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1、ExponentialGrowthandDecaySection5.7Problem:AsinglebacteriumisinaPetridish.Every3secondsthebacteriadoubles.Findtherelationshipbetweent,thenumberofseconds,andN(t),thenumberofbacteria.t(seconds)N(t)(#ofbacteria)036912…Whenwillbacteriapopulationreach1000?Varia
2、bleinexponentTakelog(base2)ofbothsidestoundotheexponentialfunction.ChangeofbaseThebacteriawillincreaseitspopulationfrom1to1000in29.90seconds.Wesaythebacteriaobeythelawofuninhibitedgrowth.Thismeansthenumberofbacteriagrowsexponentially,therelationshipbetween
3、thenumberofbacteriaandtimeisgivenbyanexponentialfunction.Formulaforuninhibitedgrowth/decayA0=initialamount(attime0)A(t)=amountaftertyears,days,etct=time(years,days,etc)k=growth/decayconstantYoudonotneedthisformula.Youcanderiveaformulalikewedidusingthetable
4、.kisspecifictosubstanceifk<0,decayifk>0,growthTousetheformula,youneedtofindkfirst.Thenyoucanusetheformulatoansweranyquestions,likewhenwillthepopulationreach1000.Let’sredotheproblemthisway.WeknowA0,theinitialamount,is1.Wealsoknowthenumberofbacteriawillbe2at
5、t=3;meaningA(3)=2.VariableinexponentTakelog(basee)ofbothsidestoundotheexponentialfunctionSubstitutingthisvalueofkandA0=1intotheequation,weget.ThisistheequationthatrelatestimettothenumberofbacteriapresentA(t).Whenwillthebacteriapopulationreach1000?Variablei
6、nexponentTakelog(basee)ofbothsidestoundotheexponentialfunction.(inseconds)Halflifeofradioactivesubstancestheamountoftimeittakesforonehalfofthesubstancepresenttodecayexpl:halflifeofradiumis1690yearst(years)amtofradium(grams)0100169050338025507012.567606.25R
7、ecalltheformulabelow.A0=initialamount(attime0)A=amountaftertyears,days,etct=timeinyears,days,etck=growth/decayconstantexpl:#4Iodine131isaradioactivematerialthatdecaysaccordingtothefunctionwhereA0istheinitialamountpresentandA(t)istheamountpresentattimet(ind
8、ays).Whatisthehalflifeofiodine131?Whenwill?VariableinexponentTakelog(basee)ofbothsidestoundotheexponentialfunction.Sothehalf-lifeofiodine131is7.97days.expl:#14Afossilizedleafcontains70%ofitscarbon14.Howoldisi
