntiderivtive of f() where f() is ontinuous on the intervl 6 6 then the definite integrl of f() on this intervl is Z f() d = F () F (). Z f() d reds the integrl from = to = of f() with respet to. It is lled definite integrl euse there re lower nd upper limits for the integrtion nd it therefore results in numeril nswer. When lulting definite integrls we n omit the onstnt of integrtion s this will lws nel out in the sutrtion proess. It is ommon to write F () F () s [F ()], nd so Z f()d =[F ()] : Sine definite integrls result in numeril nswers, we re le to evlute them using tehnolog. Instrutions for this n e found in the grphis lultor instrutions t the strt of the ook.

139 INTEGRATION (Chpter ) 599 Emple Self Tutor Find Z ( +)d: Z ( +)d = + = ³ + () =(9+6) ( +) = ³ + () Emple 5 Evlute: Z ¼ sin d Z µ + d Self Tutor Z ¼ sin d =[ os ] ¼ =( os ¼ ) ( os ) = + Z µ + d = +ln fsine >g = (6 + ln ) ( + ln ) =5+6ln = Some integrls re diffiult or even impossile to evlute. In these ses ou re epeted to use grphis lultor to evlute the integrl. Emple 6 Self Tutor Evlute Z 5 e d to n ur of signifint figures. Z 5 e d ¼ 586:

140 6 INTEGRATION (Chpter ) EXERCISE E. Evlute the following nd hek with our grphis lultor: d g j Z Z ¼ 6 Z Z 6 d os d d p d e h k Z Z Z ¼ ( ) d µ p d ¼ sin d i Z e d f l Z Z 9 Z Z ¼ 6 e d p d (e +) d sin() d m Z ¼ os d n Z ¼ sin d Evlute using tehnolog: Z ln d Z e d Z ¼ 6 ¼ sin( p ) d PROPERTIES OF DEFINITE INTEGRALS Erlier in the hpter we proved the following properties of definite integrls using the fundmentl theorem of lulus: ² ² ² ² Z Z Z Z Z [ f()] d = Z f()d = f() d + Z [f()+g()]d = f() d f()d, f() d = Z Z is n onstnt f()d + f()d Z g()d EXERCISE E. Use questions to to hek the properties of definite integrls. Find: Find: Find: Z Z Z Z p d nd ( p ) d d Z ( ) d d Z Z Z 7 d d ( ) d nd d Z Z Z ( 7 ) d d ( ) d

141 Find: Z d Z 5 Evlute the following integrls using re interprettion: Z Z f() d f() d d Z 7 Z 7 f() d f() d INTEGRATION (Chpter ) 6 Z p d ( + p ) d ( ) 6 6 Evlute the following integrls using re interprettion: Z Z 8 6 f() d f() d d Z 6 Z 8 f() d f() d ( ) Write s single integrl: 8 If Z Z If Z f() d + g() d + Z 9 Given tht d Z Z 7 Z 8 f() d f() d = nd f() d =5, Z f() d k suh tht Z 9 g() d + g() d 8 Z 6 f() d =, Z Z 6 f() d =, f() d = nd find Z 6 determine the vlue of: Z kf() d =7 If g() = nd g()=5, lulte Z 6 ( + f()) d Z g () d. f() d: f() d =7, find Z Z f() d f() d:

142 6 INTEGRATION (Chpter ) REVIEW SET A Integrte with respet to : p Find the et vlue of: Z 5 B differentiting = p, find Find the vlues of suh tht Z NON-CALCULATOR sin( 5) d e p d Z ¼ Z p d: os d= p, <<¼. ³ os d ³ 5 Integrte with respet to : sin ( os ) Z 6 B differentiting ( + ), find ( + ) (6 +)d: 7 If Z Z f() d =, (f()+)d determine: Z Z f() d f() d 8 If Z e d = e, find in the form ln k. 9 Given tht f () = sin(), f ( ¼ )= nd f()=, find the et vlue of f( ¼ ). Evlute: Z ¼ 6 ³ sin d REVIEW SET B CALCULATOR Find if: d d =( ) d d = e Evlute orret to signifint figures: Find Z e d d (ln ) nd d hene find Z p + d Z ln d: A urve = f() hs f () =8 +. Find f() if f() = nd f() =.

143 INTEGRATION (Chpter ) 6 5 Evlute the following orret to 6 signifint figures: Z p + d Z e + d 6 Suppose f () = + nd f() = f()=. Find: f() the eqution of the norml to = f() t =. 7 Find (e +) using the inomil epnsion. Hene find the et vlue of Z (e +) d. Chek our nswer to using tehnolog. REVIEW SET C Find: Z µe + d Z µ p p d Z +e d Given tht f () = + nd f()=, find f(). Find the et vlue of Evlute Z ¼ Z ³ os d p d. 5 Find d d (e sin ) nd hene find Z 6 Find ( +) n d for ll integers n. Z ¼ e (os sin ) d 7 A funtion hs grdient funtion p + p nd psses through the points (, ) 8 nd (, ). Find nd hene eplin wh the funtion = f() hs no sttionr points. Z ( + +)d = 7. Find.

144 6 INTEGRATION (Chpter )

145 Chpter Applitions of integrtion Sllus referene: 7.5, 7.6 Contents: A B C D Finding res etween urves Motion prolems Prolem solving integrtion Solids of revolution

146 66 APPLICATIONS OF INTEGRATION (Chpter ) We hve lred seen how definite integrls n e relted to the res etween funtions nd the -is. In this hpter we eplore this reltionship further, nd onsider other pplitions of integrl lulus suh s motion prolems. INVESTIGATION Z f()d AND AREAS Does Z f() d lws give us n re? Wht to do: Find Z d nd Z d. Eplin wh the first integrl in gives n re, wheres the seond integrl does not. Grphil evidene is essentil. Find Z d nd eplin wh the nswer is negtive. Chek tht Z Z Z d + d = d. A FINDING AREAS BETWEEN CURVES We hve lred estlished in Chpter tht: If f() is positive nd ontinuous on the intervl 6 6, then the re ounded = f(), the -is, nd the vertil lines = nd = is given A = Z f() d or Z d: ƒ( ) AREAS BETWEEN CURVES AND THE -AXIS Emple Self Tutor Find the re of the region enlosed =, the -is, = nd = using: geometri rgument integrtion.

147 APPLICATIONS OF INTEGRATION (Chpter ) 67 Are = 8 =6units Are = Z d = = =6units Emple Self Tutor Find the re of the region enlosed = +, the -is, = nd =. X Are = = Z ( +)d + = = units It is helpful to sketh the region. We n hek this result using grphis lultor or grphing pkge. GRAPHING PACKAGE Emple Find the re enlosed one rh of the urve = sin. Self Tutor The period of = sin is ¼ = ¼. ) the first positive -interept is ¼. sin The required re = Z ¼ sin d = ( os ) ¼ ¼ os = = (os ¼ os ) =unit

148 68 APPLICATIONS OF INTEGRATION (Chpter ) EXERCISE A. Find the re of eh of the regions desried elow using: i geometri rgument ii integrtion =5, the -is, = 6 nd = =, the -is, = nd =5 =, the -is, = nd = d =, the -is, = nd = Find the et vlue of the re of the region ounded : =, the -is nd = = sin, the -is, = nd = ¼ =, the -is, = nd = d = e, the -is, the -is nd = e the -is nd the prt of =6+ ove the -is f the es nd = p 9 g =, the -is, = nd = h =, the -is, = nd = i = p, the -is nd = j = e + e, the -is, = nd = Use tehnolog to hek our nswers. Find the re enlosed one rh of the urve = os. Use tehnolog to find the re of the region ounded : =ln, the -is, = nd = = sin, the -is, = nd = ¼ = e, the -is, = nd =:8. AREA BETWEEN TWO FUNCTIONS If two funtions f() nd g() interset t = nd =, nd f() > g() for ll 6 6, then the re of the shded region etween their points of intersetion is given Z [f() g()] d. Alterntivel, if we desrie the upper nd lower funtions s = U nd = L respetivel, then the re is Z [ U L ] d. ƒ() or U g()or L

149 APPLICATIONS OF INTEGRATION (Chpter ) 69 Proof: If we trnslte eh urve vertill through k until it is ompletel ove the -is, the re does not hnge. ƒ( ) k g() k Are of shded region = = Z Z [f()+k] d [f() g()] d Z [g() +k] d We n see immeditel tht if = f() = then the enlosed re is Z or Z [ g()] d g()d: g() Emple Self Tutor Use Z [ U L ] d to find the re ounded the -is nd =. The urve uts the -is when = ) = ) ( ) = ) =or U ) the -interepts re nd. Are = = = Z Z Z [ U L ] d [ ( )] d ( ) d = = 8 () ) the re is units. LX

150 6 APPLICATIONS OF INTEGRATION (Chpter ) Emple 5 Self Tutor Find the re of the region enlosed = + nd = +. = + meets = + where + = + ) = ) ( + )( ) = ) = Are = = = = Z Z Z [ U L ] d [( +) ( + )] d ( ) d = = units Emple 6 Find the totl re of the regions ontined = f() nd the -is for f() = +. Self Tutor f() = + = ( + ) = ( )( +) ) = f() uts the -is t,, nd : Totl re = = Z ( + ) d + = 7 Z ( + ) d + CX = 5 6 units

151 APPLICATIONS OF INTEGRATION (Chpter ) 6 EXERCISE A. Find the et vlue of the re ounded : the -is nd = + the -is, = e nd = the -is nd the prt of = 8 + elow the -is d = os, the -is, = ¼ nd = ¼ e =, the -is, = nd = f = sin, the -is, = nd = ¼ g one rh of = sin nd the -is. Find the re of the region enlosed = nd =. Consider the grphs of = nd =. Sketh the grphs on the sme set of es. Find the oordintes of the points where the grphs meet. Find the re of the region enlosed the two grphs. Determine the re of the region enlosed = p nd =. 5 On the sme set of es, grph = e nd = e, showing es interepts nd smptotes. Find lgerill the points of intersetion of = e nd = e. Find the re of the region enlosed the two urves. 6 Determine etl the re of the region ounded =e, = e nd =. 7 On the sme set of es, drw the grphs of the funtions = nd =. Determine etl the re of the region enlosed these funtions. 8 Sketh the irle with eqution + =9. Eplin wh the upper hlf of the irle hs eqution = p 9. Hene, determine Z Chek our nswer using tehnolog. p 9 d without tull integrting the funtion. 9 Find the re enlosed the funtion = f() nd the -is for: f() = 9 f() = ( )( ) f() = 5 +: The illustrted urves re those of = sin nd = sin(). Identif eh urve. Find lgerill the oordintes of A. Find the totl re enlosed C nd C for 6 6 ¼. A Cz C

152 6 APPLICATIONS OF INTEGRATION (Chpter ) Eplin wh the totl re shded is not equl to Z 7 f() d. Wht is the totl shded re equl to in terms of integrls? 5 7 = f( ) The illustrted urves re those of A E = os() nd = os. Cz Identif eh urve. B C D Determine the oordintes of A, B, C, D nd E. C Show tht the re of the shded region is ¼ units. Find, orret to signifint figures, the res of the regions enlosed the urves: = e nd = = nd = The shded re is : units. Find k, orret to deiml ples. 5 The shded re is unit. Find, orret to deiml ples. k 6 The shded re is 6 units. Find the et vlue of. B MOTION PROBLEMS DISTANCES FROM VELOCITY GRAPHS Suppose r trvels t onstnt positive veloit of 6 km h for 5 minutes. We know the distne trvelled = speed time =6km h h =5km: When we grph speed ginst time, the grph is horizontl line nd it is ler tht the distne trvelled is the re shded. 6 speed ( kmh ) vt () Qr_ time ( t hours)

153 APPLICATIONS OF INTEGRATION (Chpter ) 6 So, the distne trvelled n lso e found the definite integrl Z 6 dt =5. Now suppose the speed dereses t onstnt rte so tht the r, initill trvelling t 6 km h, stops in 6 minutes. In this se the verge speed must e km h, so the distne trvelled =km h h =km But the tringle hs re = se ltitude = 6 = 6 speed ( kmh ) vt ()t qa_p_ time ( t hours) So, one gin the shded re gives us the distne trvelled, nd we n find it using the definite integrl Z (6 6t) dt =. These results suggest tht: distne trvelled = Z t t diretion. v(t) dt provided we do not hnge If we hve hnge of diretion within the time intervl then the veloit will hnge sign. In suh ses we need to dd the omponents of re ove nd elow the t-is. For veloit-time funtion v(t) where v(t) > on the v intervl t 6 t 6 t, distne trvelled = Z t v(t) dt. t tz t t For veloit-time funtion v(t) where v(t) 6 on the intervl t 6 t 6 t, distne trvelled = Z t v(t) dt. v tz t t t Emple 7 Self Tutor The veloit-time grph for trin journe is illustrted in the grph longside. Find the totl distne trvelled the trin. 6 v( kmh ) t () h

154 6 APPLICATIONS OF INTEGRATION (Chpter ) Totl distne trvelled = totl re under the grph = re A + re B + re C + re D + re E = (:)5 + (:) (:) + (:) + (:) =:5++++:5 =km 5 5 A B C D E..... re EXERCISE B. A runner hs the veloit-time grph shown. Find the totl distne trvelled the runner. 8 6 veloit ( ms) 5 5 time () s A r trvels long stright rod with the veloit 8 ( km h ) veloit-time funtion illustrted. 6 Wht is the signifine of the grph: i ove the t-is ii elow the t-is? t( h) Find the totl distne trvelled the r Find the finl displement of the r. A list rides off from rest, elerting t onstnt rte for minutes until she rehes km h. She then mintins onstnt speed for minutes until rehing hill. She slows down t onstnt rte to km h in one minute, then ontinues t this rte for minutes. At the top of the hill she redues her speed uniforml nd is sttionr minutes lter. Drw grph to show the list s motion. How fr hs the list trvelled? DISPLACEMENT AND VELOCITY FUNCTIONS In this setion we re onerned with motion in stright line,or liner motion. Rell tht for some displement funtion s(t) the veloit funtion is v(t) =s (t). So, given veloit funtion we n determine the displement funtion the integrl Z s(t) = v(t) dt

155 APPLICATIONS OF INTEGRATION (Chpter ) 65 Using the displement funtion we n quikl determine the displement in time intervl 6 t 6. Displement = s() s() = Z v(t) dt We n lso determine the totl distne trvelled in some time intervl 6 t 6. Consider the following emple: A prtile moves in stright line with veloit funtion v(t) =t m s. How fr does it trvel in the first seonds of motion? Z Now s(t) = (t ) dt = t t + for some onstnt. Clerl, the displement of the prtile in the first seonds is s() s() = = m. However, v(t) hs sign digrm: vt () t Sine the veloit funtion hnges sign t t =seonds, the prtile reverses diretion t this time. We need to rememer this reversl of diretion t t = when lulting the totl distne trvelled. We find the positions of the prtile t t =, t = nd t =: s() =, s() =, s() =. We n hene drw digrm of the motion: The totl distne trvelled is ( + ) m =5m. \Qw_ Summr: To find the totl distne trvelled given veloit funtion v(t) =s (t) on 6 t 6 : ² Drw sign digrm for v(t) so we n determine n hnges of diretion. ² Determine s(t) integrtion, inluding onstnt of integrtion. ² Find s() nd s(). Also find s(t) t eh time the diretion hnges. ² Drw motion digrm. ² Determine the totl distne trvelled from the motion digrm.

156 66 APPLICATIONS OF INTEGRATION (Chpter ) Emple 8 Self Tutor A prtile P moves in stright line with veloit funtion v(t) =t t + ms. How fr does P trvel in the first seonds in motion? Find the displement of P fter seonds. v(t) =s (t) =t t + =(t )(t ) ) sign digrm of v is: t Sine the signs hnge, P reverses diretion t t = nd t = seonds. Z Now s(t) = (t t +)dt = t t +t + Now s() = s() = ++ = s() = = + s() = = +5 Motion digrm: +\We_ +\T_ +5\Qe_ ) totl distne = = =5 m Displement = finl position originl position = s() s() = +5 =5 m So, the displement is 5 m to the right. VELOCITY AND ACCELERATION FUNCTIONS We know tht the elertion funtion is the derivtive of veloit, so (t) =v (t). So, given n elertion funtion, we n determine the veloit funtion the integrl Z v(t) = (t) dt: EXERCISE B. A prtile hs veloit funtion v(t) = t m s s it moves in stright line. Find the totl distne trvelled in the first seond of motion. Find the displement of the prtile t the end of one seond. Prtile P hs veloit v(t) =t t m s. Find the totl distne trvelled in the first seonds of motion. Find the displement of the prtile t the end of three seonds.

157 APPLICATIONS OF INTEGRATION (Chpter ) 67 A prtile moves long the -is with veloit funtion (t) =6t t units per seond. Find the totl distne trvelled in the time intervl: 6 t 6 seonds 6 t 6 seonds. A prtile moves in stright line with veloit funtion v(t) = os t ms. Show tht the prtile osilltes etween two points. Find the distne etween the two points in. 5 The veloit of prtile trvelling in stright line is given v(t) =5 e :5t ms, where t >, t in seonds. Stte the initil veloit of the prtile. Find the veloit of the prtile fter seonds. How long will it tke for the prtile s veloit to inrese to 5 ms? d Disuss v(t) s t!. e Show tht the prtile s elertion is lws positive. f Drw the grph of v(t) ginst t. g Find the totl distne trvelled the prtile in the first seonds of motion. 6 A trin moves long stright trk with elertion t ms. If the initil veloit of the trin is 5 ms, determine the totl distne trvelled in the first minute. 7 An ojet hs initil veloit ms s it moves in stright line with elertion C funtion ms. Show tht s t inreses the ojet pprohes limiting veloit. Find the totl distne trvelled in the first seonds of motion. e t PROBLEM SOLVING BY INTEGRATION When we studied differentil lulus, we sw how to find the rte of hnge of funtion differentition. In prtil situtions it is sometimes esier to mesure the rte of hnge of vrile, for emple, the rte of wter flow through pipe. In suh situtions we n use integrtion to find funtion for the quntit onerned. Emple 9 The mrginl ost of produing urns per week is given C () =:5 : +: 6 dollrs per urn provided 6 6 : The initil osts efore prodution strts re $85. Find the totl ost of produing urns per d. The mrginl ost is C () =:5 : +: 6 dollrs per urn Z ) C() = (:5 : +: 6 ) d Self Tutor

158 68 APPLICATIONS OF INTEGRATION (Chpter ) ) C() =:5 : +: 6 + =:5 : +: + But C() = 85, so = 85 ) C() =:5 : +: + 85 ) C() = :5() :() +: () + 85 = So, the totl ost is $: Emple A metl tue hs n nnulus ross-setion s shown. The outer rdius is m nd the inner rdius is m. Within the tue, wter is mintined t temperture of o C. Within the metl the temperture drops off from inside to outside ording to dt d = where is the distne from the entrl is nd 6 6. Find the temperture of the outer surfe of the tue. m m metl wter t C tue ross-setion Self Tutor dt d = Z so T = d ) T = ln + fsine >g But when =, T = ) = ln + ) = + ln Thus T = ln + + ln T = + ln When =, T = + ln ¼ 9: ) the outer surfe temperture is 9: o C. EXERCISE C The mrginl ost per d of produing gdgets is C () =:5 + : euros per gdget. Wht is the totl ost of dil prodution of 8 gdgets given tht the fied osts efore prodution ommenes re E5 per d? The mrginl profit for produing dinner pltes per week is given P () =5 : dollrs per plte. If no pltes re mde loss of $65 eh week ours. Find the profit funtion. Wht is the mimum profit nd when does it our? Wht prodution levels enle profit to e mde?

159 APPLICATIONS OF INTEGRATION (Chpter ) 69 Jon needs to ulk up for the footll seson. His energ needs t ds fter strting his weight gin progrm re given E (t) = 5(8+:5t) :8 (8+:5t) lories per d. Find Jon s totl energ needs over the first week of the progrm. The tue ross-setion shown hs inner rdius of m nd outer rdius 6 m. Within the tue, wter is mintined t temperture of o C. Within the metl the temperture flls off t the rte dt d = :6 where is the distne from the entrl is nd metl Find the temperture of the outer surfe of the tue. D SOLIDS OF REVOLUTION Consider the urve = f() for 6 6. If the shded prt is rotted out the -is through 6 o,-dimensionl solid will e formed. This solid is lled solid of revolution. ƒ( ) ƒ( ) DEMO VOLUME OF REVOLUTION We n use integrtion to find volumes of revolution etween = nd =. The solid n e thought to e mde up of n infinite numer of thin lindril diss. Sine the volume of linder = ¼r h, the left-most dis hs pproimte volume ¼[f()] h, nd the right-most dis hs pproimte volume ¼[f()] h. ƒ( ) h In generl, ¼[f()] h is the pproimte volume for the illustrted dis. As there re infinitel mn diss, we let h!. X= Z Z ) V = lim ¼[f()] h = ¼[f()] d = ¼ d h! =

160 6 APPLICATIONS OF INTEGRATION (Chpter ) When the region enlosed = f(), the -is, nd the vertil lines =, = is rotted out the -is to generte solid, the volume of the solid is given Volume of revolution = ¼ Z d. ƒ( ) Emple Use integrtion to find the volume of the solid generted when the line = for 6 6 is revolved round the -is. Self Tutor Z Volume of revolution = ¼ = ¼ = ¼ Z d d = ¼ 6 =¼ ui units The volume of one is V one = ¼r h So, in this emple V = ¼ () ¼ () = 6¼ ¼ =¼ whih heks X Emple Self Tutor Find the volume of the solid formed when the grph of the funtion = for is revolved out the -is. 5 (, ) Z Volume of revolution = ¼ = ¼ = ¼ Z 5 Z 5 d ( ) d d 5 5 = ¼ 5 = ¼(65 ) = 65¼ ui units TI-nspire TI-8 Csio

161 APPLICATIONS OF INTEGRATION (Chpter ) 6 Emple Self Tutor One rh of = sin is rotted out the -is. sin Wht is the volume of revolution? Z Volume = ¼ = ¼ = ¼ Z ¼ Z ¼ d sin d os() d = ¼ ¼ sin() = ¼ ¼ sin(¼) sin Rememer sin = os os = + os = ¼ ¼ = ¼ units EXERCISE D. Find the volume of the solid formed when the following re revolved out the -is: = for 6 6 = p for 6 6 = for 6 6 d = for 6 6 e = for 6 6 f = p 5 for g = for 6 6 h = + for 6 6 Use tehnolog to find, orret to signifint figures, the volume of the solid of revolution formed when these funtions re rotted through 6 o out the -is: = + for 6 6 = esin for 6 6. Find the volume of revolution when the shded region is revolved out the -is. e 6

162 6 APPLICATIONS OF INTEGRATION (Chpter ) 6 8 Find the volume of revolution. A hemispheril owl of rdius 8 m 5 8 ontins wter to depth of m. Wht is the volume of wter? 5 Wht is the nme of the solid of revolution when the shded region is revolved out the -is? (, r) Find the eqution of the line segment [AB] in the A form = +. Find formul for the volume of the solid using ( h,) Z B ¼ d: 6 A irle with entre (, ) nd rdius r units hs (, r) eqution + = r. If the shded region is revolved out the -is, wht solid is formed? ( r, ) ( r,) Use integrtion to show tht the volume of revolution is ¼r. 7 Find the volume of revolution when these regions re rotted out the -is: = os for 6 6 ¼ = os() for 6 6 ¼ 8 Sketh the grph of = sin + os for 6 6 ¼. Hene, find the volume of revolution of the shpe ounded = sin + os, the -is, = nd = ¼ when it is rotted out the -is. 9 Sketh the grph of = sin() from = to = ¼. Hene, find the volume of revolution of the shpe ounded = sin(), the -is, = nd = ¼ when it is rotted out the -is. VOLUMES FOR TWO DEFINING FUNCTIONS Consider the irle with entre (, ) nd rdius unit. DEMO When this irle is revolved out the -is, we otin doughnut or torus.

163 APPLICATIONS OF INTEGRATION (Chpter ) 6 In generl, if the region ounded U = f() (on top) nd L = g() nd the lines =, = is revolved out the -is, then its volume of revolution is given : V = ¼ Z Z V = ¼ [f()] d ¼ Z ³[f()] [g()] [g()] d Z d or V = ¼ U L d Rotting out the -is gives ƒ() or U g()or L Emple Find the volume of revolution generted revolving the region etween = nd = p out the -is. (, ) Self Tutor Z Volume = ¼ = ¼ = ¼ Z Z U L d ³ p ( ) d ( ) d = ¼ 5 5 = ¼ 5 = ¼ units ()

164 6 APPLICATIONS OF INTEGRATION (Chpter ) EXERCISE D. The shded region etween = nd = is revolved out the -is. Wht re the oordintes of A nd B? Find the volume of revolution. X A B The shded region is revolved out e the -is. A e Find the oordintes of A. Find the volume of revolution. The shded region etween =, = nd = is revolved out the -is. Find the oordintes of A. Find the volume of revolution. A Find etl the volume of the solid of revolution generted rotting the region enlosed = +6 nd + =6 through 6 o out the -is. 5 The shded region is revolved out the -is. Stte the oordintes of A. Find the volume of revolution. A 8 6 Prove tht the shded re from = to infinit is infinite wheres its volume of revolution is finite. (, )

165 APPLICATIONS OF INTEGRATION (Chpter ) 65 REVIEW SET A NON-CALCULATOR The funtion = f() is grphed. Find: Z Z 6 Z 6 f() d f() d f() d Write n epression for the totl shded re. 6 ƒ( ) g() Is it true tht d Z of the shded region? Eplin our nswer riefl. f() d represents the re Determine k if the enlosed region hs re 5 units. k 5 B ppeling onl to geometril evidene, eplin wh Z e d + Z e ln d= e. e X ln 6 Find the re of the region enlosed = + + nd = +. 7 A prtile moves in stright line with veloit v(t) =t 6t +8ms, t >. Drw sign digrm for v(t). Eplin etl wht hppens to the prtile in the first 5 seonds of motion. After 5 seonds, how fr is the prtile from its originl position? d Find the totl distne trvelled in the first 5 seonds of motion. 8 Determine the re enlosed the es nd =e.

166 66 APPLICATIONS OF INTEGRATION (Chpter ) 9 Find the volume of revolution generted rotting the shded region through 6 o out the -is: X REVIEW SET B CALCULATOR A prtile moves in stright line with veloit v(t) =t t ms. Find the distne trvelled in the first seond of motion. Consider f() = +. Find the position nd nture of ll turning points of = f(). Disuss f() s! nd s!. Sketh the grph of = f(). d Find, using tehnolog, the re enlosed = f(), the -is, nd the vertil line =. A prtile moves in stright line with veloit given v(t) = sin t metres per seond. Find the totl distne trvelled the prtile in the first seonds of motion. A ot trvelling in stright line hs its engine turned off t time t =. Its veloit t time t seonds therefter is given v(t) = (t +) ms. Find the initil veloit of the ot, nd its veloit fter seonds. Disuss v(t) s t!. Sketh the grph of v(t) ginst t. d Find how long it tkes for the ot to trvel metres. e Find the elertion of the ot t n time t. f Show tht dv dt = kv, nd find the vlue of the onstnt k. 5 The figure shows the grphs of = os() nd = e for ¼ 6 6 ¼. Find orret to deiml ples: the -oordintes of their points of intersetion the re of the shded region. os e

167 APPLICATIONS OF INTEGRATION (Chpter ) 67 6 The mrginl ost per d of produing items is C () =+8e pounds per item. Wht is the totl ost of dil prodution of 8 items given tht the fied osts efore prodution ommenes re pounds per d? 7 The shded region hs re unit. Find the vlue of m. 8 Find, orret to deiml ples: the vlue of the re of the shded region. X sin sin m 9 Find the volume of the solid of revolution formed when the shded region is rotted through 6 o out the -is: os ( ) e REVIEW SET C At time t = prtile psses through the origin with veloit 7 m s. Its elertion t seonds lter is 6t m s. Find the totl distne tht the prtile hs trvelled when it momentril omes to rest for the seond time. Sketh the grphs of = sin nd = sin on the sme set of es for 6 6 ¼. Find the et vlue of the re enlosed these urves for 6 6 ¼. Find given tht the re of the region etween = e nd the -is from = to = is units. Hene determine suh tht the re of the region from = to = is lso units. e Determine the re of the region enlosed =, = sin nd = ¼. 5 Determine the re enlosed = ¼ nd = sin.

168 68 APPLICATIONS OF INTEGRATION (Chpter ) 6 OABC is retngle nd the two shded regions re equl in re. Find k. A k B k C 7 Find the volume of the solid of revolution formed when the following re rotted out the -is: = etween = nd = = + etween = nd = = sin etween = nd = ¼ d = os etween = nd = ¼. 8 Find the volume of revolution if the shded region is rotted through 6 o out the sin -is: os 9 liquid t T/ C A metl tue hs n nnulus ross-setion with rdii r nd r s shown. Within the tue liquid is mintined t rz temperture T o C. Within the metl, the temperture drops from inside to outside ording to dt d = k r where metl k is negtive onstnt nd is the distne from the entrl is. µ r Show tht the outer surfe hs temperture T + k ln. r Use V = ¼r h to find the volume of this one. Chek our nswer to integrtion. 8

169 Chpter Sttistil distriutions of disrete rndom vriles Sllus referene: 6.9, 6. Contents: A B C D Disrete rndom vriles Disrete proilit distriutions Epettion The inomil distriution

170 6 STATISTICAL DISTRIBUTIONS OF DISCRETE RANDOM VARIABLES (Chpter ) A DISCRETE RANDOM VARIABLES RANDOM VARIABLES In previous work we hve desried events minl using words. Where possile, it is fr more onvenient to use numers. A rndom vrile represents in numer form the possile outomes whih ould our for some rndom eperiment. A disrete rndom vrile X hs possile vlues,,,... For emple: ² the numer of houses in our suur whih hve power sfet swith ² the numer of new iles sold eh er ile store ² the numer of defetive light uls in the purhse order of it store. To determine the vlue of disrete rndom vrile we need to ount. A ontinuous rndom vrile X ould tke possile vlues in some intervl on the numer line. For emple: ² the heights of men ould ll lie in the intervl 5 <X<5 m ² the volume of wter in rinwter tnk during given month ould lie in the intervl <X< m. To determine the vlue of ontinuous rndom vrile we need to mesure. PROBABILITY DISTRIBUTIONS For n rndom vrile there is orresponding proilit distriution whih desries the proilit tht the vrile will tke n prtiulr vlue. The proilit tht the vrile X tkes vlue is written s P(X = ): For ontinuous distriutions we n lso sometimes write proilit distriution s funtion P (). For emple, when tossing two oins, the rndom vrile X ould e heds, hed, or heds, so X =, or. The ssoited proilit distriution is P(X =)=, P(X =)=, nd P(X =)= with grph: Qw_ proilit or Qw_ proilit Qr_ Qr_ numer of heds numer of heds

171 STATISTICAL DISTRIBUTIONS OF DISCRETE RANDOM VARIABLES (Chpter ) 6 Emple Self Tutor A supermrket hs three hekout points A, B nd C. A government inspetor heks for ur of the weighing sles t eh hekout. If weighing sle is urte then Y is reorded, nd if not, N is reorded. Suppose the rndom vrile X is the numer of urte weighing sles t the supermrket. List the possile outomes. Desrie, using X, the events of there eing: i one urte sle ii t lest one urte sle. Possile outomes: A B C X N N N Y N N N Y N N N Y N Y Y Y N Y Y Y N Y Y Y i X = ii X =, or EXERCISE A Clssif the following rndom vriles s ontinuous or disrete: the quntit of ft in susge. the mrk out of 5 for geogrph test the weight of seventeen er old student d the volume of wter in up of offee e the numer of trout in lke f the numer of hirs on t g the length of hirs on horse h the height of sk-srper. For eh of the following: i identif the rndom vrile eing onsidered ii give possile vlues for the rndom vrile iii indite whether the vrile is ontinuous or disrete. To mesure the rinfll over -hour period in Singpore, the height of wter olleted in rin guge (up to mm) is used. To investigte the stopping distne for tre with new tred pttern, rking eperiment is rried out. To hek the reliilit of new tpe of light swith, swithes re repetedl turned off nd on until the fil.

172 6 STATISTICAL DISTRIBUTIONS OF DISCRETE RANDOM VARIABLES (Chpter ) A supermrket hs four hekouts A, B, C nd D. Mngement heks the weighing devies t eh hekout. If weighing devie is urte Y is reorded; otherwise, N is reorded. The rndom vrile eing onsidered is the numer of weighing devies whih re urte. Suppose X is the rndom vrile. Wht vlues n X hve? Tulte the possile outomes nd the orresponding vlues for X. Desrie, using X, the events of: i devies eing urte ii t lest two devies eing urte. Consider tossing three oins simultneousl. The rndom vrile under onsidertion is the numer of heds tht ould result. List the possile vlues of X. Tulte the possile outomes nd the orresponding vlues of X. Find the vlues of P(X = ), the proilit of eh vlue ourring. d Grph the proilit distriution P(X = ) ginst s proilit histogrm. B DISCRETE PROBABILITY DISTRIBUTIONS For eh rndom vrile there is proilit distriution. The proilit p i of n given outome lies etween nd (inlusive), so 6 p i 6. If there re n possile outomes then np p i =, i= or in other words p + p + p + ::::: + p n =. The proilit distriution of disrete rndom vrile n e given: ² in tle form ² in grphil form ² in funtionl form s proilit mss funtion P (): It provides us with ll possile vlues of the vrile nd the proilit of the ourrene of eh vlue. Emple Self Tutor A mgzine store reorded the numer of mgzines purhsed its ustomers in one week. % purhsed one mgzine, 8% purhsed two, % purhsed three, % purhsed four, nd 5% purhsed five. Wht is the rndom vrile? Mke proilit tle for the rndom vrile. Grph the proilit distriution using spike grph.

173 STATISTICAL DISTRIBUTIONS OF DISCRETE RANDOM VARIABLES (Chpter ) 6 The rndom vrile X is the numer of mgzines sold. So, X =,,, or 5. proilit 5 P(X = ) : :8 : : : Emple Self Tutor Show tht the following re proilit distriution funtions: P () = +, =,,, P () = (:6) (:), =,,, P () = P () = 5 P () = P () = 7 All of these oe 6 P ( i ) 6, nd P P ( i )= = ) P () is proilit distriution funtion. For P () = (:6) (:), P () = (:6) (:) = (:) = :6 P () = (:6) (:) = (:6) (:) = :88 P () = (:6) (:) = (:6) (:) = : P () = (:6) (:) = (:6) = :6 Totl : All proilities lie etween nd, nd P P ( i )=. ) P () is proilit distriution funtion. EXERCISE B Find k in these proilit distriutions: P () : k :5 P(X = ) k k k k The proilities of Json soring home runs in eh gme during his sell reer re given in the following tle. X is the numer of home runs per gme. 5 P () : :88 :8 :7 : Wht is the vlue of P ()? Wht is the vlue of? Eplin wht this numer mens.

174 6 STATISTICAL DISTRIBUTIONS OF DISCRETE RANDOM VARIABLES (Chpter ) Wht is the vlue of P () + P () + P () + P () + P (5)? Eplin wht this represents. d Drw proilit distriution spike grph of P () ginst. Eplin wh the following re not vlid proilit distriutions: P () : : : : 5 P () : : :5 : Sll s numer of hits in eh 5 softll mth hs the following P () :7 : k :6 :8 : proilit distriution: Stte lerl wht the rndom vrile represents. Find k. Find: i P(X > ) ii P( 6 X 6 ) 5 A die is rolled twie. Drw grid whih shows the smple spe. Suppose X denotes the sum of the results for the two rolls. Find the proilit distriution of X. Drw proilit distriution histogrm for this sitution. 6 Find k for the following proilit distriutions: P () =k( +) for =,, P () = k for =,,, + 7 A disrete rndom vrile X hs proilit distriution given : P(X = ) =k where =,,,,. Find P(X = ) for =,,, nd. Find k nd hene find P(X > ): Emple Self Tutor A g ontins red nd lue mrles. Two mrles re rndoml seleted without replement. If X denotes the numer of reds seleted, find the proilit distriution of X. st seletion R_ W_ R B Et_ Wt_ Rt_ Qt_ nd seletion R R_ Et_ RR X B R R_ Wt_ W_ Rt_ RB BR X B W_ Qt_ BB X.6.. P( X) Totl P(X = ) 6

175 STATISTICAL DISTRIBUTIONS OF DISCRETE RANDOM VARIABLES (Chpter ) 65 8 Eletril omponents re produed nd pked into oes of. It is known tht % of the omponents m e fult. The rndom vrile X denotes the numer of fult items in the o, nd hs proilit distriution P () = (:) (:96), =,,,...,. Find the proilit tht rndoml seleted o will ontin no fult omponent. Find the proilit tht rndoml seleted o will ontin t lest one fult omponent. 9 A g ontins 5 lue nd green tikets. Two tikets re rndoml seleted without replement. We let X denote the numer of lue tikets seleted. Find the proilit distriution of X. Suppose insted tht three tikets re rndoml seleted without replement. Find the proilit distriution of X for X =,,,. When pir of die is rolled, D denotes the sum of the top fes. Displ the possile results in tle. Find P(D =7). Find the proilit distriution of D. d Find P(D > 8 j D > 6). When pir of die is rolled, N denotes the differene etween the numers on the top fes. Displ the possile results in tle. Construt proilit distriution tle for the possile vlues of N. Find P(N =). d Find P(N > j N > ). C EXPECTATION Consider the following prolem: A die is to e rolled times. On how mn osions should we epet the result to e si? In order to nswer this question we must first onsider ll possile outomes of rolling the die. The possiilities re,,,, 5 nd 6, nd eh of these is equll likel to our. Therefore, we would epet 6 of them to e si. 6 of is, so we epet of the rolls of the die to ield si. However, this does not men tht we will get sies when we roll die times. If there re n trils of n eperiment, nd n event hs proilit p of ourring in eh of the trils, then the numer of times we epet the event to our is np.

176 66 STATISTICAL DISTRIBUTIONS OF DISCRETE RANDOM VARIABLES (Chpter ) We n lso tlk out the epeted outome from one tril of n eperiment. The epeted outome for the rndom vrile X is the men result ¹. In generl, the epettion of the rndom vrile X is E(X) =¹ = n P i= i p i or np i P(X = i ). i= Emple 5 Self Tutor Eh time footller kiks for gol he hs hne of eing suessful. In prtiulr gme he hs kiks for gol. How mn gols would ou epet him to kik? p = P(gol) = ) the epeted numer of gols is np = =9 Emple 6 Self Tutor Find the men of the dt of Emple. The proilit tle is: i 5 p i : :8 : : :5 Now ¹ = P i p i = (:) + (:8) + (:) + (:) + 5(:5) =:9 In the long run, the verge numer of mgzines purhsed per ustomer is :9. FAIR GAMES In gmling, we s tht the epeted gin of the pler from eh gme is the epeted return or pout from the gme, less the mount it ost them to pl. The gme will e fir if the epeted gin is zero. Suppose X represents the gin of pler from eh gme. The gme is fir if E(X) =. Emple 7 In gme of hne, pler spins squre spinner lelled,,,. The pler wins the mount of mone shown in the tle longside, depending on whih numer omes up. Determine: the epeted return for one spin of the spinner Self Tutor Numer Winnings $ $ $5 $8

177 STATISTICAL DISTRIBUTIONS OF DISCRETE RANDOM VARIABLES (Chpter ) 67 the epeted gin of the pler if it ost $5 to pl eh gme whether ou would reommend pling this gme. Let Y denote the return or pout from eh spin. As eh numer is equll likel, the proilit for eh numer is ) epeted return = E(Y )= =$. Let X denote the gin of the pler from eh gme. Sine it osts $5 to pl the gme, the epeted gin = E(X) =E(Y ) $5 =$ $5 = $ Sine E(X) 6=, the gme is not fir. In prtiulr, sine E(X) = $, we epet on verge the pler to lose $ with eh spin. We would not reommend tht person pl the gme. EXERCISE C In prtiulr region, the proilit tht it will rin on n one d is :8. On how mn ds of the er would ou epet it to rin? If oins re tossed, wht is the hne tht the ll fll heds? If the oins re tossed times, on how mn osions would ou epet them ll to fll heds? If two die re rolled simultneousl 8 times, on how mn osions would ou epet to get doule? A single oin is tossed one. If hed ppers ou win $, nd if til ppers ou lose $. How muh would ou epet to win when pling this gme three times? 5 During the snow seson there is 7 proilit of snow flling on n prtiulr d. If Udo skis for five weeks, on how mn ds ould he epet to see snow flling? 6 A golkeeper hs proilit of sving penlt ttempt. How mn gols would he epet to sve from 9 ttempts?

178 68 STATISTICAL DISTRIBUTIONS OF DISCRETE RANDOM VARIABLES (Chpter ) 7 In rndom surve of her eletorte, politiin A disovered the residents voting intentions in reltion to herself nd her two opponents B nd C. The results re indited longside: A B C Estimte the proilit tht rndoml hosen voter in the eletorte will vote for: i A ii B iii C. If there re 75 people in the eletorte, how mn of these would ou epet to vote for: i A ii B iii C? 8 A person rolls norml si-sided die nd wins the numer of euros shown on the fe. Find the epeted return from one roll of the die. Find the epeted gin of the pler if it osts E to pl the gme. Would ou dvise the person to pl severl gmes? Suppose it osts Ek to pl the gme. Wht vlue(s) of k will result in: i fir gme ii profit eing mde the vendor? 9 A hrit fundriser gets liene to run the following gmling Result Wins gme: A die is rolled nd the returns to the pler re given in the p tle longside. To pl the gme osts $. A result of getting 6 $ 6 wins $, so in ft ou re hed $6 if ou get 6 on the, 5 $ first roll.,, $ Wht re our hnes of pling one gme nd winning: i $ ii $ iii $? Your epeted return from throwing 6 is 6 $. Wht is our epeted return from throwing: i or 5 ii, or iii,,,, 5 or 6? Wht is our overll epeted result t the end of one gme? d Wht is our overll epeted result t the end of gmes? A person pls gme with pir of oins. If two heds pper then $ is won. If hed nd til pper then $ is won. If two tils pper then $5 is lost. How muh would person epet to win pling this gme one? If the orgniser of the gme is llowed to mke n verge of $ per gme, how muh should e hrged to pl the gme one? A ountr eports rfish to overses mrkets. The uers re prepred to p high pries when the rfish rrive still live. If X is the numer of deths per dozen rfish, the proilit distriution for X is given : i 5 > 5 P ( i ) :5 :6 :5 k : : : Find k. Over long period, wht is the men numer of deths per dozen rfish?

179 STATISTICAL DISTRIBUTIONS OF DISCRETE RANDOM VARIABLES (Chpter ) 69 A rndom vrile X hs proilit distriution given P () = + for =,,. Clulte the men ¹ for this distriution. A pir of die is rolled nd the rndom vrile M is the lrger of the two numers tht re shown uppermost. In tle form, otin the proilit distriution of M. Find the men of the M-distriution. At hrit event there is mone-rising gme involving pir of ordinr die. The gme osts $ to pl. When the two die re rolled, their sum is desried the vrile X. The orgnisers deide tht sum whih is less thn or etween 7 nd 9 inlusive is loss of $, result etween nd 6 inlusive gives return of $7 nd result of or more gives return of $. Determine P(X 6 ), P( 6 X 6 6), P(7 6 X 6 9) nd P(X > ). Show tht the epeted gin of pler is given 6 (5 7) dollrs. Wht vlue would need to e for the gme to e fir? d Eplin wh the orgnisers would not let e. e If the orgnisers let e 6 nd the gme ws pled 6 times, estimte the mount of mone rised this gme. D THE BINOMIAL DISTRIBUTION Thus fr in the hpter we hve onsidered properties of generl disrete rndom vriles. We now emine speil disrete rndom vrile whih is pplied to smpling with replement. The proilit distriution ssoited with this vrile is the inomil proilit distriution. For smpling without replement the hpergeometri proilit distriution is the model used, ut tht distriution is not prt of this ourse. BINOMIAL EXPERIMENTS Consider n eperiment for whih there re two possile results: suess if some event ours, or filure if the event does not our. If we repet this eperiment in numer of independent trils, we ll it inomil eperiment. The proilit of suess p must e onstnt for ll trils. Sine suess nd filure re omplementr events, the proilit of filure is p. The rndom vrile X is the totl numer of suesses in n trils.

180 6 STATISTICAL DISTRIBUTIONS OF DISCRETE RANDOM VARIABLES (Chpter ) THE BINOMIAL PROBABILITY DISTRIBUTION Suppose spinner hs three lue edges nd one white edge. Clerl, for eh spin we will get either lue or white. The hne of finishing on lue is nd on white is. If we ll lue result suess nd white result filure, then we hve inomil eperiment. We let p e the proilit of getting lue, so p =. The proilit of getting white is p =. Consider twirling the spinner n =times. Let the rndom vrile X e the numer of suesses or lue results, so X =,, or. P(X =)=P(ll re white) = = st spin nd spin rd spin P(X =)=P( lue nd white) = P(BWW or WBW or WWB) = fthe rnhes Xg P(X =)=P( lue nd white) = P(BBW or BWB or WBB) = P(X = )= P( lues) = Er_ Qr_ B W Er_ Qr_ Er_ Qr_ B W B W Er_ Qr_ Er_ Qr_ Er_ Qr_ Er_ Qr_ B W B W X B W X B W X The oloured ftor is the numer of ws of getting one suess in three trils, whih is. We note tht: P(X =)= = ¼ :56 P(X =)= = ¼ :6 P(X =)= = ¼ :9 P(X =)= = ¼ :9.... proilit numer of lues This suggests tht: P(X = ) = where =,,,. Notie tht is the inomil epnsion for +.

181 STATISTICAL DISTRIBUTIONS OF DISCRETE RANDOM VARIABLES (Chpter ) 6 Consider inomil eperiment for whih p is the proilit of suess nd p is the proilit of filure. If there re n independent trils then the proilit tht there re r suesses nd n r filures is given P(X = r) = n r p r ( p) n r where r =,,,,,..., n. P(X = r) is the inomil proilit distriution funtion. The epeted or men outome of the eperiment is ¹ = E(X) =np. If X is the rndom vrile of inomil eperiment with prmeters n nd p, then we write X» B(n, p) where» reds is distriuted s. We n quikl lulte inomil proilities using grphis lultor. For emple: ² To find the proilit P(X = r) tht the vrile tkes the vlue r, we use the inomil proilit distriution funtion. ² To find the proilit P(X 6 r) tht the vrile tkes vlue whih is t most r, we use the inomil umultive distriution funtion. For help using our lultor, onsult the hpter of instrutions t the front of the ook. Emple 8 7% of union memers re in fvour of ertin hnge to their onditions of emploment. A rndom smple of five memers is tken. Find: the proilit tht three memers re in fvour of the hnge in onditions Self Tutor the proilit tht t lest three memers re in fvour of the hnged onditions the epeted numer of memers in the smple tht re in fvour of the hnge. Let X denote the numer of memers in the smple in fvour of the hnges. n =5, so X =,,,, or 5, nd p = 7% = :7 ) X» B(5, :7). P(X =) = 5 (:7) (:8) ¼ :9 P(X > ) = P(X 6 ) = :76 ¼ :86 E(X) =np =5 :7 = :6 memers

182 6 STATISTICAL DISTRIBUTIONS OF DISCRETE RANDOM VARIABLES (Chpter ) EXERCISE D For whih of these proilit eperiments does the inomil distriution ppl? Justif our nswers, using full sentene. A oin is thrown times. The vrile is the numer of heds. One hundred oins re eh thrown one. The vrile is the numer of heds. A o ontins 5 lue nd red mrles. I drw out 5 mrles, repling the mrle eh time. The vrile is the numer of red mrles drwn. d A o ontins 5 lue nd red mrles. I drw out 5 mrles without replement. The vrile is the numer of red mrles drwn. e A lrge in ontins ten thousnd olts, % of whih re fult. I drw smple of olts from the in. The vrile is the numer of fult olts. At mnufturing plnt, 5% of the emploees work night-shift. If 7 emploees were seleted t rndom, find the proilit tht: etl of them work night-shift less thn of them work night-shift t lest of them work night-shift. Reords show tht 6% of the items ssemled on prodution line re fult. A rndom smple of items is seleted t rndom, with replement. Find the proilit tht: none will e fult t most one will e fult t lest two will e fult d less thn four will e fult. There is 5% hne tht n pple in rte will hve lemish. If smple of 5 pples is tken, find: the proilit tht etl of these hve lemishes the proilit tht t lest one hs lemish the epeted numer of pples tht will hve lemish. 5 The lol us servie does not hve good reputtion. It is known tht the 8 m us will run lte on verge two ds out of ever five. For n week of the er tken t rndom, find the proilit of the 8 m us eing on time: ll 7 ds onl on Mond on n 6 ds d on t lest ds. 6 An infetious flu virus is spreding through shool. The proilit of rndoml seleted student hving the flu net week is :. Clulte the proilit tht out of lss of 5 students, or more will hve the flu net week. If more thn % of the students re w with the flu net week, lss test will hve to e nelled. Wht is the proilit tht the test will e nelled? If the shool hs 5 students, find the epeted numer tht will e sent from shool net week.

183 STATISTICAL DISTRIBUTIONS OF DISCRETE RANDOM VARIABLES (Chpter ) 6 7 During seson, sketll pler hs 9% suess rte in shooting from the free throw line. In one mth the sketller hs shots from the free throw line. Find the proilit tht the sketller is suessful on: i ll throws ii t lest 8 throws. Find the epeted numer of suessful throws. REVIEW SET A NON-CALCULATOR f() =, =,,, is proilit distriution funtion. + Find. Hene, find P(X > ). A rndom smple of toothrushes is mde with replement from ver lrge th where % re known to e defetive. Find the men numer of defetives in the smple. A rndom vrile X hs proilit distriution funtion given : Find k. Find the men ¹ for the distriution of X. P () : : :5 : k Three green lls nd two ellow lls re pled in ht. Two lls re rndoml drwn without replement. Let X e the numer of green lls drwn. Find the proilit tht: i X = ii X = iii X = Find E(X). 5 Lkshmi rolls norml si-sided die. She wins twie the numer of pounds s the numer shown on the fe. How muh does Lkshmi epet to win from one roll of the die? If it osts $8 to pl the gme, would ou dvise Lkshmi to pl severl gmes? Eplin our nswer. REVIEW SET B CALCULATOR A disrete rndom vrile X hs proilit distriution funtion P () =k where =,,, nd k is onstnt. Find k. Find P(X > ). Find the epeted outome for the vrile X. A mnufturer finds tht 8% of the items produed from its sseml lines re defetive. During floor inspetion, the mnufturer rndoml selets ten items. Find the proilit tht the mnufturer finds: one defetive two defetive t lest two defetive items. From dt over the lst fifteen ers it is known tht the hne of netller with knee injur needing mjor knee surger in n one seson is :. In 7 there were 87 ses of knee injuries. Find the epeted numer of mjor knee surgeries.

184 6 STATISTICAL DISTRIBUTIONS OF DISCRETE RANDOM VARIABLES (Chpter ) An X-r hs proilit of :96 of showing frture in the rm. If four different X-rs re tken of prtiulr frture, find the proilit tht: ll four show the frture the frture does not show up d t lest three X-rs show the frture onl one X-r shows the frture. 5 % of visitors to museum mke voluntr dontions upon entr. On ertin d the museum hs 75 visitors. Find: the epeted numer of dontions the proilit tht less thn visitors mke dontion. 6 A shool sketll tem hs 8 plers, eh of whom hs 75% hne of turning up for n given gme. The tem needs t lest 5 plers to void forfeiting the gme. Find the proilit tht, on rndoml hosen gme, the tem will: i hve ll of its plers ii hve to forfeit the gme. The tem pls gmes for the seson. How mn gmes would ou epet the tem to forfeit? REVIEW SET C Find k for: the proilit distriution funtion P () = k, =,, the proilit distriution P () A rndom vrile X hs proilit distriution funtion P(X = ) = for =,,,,. Find P(X = ) for =,,,,. Find the men ¹ for this distriution. k : k : A die is ised suh tht the proilit of otining 6 is 5. The die is rolled times. Let X e the numer of sies otined. Find the men of X. Onl % of oung trees tht re plnted will survive the first er. The Botnil Grdens us five oung trees. Assuming independene, lulte the proilit tht during the first er: etl one tree will survive t most one tree will survive t lest one tree will survive. 5 In gme, the numers from to re written on tikets nd pled in g. A pler drws out numer t rndom. He or she wins $ if the numer is even, $6 if the numer is squre numer, nd $9 if the numer is oth even nd squre. Clulte the proilit tht the pler wins: i $ ii $6 iii $9 How muh should e hrged to pl the gme so tht it is fir gme?

185 Chpter Sttistil distriutions of ontinuous rndom vriles Sllus referene: 6. Contents: A B C D E Continuous proilit densit funtions Norml distriutions The stndrd norml distriution ( Z-distriution) Quntiles or k-vlues Applitions of the norml distriution

186 66 STATISTICAL DISTRIBUTIONS OF CONTINUOUS RANDOM VARIABLES (Chpter ) A CONTINUOUS PROBABILITY DENSITY FUNCTIONS In the previous hpter we looked t disrete rndom vriles nd emined some proilit distriutions where the rndom vrile X ould tke the non-negtive integer vlues =,,,,,... For ontinuous rndom vrile X, n tke n rel vlue. Consequentl, funtion is used to speif the proilit distriution, nd tht funtion is lled the proilit densit funtion. Proilities re found lulting res under the proilit densit funtion for prtiulr intervl. ( ) d A ontinuous proilit densit funtion (pdf) is funtion f() > on given intervl 6 6 nd Z f() d =. f() suh tht The proilit tht the vrile X lies in the intervl 6 X 6 d is P( 6 X 6 d) = Z d f() d. Sine the vrile is ontinuous, it is importnt to reognise tht the proilit of X tking n et vlue is zero. Insted, we n onl mesure the proilit of X eing in prtiulr intervl. A onsequene of this is tht P( 6 X 6 d) =P( 6 X<d)=P( <X6d) =P( <X<d) The men ¹ or E(X) of ontinuous proilit densit funtion is defined s ¹ = Z f() d. Emple Self Tutor ½ f() = elsewhere is proilit densit funtion. Chek tht the ove sttement is true. Find P( 6 X 6 ). Find the men of the distriution.

187 STATISTICAL DISTRIBUTIONS OF CONTINUOUS RANDOM VARIABLES (Chpter ) 67 Qw_ (, ) Are = = or Z X d= = X Z P( 6 X 6 ) = f() d = Z d = = 6 ¹ = = Z f() d Z d = 6 = = 6 EXERCISE A ½ f() =, 6 6 is proilit densit funtion. elsewhere Chek tht the ove sttement is true. Find the men of the distriution. ½ ( ), 6 6 f() = elsewhere is ontinuous proilit densit funtion. Find. Sketh the grph of = f(). Find P( 6 X 6 ). d Find the men of the distriution. ½ k ( 6), f() = is proilit densit funtion. Find: elsewhere k E(X) P( 6 X 6 5). The ontinuous rndom vrile X hs the proilit densit funtion f() =, 6 6 k. Given tht P X 6 =, find nd k. 5 The time tken in hours to perform prtiulr tsk hs the proilit densit funtion: f() = ½ k p 6 6 otherwise. Find k. Find the men time for performing the tsk. Find P( 6 X 6 :5). d Find suh tht P( 6 X 6 ) =P( 6 X 6 ).

188 68 STATISTICAL DISTRIBUTIONS OF CONTINUOUS RANDOM VARIABLES (Chpter ) B NORMAL DISTRIBUTIONS The norml distriution is the most importnt distriution for ontinuous rndom vrile. Mn nturll ourring phenomen hve distriution tht is norml, or pproimtel norml. Some emples re: ² phsil ttriutes of popultion suh s height, weight, nd rm length ² rop ields ² sores for tests tken lrge popultion If X is normll distriuted then its proilit densit funtion is given Ã! ¹ f() = ¾ p ¼ e ¾ for << where ¹ is the men nd ¾ is the vrine of the distriution. Eh memer of the fmil is speified the prmeters ¹ nd ¾, nd we n write X» N(¹, ¾ ). This proilit densit funtion for the norml distriution represents fmil of ell-shped norml urves. These urves re ll smmetril out the vertil line = ¹. A tpil norml urve is illustrted longside. f () &, * Notie tht f(¹) = ¾ p ¼ : CHARACTERISTICS OF THE NORMAL PROBABILITY DENSITY FUNCTION ² The urve is smmetril out the vertil line = ¹: ² As! the norml urve pprohes its smptote, the -is. ² f() > for ll. ² The re under the urve is one unit, nd so Z f() d =. This is neessr sine the totl proilit must e. ² More sores re distriuted loser to the men thn further w. This results in the tpil ell shpe.

189 STATISTICAL DISTRIBUTIONS OF CONTINUOUS RANDOM VARIABLES (Chpter ) 69 HOW THE NORMAL DISTRIBUTION ARISES Consider the ornges piked from n ornge tree. The do not ll hve the sme weight. The vrition m e due to severl ftors, inluding: ² genetis ² different times when the flowers were fertilised ² different mounts of sunlight rehing the leves nd fruit ² different wether onditions suh s the previling winds. The result is tht most of the fruit will hve weights lose to the men, while there re fr fewer ornges tht re muh hevier or muh lighter. This results in the ell-shped distriution. One norml model hs een estlished, we n use it to mke preditions out distriution nd to nswer other relevnt questions. A TYPICAL NORMAL DISTRIBUTION A lrge smple of okle shells ws olleted nd the mimum distne ross eh shell ws mesured. Clik on the video lip ion to see how histogrm of the dt is uilt up. Then lik on the demo ion to oserve the effet of hnging the lss intervl lengths for normll distriuted dt. VIDEO CLIP DEMO THE GEOMETRICAL SIGNIFICANCE OF ¹ AND ¾ Differentiting f() = ¾ p ¼ e f () = ¾ p ¼ ) f () = onl when = ¹. Ã! ¹ ¾ µ ¹ ¾ we otin e ³ ¹ ¾ This orresponds to the point on the grph when f() is mimum. f () = ³ ¹ ¾ p ¼ e ¾ ( ¹) Differentiting gin, we otin ¾ ¾ ) f ( ¹) () = when ¾ = ¾ ) ( ¹) = ¾ ) ¹ = ¾ ) = ¹ ¾

190 65 STATISTICAL DISTRIBUTIONS OF CONTINUOUS RANDOM VARIABLES (Chpter ) So, the points of infletion re t = ¹ + ¾ nd = ¹ ¾: point of infletion point of infletion For given norml urve, the stndrd devition is uniquel determined s the horizontl distne from the vertil line = ¹ to point of infletion. For norml distriution with men ¹ nd stndrd devition ¾, the proportionl rekdown of where the rndom vrile ould lie is given elow. Norml distriution urve.%.5%.%.%.5%.%.59%.59% Notie tht: ² ¼ 68:6% of vlues lie etween ¹ ¾ nd ¹ + ¾ ² ¼ 95:% of vlues lie etween ¹ ¾ nd ¹ +¾ ² ¼ 99:7% of vlues lie etween ¹ ¾ nd ¹ +¾. INVESTIGATION STANDARD DEVIATION SIGNIFICANCE The purpose of this investigtion is to hek the proportions of norml distriution dt whih lie within ¾, ¾ nd ¾ of the men. Wht to do: Clik on the ion to strt the demonstrtion in Mirosoft Eel. DEMO Tke rndom smple of size n = from norml distriution. Find: nd s s, + s s, +s d s, +s Count ll vlues etween: s nd + s s nd +s s nd +s 5 Determine the perentge of dt vlues in these intervls. Do these onfirm the theoretil perentges given ove? 6 Repet the proedure severl times.

191 STATISTICAL DISTRIBUTIONS OF CONTINUOUS RANDOM VARIABLES (Chpter ) 65 Emple Self Tutor The hest mesurements of 8 er old mle footllers re normll distriuted with men of 95 m nd stndrd devition of 8 m. Find the perentge of footllers with hest mesurements etween: i 87 m nd m ii m nd m Find the proilit tht the hest mesurement of rndoml hosen footller is etween 87 m nd m. Find the vlue of k suh tht 6% of hest mesurements re elow k m. i We need the perentge etween ¹ ¾ nd ¹ + ¾. This is ¼ 68:%. ii We need the perentge etween ¹ + ¾ nd ¹ +¾. This is ¼ :6%: This is etween ¹ ¾ nd ¹ +¾. The perentge is (:%) + :59% ¼ 8:9%: So, the proilit is ¼ :89..%.% %.%.59%.59% Approimtel 6% of dt lies more thn one stndrd devition lower thn the men. ) k is ¾ elow the men ¹ ) k =95 8 =87m 6% k EXERCISE B. Drw eh of the following norml distriutions urtel on one set of es. Distriution men (ml) stndrd devition (ml) A 5 5 B C Eplin wh it is likel tht the distriutions of the following vriles will e norml: the volume of soft drink in ns the dimeter of olts immeditel fter mnufture. Stte the proilit tht rndoml seleted, normll distriuted vlue lies etween: ¾ elow the men nd ¾ ove the men the men nd the vlue ¾ ove the men.

192 65 STATISTICAL DISTRIBUTIONS OF CONTINUOUS RANDOM VARIABLES (Chpter ) When speifi tpe of rdish is grown without fertiliser, the weights of the rdishes produed re normll distriuted with men of g nd stndrd devition of g. When the sme tpe of rdish is grown in the sme w eept for the inlusion of fertiliser, the weights of the rdishes produed re lso normll distriuted, ut with men of g nd stndrd devition of g. Determine the proportion of rdishes grown: without fertiliser with weights less thn 5 grms with fertiliser with weights less thn 6 grms i with nd ii without fertiliser with weights etween nd 6 g inlusive d i with nd ii without fertiliser with weights greter thn or equl to 6 g. 5 Given X» N(, : ), find: the proilit tht rndoml seleted vlue lies within stndrd devitions of the men the vlue of X whih is :7 stndrd devitions elow the men. 6 The weights of Json s ornges re normll distriuted. 8% of the rop weigh more thn 5 grms nd 6% weigh more thn grms. Find ¹ nd ¾ for the rop. Wht proportion of the ornges weigh etween 5 grms nd grms? 7 The height of mle students is normll distriuted with men of 7 m nd stndrd devition of 8 m. Find the perentge of mle students whose height is: i etween 6 m nd 7 m ii etween 7 m nd 86 m. Find the proilit tht rndoml hosen student from this group hs height: i etween 78 m nd 86 m ii less thn 6 m iii less thn 5 m iv greter thn 6 m. Find the vlue of k suh tht 6% of the students re tller thn k m. 8 The heights of er old os re normll distriuted. 97:7% of them re ove m nd :8% re ove 79 m. Find ¹ nd ¾ for the height distriution. A er old o is rndoml hosen. Wht is the proilit tht his height lies etween m nd 9 m? 9 A ottle filling mhine fills n verge of ottles d with stndrd devition of. Assuming tht prodution is normll distriuted nd the er omprises 6 working ds, lulte the pproimte numer of working ds on whih: under 8 ottles re filled over 6 ottles re filled etween 8 nd ottles (inlusive) re filled.

193 STATISTICAL DISTRIBUTIONS OF CONTINUOUS RANDOM VARIABLES (Chpter ) 65 PROBABILITIES BY GRAPHICS CALCULATOR We n use grphis lultor to quikl find proilities for norml distriution. Suppose X» N(, ), so X is normll distriuted with men nd stndrd devition. How do we find P(8 6 X 6 )? How do we find if P(X 6 ) =:79? 8 Instrutions for these tsks n e found in the grphis lultor hpter t the strt of the ook. Use lultor to nswer the questions in the following eerise. It is helpful to EXERCISE B. sketh the norml distriution nd X is rndom vrile tht is distriuted normll with men shde the re of 7 nd stndrd devition. Find: interest. P(7 6 X 6 7) P(68 6 X 6 7) P(X 6 65) X is rndom vrile tht is distriuted normll with men 6 nd stndrd devition 5. Find: P(6 6 X 6 65) P(6 6 X 6 67) P(X > 6) d P(X 6 68) e P(X 6 6) f P(57:5 6 X 6 6:5) Given tht X» N(, 5 ), find suh tht: P(X < )=:78 P(X > ) =:59 P( < X < +) =:7 C THE STANDARD NORMAL DISTRIBUTION (Z-DISTRIBUTION) Ever norml X-distriution n e trnsformed into the stndrd norml distriution or Z-distriution using the trnsformtion z = ¹ ¾. In the following investigtion we determine the men nd stndrd devition of this Z-distriution. INVESTIGATION PROPERTIES OF z = ¹ ¾ Suppose rndom vrile X is normll distriuted with men ¹ nd stndrd devition ¾. For eh vlue of we n lulte z-vlue using the lgeri trnsformtion z = ¹ ¾.

194 65 STATISTICAL DISTRIBUTIONS OF CONTINUOUS RANDOM VARIABLES (Chpter ) Wht to do: Consider the -vlues:,,,,,,,,,,,, 5, 5, 5, 5, 6, 6, 7. Drw grph of the distriution to hek tht it is pproimtel norml. Find the men ¹ nd stndrd devition ¾ for the distriution of -vlues. Use the trnsformtion z = ¹ to onvert eh -vlue into z-vlue. ¾ d Find the men nd stndrd devition for the distriution of z-vlues. Clik on the ion to lod lrge smple drwn from norml popultion. B liking ppropritel we n repet the four steps of question. Write rief report of our findings. For norml X-distriution we know the proilit densit funtion is f() = ¾ p ¼ e ( ¹ ¾ ). Sustituting = z, ¹ = nd ¾ =, we find: DEMO You should hve disovered tht for Z-distriution the men is nd the stndrd devition is. We n write Z» N (, ). This is true for ll Z-distriutions generted trnsformtion of norml distriution, nd this is wh we ll it the stndrd norml distriution. The proilit densit funtion for the Z-distriution is f(z) = p ¼ e z, <z<. Notie tht the norml distriution funtion f() hs two prmeters ¹ nd ¾, wheres the stndrd norml distriution funtion f(z) hs no prmeters. This mens tht unique tle of vlues n e onstruted for f(z), nd we n use this tle to ompre norml distriutions. Before grphis lultors nd omputer pkges, the stndrd norml distriution tle ws used elusivel for norml proilit lultions suh s those whih follow. The umultive distriution funtion for the Z-distriution is () =P(Z 6 ). Sine z is ontinuous, P(Z 6 ) =P(Z <)= () Z nd P(Z 6 ) = p e z dz. ¼ this re is P( Z 6) ƒ( z) The tle of urve res on pge 655 enles us to find P(Z 6 ) for >. For < we need to use the smmetr of the stndrd norml urve. In prtiulr, we notie tht P(Z 6 ) = P(Z > ) = P(Z 6 ) f smmetrg z

195 STATISTICAL DISTRIBUTIONS OF CONTINUOUS RANDOM VARIABLES (Chpter ) 655 STANDARD NORMAL CURVE AREAS ƒ( z) () P( Z6) z the seond deiml digit of USING A GRAPHICS CALCULATOR TO FIND PROBABILITIES For TI-8 plus: To find P(Z 6 ) or P(Z < ) use normldf( E99, ): To find P(Z > ) or P(Z > ) use normldf(, E99): To find P( 6 Z 6 ) or P( < Z < ) use normldf(, ): Consult the grphis lultor instrutions t the strt of the ook for help using other lultor models.

196 656 STATISTICAL DISTRIBUTIONS OF CONTINUOUS RANDOM VARIABLES (Chpter ) Emple Self Tutor If Z is stndrd norml vrile, find without using tehnolog: P(Z 6 :5) P(Z > :8) P( : 6 Z 6 :67) P(Z 6 :5) ¼ :9.5 P(Z >:8) = P(Z 6 :8) ¼ :7995 ¼ :.8 P( : 6 Z 6 :67) = P(Z 6 :67) P(Z 6 :) = P(Z 6 :67) ( P(Z 6 :)) ¼ :786 ( :659) ¼ :8..67 INTERPRETING Z-VALUES If is n oservtion from norml distriution with men ¹ nd stndrd devition ¾, the Z-sore of is the numer of stndrd devitions is from the men. For emple: ² If z =:8 then z is :8 stndrd devitions to the right of the men. ² If z = :7 then z is :7 stndrd devitions to the left of the men. Z-vlues re therefore useful when ompring results from two or more different distriutions. Emple Self Tutor Kell sored 7% in Histor where the lss men ws 68% nd the stndrd devition ws :%. In Mthemtis she sored 66%, the lss men ws 6%, nd the stndrd devition ws 6:8%. In whih sujet did Kell perform etter ompred with the rest of her lss? Assume the sores for oth sujets were normll distriuted Kell s Z-sore for Histor = ¼ :9 : 66 6 Kell s Z-sore for Mthemtis = ¼ :588 6:8

197 STATISTICAL DISTRIBUTIONS OF CONTINUOUS RANDOM VARIABLES (Chpter ) 657 So, Kell s result in Mthemtis ws :588 stndrd devitions ove the men, wheres her result in Histor ws :9 stndrd devitions ove the men. ) Kell s result in Mthemtis ws etter, even though her sore ws lower. EXERCISE C. If Z hs stndrd norml distriution, find using tles nd sketh: P(Z 6 :) P(Z > :86) P(Z 6 :5) d P(Z > :6) e P( :86 6 Z 6 :) If Z hs stndrd norml distriution, find using tehnolog: P(Z > :87) P(Z 6 :6) P(Z > :876) d P( :86 6 Z 6 :56) e P( :67 6 Z 6 :65) If Z hs stndrd norml distriution, find: P( :5 <Z<:5) P( :96 <Z<:96) Find if Z hs stndrd norml distriution nd: P(Z 6 ) =:95 P(Z > ) =:9 5 The tle longside shows Sergio s results in his mid-er emintions, long with the lss mens nd stndrd devitions. Find Sergio s Z-vlue for eh sujet. Arrnge Sergio s performnes in eh sujet in order from est to worst. Sergio ¹ ¾ Phsis 8% 78% :8% Chemistr 77% 7% :6% Mthemtis 8% 7% :% Germn 9% 86% 9:6% Biolog 7% 6% :% 6 Pedro is studing Alger nd Geometr. He sits for the mid-er ems in eh sujet. He is told tht his Alger mrk is 56%, wheres the lss men nd stndrd devition re 5:% nd 5:8% respetivel. In Geometr he is told tht the lss men nd stndrd devition re 58:7% nd 8:7% respetivel. Wht perentge does Pedro need to hve sored in Geometr to hve n equivlent result to his Alger mrk? STANDARDISING ANY NORMAL DISTRIBUTION To find proilities for normll distriuted rndom vrile X: Step : Convert -vlues to z-vlues using z = ¹ ¾ : Step : Sketh stndrd norml urve nd shde the required region. Step : Use the stndrd norml tles or grphis lultor to find the proilit.

198 658 STATISTICAL DISTRIBUTIONS OF CONTINUOUS RANDOM VARIABLES (Chpter ) Emple 5 Given tht X is norml vrile with men 6 nd stndrd devition 7, find without using tehnolog: P(X 6 69) P(58:5 6 X 6 7:8) Interpret eh result. P(X 6 69) µ X 6 = P 6 7 = P(Z 6 ) ¼ : There is n 8:% hne tht rndoml seleted X-vlue is 69 or less. P(58:5 6 X 6 7:8) µ 58:5 6 = P 6 X = P( :5 6 Z 6 :) = P(Z 6 :) P(Z 6 :5) = P(Z 6 :) ( P(Z 6 :5)) =:99 ( :695) ¼ :6 7:8 6 7 z.5. Self Tutor There is 6:% hne tht rndoml seleted X-vlue is etween 58:5 nd 7:8 inlusive. z These proilities n lso e found using grphis lultor without tull onverting to stndrd norml Z-sores. Instrutions for this n e found in the grphis lultor hpter t the front of the ook. EXERCISE C. A rndom vrile X is normll distriuted with men 7 nd stndrd devition. B onverting to the stndrd vrile Z nd then using the tled proilit vlues for Z, find: P(X > 7) P(X 6 68) P(6:6 6 X 6 68:) A rndom vrile X is normll distriuted with men 58: nd stndrd devition 8:96. B onverting to the stndrd vrile Z nd then using our grphis lultor, find: P(X > 6:8) P(X 6 5:) P(5:67 6 X 6 68:9) The length L of nil is normll distriuted with men 5: mm nd stndrd devition :9 mm. Find: P(L > 5) P(L 6 5) P(9 6 L 6 5:5)

199 STATISTICAL DISTRIBUTIONS OF CONTINUOUS RANDOM VARIABLES (Chpter ) 659 D QUANTILES OR k-values Consider popultion of rs where the length of shell, X mm, is normll distriuted with men 7 mm nd stndrd devition mm. A iologist wnts to protet the popultion llowing onl the lrgest 5% of rs to e hrvested. He therefore sks the question: 95% of the rs hve lengths less thn wht?. To nswer this question we need to find the vlue of k suh tht P(X 6 k) =:95. The numer k is known s quntile, nd in this se is the 95% quntile. When given proilit we need to find the orresponding mesurement. This is the inverse of finding proilities, nd to do this we n either: ² red the vlue from n inverse stndrd norml tle. The tle on pge 66 inludes proilities whih re > :5 onl. For other proilities we gin rel on the smmetr of the norml distriution. For the proilit p where p < :5, we look up the Z-sore orresponding to p, then tke its negtive. ² use the inverse norml funtion on our grphis lultor. For instrutions on using our lultor, see the grphis lultor hpter t the strt of the ook. Emple 6 Self Tutor Find k for whih P(X 6 k) =:95 given tht X» N(7, ) nd X is mesured in mm. P(X 6 k) =:95 µ X 7 ) P 6 k 7 =:95 µ ) P Z 6 k 7 =:95 Using the inverse norml tle we find: k 7 ¼ :69 ) k ¼ 86: So, pproimtel 95% of the vlues re epeted to e 86: mm or less. or Using tehnolog: If P(X 6 k) =:95 then k ¼ 86:5.

200 66 STATISTICAL DISTRIBUTIONS OF CONTINUOUS RANDOM VARIABLES (Chpter ) INVERSE NORMAL PROBABILITIES pp( Z6) ( z) p z

201 STATISTICAL DISTRIBUTIONS OF CONTINUOUS RANDOM VARIABLES (Chpter ) 66 EXERCISE D Z hs stndrd norml distriution. Illustrte, then find k using tled vlues given: P(Z 6 k) =:8 P(Z 6 k) =:58 P(Z 6 k) =:7 Z hs stndrd norml distriution. Illustrte, then find k using tehnolog given: P(Z 6 k) =:8 P(Z 6 k) =:878 P(Z 6 k) =:8 Show tht P( k 6 Z 6 k) =P(Z 6 k) : Suppose Z hs stndrd norml distriution. Find k if: i P( k 6 Z 6 k) =:8 ii P( k 6 Z 6 k) =:7 Find k if P(X 6 k) =:9 nd X» N(56, 8 ). Find k if P(X > k) =:8 nd X» N(8:7, 8:8 ). E APPLICATIONS OF THE NORMAL DISTRIBUTION Emple 7 Self Tutor In 97 the heights of rug plers were found to e normll distriuted with men 79 m nd stndrd devition 7 m. Find the proilit tht rndoml seleted pler in 97 ws: t lest 75 m tll etween 7 m nd 9 m. If X is the height of pler then X is normll distriuted with ¹ = 79, ¾ =7: P(X > 75) P(7 <X<9) ¼ :76 fusing tehnologg ¼ :8 fusing tehnologg Emple 8 Self Tutor A universit professor determines tht 8% of this er s Histor ndidtes should pss the finl emintion. The emintion results re epeted to e normll distriuted with men 6 nd stndrd devition. Find the epeted lowest sore neessr to pss the emintion.

202 66 STATISTICAL DISTRIBUTIONS OF CONTINUOUS RANDOM VARIABLES (Chpter ) Let the rndom vrile X denote the finl emintion result, so X» N(6, ). We need to find k suh tht P(X > k) =:8 ) P(X 6 k) =: ) k ¼ 5: k 6 So, the professor would deide etween 5 nd 5 for the minimum pss mrk. In the following emple we must onvert to Z-sores to nswer the question. We lws need to onvert to Z-sores if we re tring to find n unknown men ¹ or stndrd devition ¾. Emple 9 Find the men nd stndrd devition of normll distriuted rndom vrile X if P(X 6 ) = : nd P(X > 5) = :. Let the unknown men nd stndrd devition e ¹ nd ¾ respetivel. Self Tutor Now P(X 6 ) = : µ X ¹ nd so P 6 ¹ =: ¾ ¾ µ ) P Z 6 ¹ =: ¾ nd P(X > 5) = : ) P(X 6 5) = :8 µ X ¹ nd so P 6 5 ¹ =:8 ¾ ¾ µ ) P Z 6 5 ¹ =:8 ¾ ¹ ) ¼ :5 ¾ 5 ¹ ) ¹ ¼ :5¾... () ) ¼ :86 ¾ ) 5 ¹ ¼ :86¾... () Solving () nd () simultneousl we get ¹ ¼ :5, ¾ ¼ :.

203 STATISTICAL DISTRIBUTIONS OF CONTINUOUS RANDOM VARIABLES (Chpter ) 66 EXERCISE E A mhine produes metl olts. The lengths of these olts hve norml distriution with men 9:8 m nd stndrd devition : m. If olt is seleted t rndom from the mhine, find the proilit tht it will hve length etween 9:7 m nd m. M s ustomers put mone for hrit into olletion o on the front ounter of his shop. The verge weekl olletion is pproimtel normll distriuted with men $ nd stndrd devition $6. Wht proportion of weeks would he epet to ollet: etween $: nd $5: t lest $5:? The students of Clss X st Phsis test. The verge sore ws 6 with stndrd devition of 5. The teher deided to wrd n A to the top 7% of the students in the lss. Assuming tht the sores were normll distriuted, find the lowest sore tht student needed to otin in order to hieve n A. Eels re wshed onto eh fter storm. Their lengths hve norml distriution with men m nd vrine m. If n eel is rndoml seleted, find the proilit tht it is t lest 5 m long. Find the proportion of eels mesuring etween m nd 5 m long. How mn eels from smple of would ou epet to mesure t lest 5 m in length? 5 Find the men nd stndrd devition of normll distriuted rndom vrile X if P(X > 5) = : nd P(X 6 8) = :6. 6 A rndom vrile X is normll distriuted. Find the men nd the stndrd devition of X, given tht P(X > 8) = : nd P(X 6 ) = :5. In the Mthemtis emintion t the end of the er, it ws found tht % of the students sored t lest 8 mrks, nd no more thn 5% sored less thn mrks. Assuming the mrks re normll distriuted, wht proportion of students sored more thn 5 mrks? 7 8 The IQ of students t shool is normll distriuted with stndrd devition 5. If% of the students hve n IQ lrger thn 5, find the men IQ of students t the shool. The distne n thlete n jump is normll distriuted with men 5: m. If 5% of the jumps this thlete re less thn 5 m, wht is the stndrd devition?

204 66 STATISTICAL DISTRIBUTIONS OF CONTINUOUS RANDOM VARIABLES (Chpter ) 9 Cirulr metl tokens re used to operte wshing mhine in lundromt. The dimeters of the tokens re normll distriuted, nd onl tokens with dimeters etween :9 nd :6 m will operte the mhine. Find the men nd stndrd devition of the distriution given tht % of the tokens re too smll, nd % re too lrge. Find the proilit tht t most one token out of rndoml seleted smple of will not operte the mhine. REVIEW SET A NON-CALCULATOR The ontents of ertin rnd of soft drink n re normll distriuted with men 77 ml nd stndrd devition : ml. Find the perentge of ns with ontents: i less thn 68:6 ml ii etween 7:8 ml nd 89:6 ml. Find the proilit tht rndoml seleted n hs ontents etween 77 ml nd 8: ml. The edile prt of th of Coffin B osters is normll distriuted with men 8:6 grms nd stndrd devition 6: grms. If the rndom vrile X is the mss of Coffin B oster: find if P(8:6 6 X 6 8:6+) =:686 find if P(X > ) =:8. f() = ½ ( ), 6 6 elsewhere is ontinuous proilit distriution funtion. Find. Sketh the grph of = f(): Find the men of the distriution. d Find P( 6 X 6 ). The ontinuous rndom vrile Z is distriuted suh tht Z» N(, ). Find the vlue of k if P( k 6 Z 6 k) =:. 5 The distne tht 5 er old o n throw tennis ll is normll distriuted with men 5 m nd stndrd devition m. The distne tht er old o n throw tennis ll is normll distriuted with men 5 m nd stndrd devition m. Jrrod is 5 ers old nd n throw tennis ll m. How fr does his er old rother Pul need to throw tennis ll to perform s well s Jrrod?

205 STATISTICAL DISTRIBUTIONS OF CONTINUOUS RANDOM VARIABLES (Chpter ) Stte the proilit tht rndoml seleted, normll distriuted vlue lies etween: ¾ ove the men nd ¾ ove the men the men nd ¾ ove the men. REVIEW SET B CALCULATOR The rm lengths of 8 er old femles re normll distriuted with men 6 m nd stndrd devition m. Find the perentge of 8 er old femles whose rm lengths re: i etween 6 m nd 7 m ii greter thn 6 m. Find the proilit tht rndoml hosen 8 er old femle hs n rm length in the rnge 56 m to 6 m. The rm lengths of 7% of the 8 er old femles re more thn m. Find the vlue of. The length of steel rods produed mhine is normll distriuted with stndrd devition of mm. It is found tht % of ll rods re less thn 5 mm long. Find the men length of rods produed the mhine. The distriution urve shown orresponds to X» N(¹, ¾ ). Are A = Are B =:. Find ¹ nd ¾. Clulte: i P(X 6 5) ii P( 6 X 6 ) A 8 B The mrks of 76 ndidtes in n IB emintion re normll distriuted with men of 9 mrks nd vrine 5. If the pss mrk is 5, estimte the numer of ndidtes who pssed the emintion. If 7% of the ndidtes sored 7, find the minimum mrk required to otin 7. 5 The life of Xenon-rnd tter is known to e normll distriuted with men of : weeks nd stndrd devition of :8 weeks. Find the proilit tht rndoml seleted tter will lst t lest 5 weeks. For how mn weeks n the mnufturer epet the tteries to lst efore 8% of them fil? 6 The rndom vrile X is normll distriuted with P(X 6 ) = :8 nd P(X > 9) = :. Find the men ¹ nd stndrd devition ¾ for X, orret to deiml ples. Hene find P( 7 6 X ¹ 6 7).

206 666 STATISTICAL DISTRIBUTIONS OF CONTINUOUS RANDOM VARIABLES (Chpter ) REVIEW SET C A rndom vrile X is normll distriuted with men :5 nd stndrd devition :. Find: P(X > ) P(8 6 X 6 ) k suh tht P(X 6 k) =:. A ftor hs mhine designed to fill ottles of drink with volume of 75 ml. It is found tht the verge mount of drink in eh ottle is 76 ml, nd tht :% of the drink ottles hve volume smller thn 75 ml. Assuming tht the mount of drink in eh ottle is distriuted normll, find the stndrd devition. X is ontinuous rndom vrile where X» N(¹, ). Find P( :5 <X ¹<:5). The lengths of metl rods produed in mnufturing proess re distriuted normll with men ¹ m nd stndrd devition 6 m. It is known tht 5:6% of the rods hve length greter thn 89:5 m. Find the men length of the metl rods. 5 The rndom vrile X is distriuted normll with men 5 nd P(X <9) ¼ :975: Find the shded re in the given digrm whih illustrtes the proilit densit funtion for X. 6 The heights of 8 er old os re normll distriuted with men of 87 m. Fifteen perent of ll these os hve heights greter thn 9 m. Find the proilit tht two 8 er old os hosen t rndom will hve heights greter thn 85 m. 8

207 Chpter 5 Misellneous questions Contents: A B Non-lultor questions Clultor questions

208 668 MISCELLANEOUS QUESTIONS (Chpter 5) A NON-CALCULATOR QUESTIONS EXERCISE 5A A geometri sequene hs S = nd S =8. Find: the ommon rtio r the twentieth term u. Show tht the sum of the first fort terms of the series ln + ln + ln 8 + ln 6 + :::::: is 8 ln. If f() =e nd g() = ln(), find: (f ± g)() (g ± f)(), in terms of, suh tht (f ± g)( )=(g ± f)( ): Let f() = ( ) +, qudrti in. Stte the oordintes of the verte. Find the es interepts. The grph of funtion g is otined trnslting the grph of f vertill through units. Stte the possile vlues of suh tht the grph of g: i hs etl one -interept ii hs no -interepts iii psses through the origin. 5 Epnd ( ) : Hene, find the oeffiient of in ( 7)( ). 6 Consider f() = p. Find: f() f( ) the domin of = f() d the rnge of = f(): 7 Let = sin o nd = tn 5 o. In terms of nd, write down epressions for: sin 6 o tn( 5 o ) os 7 o d tn o 8 Given f() = os nd g() =, solve: (f ± g)() = for 6 6 ¼ (g ± f)() = for 6 6 ¼ 9 Consider the funtions f() = + + nd g() =d +e+h with grphs illustrted. Cop nd omplete the following tle inditing whether the onstnt is positive, negtive or zero: ( ) g() Constnt d e h of f() of g() Sign

209 µ If A = µ If µ µ 5 nd A = 9 µ = MISCELLANEOUS QUESTIONS (Chpter 5) 669, find., find : Suppose (A + I) = B where I is the identit mtri. Find n epression for A in terms of B. µ Find A if B =. Find det A. 5 Consider u = i j + k nd v =i + pj k. Find: i p if u is perpendiulr to v ii jujjvj Show tht no vlue of p eists suh tht v u nd u re prllel. Consider prllelogrm ABCD: Find! BA nd! BC. Hene find j! BA j nd j! BC j: d e f Wht n e dedued out prllelogrm ABCD? Clulte os(cba): A (,, ) Find, in surd form, sin(cba): Hene find, in surd form, the re of ABCD. D B (,, ) C (,, ),,, d, e, f, g, h, i, j, k, l nd m re dt vlues whih hve een rrnged in sending order. Whih vrile represents the medin? Write down n lgeri epression for: i the rnge ii the interqurtile rnge. 5 The men nd stndrd devition for the dt set f,, g re 7:5 nd : respetivel. Cop nd omplete the tle elow finding the men nd stndrd devition of eh new dt set: New dt set Men Stndrd devition f,, g f +, +, +g f +5, +5, +5g 6 A drt is thrown t the drtord shown. It is equll likel to lnd nwhere on the ord. Given tht the drt lnds on the ord, show tht the proilit of it lnding on the shded region is etl 5 ¼ sin µ ¼. 5

210 67 MISCELLANEOUS QUESTIONS (Chpter 5) 7 For the funtion = f() with grph shown, sketh the grphs of: = f () = f () ( ) 8 Consider g() = os(): Find g (): Sketh = g () for ¼ 6 6 ¼: Write down the numer of solutions to g () = for ¼ 6 6 ¼: d Mrk point M on the sketh in where g () = nd g () >. 9 A nd B re mutull elusive events where P(A) = nd P(B )=:. Write, in terms of, n epression for P(A [ B): Find given tht P(A [ B) =:7: For the funtion g(), the sign digrms for g () nd g () re shown longside. The points A(, ), B(, ) nd C(, ) ll lie on the grph of = g(): Sketh the grph of = g(), lelling ll of the sttionr points. Consider f() =e : Show tht f () =e ( ): Find the point on the grph of = f() where the tngent is horizontl. Find vlues of for whih: i f() > ii f () > A prtile moves in stright line so tht its position s t time t seonds is given s(t) = e t + kt metres, where k is onstnt. Find the veloit funtion v(t): If the prtile is sttionr when t =ln, determine k. Solve for : log 7 = log 7= e 5 =8 d ln( ) ln() = The grph shows the veloit v ms of n ojet t time t seonds, t >. Find nd interpret: v() v () Z v(t) dt: g' (): g" (): v() t 5 t

211 MISCELLANEOUS QUESTIONS (Chpter 5) 67 5 Suppose Z Z f() d =. Find the vlue of: (f() 6) d k if Z kf() d = 5: 6 Show tht (sin µ os µ) = sin µ: Hene lulte Z ¼ (sin µ os µ) dµ: 7 The following tle shows the proilit distriution for disrete rndom vrile X. P(X = ) : :5 :5 Find nd given tht the epeted vlue of X is. 8 Consider the infinite geometri sequene: e, e, e,,... Find, in terms of e: e5 the ommon rtio the st term the sequene s sum Suppose A A, B A nd C A. 5 6 Find given tht: det A = AB = C The rndom vrile X is normll distriuted with men ¹ nd stndrd devition ¾. Let k e suh tht P ( X < k ) = : 7. Cop the norml distriution urve nd mrk on it ¹ nd k. On the grph shde the region whih illustrtes P( X<k ) = : 7: Find: i P(X > k) ii P(¹ < X < k) iii P(¹ ¾ < X < k) d If P(X > t) =:, find P(k 6 X 6 t): umultive The emintion mrks for students re frequen displed on the umultive frequen grph shown. The pss mrk for the emintion 5 ws. Wht perentge of the students pssed? A o-nd-whisker-plot for the emintion dt is: 5 mrk 6 8 m n p q From the grph, estimte: i m ii n iii p iv q

212 67 MISCELLANEOUS QUESTIONS (Chpter 5) The line L hs eqution = (tn 6 o ): Find p given tht the point A(, p) lies on L. Find the eqution of the line whih psses through A nd is perpendiulr to L. Write our nswer in the form + = : The elertion m s of n ojet moving in stright line is given (t) = os t + ¼ where t is the time in seonds. The ojet s initil veloit is 5 m s. Find n epression for the ojet s veloit v in terms of t. Find the veloit of the ojet t t = ¼ seonds. Consider f() =e +: Show tht f ln( ) + () = : Clulte f (8) f (). Give our nswer in the form ln where, Q + : 5 Given tht sin A = 5 nd ¼ 6 A 6 ¼, find: tn A sin A: 6 Consider the infinite geometri sequene, p,, 5 p,... Write the th term of the sequene in the form k p where k Q : Find the et vlue of the sum of the infinite sequene in the form + p where, Z : 7 A prtile is initill loted t P(,, ), nd it moves with fied veloit in stright line. After seonds the prtile is t Q(,, ): Find:! PQ the prtile s speed the eqution of the stright line. 8 Suppose P = µ nd Q = µ Find mtri PQ in terms of nd. Given tht PQ = ki, find, nd k. Hene, find P in terms of Q. 9 Suppose g() =e where 6 6. Sketh g on the given domin. Wht is the rnge of the funtion g? On the sme set of es used in, sketh the grph of g. d Stte the domin nd rnge of g. e Find lgerill g (): [QS] is digonl of qudrilterl PQRS where PQ =m, QR =7m, PS =5m nd QRS = o. P Show tht if SPQ = Á, then QS = p os Á m. If Á =6 o nd QSR = µ, show tht sin µ = 7 p 9 :. m 5m Q S 7m R

213 MISCELLANEOUS QUESTIONS (Chpter 5) 67 d Find the et length of [RS], given tht µ is ute. Hene, find the et perimeter nd re of PQRS. Suppose f() = + +. i Write down f () in simplest form. ii Find the eqution of the norml to f t (, 9). iii If this norml uts f t nother point A, find the oordintes of A. The grph of = f() is shown longside. i Write down n epression for the re of the shded region R. ii Clulte the et re of R. iii Suppose the region R is revolved out the -is through one revolution. Write down n epression for the volume of the solid formed. R ( ) Consider the geometri sequene:,, 6, 8,... i Write down the ommon rtio. ii Find the th term. Consider the sequene:,, 7,... i Find if the sequene is geometri. ii Does the sequene hve sum to infinit? Eplin our nswer. Suppose the sequene,, 7,... is rithmeti. Find: i its th term ii the sum of its first 5 terms. A nd B hve position vetors i +j k nd i j 8k respetivel.! Find: i AB ii the unit vetor u in the diretion of! BA: Is u perpendiulr to! OA? If C hs position vetor i + j + k, nd! OC is perpendiulr to i j +k, find. d If M is the midpoint of [AB], find the position vetor of M. e Line L psses through M nd is prllel to! OA. Write down the vetor eqution r of line L. f Suppose line L hs vetor eqution r =(mi + j k)+s(i j + k): Suppose A = i Eplin wh L is not prllel to L. ii Find m if L nd L interset. iii Find the position vetor of P, the point of intersetion of L nd L. µ. Find: i A ii A iii A. µ Show tht A n = n n for n =,, nd.

214 67 MISCELLANEOUS QUESTIONS (Chpter 5) Assuming the result from is lso true for n > 5, find A. d Let S n = A + A + A + A A n. i Write down S n s sum of mtries. µ p q ii If S n =, eplin wh r = nd s = n. r s iii Show tht p = n+ nd hene find q in similr form. iv Find S using the ove results. 5 Consider the qudrti = +. i Eplin wh this qudrti hs mimum vlue. ii Wht vlue of gives this mimum vlue? iii Wht is the mimum vlue? B In ABC, AB =, BC =, AC =8, nd the perimeter of ABC is. i Write in terms of. ii Use the osine rule to write in terms of nd os µ. A iii Hene, show tht os µ = : 8 C If the re of the tringle in is A, show tht A =6 sin µ. d Show tht A = ( + ): e f Hene, find the mimum re of ABC. Comment on the shpe of the tringle when it hs mimum re. 6 Suppose f() = nd g() = +. Find f () nd g (), the inverse funtions of f nd g. d Find (f ± g )(): Find suh tht (f ± g )() =f (). Suppose H() = f() g() : i ii iii iv Sketh the grph of H(). Inlude its smptotes nd their equtions. Find onstnts A nd B suh tht + = A + B + Clulte the et vlue of Z H() d. On our sketh in i, shde the region whose re is given Z H() d: 7 Hnnh, Heidi nd Holl hve different sets of rds, ut eh set ontins rds with the numers,,, or, one per rd. Hnnh wrongl sttes tht the proilit distriution of her set of rds is: Wh is Hnnh wrong? P(X = ) : : : : :

215 MISCELLANEOUS QUESTIONS (Chpter 5) 675 Heidi orretl sttes tht her proilit distriution is: P(X = ) : : : Wht n e dedued out nd? Holl orretl sttes tht the proilit distriution for her set of rds is ( +) P(X = ) = : 5 If one rd is rndoml hosen from Holl s set, find the proilit tht it is: i ii not. 8 An ordinr 6-sided die is used to selet one of two Rt_ B gs C nd D, nd tiket is drwn from tht g. C Y If or is rolled, g C is hosen; otherwise g D is hosen. B D Bg C ontins lue nd ellow tiket. Y Bg D ontins lue nd ellow tikets. Cop nd omplete the tree digrm showing the possile outomes. Wht is the proilit tht ellow tiket is drwn from g D? Find the proilit of drwing ellow tiket from either g. d If lue tiket is hosen, find the proilit tht it me from g D. e In gmling gme, pler wins $6 for getting lue tiket nd $9 for getting ellow one. Find the pler s epeted return. 9 Two identil tetrhedrl die re rolled. Their four verties re lerl lelled,, nd s shown in the digrm. The result when one die omes to rest is the numer on the uppermost verte. This is in the digrm. Illustrte the smple spe of 6 possile results when the two die re rolled. Let X e the sum of the sores on the two die. Wht re the possile vlues of X? Find: i P(X =) ii P(X > ) iii P(X =6j X > ) d When Mimi uses the two die to pl gme she: ² wins E5 if the sum is, ² wins E if the sum is greter thn, ² loses Ed if the sum is less thn. Wht vlue should d hve if Mimi s epeted result is E? 5 A prtile moves in stright line suh tht t time t seonds, t >, the elertion is (t) =t sin t m s. Wht is the prtile s elertion t time t = nd t = ¼ seonds? If the initil veloit of the prtile is m s, find its veloit funtion v(t). Z ¼ d Find v(t) dt nd eplin wh the result is positive. Interpret the result in with regrd to the prtile s motion.