10. Water tank. Example I. Draw the graph of the amount z of water in the tank against time t.. Explain the shape of the graph.

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1 1. Waer ank The graph A cylindrical ank conains ml of waer. A = (minues) a hole is punched in he boom, and waer begins o flow ou. I akes exacly 1 seconds for he ank o empy. Example I. Draw he graph of he amoun of waer in he ank agains ime.. Explain he shape of he graph. I begin, of course, by leing he class wach he acion. I use a 1 lire waer bole or pop bole, even hough his is no quie cylindrical, wih a hole ha s 3 o mm in diameer. We don ake measuremens or record daa ha comes laer. I invie a suden o come and draw he graph on he board and ineresingly enough, wha I ge is a sraigh line. I ask he class wha hey hink and here is general cauious nodding. Wha s good abou he graph? I sars a heigh, decreases, and his he -axis a 1. Tha seems o fi he given informaion. So we ake a voe. How many agree? I ge a fores of hands. Holy cow. Tha really surprises me. Forunaely here are a couple of objecors. Why is i sraigh? Shouldn i be a curve? This suden comes up and draws a new picure, and I ge somehing like he graph a he righ. Okay. Wha's beer abou his graph? Why should i be curved? Well he flow rae is higher a he beginning han a he end. Why is ha? More waer in he ank more pressure pushing he waer ou. So how does ha relae o he geomery of he graph? Well, when he flow rae is high, goes down fas, and he graph is seep. And hen when he flow rae is less, he graph is flaer. Tha s good, and his ime i s no hard o ge he class o agree ha i s righ. Okay wha migh we have done o ge ha sraigh-line graph? We d have o ake he waer ou a a consan rae. How migh we do ha? Maybe pump i ou a a consan rae? Tha would do i. Wha consan rae would he pump use? ml in 1 seconds means ml/s. Righ. 1 1 How does he decrease in he flow rae over ime relae o he curvaure of he graph? I's imporan ha hey are able o explain ha in an inuiive manner. v 1 waer ank 1

2 The formula. Example. Finding an equaion for he graph. I ask he class if hey can produce an equaion for a graph wih he righ shape, and of course hey give me a parabola. This urns ou o be correc on heoreical grounds. Given ha i s a parabola, I challenge hem o find is equaion. They should use he given daa ha he ank sars wih ml of waer and ha i akes 1 seconds o empy. Well he equaion of a parabola is a polynomial of degree, and many of hem sar wih he algebraic form: () = a + b + c leaving hem hree parameers, a, b and c o evaluae. And hey can quie do ha because hey ve only wo condiions o work wih, ha ()= and (1)=. So hey re suck.. I sugges ha hey draw he whole parabola (of which only he lef half belongs o he waer ank problem), and hen see wha hey can say abou he equaion. When hey do ha, hey realie ha he verex of he parabola mus be a =1. Why is ha? Because he flow rae ou of he ank deermines he slope of he graph and as he ank becomes empy, he flow rae mus fall o ero, and herefore he slope a he end poin mus be ero. Okay. How do we use ha algebraically? Well, when we know he verex of a parabola, here s anoher algebraic form ha s beer o use he compleed square form because he coordinaes of he verex boh appear among he hree parameers: In his case he verex of he parabola is a (1,) and hence he equaion has he form: = a( 1). To find a plug in he iniial condiion =, =: = a( 1) = 1 a and his solves o give a=/1 =.. Thus he equaion of he parabola is =.( 1). Theoreical consideraions of how he pressure pushes he waer ou of he hole lead in fac o he equaion of a parabola. A simpleminded argumen for his is no so easy o find, bu I give some heurisic consideraions a he end. I s nea ha a poenially complex phenomenon like his leads o a simple equaion. In fac he physical world is full of similar surprising sories The compleed square form y = a(x c) + b If you are working wih a parabola, his is acually a wonderful algebraic form o be given. The parameers a, b and c all describe feaures of he graph which are geomerically and physically significan. c and b give you he locaion of he verex (x=c and y=b), and a ells you how fas he parabola goes up. 1 waer ank

3 Experimen. I s ime o fill up he ank, open he hole, ake some careful readings, and see if we can verify he parabola resul. Wha we have o keep rack of is he volume in he ank, bu for a cylindrical ank, volume is proporional o heigh, so we used heigh above he hole as our measure of amoun. In fac I have drawn horional marks a inervals of 1 mm above he hole, all he way from 1 o 13 and hen I have he sudens record he ime ha he level arrives a each of he marks. Now we have o decide wheher hese poins lie on a parabola. They cerainly have he righ shape, bu here are any number of curves around wih ha shape, so we need some mahemaical way of checking i ou. Wha we do know is he arge shape ha we re looking for: = a( c). Our job is o see if here are any values of he parameers a and c for which he curve provides a good fi o our daa, and if so, wha are he bes-fi values of hese parameers. I ask he class for suggesions. One idea is ha we ake he wo daa poins we used in he previous secion he beginning (,13) and he end (19,) and hen use hese o find he values of a and c, and hen maybe we could plo he equaion on op of he poins and see how good he fi was. Tha s an ineresing idea, bu i only uses he informaion conained in wo of our daa poins. Wha if eiher of hose poins happened o be off? The poin is ha here will always be errors of measuremen in experimenal daa, and we need curve fiing echniques ha are robus in he sense ha hey re no going o be seriously compromised by one or wo bad poins. Indeed, some of our poins will be beer han ohers, bu i s no clear a firs which he good ones are, and we need mehods which can rise above ha. Bu he mos popular idea is o jus fi a parabola o he daa. Technology has come a long way in 1 years mos spreadshees and hand calculaors will allow you o fi a rendline of a variey of algebraic forms, no only linear bu polynomial of any degree or exponenial ec. And he kids know how o use ha. Bu I wan o be able see wih my own eyes wheher hese poins lie on a parabola, and wha my eyes are good a seeing are sraigh lines. So I wan o somehow ransform he daa so i ough o lie in a sraigh line. Wha kind of ransformaion would do ha? Well is quadraic in. Wha abou looking a is square roo? The experimen I record here used a 1-lire pop bole. I avoid he non-cylindrical par a he op and he boom by no filling he bole righ o he op, and by punching he hole in he recessed cenre of he boom so ha a residue of waer sis permanenly in he boom of he bole. heigh ime (s) (mm) Careful wih he end poins. Ineresingly enough, he end poin =19, =, is ap o be even more prone o error han he ohers. When he flow is very weak, effecs like surface ension, which our heoreical model does no accoun for, can become significan, and cause he flow o sop premaurely. In fac, read on! The elemenary approach. This is of course wha we would have done when I was a school. I s unsophisicaed, bu here s a lo of good learning in i. In ransforming he daa o a linear form, wha we are doing is wha all hose sophisicaed spreadshees are doing behind he scenes. 1 waer ank 3

4 Transforming he daa. The arge equaion = a( c) is quadraic in. Bu we can ge rid of he square by aking square roos of boh sides: 1/ = ± a 1/ ( c). I s imporan o remember he plus and minus signs, because i s no a firs clear which of he wo we wan. Le's hink abou his. There are only cerain imes ha are relevan and hose are he imes beween and c. If we look a he sign of boh sides for hese imes, we see ha everyhing s posiive excep he ( c), which is negaive. So o make boh sides posiive in ha inerval, we need o choose he minus sign: 1/ = a 1/ ( c). Acually i would read beer if we urned he ( c) erm around o read (c ) and hen ha erm would be posiive for he relevan -values: 1/ = a 1/ (c ). Now all hree erms are posiive for he relevan values of. So where have we go o? Do we have a sraigh line ye or no? Well he expression is now linear in, bu no in we don have he square anymore, bu we have a square roo insead. However, he expression is linear in 1/. Is ha any use? Yes i is. Tha ells us ha if we were o abulae he values of he square roo of, and plo hose agains, we should ge a sraigh line. Well, well. And we do! There s a real sense of pleasure and exciemen in waching ha sraigh line of poins emerge. Tha s he proof we were looking for ha he original daa poins lie along a parabola. Now we can ask for a linear rend line. We ge he equaion. 1/ = A suden remarks ha he las daa poin is below he line and farher from i han any oher. I would appear ha our experimen ended (i.e. he flow sopped) a few seconds before i should have. Does ha make any sense? I hrow his quesion ou and ge a good discussion. This is a graphic illusraion of he surface ension effec, he low pressure a he end being insufficien o overcome he viscosiy. So ha in fac a small amoun of waer remained above he hole a he end. This is he firs iny opening of a window ono a sophisicaed and powerful idea. We are changing variable. The new variable, 1/, is much friendlier for our purposes because if has he arge algebraic form, hen he 1/ -daa will be linear ^1/ 1/ ^ 1/ waer ank

5 Analysis So wha do you hink of he line? Is everyhing okay? Mos sudens seem happy enough. Bu one or wo are uneasy. The poins beween =1 and =15 are all above he line. The problem is clearly caused by he las poin. Being so low i has pulled he line down a he end. I would be beer if i weren here! Quie so. Indeed we ve already idenified ha las poin as anomalous, and we even know he reason viscosiy effecs. So we have a good reason o simply exclude i from he daa se. Le s do i. So we order up anoher rend-line using only he firs 13 poins: 1/ = and indeed i has a slighly flaer slope. And i gives a much beer fi. So his is he line we will use. Wha we can do now is "unransform" his equaion back ino an equaion for. Square boh sides: = ( ). There i is. Acually, he form we prefer i o be in is he above form = a( c) because hen he ime c o empy is one of he parameers. How do we pu i in ha form? Well we wan o have coefficien 1, so we ake he coefficien of ouside he bracke. = = (.55).55 = (.55) (5 ) ^ 1/ ^ 1/ This complees he proof ha he -poins lie in a parabola. The 1/ - daa was a sraigh line, and by squaring he equaion of ha line, we ge a parabola. Mos of my sudens have rouble wih hese manipulaions more rouble han I would like o hink hey would have. I feel ha i s no so much ha hey don know how o do he seps, as hey have no learned ha grea care is needed wih such calculaions According o he equaion as i is now wrien, he heoreical ime ha he ank should ve been empy (if here had been no surface ension effecs) is =5. This is he poin where he regression line mees he -axis (jus off he righ end of he diagram). I appears ha he flow sopped 13 seconds early. Evidenly here was some waer a he end ha he viscosiy rapped inside. Also by seing = we find ha he saring waer level is = (.55) (5) = 1 mm. Tha s o he = 13 ha we aimed for. Maybe we were a ad shor a he beginning. Or maybe no 1 waer ank 5

6 Problems. 1. A parabola has verex a x=, y=. I inersecs he y-axis a y=3. Make a skech of is graph and find is equaion.. A = (minues) a ank holds 1 L of waer. Over he nex 5 minues a oal of 5 L is pumped ou of he ank a a variable rae so ha a =5 he ank holds only 5 L. The equaion of he amoun (lires) in he ank a ime (minues) has he form = a(5 ) 3 + b. Find he parameers a and b. 3. A ank conains has lires of waer. Suppose we pump he waer ou of he ank a he consan rae of L/min. Draw he graph of he amoun in he ank agains ime over he 1-minue period unil he ank is empy and find he equaion of his line. Take he ime origin = o be he momen a which he pump begins o operae.. A cylindrical ank sars wih lires of waer. Suppose we pump waer ou of he ank for 1 minues a he consan rae of L/min and hen for anoher 1 minues a he consan rae of L/min. (a) How much waer will here be in he ank a he end of he -minue period? (b) Draw he graph of he amoun in he ank agains ime over he -minue period. (c) Find he equaion of he line segmens. (d) A wha ime is he ank half full (i.e. lires)? Illusrae his on a copy of your graph. 5. A cylindrical ank has 9 lires of waer. A = (minues) a hole punched a he base, and waer pours ou. I akes exacly minues for he ank o empy. Use he fac ha he - graph is a parabola o answer he following quesions. (a) Find a formula for as a funcion of. (b) How much waer is in he ank a =3 minues? (c) A wha ime are here L in he ank? (d) When is he ank half-full?. A he righ is a copy of he parabola ha resuls from Example 1 of his secion. Suppose he waer which runs ou of he hole is direced ino a bucke which sars empy. Draw he graph of he amoun of waer in he bucke as a funcion of ime. Wha is he geomeric relaionship beween he curve ha you have drawn and he graph a he righ? Be as precise as you can. Illusrae by including boh curves in he same diagram. 7. In Example 1 we preferred o work wih an equaion of he form = a(c ) 1/ because he parameer c has a nice physical inerpreaion as he ime for he ank o empy. Bu wha abou he parameer a? Well here isn a nice inerpreaion for i, bu here is anoher physically significan parameer, and ha s he saring heigh of he waer, call i h. Here s he quesion: find a form of he general equaion which uses hese wo nice parameers c and h, insead of a and c. [And ha s acually he naural form o work wih.] 1. A ank wih a hole in he base is empy. Saring a ime =, waer is run ino he ank a a consan rae. Afer a reasonable ime has passed, he waer level in he ank seems o be consan a deph 1 mm. Draw he graph of he waer level in he ank as a funcion of ime for. 1 waer ank

7 9. Two idenical cylindrical anks, A and B each conain ml of waer. One hole is punched a he base of A and wo holes are punched a he base of B, all hree holes of he same sie. On a single se of axes, draw graphs A and B of he amouns of waer in each ank agains ime. Wha relaionship exiss beween he slopes of graphs A and B? Illusrae his relaionship by referring o specific poins on he graphs. 1. I have wo cylindrical anks, A and B, he firs wih wice he cross secional area of he second. The anks have idenical holes a he base, and each sars wih ml of waer. On a single se of axes, draw graphs A and B of he amouns of waer in each ank agains ime. Wha relaionship exiss beween he slopes of graphs A and B? Illusrae his relaionship by referring o specific poins on he graphs. 11. The daa se a he righ is supposed o fi a model of he form = a(c ) 1/ bu here are random errors in he collecion and no reliable figure was obained for he -value when =. Transform he daa o fi a sraighline model, and use ruler and eye or a rend-line rouine o plo he daa, obain he bes-fi sraigh line, and esimae he parameers a and c. Wha s your bes guess for he ime a which =? 1. I have some x, y daa which are supposed o fi an equaion of he form y = a(x b) for some suiable values of he parameers a and b. So I plo y 1/ agains x and I ge a fairly good sraigh line. The regression equaion is y 1/ =.3.x. Wha esimaes of a and b does his line gives us? 13. I have some x, y daa which are supposed o fi an equaion of he form y = a(x+b) 3 for some suiable values of he parameers a and b. So I plo y 1/3 agains x and I ge a fairly good sraigh line. The regression equaion is y 1/3 =.5+.5x. Wha esimaes of a and b does his line gives us? 1. A large cylindrical ank conaining 1 lires of waer empies hrough a hole a he base. The daa se a he righ shows he ime (minues) a which he waer level is a differen heighs (cm). Thus he daa are supposed o fi a model of he form = a(c ). Transform he daa o fi a sraigh-line model, and use ruler and eye or a spread shee and a regression rouine o plo he daa, obain he bes-fi sraigh line, and esimae he parameers a and c. 15. The graph of.7 agains is a sraigh line wih slope 1.3 and - inercep =. Find an equaion for in erms of ? I have hree cylindrical anks, one wih 1 L, one wih L, and one wih 3 L. Now I punch holes a he base of each of hem and le hem all sar o empy a he same ime, he firs wo empy ino pool A and he hird empies ino pool B. Remarkably enough, all hree anks becomes empy a he same ime. If boh pool A and B sar empy, do hey always have he same amoun of waer? 1 waer ank 7