4.1 Water tank. height z (mm) time t (s)

Transcription

1 4.1 Waer ank (a) A cylindrical ank conains 8 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. Draw he graph of he amoun of waer in he ank agains ime. Explain he shape of he graph. Begin, of course, by leing he class wach he acion, bu wihou aking measuremens. A fine ank can be made from a clear 1 lire pop bole. I colour he waer wih food colouring, so i s easier o see from he back. I avoid he noncylindrical 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, acing like a fake fla boom. I drill a hole some 3 o 4 mm (say 1/16 h inch) in diameer. I akes some 3 minues o empy. v (b) The form of he equaion of agains depends on he shape of he ank, bu for a cylindrical ank, heoreical consideraions of how he pressure pushes he waer ou of he hole lead in fac o he equaion of a parabola. Given his, ha he - graph is a parabola, find is equaion. (c) I s ime o 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. Daa from an experimen I did wih a KCVI class is recorded a he righ. We used a large pop bole wih horional marks a inervals of 1 mm above he hole. The sudens recorded he ime ha he waer level arrived a each of he marks. I s bes o do your own experimen. (d) Use he daa you colleced o find an equaion for in erms of. Alernaively saed, find he bes fi parabola. Discuss your resuls, paying aenion o possible sources of error in your daa collecion. In paricular, check ou he wo endpoins, he saring heigh and he ending ime. heigh (mm) ime (s) Remarks. There are differen ways o find he bes parabola, and some are more insighful han ohers. One way is o ener he daa ino a compuer or graphical calculaor and ask he echnology o fi he bes quadraic polynomial. Indeed, ha s wha my sudens ell me o do, and in he wrie-up below ha s he firs roue we ll follow. Bu i s no a roue I m comforable wih and ha s no only because here s weren calculaors around when I was a kid. The ge a bes-fi curve of any kind approach makes i much harder o evaluae he experimen, o cach anomalies, o see where hings migh have gone wrong. In siuaions like his I always ry o ransform my daa so ha i ough o plo as a sraigh line. The poin is ha I can always see wheher a line is sraigh or no (whereas I can easily see wheher i s a parabola). If you do ha, in his case, here s an anomaly ha pops righ ou a you. And aking accoun of i leads you o a new esimae of when he ank was really empy. waer ank 9/13/7 1

2 The graph (a) A cylindrical ank conains 8 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. Draw he graph of he amoun of waer in he ank agains ime. Explain he shape of he graph. Soluion. 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 8, 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 Yes i did surprise me, bu ha s really wha happened. However his was 1998 and I have a feeling ha oday I d ge a curve righ off he ba. Am I righ? Do grade 1 sudens oday no have a much beer feeling for slope and rae of change even hough chaper 5 sill awais us? 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? 8 ml in 1 seconds means 8 ml/s. Righ 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. waer ank 9/13/7

3 The formula. (b) Theoreical consideraions of how he pressure pushes he waer ou of he hole lead in fac o he equaion of a parabola. Given his, ha he - graph is a parabola, find is equaion. Soluion The daa ha we have is ha he ank sars wih 8 ml of waer and ha i akes 1 seconds o empy. Well he equaion of a parabola is a polynomial of degree, and mos of my sudens 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 ()=8 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 = 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 =, =8: 8 = a( 1) = 1 a and his solves o give a=8/1 =.8. Thus he equaion of he parabola is =.8( 1) 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. waer ank 9/13/7 3

4 Experimen. (c) I s ime o 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. Daa from an experimen I did wih a KCVI class is recorded a he righ and graphed below. We used a large pop bole wih horional marks a inervals of 1 mm above he hole. The sudens recorded he ime ha he waer level arrived a each of he marks. I s bes o do your own experimen heigh (mm) ime (s) Analysis of he daa (d) Use he daa you colleced o find an equaion for in erms of. Alernaively saed, find he bes fi parabola. Discuss your resuls, paying aenion o possible sources of error in your daa collecion. In paricular, check ou he wo endpoins, he saring heigh and he ending ime. There are differen ways o do his, and as we shall see, some are more insighful han ohers. Fiing a quadraic polynomial. My sudens ell me o use echnology o direcly fi he bes parabola. For example, my Excel spreadshee gives me he graph a he righ wih equaion: = From he diagram i seems o give a prey good fi. Le s check he endpoins. The official saring heigh is 13 and he equaion gives us () = Tha s no a bad mach. To ge he ime a which he ank is empy, we se = and we ge: = 1.79 ± (.31)(18.84) = = (.31) 6.36 We ge wo soluions. Which one do we ake? y =.31x x waer ank 9/13/7 4

5 Think abou wha he parabola mus look like around hose wo roos. Since he graph is coming from he lef, i s he lef-hand ero ha i will hi. Tha gives an empy ank a = Tha s quie a bi shor of our measured ime of 19. Bu somehing else isn righ he graph also gives us a negaive slope a ha ero, suggesing ha waer should be flowing ou. Is ha wha we expec? No. In fac we expec he flow o become ero as he ank becomes empy. Indeed we expec he parabola o have he form: = a( c) wih is verex on he axis. Bu his parabola does no have ha form. For example, if we complee he square of he equaion we ge: =.31( 5) 1.44 and he verex is 1.44 unis below he axis. This is no a grea error, and we migh well live wih i. We could ake he ime o empy o be he -value of he verex which is = 5, half-way beween he roos. And call i a good day s work. Looking for a sraigh line. Bu here s anoher way o do he analysis, definiely my favorie, and ha s o shoo direcly a he arge form: = a( c). How do we make echnology resric is aenion o parabolas of he above form? The secre is o ransform he equaion ino a sraigh line. 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. There s definiely somehing wrong here. Bu given wha we ve go so far, i s no easy o see how o fix i. How do we ge a bes-fi parabola of he form = a( c)? The answer migh come from noing ha his family of parabolas has only wo parameers, a and c, whereas he general parabola has hree. Wha else has wo parameers: a line! Can we ransform he equaion ino a line? This is he daa we will now work wih: he square roo of agains. The equaion is linear in, so we should ge a sraigh line. waer ank 9/13/7 5

6 So where have we go o? Do we have a sraigh line ye or no? Well he righ-hand side is linear in, bu he lef-hand side is no linear in. Bu i is linear in he square roo of. 1/ = a 1/ (c ) In oher words, if we were o abulae he values of he square roo of, and plo hose agains, we should ge a sraigh line. Well, well ^1/ And we do! There s a real sense of pleasure and exciemen in waching ha sraigh line of poins emerge. Tha s acually a beauiful visual proof ha he original daa poins lie along a parabola. Tha s jus one of he several reasons I like his approach. Now we can ask for a linear rend line. We ge he equaion. 1/ = and i is graphed a he righ ^1/ 1/ y = -.573x A suden remarks ha he las daa poin is below he line and farher from i han any oher. This poin definiely looks like i doesn belong. [Such a poin is called an oulier.] I looks as if i ough o be moved o he righ, or, saed anoher way, i looks like 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. I urns ou ha his is a graphic illusraion of he surface ension effec, he low pressure a he end being insufficien o overcome he viscosiy or he liquid. So ha in fac a small amoun of waer remains above he hole a he end Tha las daa poin is definiely off he line, almos flagranly so. Is i rying o ell us somehing? Yes i is! I urns ou ha we can find a physical reason o expec i o be off he line. Wha his approach does, however is give us a way of deermining when he flow ough o have ended. waer ank 9/13/7 6

7 Bu firs a bi of reanalysis. The las daa poin clearly doesn belong. One of my sudens is percepive enough o observe ha he bes-fi line ough o be consruced wihou aking accoun of ha poin. Tha is, a presen, ha bad poin has he effec of pulling he line down a bi. Remove i and we d ge a beer line. Indeed we even have a good physical raionale for excluding i viscosiy effecs. Le s do i. 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 ^ 1/ y = -.55x = = (.55).55 =.35 (5 ). 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 experimenal 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 =.35(5) = 18 mm. Tha compares well wih he = 13 ha we aimed for. Maybe we were a ad shor a he beginning. waer ank 9/13/7 7

8 Problems. 1. A parabola has verex a x=4, y=. I inersecs he y-axis a y=34. 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. 4. A cylindrical ank sars wih 8 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 4 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. 4 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 6 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 4 L in he ank? (d) When is he ank half-full? 6. 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.] 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. waer ank 9/13/7 8

9 9. Two idenical cylindrical anks, A and B each conain 8 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 8 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/ = 8.3.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 =.45+.5x. Wha esimaes of a and b does his line gives us? 14. 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 =4. 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? waer ank 9/13/7 9