E. R. Huggins. Dartmouth College. Physics Calculus. Chapter 1

Transcription

1 E. R. Huggins Darmouh College Physics 2 Calculus Chaper 1

2 Copyrigh Moose Mounain Digial Press New Hampshire 375 All righs reserved

3 able of Conens CHAPER 1 INRODUCION O CALCULUS Limiing Process Cal 1-3 he Uncerainy Principle Cal 1-3 Calculus Definiion of Velociy Cal 1-5 Acceleraion Cal 1-7 Componens Cal 1-7 Inegraion Cal 1-8 Predicion of Moion Cal 1-9 Calculaing Inegrals Cal 1-11 he Process of Inegraing Cal 1-13 Indefinie Inegrals Cal 1-14 Inegraion Formulas Cal 1-14 New Funcions Cal 1-15 Logarihms Cal 1-15 he Exponenial Funcion Cal 1-16 Exponens o he Base 1 Cal 1-16 he Exponenial Funcion y x Cal 1-16 Euler's Number e = Cal 1-17 Differeniaion and Inegraion Cal 1-18 A Fas Way o go Back and Forh Cal 1-2 Consan Acceleraion Formulas Cal 1-2 Consan Acceleraion Formulas in hree Dimensions Cal 1-22 More on Differeniaion Cal 1-23 Series Expansions Cal 1-23 Derivaive of he Funcion x n Cal 1-24 he Chain Rule Cal 1-25 Remembering he Chain Rule Cal 1-25 Parial Proof of he Chain Rule (opional) Cal 1-26 Inegraion Formulas Cal 1-27 Derivaive of he Exponenial FuncionCal 1-28 Inegral of he Exponenial Funcion Cal 1-29 Derivaive as he Slope of a Curve Cal 1-3 Negaive Slope Cal 1-31 he Exponenial Decay Cal 1-32 Muon Lifeime Cal 1-32 Half Life Cal 1-33 Measuring he ime Consanrom a Graph Cal 1-34 he Sine and Cosine Funcions Cal 1-35 Radian Measure Cal 1-35 he Sine Funcion Cal 1-36 Ampliude of a Sine Wave Cal 1-37 Derivaive of he Sine Funcion Cal 1-38 i

4 Cal 1-2 Calculus 2 V V 1 1 =.4 Sec =.1 Sec (a) (c) 1 V V i ~ =.25 Sec insananeous velociy (b) (d) Figure 1 ransiion o insananeous velociy.

5 Inroducion o Calculus Cal 1-3 Calculus Chaper 1 Inroducion o Calculus CHAPER 1 CALCULUS INRODUCION O his chaper, which replaces Chaper 4 in Physics 2, is inended for sudens who have no had calculus, or as a calculus review for hose whose calculus is no well remembered. If, afer reading par way hrough his chaper, you feel your calculus background is no so bad afer all, go back o Chaper 4 in Physics 2, sudy he derivaion of he consan acceleraion formulas beginning on page 4-8, and work he projecile moion problems in he appendix o Chaper 4. hose who sudy all of his inroducion o calculus should hen proceed o he projecile moion problems in he appendix o Chaper 4 of he physics ex. In Chaper 3 of Physics 2, we used srobe phoographs o define velociy and acceleraion vecors. he basic approach was o urn up he srobe flashing rae as we did in going from Figure (3-3) o (3-4) unil all he kinks are clearly visible and he successive displacemen vecors give a reasonable descripion of he moion. We did no urn he flashing rae oo high, for he pracical reason ha he displacemen vecors became oo shoror accurae work. LIMIING PROCESS In our discussion of insananeous velociy we concepually urned he srobe all he way up as illusraed in Figures (2-22a) hrough (2-22d), redrawn here in Figure (1). In hese figures, we iniially see a fairly large change in v as he srobe rae is increased and reduced. Bu he change becomes smaller and i looks as if we are approaching some final value of v ha does no depend on he size of, provided s small enough. I looks as if we have come close o he final value in Figure (1c). he progression seen in Figure (1) is called a limiing process. he idea is ha here really is some rue value of v which we have called he insananeous velociy, and ha we approach his rue value for sufficienly small values of. his is a calculus concep, and in he language of calculus, we are aking he limi as goes o zero. he Uncerainy Principle For over 2 years, from he invenion of calculus by Newon and Leibniz unil 1924, he limiing process and he resuling concep of insananeous velociy was one of he cornersones of physics. hen in 1924 Werner Heisenberg discovered wha he called he uncerainy principle which places a limi on he accuracy of experimenal measuremens.

6 Cal 1-4 Calculus 2 Heisenberg discovered somehing very new and unexpeced. He found ha he ac of making an experimenal measuremen unavoidably affecs he resuls of an experimen. his had no been known previously because he effec on large objecs like golf balls is undeecable. Bu on an aomic scale where we sudy small sysems like elecrons moving inside an aom, he effecs no only observable, i can dominae our sudy of he sysem. One paricular consequence of he uncerainly principle is ha he more accuraely we measure he posiion of an objec, he more we disurb he moion of he objec. his has an immediae impac on he concep of insananeous velociy. If we urn he srobe all he way up, reduce o zero, we are in effec rying o measure he posiion of he objec wih infinie precision. he consequence would be an infiniely big disurbance of he moion of he objec we are sudying. If we acually could urn he srobe all he way up, we would desroy he objec we were rying o sudy. I urns ou ha he uncerainy principle can have a significanmpac on a larger scale of disance han he aomic scale. Suppose, for example, ha we consruced a chamber 1 cm on a side, and wished o sudy he projecile moion of an elecron inside. Using Galileo s idea ha objecs of differen mass fall a he same rae, we would expec ha he moion of he elecron projecile should be he same as more massive objecs. If we ook a srobe phoograph of he elecron s moion, we would expec ge resuls like hose shown in Figure (2). his figure represens projecile moion wih an acceleraion g = 98 cm/sec 2 and =.1sec, as he reader can easily check. 1 cenimeer -1 1 v When we sudy he uncerainy principle in Chaper 4 of he physics ex, we will see ha a measuremen which is accurae enough o show ha Posiion (2) is below Posiion (1), could disurb he elecron enough o reverse is direcion of moion. he nex posiion measuremen could find he elecron over where we drew Posiion (3), or back where we drew Posiion (), or anywhere in he region in beween. As a resul we could no even deermine wha direcion he elecron is moving. his uncerainy would no be he resul of a sloppy experimen, is he bes we can do wih he mos accurae and delicae measuremens possible. he uncerainy principle has had a significanmpac on he way physiciss hink abou moion. Because we now know ha he measuring process affecs he resuls of he measuremen, we see has essenial o provide experimenal definiions o any physical quaniy we wish o sudy. A concepual definiion, like urning he srobe all he way up o define insananeous velociy, can lead o fundamenal inconsisencies. Even an experimenal definiion like our srobe definiion of velociy can lead o inconsisen resuls when applied o somehing like he elecron in Figure (2). Bu hese inconsisencies are real. heir exisence is elling us ha he very concep of velociy is beginning o lose meaning for hese small objecs. On he oher hand, he idea of he limiing process and insananeous velociy is very convenien when applied o larger objecs where he effecs of he uncerainy principle are no deecable. In his case we can apply all he mahemaical ools of calculus developed over he pas 25 years. he saus of insananeous velociy has changed from a basic concep o a useful mahemaical ool. hose problems for which his mahemaical ool works are called problems in classical physics; and hose problems for which he uncerainy principle is imporan, are in he realm of wha we call quanum physics. 1 cenimeer Figure 2 Hypoheical elecron projecile moion experimen.

7 Inroducion o Calculus Cal 1-5 CALCULUS DEFINIION OF VELOCIY Wih he above perspecive on he physical limiaions on he limiing process, we can now reurn o he main opic of his chaper he use of calculus in defining and working wih velociy and acceleraion. In discussing he limiing process in calculus, one radiionally uses a special se of symbols which we can undersand if we adop he noaion shown in Figure (3). In haigure we have drawn he coordinae vecors R i and R i+1 for he i h and (i + 1) h posiions of he objec. We are now using he symbol R i o represen he displacemen of he ball during he i o i+1 inerval. he vecor equaion for R i is R i =R i+1 R i (1) In words, Equaion (1) ells us ha R i is he change, during he ime, of he posiion vecor R describing he locaion of he ball. he velociy vecor v i is now given by v i = R i (2) his is jus our old srobe definiion v i =s i /, bu using a noaion which emphasizes ha he displacemen s i = R i is he change in posiion ha occurs during he ime. he Greek leer (dela) is used boh o represen he idea ha he quaniy R i or is small, and o emphasize ha boh of hese quaniies change as we change he srobe rae. he limiing process in Figure (1) can be wrien in he form v i limi R i (3) where he word limi wih underneah, is o be read as limi as goes o zero. For example we would read Equaion (3) as he insananeous velociy v i a posiion i is he limi, as goes o zero, of he raio R i /. R i i R i + 1 R i i + 1 For wo reasons, Equaion (3) is no quie yen sandard calculus noaion. One is han calculus, only he limiing value, in his case, he insananeous velociy, is considered o be imporan. Our srobe definiion v i = R i / s only a sep in he limiing process. herefore when we see he vecor v i, we should assume has he limiing value, and no special symbol like he underline is used. For his reason we will drop he underline and wrie v i = limi R i (3a) R i = R i + 1 R i V i = R i Figure 3 Definiions of R i and v i.

8 Cal 1-6 Calculus 2 he second change deals wih he fac ha when goes o zero we need an infinie number o ime seps o ge hrough our srobe phoograph, and hus is no possible o locae a posiion by couning ime seps. Insead we measure he ime ha has elapsed since he beginning of he phoograph, and use ha ime o ell us where we are, as illusraed in Figure (4). hus insead of using v i o represen he velociy a posiion i, we wrie v() o represen he velociy a ime. Equaion (3) now becomes v() = limi R() (3b) where we also replaced R i by is value R() a ime. Alhough Equaion (3b) is in more or less sandard calculus noaion, he noaion is clumsy. Is a pain o keep wriing he word limi wih a underneah. o sreamline he noaion, we replace he Greek leer wih he English leer d as follows limi R() dr() (4) (he symbol means defined equal o.) o a mahemaician, he symbol dr()/ is jus shorhand =.1sec =.2sec noaion for he limiing process we have been describing. Bu o a physicis, here is a differen, more pracical meaning. hink of as a shor, shor enough so ha he limiing process has essenially occurred, bu no oo shor o see whas going on. In Figure (1), a value of less han.25 seconds is probably good enough. If is small buinie, hen we know exacly wha he dr() is. Is he small buinie displacemen vecor a he ime. Is our old srobe definiion of velociy, wih he added condiion ha is such a shor ime inerval ha he limiing process has occurred. From his poin of view, is a real ime inerval and dr() a real vecor, which we can work wih in a normal way. he only hing special abou hese quaniies is ha when we see he leer d insead of, we mus remember ha a limiing process is involved. In his noaion, he calculus definiion of velociy is v() = dr() (5) where R() and v() are he paricle s coordinae vecor and velociy vecor respecively, as shown in Figure (5). Remember ha his is jusancy shorhand noaion for he limiing process we have been describing. = sec =.3sec V() =.4sec R() a =.3 sec R() =.5sec Figure 4 Raher han couning individual images, we can locae a posiion by measuring he elapsed ime. In his figure, we have drawn he displacemen vecor R() a ime =.3 sec. Figure 5 Insananeous posiion and velociy a ime.

9 Inroducion o Calculus Cal 1-7 ACCELERAION In he analysis of srobe phoographs, we defined boh a velociy vecor v and an acceleraion vecor a. he definiion of a, shown in Figure (2-12) reproduced here in Figure (6) was a i v i+1 v i (6) In our graphical work we replaced v i by s i / so ha we could work direcly wih he displacemen vecors s i and experimenally deermine he behavior of he acceleraion vecor for several kinds of moion. Le us now change his graphical definiion of acceleraion over o a calculus definiion, using he ideas jus applied o he velociy vecor. Firs, assume ha he ball reached posiion i a ime as shown in Figure (6). hen we can wrie v i =v() v i+1 =v(+ ) o change he ime dependence from a coun of srobe flashes o he coninuous variable. Nex, define he vecor v() by v() v(+ ) v() = v i+1 v i (7) We see ha v()is he change in he velociy vecor as he ime advances from o +. he srobe definiion of a i can now be wrien a() srobe definiion v( + ) v() = v() (8) Now go hrough he limiing process, urning he srobe up, reducing unil he value of a() seles down o is limiing value. We have a() calculus definiion = limi v + v = limi v() (9) Finally use he shorhand noaion d/ for he limiing process: a() = dv() (1) Equaion (1) does no make sense unless you remember has noaion for all he ideas expressed above. Again, physiciss hink of as a shor buinie ime inerval, and dv() as he small buinie change in he velociy vecor during he ime inerval. I s our srobe definiion of acceleraion wih he added requiremen ha s shor enough ha he limiing process has already occurred. Componens Even if you have sudied calculus, you may no recall encounering formulas for he derivaives of vecors, like dr()/ and dv()/ which appear in Equaions (5) and (1). o bring hese equaions ino a more familiar form where you can apply sandard calculus formulas, we will break he vecor Equaions (5) and (1) down ino componen equaions. posiion a ime V i posiion a ime + In he chaper on vecors, we saw ha any vecor equaion like A =B+C (11) is equivalen o he hree componen equaions a i = ( V i + 1 V i ) ( V i + 1 V i ) V i V i + 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 equaions was ha hey are simply numerical equaions and no graphical work or rigonomery is required. Figure 6 Experimenal definiion of he acceleraion vecor.

10 Cal 1-8 Calculus 2 he limiing process in calculus does no affec he decomposiion of a vecor ino componens, hus Equaion (5) for v() and Equaion (1) for a() become v() = dr()/ (5) v x () = dr x ()/ (5a) v y () = dr y ()/ (5b) v z () = dr z ()/ (5c) and a() = dv()/ (1) a x () = dv x ()/ a y () = dv y ()/ a z () = dv z ()/ (1a) (1b) (1c) Ofen we use he leer x for he x coordinae of he vecor R and we use y for R y and z for R z. Wih his noaion, Equaion (5) assumes he shorer and perhaps more familiar form v x () = ()/ v y () = dy()/ v z () = dz()/ y (5a ) (5b ) (5c ) A his poin he noaion has become decepively shor. You now have o remember ha x() sands for he x coordinae of he paricle a a ime. We have finally boiled he noaion down o he poin where i would be familiar from any calculus course. If we resric our aenion o one dimensional moion along he x axis. hen all we have o concern ourselves wih are he x componen equaions R Figure 7 x INEGRAION When we worked wih srobe phoographs, he phoograph old us he posiion R() of he ball as ime passed. Knowing he posiion, we can hen use Equaion (5) o calculae he ball's velociy v() and hen Equaion (1) o deermine he acceleraion a(). In general, however, we wan o go he oher way, and predic he moion from a knowledge of he acceleraion. For example, imagine ha you were in Galileo's posiion, hired by a prince o predic he moion of cannonballs. You know ha a cannonball should no be much affeced by air resisance, hus he acceleraion hroughous rajecory should be he consan graviaional acceleraion g. You know ha a() = g ; how hen do you use ha knowledge in Equaions (5) and (1) o predic he moion of he ball? he answer is ha you canno wih he equaions in heir presenorm. he equaions ell you how o go from R() o a(), while o predic moion you need o go he oher way, from a() o R(). he opic of his secion is o see how o reverse he direcions in which we use our calculus equaions. Equaions (5) and (1) involve he process called differeniaion. We will see ha when we go he oher way he reverse of differeniaion is a process called inegraion. We will see ha inegraion is a simple concep, bu a process has someimes hard o perform wihou he aid of a compuer. v x () = () a x () = dv x () (1a)

11 Inroducion o Calculus Cal 1-9 Predicion of Moion In our earlier discussion, we have used srobe phoographs o analyze moion. Le us see wha we can learn from such a phoograph for predicing moion. Figure (8) is our familiar projecile moion phoograph showing he displacemen s of a ball during he ime he ball raveled from a posiion labeled () o he posiion labeled (4). If he ball is now a posiion () and each of he images is.1 seconds apar, hen he vecor s ells us where he ball will be a a ime of.4 seconds from now. If we can predic s, we can predic he moion of he ball. he general problem of predicing he moion of he ball is o be able o calculae s() for any ime. From Figure (8) we see ha s is he vecor sum of he individual displacemen 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 wriing ou each erm, we can use he summaion sign Σ o wrie s = 4 Σ i=1 v i (12a) Equaion (12) is approximae in ha he v i are approximae (srobe) velociies, no he insananeous velociies we wanor a calculus discussion. In Figure (9) we improved he siuaion by cuing o 1/4 of is previous value, giving us four imes as many images and more accurae velociies v i. We see ha he displacemen s is now he sum of 16 vecors s = s 1 +s 2 +s s 15 +s 16 (13) Expressing his in erms of he velociy vecors v 1 o v 16 we have s =v 1 +v 2 +v v 15 +v 16 (14) or using our more compac noaion s = 16 Σ i=1 v i (14a) While Equaion (14) for s looks quie differen han Equaion (12), he sum of sixeen vecors insead of four, he displacemen vecors s in he wo cases are exacly he same. Adding more inermediae images did no change where he ball was locaed a he ime of =.4 seconds. In going from Equaion (12) o (14), wha has changed in shorening he ime sep, is ha he individual velociy vecors v i become more nearly equal o he insananeous velociy of he ball a each image. = 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 Figure 8 o predic he oal displacemen s, we add up he individual displacemens s i. Figure 9 Wih a shorer ime inerval, we add up more displacemen vecors o ge he oal displacemen s.

12 Cal 1-1 Calculus 2 If we reduced again by anoher facor of 1/4, so ha we had 64 images in he inerval = o =.4 sec, he formula for s would become s = 64 Σ i=1 v i (15a) where now he v i are sill closer o represening he ball's insananeous velociy. he more we reduce, he more images we include, he closer each v i comes o he insananeous velociy v. While adding more images gives us more vecors we have o add up o ge he oal displacemen s, here is very lile change in our formula for s. If we had a million images, we would simply wrie s = 1 Σ i=1 v i (16a) In his case he v i would be physically indisinguishable from he insananeous velociy v(). We have essenially reached a calculus limi, bu we have problems wih he noaion. Is clearly inconvenien o label each v i and hen coun he images. Insead we would like noaion hanvolves he insananeous velociy v() and expresses he beginning and end poins in erms of he iniial ime 1 and final ime, raher han he iniial and final image numbers i. In he calculus noaion, we replace he summaion sign Σ by somehing ha looks almos like he summaion sign, namely he inegral sign. (he French word for inegraion is he same as heir word for summaion.) Nex we replaced he individual v i by he coninuous variable v() and finally express he end poins by he iniial ime and he final ime. he resuls s = n Σ i=1 v i as he number n becomes infiniely large v() (17) Calculus noaion is more easily handled, or is a leas more familiar, if we break vecor equaions up ino componen equaions. Assume ha he ball sared a posiion i which has componens x i =x( ) [read x( ) as x a ime ] and y i =y( ) as shown in Figure (1). he final posiion f is a x f =x( ) and y f =y( ). hus he displacemen s has x and y componens s x =x( ) x( ) s y =y( ) y( ) Breaking Equaion (17) ino componen equaions gives s x =x( ) x( )= s y =y( ) y( )= v x () (18a) v y () (18b) Here we will inroduce one more piece of noaion ofen used in calculus courses. On he lef hand side of Equaion (18a) we have x( ) x( ) which we can hink of as he variable x() evaluaed over he inerval of ime from o. We will ofen deal wih variables evaluaed over some inerval and have a special noaion for ha. We will wrie x( ) x( ) x() (19) You are o read he symbol x() f i as "x of evaluaed from o ". We wrie he iniial ime a he boom of he verical bar, he final ime a he op. y i y f (y y ) f i i (x x ) x i x f x() x() Figure 1 Breaking he vecor s ino componens. S f i f

13 Inroducion o Calculus Cal 1-11 We use similar noaion for any kind of variable, for example x 2 f(x) f(x 2 ) f(x 1 ) (19a) x1 Remember o subrac he variable when evaluaed a he value a he boom of he verical bar. Wih his noaion, our Equaion (18) can be wrien s x =x() = i s y =y() i = v x () (18 a ) v y () (18 b ) Calculaing Inegrals Equaion (2) is nice and compac, bu how do you use i? How do you calculae inegrals? he key is o remember ha an inegral is jus a fancy noaion for a sum of erms, where we make he ime sep () very small. Keeping his in mind, we will see ha here is a very easy way o inerpre an inegral. Figure 11a Srobe phoograph of ball moving a consan velociy in x direcion. v x() v x Figure 11b Graph of v x () versus or he ball of Figure 11a. v () x v x i v x7 Figure 11c Each v x s he area of a recangle. f x o ge his inerpreaion, le us sar wih he simple case of a ball moving in a sraigh line, for insance, he x direcion, a a consan velociy v x. A srobe picure of his moion would look like ha shown in Figure (11a). Figure (11b) is a graph of he ball's velociy v x () as a funcion of he ime. he verical axis is he value of v x, he horizonal axis is he ime. Since he ball is raveling a consan velociy, v x has a consan value and is hus represened by a sraigh horizonal line. In order o calculae he disance ha he ball has raveled during he ime inerval from o, we need o evaluae he inegral s x = v x () disance ball ravels in ime inerval o (18a) o acually evaluae he inegral, we will go back o our summaion noaion s x = i final v xi (2) Σ i iniial and show individual ime seps n he graph of v x versus, as in Figure (11c). We see ha each erm in Equaion (2) is represened in Figure (11c) by a recangle whose heighs v x and whose wih is. We have shaded in he recangle represening he 7h erm v x7. We see ha v x7 s jus he area of he shaded recangle, and is clear ha he sum of all he areas of he individual recangles is he oal area under he curve, saring a ime and ending a ime. Here we are beginning o see ha he process of inegraion is equivalen o finding he area under a curve. Wih a simple curve like he consan velociy v x () in Figure (11c), we see by inspecion ha he oal area from o is jus he area of he complee recangle of heigh v x and wih ( ). hus s x =v x ( ) (21) his is he expeced resulor consan velociy, namely disance raveled = velociy ime for consan velociy (21a)

14 Cal 1-12 Calculus 2 o see ha you are no resriced o he case of consan velociy, suppose you drove on a freeway due eas (he x direcion) saring a 9: AM and sopping for lunch a 12 noon. Every minue during your rip you wroe down he speedomeer reading so ha you had an accurae plo of v x () for he enire morning, a plo like ha shown in Figure (12). From such a plo, could you deermine he disance s x ha you had ravelled? Your bes answer is o muliply each value v i of your velociy by he ime o calculae he average disance raveled each minue. Summing hese up from he iniial ime = 9:AM o he final ime = noon, you have as your esimae s x Σ i v xi (he symbol means approximaely equal.) o ge a more accurae value for he disance raveled, you should measure your velociy a shorer ime inervals and add up he larger number of smaller recangles. he precise answer should be obained in he limi as goes o zero s x = limi Σ v xi = v x () (22) i his limis jus he area under he curve has supposed o represen he insananeous velociy v x (). v () x v x7 9am Figure 12 Plo of v x () for a rip saring a 9: AM and finishing a noon. he disance raveled is he area under he curve. noon hus we can inerpre he inegral of a curve as he area under he curve even when he curve is no consan or fla. Mahemaicians concern hemselves wih curves ha are so wild has difficul or impossible o deermine he area under hem. Such curves seldom appear in physics problems. While he basic idea of inegraion is simple jus finding he area under a curve in pracice i can be quie difficul o calculae he area. Much of an inroducory calculus course is devoed o finding he formulas for he areas under various curves. here are also books called ables of inegrals 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 physics ex, we will discuss a mahemaical echnique called Fourier analysis. his is a echnique in which we can describe he shape of any coninuous curve in erms of a sum of sin waves. (Why we wan o do ha will become clear hen.) he process of Fourier analysis involves finding he area under some very complex curves, curves ofen involving experimenal daa for which we have no formula, only graphs. Such curves canno be inegraed by using a able of inegrals, wih he resul ha Fourier analysis was no widely used unil he adven of he modern digial compuer. he compuer made a difference, because we can find he area under almos any curve by breaking he curve ino shor pieces of lengh, calculaing he area v i of each narrow recangle, and adding up he area of he recangles o ge he oal area. If he curve is so wild ha we have o break ino a million segmens o ge an accurae answer, ha migh be oo hard o do by hand, bu usually a very simple and rapid job for a compuer. Compuers can be much more efficien han people anegraion.

15 Inroducion o Calculus Cal 1-13 he Process of Inegraing here is a language for he process of inegraion which we will now ake you hrough. In each case we will check ha he resuls are wha we would expecrom our summaion definiion, or he idea ha an inegral is he area under a curve. he simplesnegral we will encouner in he calculaion of he area under a curve of uni heigh as shown in Figure (13). We have he area of a recangle of heigh 1 and lengh ( ) 1 1 = =( ) i area = 1( ) Figure 13 Area under a curve of uni heigh. f i (22) We will use some special language o describe his inegraion. We will say ha he inegral of is simply he ime, and ha he inegral of from o is equal o evaluaed from o. In symbols his is wrien as = =( ) (23) Recall ha he verical line afer a variable means o evaluae ha variable a he final posiion (upper value), minus ha variable evaluaed a he iniial posiion (lower value). Noice ha his prescripion gives he correc answer. he nex simplesnegral is he inegral of a consan, like a consan velociy v x over he inerval o v x =v x ( ) (24) Since ( )= f, we can replace ( f ) in Equaion (24) by he inegral o ge v x =v x v x a consan (25) and we see ha a consan like v x can be aken ouside he inegral sign. Le us ry he simples case we can hink of where v x is no consan. Suppose v x sars a zero a ime = and increases linearly according o he formula v x =a (26) a v x Figure 15 v = a x When we ge up o he ime he velociy will be (a ) as shown in Figure (15). he area under he curve v x =as a riangle whose base is of lengh and heighs a. he area of his riangle is one half he base imes he heigh, hus we geor he disance s x raveled by an objec moving wih his velociy s x = v x = 1 (base) (aliude) 2 = 1 (27) 2 ( )(a )=1 2 a 2 Now le us repea he same calculaion using he language one would find in a calculus book. We have s x = v x = (a) (28) he consan (a) can come ouside, and we know ha he answer is 1/2a 2, hus we can wrie v x i area = v ( ) Figure 14 Area under he consan v x curve. x f i s x =a = 1 2 a 2 (29) In Equaion (29) we can cancel he a's o ge he resul x = (3)

16 Cal 1-14 Calculus 2 In a calculus ex, you would find he saemen ha he inegral s equal o 2 /2 and ha he inegral should be evaluaed as follows = 2 2 = f = 2 2 (31) Indefinie 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 limis on he inegral sign, like and, we have whas called a definie inegral. However here are imes where we jus wan o know wha he form of he inegral is, wih he idea ha we will pun he limis laer. In his case we have whas called an indefinie inegral, such as = 2 2 indefinie inegral (32) he difference beween our definie inegral in Equaion (31) and he indefinie one in Equaion (32) is ha we have no chosen he limis yen Equaion (32). If possible, a able of inegrals will give you a formula for he indefinie inegral and le you pun whaever limis you wan. Inegraion Formulas For some ses of curves, here are simple formulas for he area under hem. One example is he se of curves of he form n. We have already considered he cases where n = and n = 1. n = = = n = 1 1 = = 2 2 Some resuls we will prove laer are n = n = 3 3 = 4 4 (33a,b,c,d) Looking a he way hese inegrals are urning ou, we suspec ha he general rule is n = n+1 n+1 (34) I urns ou ha Equaion (34) is a general resulor any value of n excep n = 1. If n = 1, hen you would have division by zero, which canno be he answer. (We will shorly discuss he special case where n = 1.) As long as we say away from he n = 1 case, he formula works for negaive numbers. For example 2 = 2 = 1 2 = ( 2+1) 2+1 = 1 ( 1) (35) In our discussion of graviaional and elecrical poenial energy, we will encouner inegrals of he form seen in Equaion (35). Exercise 1 Using Equaion (34) and he fac ha consans can come ouside he inegral, evaluae he following inegrals: (a) x i does no maer wheher we call he variable or x (b) (c) (d) x=2 x 5 x=1 =2 =1 2 3 also skechhe area being evaluaed Show ha you ge a posiivearea. GmM r 2 dr whereg, m, and M are consans 2 = 3 3 (e) a y 3/2dy (a) is a consan

17 Inroducion o Calculus Cal 1-15 NEW FUNCIONS Logarihms We have seen ha when we inegrae a curve or funcion like 2, we ge a new funcion 3 /3. he funcions 2 and 3 appear o be fairly similar; he inegraion did no creae somehing radically differen. However, he process of inegraion can lead o some curves wih enirely differen behavior. his happens, for example, in ha special case n = 1 when we ry o do he inegral of 1. Is cerainly no hard o plo 1, he resuls shown in Figure (16). Also here is nohing fundamenally difficul or peculiar abou measuring he area under he 1 curve from some o, as long as we say away from he origin = where 1 blows up. he formula for his area urns ou, however, o be he new funcion called he naural logarihm, abbreviaed by he symbol ln. he area in Figure (16) is given by he formula 1 =ln( ) ln( ) (36) 1 curve 1 wo of he imporan bu peculiar feaures of he naural logarihm are ln(ab) = ln(a) + ln(b) (37) ln( 1 a )= ln(a) (38) hus we ge, for example ln( ) ln( )=ln( )+ln 1 =ln (39) hus he area under he curve in Figure (16) is =ln (4) While he naural logarihm has some raher peculiar properies is easy o evaluae because is available on all scienific calculaors. For example, if =.5 seconds and = 4 seconds, hen we have ln =ln 4 = ln (8) (41) i.5 Enering he number 8 on a scienific calculaor and pressing he buon labeled ln, gives ln (8) = 2.79 (42) which is he answer. i Figure 16 Plo of 1. he area under his curve is he naural logarihm ln. f Exercise 2 Evaluae he inegrals 1.1 x 1.1 x Why are he answers he same?

18 Cal 1-16 Calculus 2 he Exponenial Funcion We have jus seen ha, while he logarihm funcion may have some peculiar properies, is easy o evaluae using a scienific calculaor. he quesion we now wan o consider is wheher here is some funcion ha undoes he logarihm. When we ener he number 8 ino he calculaor and press ln, we ge he number Now we are asking if, when we ener he number 2.79, can we press some key and ge back he number 8? he answer is, you press he key labeled e x. he e x key performs he exponenial funcion which undoes he logarihm funcion. We say ha he exponenial funcion e x is he inverse of he logarihm funcion ln. Exponens o he Base 1 You are already familiar wih exponens o he base 1, as in he following examples 1 =1 1 1 =1 1 2 = = 1,, 1 1 = 1/1 = = 1/1 = =.1 (43) he exponen, he number wrien above he 1, ells us how many facors of 1 are involved. A minus sign means how many facors of 1 we divide by. From his alone we deduce he following rules for he exponen o he base 1. 1 a = 1 1 a (44) he inverse of he exponen o he base 1 is he funcion called logarihm o he base 1 which is denoed by he key labeled log on a scienific calculaor. Formally his means ha log 1 y =y (46) Check his ou on your scienific calculaor. For example, ener he number 1,, and press he log buon and see if you ge he number 6. ry several examples so ha you are confiden of he resul. he Exponenial Funcion y x Anoher key on your scienific calculaor is labeled y x. his allows you o deermine he value of any number y raised 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 Is quie 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,.)

19 Inroducion o Calculus Cal 1-17 Exercise 3 Use your scienific calculaor o evaluae he following quaniies. (You should ge he answers shown.) (a) 1 6 (1) (b) 2 3 (8) (c) 23 (1) (d) 1 1 (.1) (o do his calculaion, ener 1, hen press y x. hen ener 1, hen press he +/ key o change i 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 familiar wih he y x key. (You should noe ha any posiive number raised o he power is 1. Also, some calculaors, in paricular he one I am using, canno handle any negaive values of y, no even ( 2) 2 which is +4) Euler's Number e = We have seen ha he funcion log on he scienific calculaor undoes, is he inverse of, powers of 1. For example, we saw ha log 1 x =x (46 repeaed) Example: log 1 6 =6 Earlier we saw ha he exponenial funcion e x was he inverse of he naural logarihm ln. his means ha ln e x =x (49) he difference beween he logarihm log and he naural logarihm ln, is ha log undoes exponens of he number 1, while ln undoes exponens of he number e. his special number e, one of he fundamenal mahemaical consans like π, is known as Euler's number, and is always denoed by he leer e. You can find he numerical value of Euler's number e on your calculaor by evaluaing e 1 =e (5) o do his, ener 1 ino your calculaor, press he e x key, and you should see he resul e 1 = e = (51) We will run ino his number hroughou he course. You should remember ha e is abou 2.7, or you migh even remember (Only remembering e as 2.7 is as kluzy as remembering π as 3.1) he erminology in mah courses is ha he funcion log, which undoes exponens of he number 1, is he logarihm o he base 1. he funcion ln, wha we have called he naural logarihm, which undoes exponens of he number e, is he logarihm o he base e. You can have logarihms o any base you wan, bun pracice we only use base 1 (because we have 1 fingers) and he base e. he base e is special, in par because has he logarihm ha naurally arises when we inegrae he funcion 1/x. We will see shorly ha he funcions ln and e x have several more, very special feaures.

20 Cal 1-18 Calculus 2 DIFFERENIAION AND INEGRAION he scienific calculaor is a good ool for seeing how he funcions like ln and e x are inverse of each oher. Anoher example of inverse operaions is inegraion and differeniaion. We have seen hanegraion allows us o go he oher way from differeniaion [finding x() from v(), raher han v() from x()]. However is no so obvious hanegraion and differeniaion are inverse operaions when you hink of inegraion as finding he area under a curve, and differeniaion as finding limis of x/ as goes o zero. Is ime now o make his relaionship clear. Firs, le us review our concep of a derivaive. Going back o our srobe phoograph of Figure (3), replacing R i by R() and R i+1 by R(+ ), as shown in Figure (3a), our srobe velociy was hen given by R(+ ) R() v() = (52) he calculus definiion of he velociy is obained by reducing he srobe ime inerval unil we obain he insananeous velociy v. v calculus = limi R( + ) R() R() i V() = R(+ ) R = R(+ ) R() i + 1 R R(+ ) R() = (53) While Equaion (53) looks like is applied o he explici case of he srobe phoograph of projecile moion, is easily exended o cover any process of differeniaion. Whaever funcion we have [we had R(),suppose is now f()], evaluae i a wo closely spaced imes, subrac he older value from he newer one, and divide by he ime difference. aking he limi as becomes very small gives us he derivaive df() limi( + ) f() (54) he variable wih which we are differeniaing does no have o be ime. I can be any variable ha we can divide ino small segmens, such as x; d f(x) limi(x + x) f(x) x x (55) Le us see how he operaion defined in Equaion (55) is he inverse of finding he area under a curve. Suppose we have a curve, like our old v x () graphed as a funcion of ime, as shown in Figure (17). o find ou how far we raveled in a ime inerval from o some laer ime, we would do he inegral x() = v x () (56) he inegral in Equaion (56) ells us how far we have gone a any ime during he rip. he quaniy x() is a funcion of his ime. v () x i x() Figure 17 he disance raveled by he ime is he area under he velociy curve up o he ime. Figure 3a Defining he srobe velociy.

21 Inroducion o Calculus Cal 1-19 Now le us differeniae he funcion x() wih respec o he variable. By our definiion of differeniaion we have d d x() = limi x( + ) x() (57) Figure (17) shows us he funcion x(). Is he area under he curve v() saring a and going up o ime =. Figure (18) shows us he funcion x( + ). I is he area under he same curve, saring a bu going up o = +. When we subrac hese wo areas, all we have lefs he area of he slender recangle shown in Figure (19). v () x v () x x() i Figure 17 repeaed he disance x() raveled by he ime x(+ ) + Figure 18 he disance x (+ ) raveled by he ime +. v () x v () x v () x he recangle has a heigh approximaely v() and a wih for an area x( + ) x() = v x () (58) Dividing hrough by gives x( + ) x() v x () = (59) he only approximaion in Equaion (59) is a he op of he recangle. If he curve is nola, v x ( + ) will be differenrom v x () and he area of he sliver will have a value somewhere beween v x () and v x ( + ). Buf we ake he limi as goes o zero, he value of v x ( + ) mus approach v x (), and we end up wih he exac resul v x () = limi x( + ) x() (6) his is jus he derivaive ()/ evaluaed a =. v x () = () where we sared from x() = = v x () (61a) (61b) Equaions (61a) and (61b) demonsrae explicily how differeniaion and inegraion are inverse operaions. he derivaive allowed us o go from x() o v x () while he inegral ook us from v x () o x(). his inverse is no as simple as pushing a buon on a calculaor o go from ln o e x. Here we have o deal wih limis on he inegraion and a shif of variables from o. Bu hese wo processes do allow us o go back and forh. Figure 18 he disance x (+ ) x() raveled during he ime. +

22 Cal 1-2 Calculus 2 A Fas Way o go Back and Forh We inroduced our discussion of inegraion by poining ou ha equaions v x () = () ; a x () = dv x () (62a,b) wen he wrong way in ha we were more likely o know he acceleraion a x () and from ha wan o calculae he velociy v x () and disance raveled x(). Afer many seps, we found hanegraion was wha we needed. We do no wan o repea all hose seps. Insead we would like a quick and simple way o go he oher way around. Here is how you do i. hink of he in (62a) as a small buinie ime inerval. ha means we can rea like any oher number and muliply boh sides of Equaion (62a) hrough by i. v x () = () () = v x () (63) Now inegrae boh sides of Equaion (63) from some iniial ime o a final ime. (If you do he same hing o boh sides of an equaion, boh sides should sill be equal o each oher.) () = v x () (64) i If is o be hough of as a small buinie ime sep, hen () is he small buinie disance we moved in he ime. he inegral on he lef side of Equaion (64) is jus he sum of all hese shor disances moved, which is jus he oal disance moved during he ime from o. () =x() =x() x (i ) (65) i hus we end up wih he resul x() = vx () (66) Equaion (66) is a lile more general han (62b) for i allows for he fac ha x( ) migh no be zero. If, however, we say ha we sared our rip a x( )=, hen we ge he resul x() = v x () (67) i represening he disance raveled since he sar of he rip. Consan Acceleraion Formulas he consan acceleraion formulas, so well known from high school physics courses, are an excellen applicaion of he procedures we have jus described. We will begin wih moion in one dimension. Suppose a car is raveling due eas, in he x direcion, and for a while has a consan acceleraion a x. he car passes us a a ime =, raveling a a speed v x. A some laer ime, if he acceleraion a x remains consan, how far away from us will he car be? We sar wih he equaion a x () = dv x () (68) Muliplying hrough by o ge dv x () = a x () hen inegraing from ime = o ime =, we ge dv x () = a x () Since he inegral dv x () =v x (), we have dv x () (69) =v x () =vx () v x () (7) where v x () is he velociy v x of he car when i passed us a ime =. While we can always do he lef hand inegral in Equaion (69), we canno do he righ hand inegral unil we know a x (). For he consan acceleraion problem, however, we know ha a x () = a x is consan, and we have a x () = a x (71)

23 Inroducion o Calculus Cal 1-21 Since consans can come ouside he inegral sign, we ge a x =a x =a x =ax (72) where we used =. Subsiuing Equaions (7) and (72) in (69) gives v x v x =a x (73) Since Equaion (73) applies for any ime, we can replace by o ge he well known resul v x () = v x +a x (a x consan) (74) Equaion (74) ells us he speed of he car a any ime afer i passed us, as long as he acceleraion remains consan. o find ou how far away he car is, we sar wih he equaion v x () = () (62a) Muliplying hrough by o ge () = v x () hen inegraing from ime = o ime = gives (as we saw earlier) () he lef hand side is () = v x () (75) =x() =x() x() (76) If we measure along he x axis, saring from where we are (where he car was a = ) hen x() =. In order o do he righ hand inegral in Equaion (75), we have o know wha he funcion v x () is. Buor consan acceleraion, we have from Equaion (74) v x () = v x +a x, hus v x () = (v x +a x ) (77) One of he resuls of inegraion ha you should prove for yourself (jus skech he areas) is he rule i f hus we ge a(x) + b(x) (v x +a x ) f = a(x) i = v x f + b(x) i + a x (78) (79) Since consans can come ouside he inegrals, his is equal o (v x +a x ) Earlier we saw ha = = 2 2 =v x +a x (8) = = (23) = 2 2 =2 2 (3) hus we ge (v x +a x ) =v x a x 2 (81) Using Equaions (76) and (81) in (75) gives x() x =v x a x 2 aking x = and replacing by gives he oher consan acceleraion formula x() = v x a x 2 (a x consan) (82) You can now see ha he facor of 2 /2 in he consan acceleraion formulas comes from he inegral.

24 Cal 1-22 Calculus 2 Exercise 4 Find he formula for he velociy v() and posiion x() for a car moving wih consan acceleraion a x, ha was locaed a posiion x i a some iniial ime. Sar your calculaion from he equaions v x () = () a x () = dv x() and go hrough all he seps ha we did o ge Equaions (74) and (82). See if you can do his wihou looking a he ex. If you have o look back o see wha some seps are, hen finish he derivaion looking a he ex. hen a day or so laer, clean off your desk, ge ou a blank shee of paper, wrie down his problem, pu he book away and do he derivaion. Keep doing his unil you can do he derivaion of he consan acceleraion formulas wihou looking a he ex. Consan Acceleraion Formulas in hree Dimensions o handle he case of moion wih consan acceleraion in hree dimensions, you sar wih he separae equaions v x () = () v y () = dy() v z () = dz() a x () = dv x () a y () = dv y () a z () = dv z () (83) hen repea, for each pair of equaions, he seps ha led o he consan acceleraion formulas for moion in he x direcion. he resuls will 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 final sep is o combine hese six equaions ino he wo vecor equaions x() = v a2 ; v() = v +a (85) hese are he equaions we analyzed graphically in Chaper 3 of he physics ex, in Figure (3-34) and Exercise (3-9). (here we wroe s insead of x(), and v i raher han v.) In many inroducory physics courses, considerable emphasis is placed on solving consan acceleraion problems. You can spend weeks pracicing on solving hese problems, and become very good a. However, when you have done his, you have no learned very much physics because mosorms of moion are no wih consan acceleraion, and hus he formulas do no apply. he formulas were imporan hisorically, for hey were he firs o allow he accurae predicion of moion (of cannonballs). Buf oo much emphasis is placed on hese problems, sudens end o use hem where hey do no apply. For his reason we have placed he exercises using he consan acceleraion equaions in an appendix a he end of chaper 4 of he physics ex. here are pleny of problems here for all he pracice you will need wih hese equaions. Doing hese exercises requires only algebra, here is no pracice wih calculus. o ge some experience wih calculus, be sure ha you can confidenly do Exercise 4.

25 Inroducion o Calculus Cal 1-23 MORE ON DIFFERENIAION In our discussion of inegraion, we saw ha he basic idea was ha he inegral of some curve or funcion f() was equal o he area under ha curve. has an easy enough concep. he problems arose when we acually ried o find he formulas for he areas under various curves. he only areas we acually calculaed were he recangular area under f() = consan and he riangular area under f() = a. I was perhaps a surprise ha he area under he simple curve 1/ should urn ou o be a logarihm. For differeniaion, he basic idea of he process is given by he formula df() = limi f+ f() (54 repeaed) Equaion (54) is shor hand noaion for a whole series of seps which we inroduced hrough he use of srobe phoographs. he basic idea of differeniaion is more complex han inegraion, bu, as we will now see, is ofen a lo easier o find he derivaive of a curve han is inegral. Series Expansions An easy way o find he formula for he derivaive of a curve is o use a series expansion. We will illusrae he process by using he binomial expansion o calculae he derivaive of he funcion x n where n is any consan. We used he binomial expansion, or a leas he firs wo erms, in Chaper 1 of he physics ex. ha was during our discussion of he approximaion formulas ha are useful in relaivisic calculaions. As we menioned in Exercise (1-5), he binomial expansion is (x + α) n =x n +nαx n 1 n(n 1) + α 2! 2 x n 2 (86) When α is a number much smaller han 1 (α <<1), we can neglec α 2 compared o α (if α =.1, α 2 =.1 ), wih he resul ha we can accuraely approximae (x + α) n by (x + α) n x n +nαx n 1 α << 1 (87) Equaion (87) gives us all he approximaion formulas found in Equaions (1-2) hrough (1-25) on page 1-28 of he physics ex. As an example of Equaion (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 his ener 5.1, press he y x buon, hen ener 7 and press he = buon.) Le us now see how his resul compares wih (x + α) n x n +nαx n 1 (89) (5 +.1) (.1)5 6 We have 5 7 = (9) = = (91) Adding he numbers in (9) and (91) ogeher gives (.1)5 6 = (92) hus we end up wih insead of 79225, which is no oo bad a resul. he smaller α is compared o one, he beer he approximaion.

26 Cal 1-24 Calculus 2 Derivaive of he Funcion x n We are now ready o use our approximaion formula (87) o calculae he derivaive of he funcion x n. From he definiion of he derivaive we have d(x n ) = limi x (x + x) n x n x (93) Since x is o become infiniesimally small, we can use our approximaion 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) Using his in Equaion (93) gives d(x n ) = limi x x n +n( x)x n 1 x n x (95) We used an equal sign raher han an approximaely equal sign in Equaion (95) because our approximaion formula (94) becomes exac when x becomes infiniesimally small. In Equaion (95), he erms x n cancel and we are lef wih d(x n ) = limi x n( x)x n 1 x A his poin, he facors x cancel and we have d(x n ) (96) = limi x nxn 1 (97) Since no x's remain in our formula, we end up wih he exac resul d(x n ) =nx n 1 (98) Equaion (98) is he general formula for he derivaive of he funcion x n. In our discussion of inegraion, we saw ha a consan could come ouside he inegral. he same hing happens wih a derivaive. Consider, for example, d af(x) = limi x af(x + x) af(x) x Since he consan a has nohing o do wih he limiing process, his can be wrien d af(x) = a limi x =a df(x) f(x + x) f(x) x (99) Exercise 5 Calculae he derivaive wih respec o x (i.e., d/) of he following funcions. (When negaive powers of x are involved, assume x is no equal o zero.) (a) x (b) x 2 (c) x 3 (d) 5x 2 3x (Before you do par (d), use he definiion of he derivaive o prove ha d df(x) f(x) + g(x) = + dg(x) ) (e) x 1 (f) x 2 (g) (h) x 1/ x (i) 3x.73 (j) 7x.2 (k) 1 (In par (k) firs show ha his should be zero from he definiion of he derivaive. hen wrie 1=x and show ha Equaion (98) also works, as long as x is no zero.) (l) 5

27 Cal 1-25 he Chain Rule here is a simple rick called he chain rule ha makes i easy o differeniae a wide variey of funcions. he rule is df y(x) = df(y) dy dy chain rule (1) o see how his rule works, consider he funcion f(x) = x 2 n (11) We know ha his is jus(x) = x 2n, and he derivaive is df(x) = d x2n = 2nx 2n 1 (12) Bu suppose ha we did no know his rick, and herefore did no know how o differeniae (x 2 ) n. We do, however, know how o differeniae powers like x 2 and y n. he chain rule allows us o use his knowledge in order o figure ou how o differeniae he more complex funcion(x 2 ) n. We begin by defining y(x) as y(x) = x 2 (13) hen our funcion f(x) = (x 2 ) n can be wrien in erms of y as follows f(x) = (x 2 ) n = y(x) n =(y) n = f(y) f(y) = (y) n (14) Using (14) and (15) in he chain rule (1) gives 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 which is he answer we expec. (17) In our example, using he chain rule was more difficul han differeniaing direcly because we already knew how o differeniae x 2n. Bu we will shorly encouner examples of new funcions ha we do no know how o differeniae direcly, bu which can be wrien in he form f[y(x)]; and where we know df/dy and dy/. We can hen use he chain rule o evaluae he derivaive df/. We will give you pracice wih he chain rule when we encouner hese funcions. Remembering he Chain Rule he chain rule can be remembered by hinking of he dy's as cancelling as shown. Differeniaing (13) and (14) gives dy(x) = d x2 =2x (15) df(y) dy dy = df(y) remembering he chain rule (18) df(y) dy = d dy yn =ny n 1 (16)

28 Cal 1-26 Calculus 2 Parial Proof of he Chain Rule (opional) he proof of he chain rule is closely relaed o cancellaion we showed in Equaion (18). A parial proof of he rule proceeds as follows. Suppose we have some funcion f(y) where y is a funcion of he variable x. As a resul[y(x)] is iself a funcion of x and can be differeniaed wih respec o x. d fy(x) = limi fy(x+ x) fy(x) x x Now define he quaniy 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 Equaion (123) becomes d fy(x) = limi x f(y + y) f(y) x Now muliply (125) hrough by 1= y y(x + x) y(x) = y y o ge (125) (126) (We call his a parial proof for he following reason. For some funcions y(x), he quaniy y =yx+ x y(x) may be idenically zero for a small range of x. In ha case we would be dividing by zero (he 1/ y ) even before we ook he limi as x goes o zero. A more complee proof handles he special cases separaely. he resuling chain rule sill works however, even for hese special cases.) Since y =y(x+ x) y(x) goes o zero as x goes o zero, we can wrie Equaion (127) as d fy(x) = limi(y + y) f(y) y y limi y(x + x) y(x) x x = df(y) dy dy (1 repeaed) his rule works as long as he derivaives df/dy and dy/ are meaningful, i.e., we say away from kinks or disconinuiies in f and y. d fy(x) = limi(y + y) f(y) y(x + x) y(x) x x y = limi(y + y) f(y) y(x + x) y(x) x y x (127) where we inerchanged x and y in he denominaor.

29 Cal 1-27 INEGRAION FORMULAS Knowing he formula for he derivaive of he funcion x n, and knowing hanegraion undoes differeniaion, we can now use Equaion (98) n =nxn 1 (98 repeaed) o find he inegral of he funcion x n. We will see ha his rick works for all cases excep he special case where n = 1, i.e., he special case where he inegral is a naural logarihm. o inegrae x n, le us go back o our calculaion of he disance s x or x() raveled by an objec moving in he x direcion a a velociy v x. his was given by Equaions (19) or (56) as x() = vx () (128) where he insananeous velociy v x () is defined as v x () = () (129) Suppose x() had he special form x() = n+1 (a special case) (13) hen we know from our derivaive formulas ha v() = () = (n+1) = (n+1) n (131) Subsiuing x() = n+1 and v() = (n+1) n ino Equaion (128) gives x() = vx () (128) i Dividing hrough by (n+1) gives n = 1 n+1 n+1 If we choose =, we ge he simpler resul n = n+1 n+1 and he indefinie inegral can be wrien n = n+1 n+1 (133) (134) (135) (also 34) his is he general rule we saed wihou proof back in Equaion (34). Noe ha his formula says nohing abou he case n = 1, i.e., when we inegrae 1 =1/, because n +1 = 1 +1 = and we end up wih division by zero. Buor all oher values of n, we now have derived a general formula for finding he area under any curve of he form x n (or n ). his is a raher powerful resul considering he problems one encouners acually finding areas under curves. (If you did no do Exercise 1, he inegraion exercises on page 14, or had difficuly wih hem, go back and do hem now.) n+1 (132) = (n+1) n = (n+1) n

30 Cal 1-28 Calculus 2 Derivaive of he Exponenial Funcion he previous work shows us haf we have a series expansion for a funcion, is easy o obain a formula for he derivaive of he funcion. We will now apply his echnique o calculae he derivaive and inegral of he exponenial funcion e x. here is a series expansion for he funcion e x ha works for any value of α in 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 quaniies 2!, 3! are called facorials. For example 3! is called hree facorial.) o see how well he series (136) works, consider he case α =.1. From he series we have, up o he α 3 erm α =.1 α 2 =.1 ; α 2 /2 =.5 α 3 =.1 ; α 3 / 6 =.167 Giving us he approximae value 1+α + α2 2! + α3 = (137) 3! When we ener.1 ino a scienific calculaor and press he e x buon, we ge exacly he same resul. hus he calculaor is no more accurae han including he α 3 erm in he series, for values of α equal o.1 or less. Le us now see how o use he series 136 for calculaing he derivaive of e x. We have, from he definiion of a derivaive, d f(x) limi(x + x) f(x) (56 repea) x x If f(x) = e x, we ge d(e x ) = limi x e x+ x e x x (138) o do his calculaion, we have o evaluae he quaniy e x+ x. Firs, we use he fac haor exponenials e a+b =e a e b (Remember ha = =1 5.) hus e x+ x =e x e x (139) Now use he approximaion formula (136), seing α = x and hrowing ou he α 2 and α 3 and higher erms because we are going o le x go o zero e x 1+ x (14) Subsiuing (14) in (139) gives e x+ x e x (1 + x) =e x +e x x (141) Nex use (141) in (138) o ge de x = limi e x +e x x e x (142) x x he e x erms cancel and we are lef wih de x = limi e x x = x x limi x ex (143) Since he x s cancelled, we are lef wih he exac resul de x =e x (144) We see ha he exponenial funcion e x has he special propery has is own derivaive.

31 Cal 1-29 We will ofen wan o know he derivaive, no jus of he funcion e x bu of he slighly more general resul e ax where a is a consan. has, we wan o find d eax (a = consan) (145) Solving his problem provides us wih our firs meaningful applicaion of he chain 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) Using (148) and (149) in (147) gives de ax = ey (a) = e ax (a) = ae ax hus we have d eax =ae ax (15) his resul will be used so ofen is worh memorizing. Exercise 6 For furher pracice wih he chain rule, show ha de ax2 = 2axe ax2 Do his by choosing y=ax 2, and hen do i again by choosing y=x 2. Inegral of he Exponenial Funcion o calculae he inegral of e ax, we will use he same rick as we used for he inegral of x n, bu we will be a bi more formal his ime. Le us sar wih Equaion (128) relaing posiion x() and velociy v() = ()/ go ge x() = v x () = () (128) Since Equaion (128) holds for any funcion x() [we did no pu any resricions on x()], we can wrie Equaion (128) in a more absrac way relaing any funcion f(x) o is derivaive df(x)/; f(x) xi x f = x f df(x) (151) x i o calculae he inegral of e ax, we se(x) = e ax and df(x)/ = ae ax o ge x f x f = ae ax (152) e ax x i x i Dividing (157) hrough by (a) gives us he definie inegral x f x i e ax = 1 a eax x f (a = consan) (153) he corresponding indefinie inegral is x i e ax = eax a (a = consan) (154) Exercise 7 he naural logarihm is defined by he equaion ln (x) = 1 x (see Equaions 33-4) Use Equaion (151) o show ha d (ln x) = 1 x (155) (Hin inegrae boh sides of Equaion (155) wih respec o x.)

32 Cal 1-3 Calculus 2 DERIVAIVE AS HE SLOPE OF A CURVE Up o now, we have emphasized he idea ha he derivaive of a funcion f(x) is given by he limiing process df(x) = limi x f(x + x) f(x) x (55 repeaed) We saw ha his form was convenien when we had an explici way of calculaing f(x + x), as we did by using a series expansion. However, a lo of words are required o explain he seps involved in doing he limiing process indicaed in Equaion (55). In conras, he idea of an inegral as being he area under a curve is much easier o sae and visualize. Now we will provide an easy way o sae and inerpre he derivaive of a curve. Consider he funcion f(x) graphed in Figure (2). A a disance x down he x axis, he curve had a heigh f(x) as shown. Slighly farher down he x axis, a x+ x, he curve has risen o a heigh(x + x). f(x+ x) f(x) f(x) x x x+ x Figure 2 wo poins on a curve, a disance x apar. x Figure (2a) is a blowup of he curve in he region beween x and x+ x. If he disance x is sufficienly small, he curve beween x and x + x should be approximaely a sraigh line and ha par of he curve should be approximaely he hypoenuse of he righ riangle abc seen in Figure (2a). Since he side opposie o he angle θ * is f(x + x) f(x), and he adjacen side is x, we have he resul ha he angen of he angle θ * is an θ * f(x + x) f(x) = (156) x When we make x smaller and smaller, ake he limi as x, we see ha he angle θ * becomes more nearly equal o he angle θ shown in Figure (21), he angle of he curve when i passes hrough he poin x. hus an θ = limi(x + x) f(x) (157) x x he angen of he angle a which he curve passes hrough he poin x is called he slope of he curve a he poin x. hus from Equaion (157) we see ha he slope of he curve is equal o he derivaive of he curve a ha poin. We now have he inerpreaion ha he derivaive of a curve a some poins equal o he slope of he curve a ha poin, while he inegral of a curve is equal o he area under he curve up o ha poin. f(x) θ f(x+ x) f(x) a θ* x Figure 2a A his poin, he curve is iled by approximaely an angle θ *. c } b f(x+ x) f(x) x Figure 21 he angen of he angle θ a which he curve passes hrough he poin x is called he slope of he curve a ha poin.

33 Cal 1-31 Negaive Slope In Figure (22) we compare he slopes of a rising and a falling curve. In (22a), where he curve is rising, he quaniy f(x + x) is greaer han f(x) and he derivaive or slope df(x) = limi x is a posiive number. f(x + x) f(x) x In conras, for he downward curve of Figure (22b), f(x + x) is less han f(x) and he slope is negaive. For a curve headed downward, we have df(x) = an(θ) downward heading curve (158) (For his case you can hink of θ as a negaive angle, so ha an(θ) would auomaically come ou negaive. However is easier simply o remember ha he slope of an upward direced curve is posiive and ha of a downward direced cure is negaive.) Exercise 8 Esimae he numerical value of he slope of he curve shown in Figure (23) a poins (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 raio of f(x + x) f(x) o x. Your answers should be roughly 1,, 1, +,. f(x) a b c Figure 23 Esimae he slope a he various poins indicaed. d e x posiive slope θ f(x) f(x+ x) x x+ x f(x+ x) f(x) x is posiive negaive slope θ f(x+ x) f(x) is negaive x f(x) f(x+ x) x x+ x Figure 22 Going uphill is a posiive slope, downhill is a negaive slope.

34 Cal 1-32 Calculus 2 HE EXPONENIAL DECAY A curve ha we will encouner several imes during he course is he funcion e ax shown in Figure (24), which we call an exponenial decay. Since exponens always have o be dimensionless numbers, we are wriing he consan (a) in he form 1/x so ha he exponen x/x is more obviously dimensionless. he funcion e x/x has several very special properies. A x =, i has he numerical 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). his decrease coninues indefiniely. Is he characerisic feaure of an exponenial decay. Muon Lifeime In he muon lifeime experimen, we saw ha he number of muons surviving decreased wih ime. A he end of wo microseconds, more han half of he original 648 muons were sill presen. By 6 microseconds, only 27 remained. he decay of hese muons is an example of an exponenial decay of he form number of surviving muons = number of muons a ime = e / (161) where is he ime i akes for he number of muons remaining o drop by a facor of 1/e = 1/2.7. ha ime is called he muon lifeime. We can use Equaion (161) o esimae he muon lifeime. In he movie, he number of mesons a he op of he graph, reproduced in Figure (25), is 648. ha is a ime =. Down a ime = 6 microseconds, he number surviving is 27. Puing hese numbers ino Equaion (161) gives 27 surviving muons = 648 iniial muons e 6/ e 6/ = =.42 (162) ake he naural logarihm ln of boh sides of Equaion (162), [remembering ha ln e x =x] gives ln e 6/ = 6 = ln.42 = 3.17 where we enered.42 on a scienific calculaor and pressed he ln key. Solving for we ge = 6 = 1.9 microseconds (163) 3.17 his is close o he acceped value of = 2.2 microseconds which has been deermined from he sudy of many housands of muon decays. 1 e x/x 1/e 1/e 2 x 2x 3x Figure 24 As we go ou an addiional disance x, he exponenial curve drops by anoher facor of 1/e. x Figure 25 he lifeime of each deeced muon is represened by he lengh of a verical line. We can see ha many muons live as long as 2 microseconds (2µs), buew live as long as 6 microseconds.