Calculus Chapter 1 Introduction to Calculus

Transcription

1 Inroducon o Calculus Cal 1-3 Calculus Chaper 1 Inroducon o Calculus CHAPER 1 CALCULUS INRODUCION O hs chaper, whch replaces Chaper 4 n Physcs 2, s nended for sudens who have no had calculus, or as a calculus revew for hose whose calculus s no well remembered. If, afer readng par way hrough hs chaper, you feel your calculus background s no so bad afer all, go back o Chaper 4 n Physcs 2, sudy he dervaon of he consan acceleraon formulas begnnng on page 4-8, and work he projecle moon problems n he appendx o Chaper 4. hose who sudy all of hs nroducon o calculus should hen proceed o he projecle moon problems n he appendx o Chaper 4 of he physcs ex. In Chaper 3 of Physcs 2, we used srobe phoographs o defne velocy and acceleraon vecors. he basc approach was o urn up he srobe flashng rae as we dd n gong from Fgure (3-3) o (3-4) unl all he knks are clearly vsble and he successve dsplacemen vecors gve a reasonable descrpon of he moon. We dd no urn he flashng rae oo hgh, for he praccal reason ha he dsplacemen vecors became oo shor for accurae work. LIMIING PROCESS In our dscusson of nsananeous velocy we concepually urned he srobe all he way up as llusraed n Fgures (2-22a) hrough (2-22d), redrawn here n Fgure (1). In hese fgures, we nally see a farly large change n v as he srobe rae s ncreased and reduced. Bu he change becomes smaller and looks as f we are approachng some fnal value of v ha does no depend on he sze of, provded s small enough. I looks as f we have come close o he fnal value n Fgure (1c). he progresson seen n Fgure (1) s called a lmng process. he dea s ha here really s some rue value of v whch we have called he nsananeous velocy, and ha we approach hs rue value for suffcenly small values of. hs s a calculus concep, and n he language of calculus, we are akng he lm as goes o zero. he Uncerany Prncple For over 2 years, from he nvenon of calculus by Newon and Lebnz unl 1924, he lmng process and he resulng concep of nsananeous velocy was one of he cornersones of physcs. hen n 1924 Werner Hesenberg dscovered wha he called he uncerany prncple whch places a lm on he accuracy of expermenal measuremens.

2 Cal 1-4 Calculus 2 Hesenberg dscovered somehng very new and unexpeced. He found ha he ac of makng an expermenal measuremen unavodably affecs he resuls of an expermen. hs had no been known prevously because he effec on large objecs lke golf balls s undeecable. Bu on an aomc scale where we sudy small sysems lke elecrons movng nsde an aom, he effec s no only observable, can domnae our sudy of he sysem. One parcular consequence of he unceranly prncple s ha he more accuraely we measure he poson of an objec, he more we dsurb he moon of he objec. hs has an mmedae mpac on he concep of nsananeous velocy. If we urn he srobe all he way up, reduce o zero, we are n effec ryng o measure he poson of he objec wh nfne precson. he consequence would be an nfnely bg dsurbance of he moon of he objec we are sudyng. If we acually could urn he srobe all he way up, we would desroy he objec we were ryng o sudy. I urns ou ha he uncerany prncple can have a sgnfcan mpac on a larger scale of dsance han he aomc scale. Suppose, for example, ha we consruced a chamber 1 cm on a sde, and wshed o sudy he projecle moon of an elecron nsde. Usng Galleo s dea ha objecs of dfferen mass fall a he same rae, we would expec ha he moon of he elecron projecle should be he same as more massve objecs. If we ook a srobe phoograph of he elecron s moon, we would expec ge resuls lke hose shown n Fgure (2). hs fgure represens projecle moon wh an acceleraon g = 98 cm/sec 2 and =.1sec, as he reader can easly check. 1 cenmeer -1 1 v When we sudy he uncerany prncple n Chaper 4 of he physcs ex, we wll see ha a measuremen whch s accurae enough o show ha Poson (2) s below Poson (1), could dsurb he elecron enough o reverse s drecon of moon. he nex poson measuremen could fnd he elecron over where we drew Poson (3), or back where we drew Poson (), or anywhere n he regon n beween. As a resul we could no even deermne wha drecon he elecron s movng. hs uncerany would no be he resul of a sloppy expermen, s he bes we can do wh he mos accurae and delcae measuremens possble. he uncerany prncple has had a sgnfcan mpac on he way physcss hnk abou moon. Because we now know ha he measurng process affecs he resuls of he measuremen, we see ha s essenal o provde expermenal defnons o any physcal quany we wsh o sudy. A concepual defnon, lke urnng he srobe all he way up o defne nsananeous velocy, can lead o fundamenal nconssences. Even an expermenal defnon lke our srobe defnon of velocy can lead o nconssen resuls when appled o somehng lke he elecron n Fgure (2). Bu hese nconssences are real. her exsence s ellng us ha he very concep of velocy s begnnng o lose meanng for hese small objecs. On he oher hand, he dea of he lmng process and nsananeous velocy s very convenen when appled o larger objecs where he effecs of he uncerany prncple are no deecable. In hs case we can apply all he mahemacal ools of calculus developed over he pas 25 years. he saus of nsananeous velocy has changed from a basc concep o a useful mahemacal ool. hose problems for whch hs mahemacal ool works are called problems n classcal physcs; and hose problems for whch he uncerany prncple s mporan, are n he realm of wha we call quanum physcs. 1 cenmeer Fgure 2 Hypohecal elecron projecle moon expermen.

3 Inroducon o Calculus Cal 1-5 CALCULUS DEFINIION OF VELOCIY Wh he above perspecve on he physcal lmaons on he lmng process, we can now reurn o he man opc of hs chaper he use of calculus n defnng and workng wh velocy and acceleraon. In dscussng he lmng process n calculus, one radonally uses a specal se of symbols whch we can undersand f we adop he noaon shown n Fgure (3). In ha fgure we have drawn he coordnae vecors R and R +1 for he h and ( + 1) h posons of he objec. We are now usng he symbol R o represen he dsplacemen of he ball durng he o +1 nerval. he vecor equaon for R s R =R +1 R (1) In words, Equaon (1) ells us ha R s he change, durng he me, of he poson vecor R descrbng he locaon of he ball. he velocy vecor v s now gven by v = R (2) hs s jus our old srobe defnon v =s /, bu usng a noaon whch emphaszes ha he dsplacemen s = R s he change n poson ha occurs durng he me. he Greek leer (dela) s used boh o represen he dea ha he quany R or s small, and o emphasze ha boh of hese quanes change as we change he srobe rae. he lmng process n Fgure (1) can be wren n he form v lm R (3) where he word lm wh underneah, s o be read as lm as goes o zero. For example we would read Equaon (3) as he nsananeous velocy v a poson s he lm, as goes o zero, of he rao R /. R R + 1 R + 1 For wo reasons, Equaon (3) s no que ye n sandard calculus noaon. One s ha n calculus, only he lmng value, n hs case, he nsananeous velocy, s consdered o be mporan. Our srobe defnon v = R / s only a sep n he lmng process. herefore when we see he vecor v, we should assume ha s he lmng value, and no specal symbol lke he underlne s used. For hs reason we wll drop he underlne and wre v = lm R (3a) R = R + 1 R V = R Fgure 3 Defnons of R and v.

4 Cal 1-6 Calculus 2 he second change deals wh he fac ha when goes o zero we need an nfne number o me seps o ge hrough our srobe phoograph, and hus s no possble o locae a poson by counng me seps. Insead we measure he me ha has elapsed snce he begnnng of he phoograph, and use ha me o ell us where we are, as llusraed n Fgure (4). hus nsead of usng v o represen he velocy a poson, we wre v() o represen he velocy a me. Equaon (3) now becomes v() = lm R() (3b) where we also replaced R by s value R() a me. Alhough Equaon (3b) s n more or less sandard calculus noaon, he noaon s clumsy. I s a pan o keep wrng he word lm wh a underneah. o sreamlne he noaon, we replace he Greek leer wh he Englsh leer d as follows lm R() dr() d (4) (he symbol means defned equal o.) o a mahemacan, he symbol dr()/d s jus shorhand =.1sec =.2sec noaon for he lmng process we have been descrbng. Bu o a physcs, here s a dfferen, more praccal meanng. hnk of d as a shor, shor enough so ha he lmng process has essenally occurred, bu no oo shor o see wha s gong on. In Fgure (1), a value of d less han.25 seconds s probably good enough. If d s small bu fne, hen we know exacly wha he dr() s. I s he small bu fne dsplacemen vecor a he me. I s our old srobe defnon of velocy, wh he added condon ha d s such a shor me nerval ha he lmng process has occurred. From hs pon of vew, d s a real me nerval and dr() a real vecor, whch we can work wh n a normal way. he only hng specal abou hese quanes s ha when we see he leer d nsead of, we mus remember ha a lmng process s nvolved. In hs noaon, he calculus defnon of velocy s v() = dr() d (5) where R() and v() are he parcle s coordnae vecor and velocy vecor respecvely, as shown n Fgure (5). Remember ha hs s jus fancy shorhand noaon for he lmng process we have been descrbng. = sec =.3sec V() =.4sec R() a =.3 sec R() =.5sec Fgure 4 Raher han counng ndvdual mages, we can locae a poson by measurng he elapsed me. In hs fgure, we have drawn he dsplacemen vecor R() a me =.3 sec. Fgure 5 Insananeous poson and velocy a me.

5 Inroducon o Calculus Cal 1-7 ACCELERAION In he analyss of srobe phoographs, we defned boh a velocy vecor v and an acceleraon vecor a. he defnon of a, shown n Fgure (2-12) reproduced here n Fgure (6) was a v +1 v (6) In our graphcal work we replaced v by s / so ha we could work drecly wh he dsplacemen vecors s and expermenally deermne he behavor of he acceleraon vecor for several knds of moon. Le us now change hs graphcal defnon of acceleraon over o a calculus defnon, usng he deas jus appled o he velocy vecor. Frs, assume ha he ball reached poson a me as shown n Fgure (6). hen we can wre v =v() v +1 =v(+ ) o change he me dependence from a coun of srobe flashes o he connuous varable. Nex, defne he vecor v() by v() v(+ ) v() = v +1 v (7) We see ha v()s he change n he velocy vecor as he me advances from o +. he srobe defnon of a can now be wren a() srobe defnon v( + ) v() = v() (8) Now go hrough he lmng process, urnng he srobe up, reducng unl he value of a() seles down o s lmng value. We have a() calculus defnon = lm v + v = lm v() (9) Fnally use he shorhand noaon d/d for he lmng process: a() = dv() d (1) Equaon (1) does no make sense unless you remember ha s noaon for all he deas expressed above. Agan, physcss hnk of d as a shor bu fne me nerval, and dv() as he small bu fne change n he velocy vecor durng he me nerval d. I s our srobe defnon of acceleraon wh he added requremen ha s shor enough ha he lmng process has already occurred. Componens Even f you have suded calculus, you may no recall encounerng formulas for he dervaves of vecors, lke dr()/ and dv()/ whch appear n Equaons (5) and (1). o brng hese equaons no a more famlar form where you can apply sandard calculus formulas, we wll break he vecor Equaons (5) and (1) down no componen equaons. poson a me V poson a me + In he chaper on vecors, we saw ha any vecor equaon lke A =B+C (11) s equvalen o he hree componen equaons a = ( V + 1 V ) ( V + 1 V ) V V + 1 A x =B x +C x A y =B y +C y (12) A z =B z +C z he advanage of he componen equaons was ha hey are smply numercal equaons and no graphcal work or rgonomery s requred. Fgure 6 Expermenal defnon of he acceleraon vecor.

6 Cal 1-8 Calculus 2 he lmng process n calculus does no affec he decomposon of a vecor no componens, hus Equaon (5) for v() and Equaon (1) for a() become v() = dr()/d (5) v x () = dr x ()/d (5a) v y () = dr y ()/d (5b) v z () = dr z ()/d (5c) and a() = dv()/d (1) a x () = dv x ()/d a y () = dv y ()/d a z () = dv z ()/d (1a) (1b) (1c) Ofen we use he leer x for he x coordnae of he vecor R and we use y for R y and z for R z. Wh hs noaon, Equaon (5) assumes he shorer and perhaps more famlar form v x () = ()/d v y () = dy()/d v z () = dz()/d y (5a ) (5b ) (5c ) A hs pon he noaon has become decepvely shor. You now have o remember ha x() sands for he x coordnae of he parcle a a me. We have fnally boled he noaon down o he pon where would be famlar from any calculus course. If we resrc our aenon o one dmensonal moon along he x axs. hen all we have o concern ourselves wh are he x componen equaons R Fgure 7 x INEGRAION When we worked wh srobe phoographs, he phoograph old us he poson R() of he ball as me passed. Knowng he poson, we can hen use Equaon (5) o calculae he ball's velocy v() and hen Equaon (1) o deermne he acceleraon a(). In general, however, we wan o go he oher way, and predc he moon from a knowledge of he acceleraon. For example, magne ha you were n Galleo's poson, hred by a prnce o predc he moon of cannonballs. You know ha a cannonball should no be much affeced by ar ressance, hus he acceleraon hroughou s rajecory should be he consan gravaonal acceleraon g. You know ha a() = g ; how hen do you use ha knowledge n Equaons (5) and (1) o predc he moon of he ball? he answer s ha you canno wh he equaons n her presen form. he equaons ell you how o go from R() o a(), whle o predc moon you need o go he oher way, from a() o R(). he opc of hs secon s o see how o reverse he drecons n whch we use our calculus equaons. Equaons (5) and (1) nvolve he process called dfferenaon. We wll see ha when we go he oher way he reverse of dfferenaon s a process called negraon. We wll see ha negraon s a smple concep, bu a process ha s somemes hard o perform whou he ad of a compuer. v x () = () d a x () = dv x () d (1a)

7 Inroducon o Calculus Cal 1-9 Predcon of Moon In our earler dscusson, we have used srobe phoographs o analyze moon. Le us see wha we can learn from such a phoograph for predcng moon. Fgure (8) s our famlar projecle moon phoograph showng he dsplacemen s of a ball durng he me he ball raveled from a poson labeled () o he poson labeled (4). If he ball s now a poson () and each of he mages s.1 seconds apar, hen he vecor s ells us where he ball wll be a a me of.4 seconds from now. If we can predc s, we can predc he moon of he ball. he general problem of predcng he moon of he ball s o be able o calculae s() for any me. From Fgure (8) we see ha s s he vecor sum of he ndvdual dsplacemen vecors s 1, s 2, s 3 and s 4 s = s 1 +s 2 +s 3 +s 4 (11) We can hen use he fac ha s 1 =v 1, s 2 =v 2, ec. o ge s =v 1 +v 2 +v 3 +v 4 (12) Raher han wrng ou each erm, we can use he summaon sgn Σ o wre s = 4 Σ =1 v (12a) Equaon (12) s approxmae n ha he v are approxmae (srobe) veloces, no he nsananeous veloces we wan for a calculus dscusson. In Fgure (9) we mproved he suaon by cung o 1/4 of s prevous value, gvng us four mes as many mages and more accurae veloces v. We see ha he dsplacemen s s now he sum of 16 vecors s = s 1 +s 2 +s s 15 +s 16 (13) Expressng hs n erms of he velocy vecors v 1 o v 16 we have s =v 1 +v 2 +v v 15 +v 16 (14) or usng our more compac noaon s = 16 Σ =1 v (14a) Whle Equaon (14) for s looks que dfferen han Equaon (12), he sum of sxeen vecors nsead of four, he dsplacemen vecors s n he wo cases are exacly he same. Addng more nermedae mages dd no change where he ball was locaed a he me of =.4 seconds. In gong from Equaon (12) o (14), wha has changed n shorenng he me sep, s ha he ndvdual velocy vecors v become more nearly equal o he nsananeous velocy of he ball a each mage. = S 1 1 S 2 2 S = 4 8 S 3 S 3 S 12 S 4 S = S + 1 S S 16 S = S + 1 S S 16 S 16 =.4 sec 4 =.4 sec 16 Fgure 8 o predc he oal dsplacemen s, we add up he ndvdual dsplacemens s. Fgure 9 Wh a shorer me nerval, we add up more dsplacemen vecors o ge he oal dsplacemen s.

8 Cal 1-1 Calculus 2 If we reduced agan by anoher facor of 1/4, so ha we had 64 mages n he nerval = o =.4 sec, he formula for s would become s = 64 Σ =1 v (15a) where now he v are sll closer o represenng he ball's nsananeous velocy. he more we reduce, he more mages we nclude, he closer each v comes o he nsananeous velocy v. Whle addng more mages gves us more vecors we have o add up o ge he oal dsplacemen s, here s very lle change n our formula for s. If we had a mllon mages, we would smply wre s = 1 Σ =1 v (16a) In hs case he v would be physcally ndsngushable from he nsananeous velocy v(). We have essenally reached a calculus lm, bu we have problems wh he noaon. I s clearly nconvenen o label each v and hen coun he mages. Insead we would lke noaon ha nvolves he nsananeous velocy v() and expresses he begnnng and end pons n erms of he nal me 1 and fnal me f, raher han he nal and fnal mage numbers. In he calculus noaon, we replace he summaon sgn Σ by somehng ha looks almos lke he summaon sgn, namely he negral sgn. (he French word for negraon s he same as her word for summaon.) Nex we replaced he ndvdual v by he connuous varable v() and fnally express he end pons by he nal me and he fnal me f. he resul s s = n Σ =1 v as he number n becomes nfnely large f v()d (17) Calculus noaon s more easly handled, or s a leas more famlar, f we break vecor equaons up no componen equaons. Assume ha he ball sared a poson whch has componens x =x( ) [read x( ) as x a me ] and y =y( ) as shown n Fgure (1). he fnal poson f s a x f =x( f ) and y f =y( f ). hus he dsplacemen s has x and y componens s x =x( f ) x( ) s y =y( f ) y( ) Breakng Equaon (17) no componen equaons gves s x =x( f ) x( )= s y =y( f ) y( )= v x ()d f (18a) v y ()d f (18b) Here we wll nroduce one more pece of noaon ofen used n calculus courses. On he lef hand sde of Equaon (18a) we have x( f ) x( ) whch we can hnk of as he varable x() evaluaed over he nerval of me from o f. We wll ofen deal wh varables evaluaed over some nerval and have a specal noaon for ha. We wll wre x( f ) x( ) x() f (19) You are o read he symbol x() f as "x of evaluaed from o f ". We wre he nal me a he boom of he vercal bar, he fnal me f a he op. y y f (y y ) f (x x ) x x f x( ) x( f) Fgure 1 Breakng he vecor s no componens. S f f

9 Inroducon o Calculus Cal 1-11 We use smlar noaon for any knd of varable, for example x 2 f(x) f(x 2 ) f(x 1 ) (19a) x1 Remember o subrac he varable when evaluaed a he value a he boom of he vercal bar. Wh hs noaon, our Equaon (18) can be wren f s x =x() = s y =y() f = f v x ()d (18 a ) f v y ()d (18 b ) Calculang Inegrals Equaon (2) s nce and compac, bu how do you use? How do you calculae negrals? he key s o remember ha an negral s jus a fancy noaon for a sum of erms, where we make he me sep () very small. Keepng hs n mnd, we wll see ha here s a very easy way o nerpre an negral. Fgure 11a Srobe phoograph of ball movng a consan velocy n x drecon. v x() v x Fgure 11b Graph of v x () versus for he ball of Fgure 11a. v () x v x v x7 Fgure 11c Each v x s he area of a recangle. f f f x o ge hs nerpreaon, le us sar wh he smple case of a ball movng n a sragh lne, for nsance, he x drecon, a a consan velocy v x. A srobe pcure of hs moon would look lke ha shown n Fgure (11a). Fgure (11b) s a graph of he ball's velocy v x () as a funcon of he me. he vercal axs s he value of v x, he horzonal axs s he me. Snce he ball s ravelng a consan velocy, v x has a consan value and s hus represened by a sragh horzonal lne. In order o calculae he dsance ha he ball has raveled durng he me nerval from o f, we need o evaluae he negral s x = f v x ()d dsance ball ravels n me nerval o f (18a) o acually evaluae he negral, we wll go back o our summaon noaon s x = fnal v x (2) Σ nal and show ndvdual me seps n he graph of v x versus, as n Fgure (11c). We see ha each erm n Equaon (2) s represened n Fgure (11c) by a recangle whose hegh s v x and whose wdh s. We have shaded n he recangle represenng he 7h erm v x7. We see ha v x7 s jus he area of he shaded recangle, and s clear ha he sum of all he areas of he ndvdual recangles s he oal area under he curve, sarng a me and endng a me f. Here we are begnnng o see ha he process of negraon s equvalen o fndng he area under a curve. Wh a smple curve lke he consan velocy v x () n Fgure (11c), we see by nspecon ha he oal area from o f s jus he area of he complee recangle of hegh v x and wdh ( f ). hus s x =v x ( f ) (21) hs s he expeced resul for consan velocy, namely dsance raveled = velocy me for consan velocy (21a)

10 Cal 1-12 Calculus 2 o see ha you are no resrced o he case of consan velocy, suppose you drove on a freeway due eas (he x drecon) sarng a 9: AM and soppng for lunch a 12 noon. Every mnue durng your rp you wroe down he speedomeer readng so ha you had an accurae plo of v x () for he enre mornng, a plo lke ha shown n Fgure (12). From such a plo, could you deermne he dsance s x ha you had ravelled? Your bes answer s o mulply each value v of your velocy by he me o calculae he average dsance raveled each mnue. Summng hese up from he nal me = 9:AM o he fnal me f = noon, you have as your esmae s x Σ v x (he symbol means approxmaely equal.) o ge a more accurae value for he dsance raveled, you should measure your velocy a shorer me nervals and add up he larger number of smaller recangles. he precse answer should be obaned n he lm as goes o zero s x = lm f Σ v x = v x ()d (22) hs lm s jus he area under he curve ha s supposed o represen he nsananeous velocy v x (). v () x v x7 9am Fgure 12 Plo of v x () for a rp sarng a 9: AM and fnshng a noon. he dsance raveled s he area under he curve. noon hus we can nerpre he negral of a curve as he area under he curve even when he curve s no consan or fla. Mahemacans concern hemselves wh curves ha are so wld ha s dffcul or mpossble o deermne he area under hem. Such curves seldom appear n physcs problems. Whle he basc dea of negraon s smple jus fndng he area under a curve n pracce can be que dffcul o calculae he area. Much of an nroducory calculus course s devoed o fndng he formulas for he areas under varous curves. here are also books called ables of negrals where you look up he formula for a curve and he able ells you he formula for he area under ha curve. In Chaper 16 of he physcs ex, we wll dscuss a mahemacal echnque called Fourer analyss. hs s a echnque n whch we can descrbe he shape of any connuous curve n erms of a sum of sn waves. (Why we wan o do ha wll become clear hen.) he process of Fourer analyss nvolves fndng he area under some very complex curves, curves ofen nvolvng expermenal daa for whch we have no formula, only graphs. Such curves canno be negraed by usng a able of negrals, wh he resul ha Fourer analyss was no wdely used unl he adven of he modern dgal compuer. he compuer made a dfference, because we can fnd he area under almos any curve by breakng he curve no shor peces of lengh, calculang he area v of each narrow recangle, and addng up he area of he recangles o ge he oal area. If he curve s so wld ha we have o break no a mllon segmens o ge an accurae answer, ha mgh be oo hard o do by hand, bu usually a very smple and rapd job for a compuer. Compuers can be much more effcen han people a negraon.

11 Inroducon o Calculus Cal 1-13 he Process of Inegrang here s a language for he process of negraon whch we wll now ake you hrough. In each case we wll check ha he resuls are wha we would expec from our summaon defnon, or he dea ha an negral s he area under a curve. he smples negral we wll encouner n he calculaon of he area under a curve of un hegh as shown n Fgure (13). We have he area of a recangle of hegh 1 and lengh ( f ) 1 f f 1d = d =( f ) area = 1( ) Fgure 13 Area under a curve of un hegh. f f (22) We wll use some specal language o descrbe hs negraon. We wll say ha he negral of d s smply he me, and ha he negral of d from o f s equal o evaluaed from o f. In symbols hs s wren as f f d = =( f ) (23) Recall ha he vercal lne afer a varable means o evaluae ha varable a he fnal poson f (upper value), mnus ha varable evaluaed a he nal poson (lower value). Noce ha hs prescrpon gves he correc answer. he nex smples negral s he negral of a consan, lke a consan velocy v x over he nerval o f f v x d =v x ( f ) (24) Snce ( f )= f d, we can replace ( f ) n Equaon (24) by he negral o ge f v x d =v x d f v x a consan (25) and we see ha a consan lke v x can be aken ousde he negral sgn. Le us ry he smples case we can hnk of where v x s no consan. Suppose v x sars a zero a me = and ncreases lnearly accordng o he formula v x =a (26) a f v x Fgure 15 v = a x When we ge up o he me f he velocy wll be (a f ) as shown n Fgure (15). he area under he curve v x =a s a rangle whose base s of lengh f and hegh s a f. he area of hs rangle s one half he base mes he hegh, hus we ge for he dsance s x raveled by an objec movng wh hs velocy f s x = v x d = 1 (base) (alude) 2 = 1 (27) 2 ( f )(a f )=1 2 a f 2 Now le us repea he same calculaon usng he language one would fnd n a calculus book. We have s x = f f v x d = (a)d (28) he consan (a) can come ousde, and we know ha he answer s 1/2a f 2, hus we can wre f v x area = v ( ) Fgure 14 Area under he consan v x curve. x f f s x =a f d= 1 2 a f 2 (29) In Equaon (29) we can cancel he a's o ge he resul x d = 1 2 f 2 (3)

12 Cal 1-14 Calculus 2 In a calculus ex, you would fnd he saemen ha he negral d s equal o 2 /2 and ha he negral should be evaluaed as follows f d = 2 f 2 = f = f 2 2 (31) Indefne Inegrals When we wan o measure an acual area under a curve, we have o know where o sar and sop. When we pu hese lms on he negral sgn, lke and f, we have wha s called a defne negral. However here are mes where we jus wan o know wha he form of he negral s, wh he dea ha we wll pu n he lms laer. In hs case we have wha s called an ndefne negral, such as d = 2 2 ndefne negral (32) he dfference beween our defne negral n Equaon (31) and he ndefne one n Equaon (32) s ha we have no chosen he lms ye n Equaon (32). If possble, a able of negrals wll gve you a formula for he ndefne negral and le you pu n whaever lms you wan. Inegraon Formulas For some ses of curves, here are smple formulas for he area under hem. One example s he se of curves of he form n. We have already consdered he cases where n = and n = 1. n = d = d = n = 1 1 d = d = 2 2 Some resuls we wll prove laer are n = n = 3 3 d = 4 4 (33a,b,c,d) Lookng a he way hese negrals are urnng ou, we suspec ha he general rule s n d = n+1 n+1 (34) I urns ou ha Equaon (34) s a general resul for any value of n excep n = 1. If n = 1, hen you would have dvson by zero, whch canno be he answer. (We wll shorly dscuss he specal case where n = 1.) As long as we say away from he n = 1 case, he formula works for negave numbers. For example 2 d = d 2 = 1 d 2 = ( 2+1) 2+1 = ( 1) 1 (35) In our dscusson of gravaonal and elecrcal poenal energy, we wll encouner negrals of he form seen n Equaon (35). Exercse 1 Usng Equaon (34) and he fac ha consans can come ousde he negral, evaluae he followng negrals: (a) x does no maer wheher we call he varable or x (b) (c) (d) x=2 x 5 x=1 =2 =1 d 2 3 also skechhe area beng evaluaed Show ha you ge a posvearea. GmM r 2 dr whereg, m, and M are consans 2 d = 3 3 (e) a y 3/2dy (a) s a consan

13 Inroducon o Calculus Cal 1-15 NEW FUNCIONS Logarhms We have seen ha when we negrae a curve or funcon lke 2, we ge a new funcon 3 /3. he funcons 2 and 3 appear o be farly smlar; he negraon dd no creae somehng radcally dfferen. However, he process of negraon can lead o some curves wh enrely dfferen behavor. hs happens, for example, n ha specal case n = 1 when we ry o do he negral of 1. I s ceranly no hard o plo 1, he resul s shown n Fgure (16). Also here s nohng fundamenally dffcul or pecular abou measurng he area under he 1 curve from some o f, as long as we say away from he orgn = where 1 blows up. he formula for hs area urns ou, however, o be he new funcon called he naural logarhm, abbrevaed by he symbol ln. he area n Fgure (16) s gven by he formula f 1 d =ln( f ) ln( ) (36) 1 curve 1 wo of he mporan bu pecular feaures of he naural logarhm are ln(ab) = ln(a) + ln(b) (37) ln( 1 a )= ln(a) (38) hus we ge, for example ln( f ) ln( )=ln( f )+ln 1 =ln f (39) hus he area under he curve n Fgure (16) s f d =ln f (4) Whle he naural logarhm has some raher pecular properes s easy o evaluae because s avalable on all scenfc calculaors. For example, f =.5 seconds and f = 4 seconds, hen we have ln f =ln 4 = ln (8) (41).5 Enerng he number 8 on a scenfc calculaor and pressng he buon labeled ln, gves ln (8) = 2.79 (42) whch s he answer. Fgure 16 Plo of 1. he area under hs curve s he naural logarhm ln. f Exercse 2 Evaluae he negrals 1.1 x 1.1 x Why are he answers he same?

14 Cal 1-16 Calculus 2 he Exponenal Funcon We have jus seen ha, whle he logarhm funcon may have some pecular properes, s easy o evaluae usng a scenfc calculaor. he queson we now wan o consder s wheher here s some funcon ha undoes he logarhm. When we ener he number 8 no he calculaor and press ln, we ge he number Now we are askng f, when we ener he number 2.79, can we press some key and ge back he number 8? he answer s, you press he key labeled e x. he e x key performs he exponenal funcon whch undoes he logarhm funcon. We say ha he exponenal funcon e x s he nverse of he logarhm funcon ln. Exponens o he Base 1 You are already famlar wh exponens o he base 1, as n he followng examples 1 =1 1 1 =1 1 2 = = 1,, 1 1 = 1/1 = = 1/1 = =.1 (43) he exponen, he number wren above he 1, ells us how many facors of 1 are nvolved. A mnus sgn means how many facors of 1 we dvde by. From hs alone we deduce he followng rules for he exponen o he base 1. 1 a = 1 1 a (44) he nverse of he exponen o he base 1 s he funcon called logarhm o he base 1 whch s denoed by he key labeled log on a scenfc calculaor. Formally hs means ha log 1 y =y (46) Check hs ou on your scenfc calculaor. For example, ener he number 1,, and press he log buon and see f you ge he number 6. ry several examples so ha you are confden of he resul. he Exponenal Funcon y x Anoher key on your scenfc calculaor s labeled y x. hs allows you o deermne he value of any number y rased o he power (or exponen) x. For example, ener he number y = 1, and press he y x key. hen ener he number x = 6 and press he = key. You should see he answer y x =1 6 = 1 I s que clear ha all exponens obey he same rules we saw for powers of 1, namely y a y b =y a+b (47) (Example y 2 y 3 = y y y y y =y 5.) And as before y a 1 y a (48) 1 a 1 b =1 a+b (45) (Example = 1 1 = 1,.)

15 Inroducon o Calculus Cal 1-17 Exercse 3 Use your scenfc calculaor o evaluae he followng quanes. (You should ge he answers shown.) (a) 1 6 (1) (b) 2 3 (8) (c) 23 (1) (d) 1 1 (.1) (o do hs calculaon, ener 1, hen press y x. hen ener 1, hen press he +/ key o change o 1, hen press = o ge he answer.1) (e) 2.5 (f) log (1) (g) ln (2.7183) (1/ 2=.77) (1) (1) (very close o 1) ry some oher examples on your own o become compleely famlar wh he y x key. (You should noe ha any posve number rased o he power s 1. Also, some calculaors, n parcular he one I am usng, canno handle any negave values of y, no even ( 2) 2 whch s +4) Euler's Number e = We have seen ha he funcon log on he scenfc calculaor undoes, s he nverse of, powers of 1. For example, we saw ha log 1 x =x (46 repeaed) Example: log 1 6 =6 Earler we saw ha he exponenal funcon e x was he nverse of he naural logarhm ln. hs means ha ln e x =x (49) he dfference beween he logarhm log and he naural logarhm ln, s ha log undoes exponens of he number 1, whle ln undoes exponens of he number e. hs specal number e, one of he fundamenal mahemacal consans lke π, s known as Euler's number, and s always denoed by he leer e. You can fnd he numercal value of Euler's number e on your calculaor by evaluang e 1 =e (5) o do hs, ener 1 no your calculaor, press he e x key, and you should see he resul e 1 = e = (51) We wll run no hs number hroughou he course. You should remember ha e s abou 2.7, or you mgh even remember (Only rememberng e as 2.7 s as kluzy as rememberng π as 3.1) he ermnology n mah courses s ha he funcon log, whch undoes exponens of he number 1, s he logarhm o he base 1. he funcon ln, wha we have called he naural logarhm, whch undoes exponens of he number e, s he logarhm o he base e. You can have logarhms o any base you wan, bu n pracce we only use base 1 (because we have 1 fngers) and he base e. he base e s specal, n par because ha s he logarhm ha naurally arses when we negrae he funcon 1/x. We wll see shorly ha he funcons ln and e x have several more, very specal feaures.

16 Cal 1-18 Calculus 2 DIFFERENIAION AND INEGRAION he scenfc calculaor s a good ool for seeng how he funcons lke ln and e x are nverse of each oher. Anoher example of nverse operaons s negraon and dfferenaon. We have seen ha negraon allows us o go he oher way from dfferenaon [fndng x() from v(), raher han v() from x()]. However s no so obvous ha negraon and dfferenaon are nverse operaons when you hnk of negraon as fndng he area under a curve, and dfferenaon as fndng lms of x/ as goes o zero. I s me now o make hs relaonshp clear. Frs, le us revew our concep of a dervave. Gong back o our srobe phoograph of Fgure (3), replacng R by R() and R +1 by R(+ ), as shown n Fgure (3a), our srobe velocy was hen gven by R(+ ) R() v() = (52) he calculus defnon of he velocy s obaned by reducng he srobe me nerval unl we oban he nsananeous velocy v. v calculus = lm R( + ) R() R() V() = R(+ ) R = R(+ ) R() + 1 R R(+ ) R() = (53) Whle Equaon (53) looks lke s appled o he explc case of he srobe phoograph of projecle moon, s easly exended o cover any process of dfferenaon. Whaever funcon we have [we had R(),suppose s now f()], evaluae a wo closely spaced mes, subrac he older value from he newer one, and dvde by he me dfference. akng he lm as becomes very small gves us he dervave df() d lm f( + ) f() (54) he varable wh whch we are dfferenang does no have o be me. I can be any varable ha we can dvde no small segmens, such as x; d f(x) lm f(x + x) f(x) x x (55) Le us see how he operaon defned n Equaon (55) s he nverse of fndng he area under a curve. Suppose we have a curve, lke our old v x () graphed as a funcon of me, as shown n Fgure (17). o fnd ou how far we raveled n a me nerval from o some laer me, we would do he negral x() = v x ()d (56) he negral n Equaon (56) ells us how far we have gone a any me durng he rp. he quany x() s a funcon of hs me. v () x x() Fgure 17 he dsance raveled by he me s he area under he velocy curve up o he me. Fgure 3a Defnng he srobe velocy.

17 Inroducon o Calculus Cal 1-19 Now le us dfferenae he funcon x() wh respec o he varable. By our defnon of dfferenaon we have d d x() = lm x( + ) x() (57) Fgure (17) shows us he funcon x(). I s he area under he curve v() sarng a and gong up o me =. Fgure (18) shows us he funcon x( + ). I s he area under he same curve, sarng a bu gong up o = +. When we subrac hese wo areas, all we have lef s he area of he slender recangle shown n Fgure (19). v () x v () x x() Fgure 17 repeaed he dsance x() raveled by he me Fgure 18 x(+ ) + he dsance x (+ ) raveled by he me +. v () x v () x v () x he recangle has a hegh approxmaely v() and a wdh for an area x( + ) x() = v x () (58) Dvdng hrough by gves x( + ) x() v x () = (59) he only approxmaon n Equaon (59) s a he op of he recangle. If he curve s no fla, v x ( + ) wll be dfferen from v x () and he area of he slver wll have a value somewhere beween v x () and v x ( + ). Bu f we ake he lm as goes o zero, he value of v x ( + ) mus approach v x (), and we end up wh he exac resul v x () = lm x( + ) x() (6) hs s jus he dervave ()/d evaluaed a =. v x () = () d where we sared from x() = = v x ()d (61a) (61b) Equaons (61a) and (61b) demonsrae explcly how dfferenaon and negraon are nverse operaons. he dervave allowed us o go from x() o v x () whle he negral ook us from v x () o x(). hs nverse s no as smple as pushng a buon on a calculaor o go from ln o e x. Here we have o deal wh lms on he negraon and a shf of varables from o. Bu hese wo processes do allow us o go back and forh. Fgure 18 he dsance x (+ ) x() raveled durng he me. +

18 Cal 1-2 Calculus 2 A Fas Way o go Back and Forh We nroduced our dscusson of negraon by ponng ou ha equaons v x () = () ; a d x () = dv x () (62a,b) d wen he wrong way n ha we were more lkely o know he acceleraon a x () and from ha wan o calculae he velocy v x () and dsance raveled x(). Afer many seps, we found ha negraon was wha we needed. We do no wan o repea all hose seps. Insead we would lke a quck and smple way o go he oher way around. Here s how you do. hnk of he d n (62a) as a small bu fne me nerval. ha means we can rea lke any oher number and mulply boh sdes of Equaon (62a) hrough by. v x () = () d () = v x ()d (63) Now negrae boh sdes of Equaon (63) from some nal me o a fnal me. (If you do he same hng o boh sdes of an equaon, boh sdes should sll be equal o each oher.) () = v x ()d (64) If d s o be hough of as a small bu fne me sep, hen () s he small bu fne dsance we moved n he me d. he negral on he lef sde of Equaon (64) s jus he sum of all hese shor dsances moved, whch s jus he oal dsance moved durng he me from o. () =x() =x() x ( ) (65) hus we end up wh he resul x() = vx ()d (66) Equaon (66) s a lle more general han (62b) for allows for he fac ha x( ) mgh no be zero. If, however, we say ha we sared our rp a x( )=, hen we ge he resul x() = v x ()d (67) represenng he dsance raveled snce he sar of he rp. Consan Acceleraon Formulas he consan acceleraon formulas, so well known from hgh school physcs courses, are an excellen applcaon of he procedures we have jus descrbed. We wll begn wh moon n one dmenson. Suppose a car s ravelng due eas, n he x drecon, and for a whle has a consan acceleraon a x. he car passes us a a me =, ravelng a a speed v x. A some laer me, f he acceleraon a x remans consan, how far away from us wll he car be? We sar wh he equaon a x () = dv x () (68) d Mulplyng hrough by d o ge dv x () = a x ()d hen negrang from me = o me f =, we ge dv x () = a x ()d Snce he negral dv x () =v x (), we have dv x () =v x () (69) =vx () v x () (7) where v x () s he velocy v x of he car when passed us a me =. Whle we can always do he lef hand negral n Equaon (69), we canno do he rgh hand negral unl we know a x (). For he consan acceleraon problem, however, we know ha a x () = a x s consan, and we have a x ()d = a x d (71)

19 Inroducon o Calculus Cal 1-21 Snce consans can come ousde he negral sgn, we ge a x d =a x d =a x =ax (72) where we used d =. Subsung Equaons (7) and (72) n (69) gves v x v x =a x (73) Snce Equaon (73) apples for any me, we can replace by o ge he well known resul v x () = v x +a x (a x consan) (74) Equaon (74) ells us he speed of he car a any me afer passed us, as long as he acceleraon remans consan. o fnd ou how far away he car s, we sar wh he equaon v x () = () (62a) d Mulplyng hrough by d o ge () = v x () d hen negrang from me = o me = gves (as we saw earler) () he lef hand sde s () = v x ()d =x() (75) =x() x() (76) If we measure along he x axs, sarng from where we are (where he car was a = ) hen x() =. In order o do he rgh hand negral n Equaon (75), we have o know wha he funcon v x () s. Bu for consan acceleraon, we have from Equaon (74) v x () = v x +a x, hus v x ()d = (v x +a x )d (77) One of he resuls of negraon ha you should prove for yourself (jus skech he areas) s he rule f hus we ge a(x) + b(x) (v x +a x )d f = a(x) = v x d f + b(x) + a x d (78) (79) Snce consans can come ousde he negrals, hs s equal o (v x +a x )d Earler we saw ha d = d = 2 2 =v x d +a x d (8) = = (23) = 2 2 =2 2 (3) hus we ge (v x +a x )d =v x a x 2 (81) Usng Equaons (76) and (81) n (75) gves x() x =v x a x 2 akng x = and replacng by gves he oher consan acceleraon formula x() = v x a x 2 (a x consan) (82) You can now see ha he facor of 2 /2 n he consan acceleraon formulas comes from he negral d.

20 Cal 1-22 Calculus 2 Exercse 4 Fnd he formula for he velocy v() and poson x() for a car movng wh consan acceleraon a x, ha was locaed a poson x a some nal me. Sar your calculaon from he equaons v x () = () d a x () = dv x() d and go hrough all he seps ha we dd o ge Equaons (74) and (82). See f you can do hs whou lookng a he ex. If you have o look back o see wha some seps are, hen fnsh he dervaon lookng a he ex. hen a day or so laer, clean off your desk, ge ou a blank shee of paper, wre down hs problem, pu he book away and do he dervaon. Keep dong hs unl you can do he dervaon of he consan acceleraon formulas whou lookng a he ex. Consan Acceleraon Formulas n hree Dmensons o handle he case of moon wh consan acceleraon n hree dmensons, you sar wh he separae equaons v x () = () d v y () = dy() d v z () = dz() d a x () = dv x () d a y () = dv y () d a z () = dv z () d (83) hen repea, for each par of equaons, he seps ha led o he consan acceleraon formulas for moon n he x drecon. he resuls wll be x() = v x a x 2 y() = v y a y 2 z() = v z a z 2 v x () = v x +a x v y () = v y +a y v z () = v z +a z (84) he fnal sep s o combne hese sx equaons no he wo vecor equaons x() = v a2 ; v() = v +a (85) hese are he equaons we analyzed graphcally n Chaper 3 of he physcs ex, n Fgure (3-34) and Exercse (3-9). (here we wroe s nsead of x(), and v raher han v.) In many nroducory physcs courses, consderable emphass s placed on solvng consan acceleraon problems. You can spend weeks praccng on solvng hese problems, and become very good a. However, when you have done hs, you have no learned very much physcs because mos forms of moon are no wh consan acceleraon, and hus he formulas do no apply. he formulas were mporan hsorcally, for hey were he frs o allow he accurae predcon of moon (of cannonballs). Bu f oo much emphass s placed on hese problems, sudens end o use hem where hey do no apply. For hs reason we have placed he exercses usng he consan acceleraon equaons n an appendx a he end of chaper 4 of he physcs ex. here are pleny of problems here for all he pracce you wll need wh hese equaons. Dong hese exercses requres only algebra, here s no pracce wh calculus. o ge some experence wh calculus, be sure ha you can confdenly do Exercse 4.

21 Inroducon o Calculus Cal 1-23 MORE ON DIFFERENIAION In our dscusson of negraon, we saw ha he basc dea was ha he negral of some curve or funcon f() was equal o he area under ha curve. ha s an easy enough concep. he problems arose when we acually red o fnd he formulas for he areas under varous curves. he only areas we acually calculaed were he recangular area under f() = consan and he rangular area under f() = a. I was perhaps a surprse ha he area under he smple curve 1/ should urn ou o be a logarhm. For dfferenaon, he basc dea of he process s gven by he formula df() d = lm f+ f() (54 repeaed) Equaon (54) s shor hand noaon for a whole seres of seps whch we nroduced hrough he use of srobe phoographs. he basc dea of dfferenaon s more complex han negraon, bu, as we wll now see, s ofen a lo easer o fnd he dervave of a curve han s negral. Seres Expansons An easy way o fnd he formula for he dervave of a curve s o use a seres expanson. We wll llusrae he process by usng he bnomal expanson o calculae he dervave of he funcon x n where n s any consan. We used he bnomal expanson, or a leas he frs wo erms, n Chaper 1 of he physcs ex. ha was durng our dscusson of he approxmaon formulas ha are useful n relavsc calculaons. As we menoned n Exercse (1-5), he bnomal expanson s (x + α) n =x n +nαx n 1 n(n 1) + α 2! 2 x n 2 (86) When α s a number much smaller han 1 (α <<1), we can neglec α 2 compared o α (f α =.1, α 2 =.1 ), wh he resul ha we can accuraely approxmae (x + α) n by (x + α) n x n +nαx n 1 α << 1 (87) Equaon (87) gves us all he approxmaon formulas found n Equaons (1-2) hrough (1-25) on page 1-28 of he physcs ex. As an example of Equaon (87), jus o see ha works, le us ake x = 5, n = 7 and α =.1 o calculae (5.1) 7. From he calculaor we ge (5.1) 7 = (88) (o do hs ener 5.1, press he y x buon, hen ener 7 and press he = buon.) Le us now see how hs resul compares wh (x + α) n x n +nαx n 1 (89) (5 +.1) (.1)5 6 We have 5 7 = (9) = = (91) Addng he numbers n (9) and (91) ogeher gves (.1)5 6 = (92) hus we end up wh nsead of 79225, whch s no oo bad a resul. he smaller α s compared o one, he beer he approxmaon.

22 Cal 1-24 Calculus 2 Dervave of he Funcon x n We are now ready o use our approxmaon formula (87) o calculae he dervave of he funcon x n. From he defnon of he dervave we have d(x n ) = lm x (x + x) n x n x (93) Snce x s o become nfnesmally small, we can use our approxmaon formula for (x + α) n. We ge (x + α ) n x n +n(α)x n 1 (α << 1) (x + x) n x n +n( x)x n 1 ( x<<1)(94) Usng hs n Equaon (93) gves d(x n ) = lm x x n +n( x)x n 1 x n x (95) We used an equal sgn raher han an approxmaely equal sgn n Equaon (95) because our approxmaon formula (94) becomes exac when x becomes nfnesmally small. In Equaon (95), he erms x n cancel and we are lef wh d(x n ) = lm x n( x)x n 1 x A hs pon, he facors x cancel and we have d(x n ) (96) = lm x nxn 1 (97) Snce no x's reman n our formula, we end up wh he exac resul d(x n ) =nx n 1 (98) Equaon (98) s he general formula for he dervave of he funcon x n. In our dscusson of negraon, we saw ha a consan could come ousde he negral. he same hng happens wh a dervave. Consder, for example, d af(x) = lm x af(x + x) af(x) x Snce he consan a has nohng o do wh he lmng process, hs can be wren d af(x) = a lm x =a df(x) f(x + x) f(x) x (99) Exercse 5 Calculae he dervave wh respec o x (.e., d/) of he followng funcons. (When negave powers of x are nvolved, assume x s no equal o zero.) (a) x (b) x 2 (c) x 3 (d) 5x 2 3x (Before you do par (d), use he defnon of he dervave o prove ha d df(x) f(x) + g(x) = + dg(x) ) (e) x 1 (f) x 2 (g) (h) x 1/ x () 3x.73 (j) 7x.2 (k) 1 (In par (k) frs show ha hs should be zero from he defnon of he dervave. hen wre 1=x and show ha Equaon (98) also works, as long as x s no zero.) (l) 5

23 Cal 1-25 he Chan Rule here s a smple rck called he chan rule ha makes easy o dfferenae a wde varey of funcons. he rule s df y(x) = df(y) dy dy chan rule (1) o see how hs rule works, consder he funcon f(x) = x 2 n (11) We know ha hs s jus f(x) = x 2n, and he dervave s df(x) = d x2n = 2nx 2n 1 (12) Bu suppose ha we dd no know hs rck, and herefore dd no know how o dfferenae (x 2 ) n. We do, however, know how o dfferenae powers lke x 2 and y n. he chan rule allows us o use hs knowledge n order o fgure ou how o dfferenae he more complex funcon(x 2 ) n. We begn by defnng y(x) as y(x) = x 2 (13) hen our funcon f(x) = (x 2 ) n can be wren n erms of y as follows f(x) = (x 2 ) n = y(x) n =(y) n = f(y) f(y) = (y) n (14) Usng (14) and (15) n he chan rule (1) gves df(y) = df dy dy = nyn 1 2x = 2ny n 1 x =2nx 2 n 1 x =2nx 2(n 1) x =2nx (2n 2) x (2n 2) + 1 =2nx = 2nx 2n 1 whch s he answer we expec. (17) In our example, usng he chan rule was more dffcul han dfferenang drecly because we already knew how o dfferenae x 2n. Bu we wll shorly encouner examples of new funcons ha we do no know how o dfferenae drecly, bu whch can be wren n he form f[y(x)]; and where we know df/dy and dy/. We can hen use he chan rule o evaluae he dervave df/. We wll gve you pracce wh he chan rule when we encouner hese funcons. Rememberng he Chan Rule he chan rule can be remembered by hnkng of he dy's as cancellng as shown. Dfferenang (13) and (14) gves dy(x) = d x2 =2x (15) df(y) dy dy = df(y) rememberng he chan rule (18) df(y) dy = d dy yn =ny n 1 (16)

24 Cal 1-26 Calculus 2 Paral Proof of he Chan Rule (oponal) he proof of he chan rule s closely relaed o cancellaon we showed n Equaon (18). A paral proof of he rule proceeds as follows. Suppose we have some funcon f(y) where y s a funcon of he varable x. As a resul f[y(x)] s self a funcon of x and can be dfferenaed wh respec o x. d fy(x) = lm fy(x+ x) fy(x) x x Now defne he quany y by so ha (123) y y(x + x) y(x) (124) y(x + x) = y(x) + y f[y(x + x)] = f(y + y) and Equaon (123) becomes d fy(x) = lm x f(y + y) f(y) x Now mulply (125) hrough by 1= y y(x + x) y(x) = y y o ge (125) (126) (We call hs a paral proof for he followng reason. For some funcons y(x), he quany y =yx+ x y(x) may be dencally zero for a small range of x. In ha case we would be dvdng by zero (he 1/ y ) even before we ook he lm as x goes o zero. A more complee proof handles he specal cases separaely. he resulng chan rule sll works however, even for hese specal cases.) Snce y = y(x + x) y(x) goes o zero as x goes o zero, we can wre Equaon (127) as d fy(x) = lm f(y + y) f(y) y y lm y(x + x) y(x) x x = df(y) dy dy (1 repeaed) hs rule works as long as he dervaves df/dy and dy/ are meanngful,.e., we say away from knks or dsconnues n f and y. d fy(x) = lm f(y + y) f(y) y(x + x) y(x) x x y = lm f(y + y) f(y) y(x + x) y(x) x y x where we nerchanged x and yn he denomnaor. (127)

25 Cal 1-27 INEGRAION FORMULAS Knowng he formula for he dervave of he funcon x n, and knowng ha negraon undoes dfferenaon, we can now use Equaon (98) n =nxn 1 (98 repeaed) o fnd he negral of he funcon x n. We wll see ha hs rck works for all cases excep he specal case where n = 1,.e., he specal case where he negral s a naural logarhm. o negrae x n, le us go back o our calculaon of he dsance s x or x() raveled by an objec movng n he x drecon a a velocy v x. hs was gven by Equaons (19) or (56) as x() = vx ()d (128) where he nsananeous velocy v x () s defned as v x () = () (129) d Suppose x() had he specal form x() = n+1 (a specal case) (13) hen we know from our dervave formulas ha v() = () d = d(n+1) d = (n+1) n (131) Subsung x() = n+1 and v() = (n+1) n no Equaon (128) gves x() = vx ()d (128) Dvdng hrough by (n+1) gves n d = 1 n+1 n+1 If we choose =, we ge he smpler resul n d = n+1 n+1 and he ndefne negral can be wren n d = n+1 n+1 (135) (also 34) (133) (134) hs s he general rule we saed whou proof back n Equaon (34). Noe ha hs formula says nohng abou he case n = 1,.e., when we negrae 1 =1/, because n +1 = 1 +1 = and we end up wh dvson by zero. Bu for all oher values of n, we now have derved a general formula for fndng he area under any curve of he form x n (or n ). hs s a raher powerful resul consderng he problems one encouners acually fndng areas under curves. (If you dd no do Exercse 1, he negraon exercses on page 14, or had dffculy wh hem, go back and do hem now.) n+1 (132) = (n+1) n d = (n+1) n d

26 Cal 1-28 Calculus 2 Dervave of he Exponenal Funcon he prevous work shows us ha f we have a seres expanson for a funcon, s easy o oban a formula for he dervave of he funcon. We wll now apply hs echnque o calculae he dervave and negral of he exponenal funcon e x. here s a seres expanson for he funcon e x ha works for any value of α n he range 1 o +1. e α 1+α + α2 2! + α3 + (136) 3! where 2! = 2 1, 3! = 3 2 1=6, ec. (he quanes 2!, 3! are called facorals. For example 3! s called hree facoral.) o see how well he seres (136) works, consder he case α =.1. From he seres we have, up o he α 3 erm α =.1 α 2 =.1 ; α 2 /2 =.5 α 3 =.1 ; α 3 / 6 =.167 Gvng us he approxmae value 1+α + α2 2! + α3 = (137) 3! When we ener.1 no a scenfc calculaor and press he e x buon, we ge exacly he same resul. hus he calculaor s no more accurae han ncludng he α 3 erm n he seres, for values of α equal o.1 or less. Le us now see how o use he seres 136 for calculang he dervave of e x. We have, from he defnon of a dervave, d f(x) lm f(x + x) f(x) (56 repea) x x If f(x) = e x, we ge d(e x ) = lm x e x+ x e x x (138) o do hs calculaon, we have o evaluae he quany e x+ x. Frs, we use he fac ha for exponenals e a+b =e a e b (Remember ha = =1 5.) hus e x+ x =e x e x (139) Now use he approxmaon formula (136), seng α = x and hrowng ou he α 2 and α 3 and hgher erms because we are gong o le x go o zero e x 1+ x (14) Subsung (14) n (139) gves e x+ x e x (1 + x) =e x +e x x (141) Nex use (141) n (138) o ge de x = lm e x +e x x e x (142) x x he e x erms cancel and we are lef wh de x = lm e x x = x x lm x ex (143) Snce he x s cancelled, we are lef wh he exac resul de x =e x (144) We see ha he exponenal funcon e x has he specal propery ha s s own dervave.

27 Cal 1-29 We wll ofen wan o know he dervave, no jus of he funcon e x bu of he slghly more general resul e ax where a s a consan. ha s, we wan o fnd d eax (a = consan) (145) Solvng hs problem provdes us wh our frs meanngful applcaon of he chan rule df(y) If we se = df(y) dy dy (1 repeaed) y = ax (146) hen we have de ax = dey dy dy (147) Now de y dy =ey (148) dy = d (ax) = a =a 1=a (149) Usng (148) and (149) n (147) gves de ax = ey (a) = e ax (a) = ae ax hus we have d eax =ae ax (15) hs resul wll be used so ofen s worh memorzng. Exercse 6 For furher pracce wh he chan rule, show ha de ax2 = 2axe ax2 Do hs by choosng y=ax 2, and hen do agan by choosng y=x 2. Inegral of he Exponenal Funcon o calculae he negral of e ax, we wll use he same rck as we used for he negral of x n, bu we wll be a b more formal hs me. Le us sar wh Equaon (128) relang poson x() and velocy v() = ()/d go ge x() f = v x ()d f = f () d d (128) Snce Equaon (128) holds for any funcon x() [we dd no pu any resrcons on x()], we can wre Equaon (128) n a more absrac way relang any funcon f(x) o s dervave df(x)/; f(x) x x f = x f df(x) (151) x o calculae he negral of e ax, we se f(x) = e ax and df(x)/ = ae ax o ge x f x f = ae ax (152) e ax x x Dvdng (157) hrough by (a) gves us he defne negral x f x e ax = 1 a eax x f (a = consan) (153) he correspondng ndefne negral s x e ax = eax a (a = consan) (154) Exercse 7 he naural logarhm s defned by he equaon ln (x) = 1 x (see Equaons 33-4) Use Equaon (151) o show ha d (ln x) = 1 x (155) (Hn negrae boh sdes of Equaon (155) wh respec o x.)

28 Cal 1-3 Calculus 2 DERIVAIVE AS HE SLOPE OF A CURVE Up o now, we have emphaszed he dea ha he dervave of a funcon f(x) s gven by he lmng process df(x) = lm x f(x + x) f(x) x (55 repeaed) We saw ha hs form was convenen when we had an explc way of calculang f(x + x), as we dd by usng a seres expanson. However, a lo of words are requred o explan he seps nvolved n dong he lmng process ndcaed n Equaon (55). In conras, he dea of an negral as beng he area under a curve s much easer o sae and vsualze. Now we wll provde an easy way o sae and nerpre he dervave of a curve. Consder he funcon f(x) graphed n Fgure (2). A a dsance x down he x axs, he curve had a hegh f(x) as shown. Slghly farher down he x axs, a x+ x, he curve has rsen o a hegh f(x + x). f(x+ x) f(x) Fgure 2 f(x) x x x+ x wo pons on a curve, a dsance x apar. x Fgure (2a) s a blowup of he curve n he regon beween x and x+ x. If he dsance x s suffcenly small, he curve beween x and x + x should be approxmaely a sragh lne and ha par of he curve should be approxmaely he hypoenuse of he rgh rangle abc seen n Fgure (2a). Snce he sde oppose o he angle * s f(x + x) f(x), and he adjacen sde s x, we have he resul ha he angen of he angle * s an * f(x + x) f(x) = (156) x When we make x smaller and smaller, ake he lm as x, we see ha he angle * becomes more nearly equal o he angle shown n Fgure (21), he angle of he curve when passes hrough he pon x. hus an = lm f(x + x) f(x) (157) x x he angen of he angle a whch he curve passes hrough he pon x s called he slope of he curve a he pon x. hus from Equaon (157) we see ha he slope of he curve s equal o he dervave of he curve a ha pon. We now have he nerpreaon ha he dervave of a curve a some pon s equal o he slope of he curve a ha pon, whle he negral of a curve s equal o he area under he curve up o ha pon. f(x) f(x+ x) f(x) a * x Fgure 2a A hs pon, he curve s led by approxmaely an angle *. c } b f(x+ x) f(x) x Fgure 21 he angen of he angle a whch he curve passes hrough he pon x s called he slope of he curve a ha pon.

29 Cal 1-31 Negave Slope In Fgure (22) we compare he slopes of a rsng and a fallng curve. In (22a), where he curve s rsng, he quany f(x + x) s greaer han f(x) and he dervave or slope df(x) = lm x s a posve number. f(x + x) f(x) x In conras, for he downward curve of Fgure (22b), f(x + x) s less han f(x) and he slope s negave. For a curve headed downward, we have df(x) = an() downward headng curve (158) (For hs case you can hnk of as a negave angle, so ha an() would auomacally come ou negave. However s easer smply o remember ha he slope of an upward dreced curve s posve and ha of a downward dreced cure s negave.) Exercse 8 Esmae he numercal value of he slope of he curve shown n Fgure (23) a pons (a), (b), (c), (d) and (e). In each case do a skech of f(x + x) f(x) for a small x, and le he slope be he rao of f(x + x) f(x) o x. Your answers should be roughly 1,, 1, +,. f(x) a b c Fgure 23 Esmae he slope a he varous pons ndcaed. d e x posve slope f(x) f(x+ x) x x+ x f(x+ x) f(x) x s posve negave slope f(x+ x) f(x) s negave x f(x) f(x+ x) x x+ x Fgure 22 Gong uphll s a posve slope, downhll s a negave slope.

30 Cal 1-32 Calculus 2 HE EXPONENIAL DECAY A curve ha we wll encouner several mes durng he course s he funcon e ax shown n Fgure (24), whch we call an exponenal decay. Snce exponens always have o be dmensonless numbers, we are wrng he consan (a) n he form 1/x so ha he exponen x/x s more obvously dmensonless. he funcon e x/x has several very specal properes. A x =, has he numercal value 1 (e =1). When we ge up o x=x, he curve has dropped o a value e x/x =e 1 = 1 e (a x = x ) (159) When we go ou o x=2x o, he curve has dropped o e 2x /x =e 2 = 1 e 2 (16) Ou a x=3x, he curve has dropped by anoher facor of e o (1/e)(1/e)(1/e). hs decrease connues ndefnely. I s he characersc feaure of an exponenal decay. Muon Lfeme In he muon lfeme expermen, we saw ha he number of muons survvng decreased wh me. A he end of wo mcroseconds, more han half of he orgnal 648 muons were sll presen. By 6 mcroseconds, only 27 remaned. he decay of hese muons s an example of an exponenal decay of he form number of survvng muons = number of muons a me = e / (161) where s he me akes for he number of muons remanng o drop by a facor of 1/e = 1/2.7. ha me s called he muon lfeme. We can use Equaon (161) o esmae he muon lfeme. In he move, he number of mesons a he op of he graph, reproduced n Fgure (25), s 648. ha s a me =. Down a me = 6 mcroseconds, he number survvng s 27. Pung hese numbers no Equaon (161) gves 27 survvng muons = 648 nal muons e 6/ e 6/ = =.42 (162) ake he naural logarhm ln of boh sdes of Equaon (162), [rememberng ha ln e x =x] gves ln e 6/ = 6 = ln.42 = 3.17 where we enered.42 on a scenfc calculaor and pressed he ln key. Solvng for we ge = 6 = 1.9 mcroseconds (163) 3.17 hs s close o he acceped value of = 2.2 mcroseconds whch has been deermned from he sudy of many housands of muon decays. 1 e x/x 1/e 1/e 2 x 2x 3x Fgure 24 As we go ou an addonal dsance x, he exponenal curve drops by anoher facor of 1/e. x Fgure 25 he lfeme of each deeced muon s represened by he lengh of a vercal lne. We can see ha many muons lve as long as 2 mcroseconds (2µs), bu few lve as long as 6 mcroseconds.