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<ul><li><p>RockHoffmanMinhPham</p><p>MeganSeymourTheCarandtheTruckWriteUp</p><p>TheproblemcalledTheCarandtheTruckdealswithacarandtrucktravellingalonga</p><p>straightroadatdifferenttimesanddifferentvelocities.Thecarstartsanhourearlierthanthe</p><p>truckandhasavelocitythatisindicatedbythevelocity(mph)vs.time(hour)graphonpage</p><p>317,whichstartsoutconcavedownandincreasinguntilt=2hours,whenthederivativeisequal</p><p>tozero,andtherestofthegraphisdecreasing.Anhourlater,att=1,thetruckstartstotravelat</p><p>aconstantvelocityof50mph.Theproblemfirstasksthedistancethatthecarhastravelledin</p><p>thefirsthour.Thenitaskstherateofchange,orthederivative,ofthedistancebetweenthecar</p><p>andthetruckatt=3hours,inadditiontothereallifesignificanceofthefactthatthecars</p><p>velocityismaximizedatt=2hours.Next,thequestionaskstofindwhenthedistancebetween</p><p>thecarandthetruckisthegreatestandwhatthatdistanceis.Theproblemthegoesontoask</p><p>whenthetruckovertakesthecar(intime)andatwhatdistancefromthestartingpoint.Finally,</p><p>theproblemhypotheticallyproposesthatbothvehiclesstartatthesametimeandasksforthe</p><p>graphofthetwofunctionswhenbothhavinginitialvaluesof(0,0)andtodeterminehowmany</p><p>timesthetruckandcarpasseachotherandwhatitmeansintermsofthedistancebetweenthe</p><p>twopassoneanother.Onemustusehisorherknowledgeofareaunderthecurveofvelocity</p><p>graphs,theFundamentalTheoremofCalculus,integrals,andcriticalthinkingtoanswerthis</p><p>problemaccurately.</p><p>Concepts:</p><p>Thewaytocalculatedistancefromavelocityequationisbytakingtheintegralofthat</p><p>equation.Thevelocityequationisthederivativeofthepositionfunction.Theintegralis</p><p>essentiallytheareaunderthegraph.Tocalculatethisalgebraically,onemusttakethe</p></li><li><p>antiderivativeoftheequation,butinthecaseofthisproblem,sincethegraphisgiven,onemust</p><p>solvetheproblemgraphically.</p><p>ThegraphicalwastocorrectlyexecutetakinganIntegraltogetthetotaldistance</p><p>traveledisbylookingattheareaunderthevelocitycurve.Onecanestimatethisthroughtaking</p><p>Riemann'ssumsorjustcountingthenumberofboxesunderthecurve.ARiemannssumcan</p><p>beanoverestimateoranunderestimatedependingonthenatureofthegraphandwhatkindof</p><p>sumthestudentchoosestouse.</p><p>TotakeaRiemannsum:</p><p>Thestudentmustfirstsplitthegraphintointervals,whichdonthavetonecessarilybe</p><p>equal,buttheyhavetobeknown.Theseintervalswillbeknownastn,andknowingthemis</p><p>essentialintakingRiemannsums.Then,forlefthandsumsonetakesthef(t)valuefromtheleft</p><p>sideoftheintervalandmultipliesitbytheinterval.Onehastoexecutethisforalloftheintervals</p><p>andthenthesumwillbecomplete.Forarighthandsumonewoulddothesamethingbutusing</p><p>thef(t)valuefromtherightsideoftheinterval.</p><p>AgeneralRiemannsumforfontheinterval[a,b]isasumoftheform:</p><p>f(ci) i,n</p><p>i=1t </p><p>Where1istheinitial(starting)pointandnistheendpoint. isthevalueofinterval,i,andf(ci)t </p><p>istheoutputvalueatthisinterval,whichcanbeanypointonthisinterval,whetherleftmost,</p><p>rightmost,orevenmidpoint.</p><p>Thisishowonesetsupanintegral:</p></li><li><p>Supposefiscontinuousfor .Thedefiniteintegralofffromatobcanbewrittena t b </p><p>as, (t)dtb</p><p>af </p><p>However,onemustremember,inordertogetthecorrectdistancethatisbeingasked</p><p>for,onemustdeterminetheboundsofthisIntegral.Or,graphicallyspeaking,onemustknowthe</p><p>beginning(leftendpoint)andtheend(rightendpoint)oftheportionofthevelocitygraphtheirare</p><p>seekingtointegrate.</p><p>Solutions:</p><p>a. Whenthetruckstarts,thecarhastraveled1houralready.Thedistancetraveledbythecarcanbefoundbythetotalareainthevelocitygraphfrom0to1hour,whichis7rectangles.Eachtriangleis5miles;therefore,thetotaldistanceis35hours.</p><p>b. Velocityofcarat3pm:67mph</p><p>Velocityoftruckat3pm:50mphWecallfunctionf(x)asthedistancebetweenthecarandthetruck:F(x)=f(car)f(truck)Therateofchangeofthedistancebetweenthecarandthetruckwillbethederivativeofthefunctionf(x),whichisthedifferenceinvelocity.F(x)=f(Car)f(truck)F(x)=v(car)v(truck)F(3)=6750=17mphTherefore,at3pm,thedistancebetweenthecarandthetruckisincreasingat17milesperhour.</p></li><li><p>Whenthecarsvelocityismaximizedat2pm,therateofthedistancebetweenthecarandthetruck(F(x)=f(Car)f(truck))willbethegreatestsincethetrucksvelocitystaysconstantandsothedifferencebetweenthemwillbethegreatest.</p><p>c. Whenthecarisaheadofthetruck,thedistancebetweenthecarandthetruckwillkeepincreasinguntilthetrucksvelocitycatchesupwiththecarsvelocity.Therefore,thegreatestdistancewilloccurwhenv(truck)=v(car).Sincev(truck)isalways50mph,v(car)willbe50mphatt=4.3hours(4:15pm).Thedistancetraveledbythecarfromt=0tot=4.3istheareaunderthev(car)graph,whichisaround50squares;thereforethedistanceis250miles</p><p>Thedistancetraveledbythetruckfromt=1tot=4.3is50*3.3=165miles(33squares)</p><p>Therefore,thegreatestdistanceis250165=85miles(17squares)</p><p>d. Thetruckovertakesthecarwhenthedistancetraveledbythecarandthetruckare</p><p>equal.Thismeansthattheareaunderthecurvev(car)andtheareaunderthecurvev(truck)areequal.Untilthetime4.3,thedistanceofthecarandthetruckarethegreatestapart.Aftert=4.3,thetruckstartscatchinguptothecar,andsothegapbetweentheirdistancegetsmaller.Underthegraphbelow,theredshadedpartrepresentsthetotaldistanceapartbetweenthecarandthetruckuntilt=4.3.Tocancel</p></li><li><p>itout,wehavetofindtheareathatisequaltothattotaldistanceaftert=4.3sothatthedifferencescancancelout.Thecorrespondingtimewillbethetimewherethecarandthetruckhavetraveledthesamedistance.Thenumberofsquaresbeforet=4.3:17squaresTherefore,thetimewillbe:t=8.3(wherethe17squaresafter4.3)=8:25pm</p><p>e. </p><p>f. Thegraphsintersecttwice,att=0.7and4.3hours. Thefirstintersectionpoint(t=0.7)meansthatv(car)=v(truck).Beforet=0.7,</p><p>v(truck)isgreaterthanv(car),whichmeansthatthetruckistravelingawayfromthe</p></li><li><p>carcar.Aftert=0.7,v(car)becomesgreaterthanv(truck),whichmeansthatthecariscatchingupwiththetruckandclosingthedistancebetweenthem.</p><p> Thesecondintersectionpoint(t=4.3)alsomeansthatv(car)=v(truck).Atthispoint,thedistanceapartbetweenthecarandthetruckarethegreatest.Beforet=4.3,v(car)isgreaterthanv(truck),whichmeansthatthecaristravelingawayfromthetruck.Aftert=0.7,v(truck)becomesgreaterthanv(car),whichmeansthatthetruckiscatchingupwiththecarandclosingthedistancebetweenthem.</p></li></ul>