Accumulion: Thoughs On () By Lin McMullin f f f d = + The gols of he AP* Clculus progrm include he semen, Sudens should undersnd he definie inegrl s he ne ccumulion of chnge. 1 The Topicl Ouline includes he opic he definie inegrl of he re of chnge of quniy over n inervl inerpreed s he chnge of he quniy over he inervl: b f d= f b f. The equion bove is simply he Fundmenl Theorem of Clculus, bu he opic sresses wh i mens rher hn is use s echnique for evluing definie inegrl. Our equion, () f f f d = +, is he FTC epressed in slighly differen form wih chnge of vribles. I becomes funcion defined using n inegrl. The ide nd he use of his equion in his form mke myrid of problems esy o undersnd nd work. Ye he word ccumulion hs, o he bes of my recollecion, never ppered on he em in his echnicl sense, nor does i pper in mos populr ebooks. While here re erlier ems on which his ide cn be used, he em quesions AB/BC (d) nd especilly AB 6 is when his pproch mde is debu. I cn be used on mny free-response quesions since nd in he muliple-choice quesions from he nd 8 relesed ems. As wih some oher opics h pper on he AP clculus ems, since he opic does no pper in he curren ediions of mos of he ebooks used in AP Clculus courses i is 1 AP Clculus Course Descripion, The College Bord, 9, p.7 AP Clculus Course Descripion, The College Bord, 9, p.7 This is probbly why i is no used on he ems. Accumulion is no menioned in he indees of Anon, Finney, Foerser, Hughes Hlle, Osebee, Rogwski or Sewr. There is single senence in he curren ediion of Lrson, wih no eercises using he ide. For emple, 1997 BC 89, see below. 6 This is good problem o sudy. Boh he old mehod (solving n iniil vlue differenil equion mehod) nd he new pproch discussed here re shown on he scoring sndrd. In similr quesions in ler yer only he new pproch is shown. 1
imporn h you use emples from recen ems nd mke up problems of your own if necessry. The equion f = f + f dis, of course, only simple resemen of he Fundmenl Theorem of Clculus. Ye in his form here re myrid of uses for he equion. On he 8 AP Clculus ems here re no less hn 7 quesions where his cn be used, ye, s menioned bove, his form nd his pproch is no menioned in he mos ebooks. I ppers 1 imes on he 1 AB scoring sndrds. 7 We will consider some of he uses of his equion, wih some emples from he 8 nd 9 AB Clculus ems shorly. The equion sys h he funcion vlue is equl o some sring vlue (which my be zero) plus he ccumuled chnge. Finl Vlue = Sring Vlue + Accumuled Chnge () f f f d = + The inegrl f d gives he ccumuled chnge (or he ne chnge) from some sring poin o of funcion, f, in erms of is derivive, f. When we dd he sring vlue, f o he inegrl we hve he vlue of he funcion. If he derivive is he velociy, v (), of moving objec hen he inegrl gives he displcemen over he ime inervl [, ]. Here he equion gives he posiion, s( ), of he moving objec ime. The objec srs some posiion s( ). The equion hen looks like his: () s s v d = + 7 1 A 1(c) is bi of overkill; he quesion is bes n Algebr quesion. However, i is good, esy, emple of he ide.
The firs ime you sw his equion ws for funcions whose re of chnge (derivive) ws consn, usully clled he slope, m. If you know one poin of he funcion, sy (, ) y-vlue nywhere on he funcion is y, hen he y = y + md = y + m = y + m m y= y + m This is, of course, he poin-slope equion of line. You cerinly didn use he inegrl in Algebr 1, bu he poin-slope form is jus specil cse of our equion. The finl vlue, y, is equl o he sring vlue, y, plus he ccumuled chnge. The ccumuled chnge, he chnge in he -vlues, is he re of chnge, chnge in y m = muliplied by, he cul chnge in,. chnge in In clculus we del wih vrible slopes, vrible res of chnge. Here re wo similr emples from he 8 AB Em. In boh he derivive is clled he velociy of moving objec. 8 AB 7: A pricle moves long he -is wih velociy given by v () = + 6for ime. If he pricle is posiion = ime =, wh is he posiion of he pricle ime = 1? (Muliple-choice nswers omied.) Using our equion, he soluion cn be wrien immediely: () 1 1 1 = + + 6d = + + = + = 6 8 AB 87: An objec rveling in srigh line hs posiion ime. If he iniil posiion is = nd he velociy of he objec is v () = 1+, wh is he posiion of he objec ime =? (Muliple-choice nswers omied.) The wording is lmos idenicl o he previous emple s is he soluion. The inegrion is done by clculor. = + 1+ d +.11 6.1
dy These wo could hve been pproched s iniil vlue problems: condiion (, f ( )) = f d wih he iniil. The old pproch is o find n niderivive including consn C, use he iniil condiion o evlue C, wrie he priculr soluion, nd finlly evlue he soluion =. The new pproch, which cn be used wih ny firs-order differenil equion where he derivive is funcion of only one vrible, is o use he equion. dy = + = d + y f d f f d Here s similr quesion from long ime go. 1997 BC 89 If f is n niderivive of 1+ such h f ( 1) =, hen f = (Muliple-choice nswers omied.) Once gin he inegrion is done by clculor especilly since he niderivive is very difficul o compue. f = f ( 1) + d.76 1 1+ Here is noher emple from he 8 AB em h forces sudens o recognize his form wih generic funcion 8 AB 81: If G is n niderivive for f ( ) nd G = 7, hen (A) f (B) 7 f (D) 7 + f () d + (C) f () d (E) () G = 7 + f d Answer (E) On he 8 AP Clculus ess he equion could be used on ll of hese quesions: o AB Muliple-choice 7, 81 nd 87 o Free-response: AB / BC (d), AB (c), AB / BC () wice nd BC (c).
While oher pproches my be possible, ny ime you re given sring vlue nd derivive (re of chnge, slope, ec.) his mehod will ge you he nswer quickly. Free-response quesions ofen cn be solved using our equion. Here is ypicl problem, one of mny. 9 AB 6 gve he derivive of funcion f s he piecewise defined funcion g for f = / e for < where g ws he semicircle shown on he grph below. The -inerceps of f re = nd = ln. The iniil condiion is f =. (All given) 1 1 1 1 In pr (b) sudens were sked o find f nd f ( ) srigh forwrd:. Using our pproch he resuls re / / / f = + e d= + 1e = 8 1e f = f + g d= f g d= 8 π = π The second definie inegrl is mos esily found by subrcing he re of he semicircle, π, from he re of recngle, 8, drwn round i. Noice h g d is bckwrds : he upper limi of inegrion is less hn he lower limi. Therefore, is vlue is he opposie of he re of he region beween he semicircle nd he -is. Inegrion quesions wih he lower limi of inegrion greer hn he upper limi, nd where he vlues mus be found from grph re confusing for sudens. A wy of looking his
siuion is o reson his wy: We re looking for f ( ). If we inegre sring = we hve ccumuled he moun ( 8 π ) by he ime we ge o = where he vlue is (given). So if we subrc he ( 8 π ) from, we will hve our sring vlue f ( ). Symboliclly, he deils look like his nd his voids he inegrl wih he lower limi greer hn he upper. ( ) + = f g d f f ( ) + ( 8 π ) = f ( ) = ( 8 π ) Pr (c) of his sme quesion sked sudens o find he vlue which f hs is bsolue mimum. The mimum occurs = ln since his is he only plce where he derivive of f chnges from posiive o negive. While he preceding senence receives full credi, here is noher pproch using he Cndide s Tes nd our equion: The cndides for he locion of he mimum re =, ln = = M, nd =. M = + Since f f M f f d. i follows h f ( M) > f ( ) on he inervl[, M ] he inegrl is posiive, / = + M f f M e d. Since on he inervl [M, ] he inegrl is negive, i follows h f ( M) > f Therefore, he mimum occurs = M. Admiedly, his is overkill, bu i shows noher use of he concep. And few sudens did use his pproch on he em! If you look hrough he AP Clculus ems, especilly from on, you will find mny emples where his equion cn be used o dvnge; if you look hrough he ebooks you will find no such emples. While here re lwys oher wys o pproch hese problems, his 6
one is srighforwrd, esy o undersnd nd pplicble in vriey of siuions. You need o supplemen your ebook so h sudens will undersnd nd be ble o use his pproch. Bu wheher you re eching AP or no his pproch mke so mny problems so much esier o undersnd nd compue. Lin McMullin Direcor of Mhemics Progrms Nionl Mh nd Science Iniiive Norh S. Pul Sree Dlls, Tes 71 lmcmullin@nionlmhndscience.org Addiionl Hndous www.linmcmullin.ne. Click on Clculus AP* is rdemrk of he College Bord. The College Bord ws no involved in he producion of his produc. 7