Lesson 14: Exponential Growth and Decay

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Transcript of Lesson 14: Exponential Growth and Decay

  • 1. Section3.4ExponentialGrowthandDecayV63.0121.006/016, CalculusI March10, 2010 AnnouncementsMidtermis(almost)graded, willbereturnedinrecitationScoresandgradeswillbemassagedWewilldropthevelowestWebAssign-ments. .... .

2. AnnouncementsMidtermis(almost)graded, willbereturnedinrecitationScoresandgradeswillbemassagedWewilldropthevelowestWebAssign-ments. .... . 3. OutlineRecallTheequation y = kyModelingsimplepopulationgrowthModelingradioactivedecay Carbon-14DatingNewtonsLawofCoolingContinuouslyCompoundedInterest. . . . . . 4. Derivativesofexponentialandlogarithmicfunctionsyyex exax(ln a)ax 1ln xx1 1 loga x ln a x . . . . . . 5. OutlineRecallTheequation y = kyModelingsimplepopulationgrowthModelingradioactivedecay Carbon-14DatingNewtonsLawofCoolingContinuouslyCompoundedInterest. . . . . . 6. Denition A differentialequation isanequationforanunknownfunction whichincludesthefunctionanditsderivatives.. .. . .. 7. Denition A differentialequation isanequationforanunknownfunction whichincludesthefunctionanditsderivatives.Example NewtonsSecondLaw F = ma isadifferentialequation, where a(t) = x (t).. .. .. . 8. Denition A differentialequation isanequationforanunknownfunction whichincludesthefunctionanditsderivatives.Example NewtonsSecondLaw F = ma isadifferentialequation, where a(t) = x (t). Inaspring, F(x) = kx, where x isdisplacementfrom equilibriumand k isaconstant. So k kx(t) = mx (t) = x (t) + x(t) = 0.m... . . . 9. Denition A differentialequation isanequationforanunknownfunction whichincludesthefunctionanditsderivatives.Example NewtonsSecondLaw F = ma isadifferentialequation, where a(t) = x (t). Inaspring, F(x) = kx, where x isdisplacementfrom equilibriumand k isaconstant. So k kx(t) = mx (t) = x (t) + x(t) = 0.m Themostgeneralsolutionis x(t) = A sin t + B cos t, where = k/m.... . . . 10. TheEquation y = 2t Example Find a solutionto y (t) = 2t. Findthe mostgeneral solutionto y (t) = 2t.. . . . . . 11. TheEquation y = 2t ExampleFind a solutionto y (t) = 2t.Findthe mostgeneral solutionto y (t) = 2t. SolutionA solutionis y(t) = t2 . . . . . . . 12. TheEquation y = 2t ExampleFind a solutionto y (t) = 2t.Findthe mostgeneral solutionto y (t) = 2t. SolutionA solutionis y(t) = t2 .Thegeneralsolutionis y = t2 + C.(checkthis) . . . . . . 13. Theequation y = ky Example Find a solutionto y (t) = y(t). Findthe mostgeneral solutionto y (t) = y(t).... . . . 14. Theequation y = ky ExampleFind a solutionto y (t) = y(t).Findthe mostgeneral solutionto y (t) = y(t). SolutionA solutionis y(t) = et . ... . . . 15. Theequation y = ky ExampleFind a solutionto y (t) = y(t).Findthe mostgeneral solutionto y (t) = y(t). SolutionA solutionis y(t) = et .Thegeneralsolutionis y = Cet , not y = et + C.(checkthis) .. . . . . 16. Kickitupanotch Example Findasolutionto y = 2y. Findthegeneralsolutionto y = 2y. . . . . . . 17. Kickitupanotch ExampleFindasolutionto y = 2y.Findthegeneralsolutionto y = 2y. Solutiony = e2ty = Ce2t. . . . . . 18. Ingeneral Example Findasolutionto y = ky. Findthegeneralsolutionto y = ky. . . . . . . 19. Ingeneral ExampleFindasolutionto y = ky.Findthegeneralsolutionto y = ky. Solutiony = ekty = Cekt. . . . . . 20. Ingeneral Example Findasolutionto y = ky. Findthegeneralsolutionto y = ky. Solution y = ekt y = Cekt RemarkWhatis C? Plugin t = 0: y(0) = Cek0 = C 1 = C, so y(0) = y0 , the initialvalue of y. .. . . . . 21. ExponentialGrowth Itmeanstherateofchange(derivative)isproportionaltothe currentvalue Examples: Naturalpopulationgrowth, compoundedinterest, socialnetworks. .... . 22. OutlineRecallTheequation y = kyModelingsimplepopulationgrowthModelingradioactivedecay Carbon-14DatingNewtonsLawofCoolingContinuouslyCompoundedInterest. . . . . . 23. Bacteria Sinceyouneedbacteriatomakebacteria, theamountofnewbacteriaatanymomentisproportionaltothetotalamountofbacteria.Thismeansbacteriapopulationsgrowexponentially.. . . . . . 24. BacteriaExample Example A colonyofbacteriaisgrownunderidealconditionsina laboratory. Attheendof3hoursthereare10,000bacteria. At theendof5hoursthereare40,000. Howmanybacteriawere presentinitially? .. .. . . 25. BacteriaExample Example A colonyofbacteriaisgrownunderidealconditionsina laboratory. Attheendof3hoursthereare10,000bacteria. At theendof5hoursthereare40,000. Howmanybacteriawere presentinitially?Solution Since y = ky forbacteria, wehave y = y0 ekt . Wehave 10, 000 = y0 ek340, 000 = y0 ek5. ... . . 26. BacteriaExample Example A colonyofbacteriaisgrownunderidealconditionsina laboratory. Attheendof3hoursthereare10,000bacteria. At theendof5hoursthereare40,000. Howmanybacteriawere presentinitially?Solution Since y = ky forbacteria, wehave y = y0 ekt . Wehave 10, 000 = y0 ek3 40, 000 = y0 ek5Dividingtherstintothesecondgives 4 = e2k = 2k = ln 4 = k = ln 2. Nowwehave 10, 000 = y0 eln 23 = y0 810, 000 So y0 = = 1250.8. . . . . . 27. Couldyoudothatagainplease? Wehave 10, 000 = y0 ek340, 000 = y0 ek5 Dividingtherstintothesecondgives 40, 000y e5k= 0 3k10, 000y0 e = 4 = e2k = ln 4 = ln(e2k ) = 2k ln 4 ln 22 2 ln 2 = k = = == ln 22 22. . . . . . 28. OutlineRecallTheequation y = kyModelingsimplepopulationgrowthModelingradioactivedecay Carbon-14DatingNewtonsLawofCoolingContinuouslyCompoundedInterest. . . . . . 29. Modelingradioactivedecay Radioactivedecayoccursbecausemanylargeatomsspontaneouslygiveoffparticles. . .. . . . 30. Modelingradioactivedecay Radioactivedecayoccursbecausemanylargeatomsspontaneouslygiveoffparticles.Thismeansthatinasample ofabunchofatoms, wecan assumeacertainpercentage ofthemwillgooffatany point. (Forinstance, ifall atomofacertainradioactive elementhavea20%chance ofdecayingatanypoint, thenwecanexpectina sampleof100that20of themwillbedecaying.) . .. . . . 31. Thustherelativerateofdecayisconstant: y =kywhere k is negative.. . . . . . 32. Thustherelativerateofdecayisconstant: y =kywhere k is negative. Soy = ky = y = y0 ektagain!. . . . . . 33. Thustherelativerateofdecayisconstant: y =kywhere k is negative. Soy = ky = y = y0 ektagain! Itscustomarytoexpresstherelativerateofdecayintheunitsof half-life: theamountoftimeittakesapuresampletodecayto onewhichisonlyhalfpure.. ..... 34. Example Thehalf-lifeofpolonium-210isabout138days. Howmuchofa 100gsampleremainsafter t years? . . .. . . 35. Example Thehalf-lifeofpolonium-210isabout138days. Howmuchofa 100gsampleremainsafter t years?Solution Wehave y = y0 ekt , where y0 = y(0) = 100 grams. Then 365 ln 2 50 = 100ek138/365 = k = .138 Therefore 365ln 2y(t) = 100e 138t= 100 2365t/138 . ... . . . 36. Example Thehalf-lifeofpolonium-210isabout138days. Howmuchofa 100gsampleremainsafter t years?Solution Wehave y = y0 ekt , where y0 = y(0) = 100 grams. Then 365 ln 250 = 100ek138/365 = k = .138 Therefore 365ln 2y(t) = 100e 138t= 100 2365t/138 .Notice y(t) = y0 2t/t1/2 , where t1/2 isthehalf-life.... . . . 37. Carbon-14Dating Theratioofcarbon-14tocarbon-12inanorganismdecaysexponentially:p(t) = p0 ekt . Thehalf-lifeofcarbon-14isabout5700years. Sotheequationfor p(t) isln2 p(t) = p0 e 5700 t Anotherwaytowritethiswouldbe p(t) = p0 2t/5700. .... . 38. Example Supposeafossilisfoundwheretheratioofcarbon-14to carbon-12is10%ofthatinalivingorganism. Howoldisthe fossil?.. ... . 39. Example Supposeafossilisfoundwheretheratioofcarbon-14to carbon-12is10%ofthatinalivingorganism. Howoldisthe fossil?Solution Wearelookingforthevalueof t forwhich p(t) = 0.1p(0) . . . . . . 40. Example Supposeafossilisfoundwheretheratioofcarbon-14to carbon-12is10%ofthatinalivingorganism. Howoldisthe fossil?Solution Wearelookingforthevalueof t forwhich p(t) = 0.1p(0)Fromtheequationwehave 2t/5700 = 0.1 t ln 2 = ln 0.1 5700 ln 0.1t= 5700 18, 940ln 2. . . . . . 41. Example Supposeafossilisfoundwheretheratioofcarbon-14to carbon-12is10%ofthatinalivingorganism. Howoldisthe fossil?Solution Wearelookingforthevalueof t forwhichp(t)= 0.1 p(0)Fromtheequationwehave 2t/5700 = 0.1 t ln 2 = ln 0.1 5700 ln 0.1t= 5700 18, 940ln 2 Sothefossilisalmost19,000yearsold. . . . . . . 42. OutlineRecallTheequation y = kyModelingsimplepopulationgrowthModelingradioactivedecay Carbon-14DatingNewtonsLawofCoolingContinuouslyCompoundedInterest. . . . . . 43. NewtonsLawofCooling NewtonsLawofCooling statesthattherateofcoolingofanobjectisproportionaltothetemperaturedifferencebetweentheobjectanditssurroundings.. . . . . . 44. NewtonsLawofCooling NewtonsLawofCooling statesthattherateofcoolingofanobjectisproportionaltothetemperaturedifferencebetweentheobjectanditssurroundings.Thisgivesusadifferentialequationoftheform dT= k (T T s ) dt(where k < 0 again). . . . . . . 45. GeneralSolutiontoNLC problems Tosolvethis, changethevariable y(t) = T(t) Ts . Then y = T and k(T Ts ) = ky. TheequationnowlookslikedT dy= k(T Ts ) = ky dt dt . ... .. 46. GeneralSolutiontoNLC problems Tosolvethis, changethevariable y(t) = T(t) Ts . Then y = T and k(T Ts ) = ky. TheequationnowlookslikedT dy= k(T Ts ) = ky dt dt Nowwecansolve!y = ky = y = Cekt= T Ts = Cekt = T = Cekt + Ts . ... .. 47. GeneralSolutiontoNLC problems Tosolvethis, changethevariable y(t) = T(t) Ts . Then y = T and k(T Ts ) = ky. TheequationnowlookslikedT dy= k(T Ts ) = ky dt dt Nowwecansolve!y = ky = y = Cekt= T Ts = Cekt = T = Cekt + TsPluggingin t = 0, wesee C = y0 = T0 Ts . So T(t) = (T0 Ts )ekt + Ts . ... .. 48. Example A hard-boiledeggat 98 C isputinasinkof 18 C water. After5 minutes, theeggstemperatureis 38 C. Assumingthewaterhas notwarmedappreciably, howmuchlongerwillittaketheeggto reach 20 C?.. ... . 49. Example A hard-boiledeggat 98 C isputinasinkof 18 C water. After5 minutes, theeggstemperatureis 38 C. Assumingthewaterhas notwarmedappreciably, howmuchlongerwillittaketheeggto reach 20 C?Solution WeknowthatthetemperaturefunctiontakestheformT(t) = (T0 Ts )ekt + Ts = 80ekt + 18Tond k, plugin t = 5: 38 = T(5) = 80e5k + 18andsolvefor k.... .. . 50. Finding k 38 = T(5) = 80e5k + 1820 = 80e5k 1 = e5k( )4 1 ln= 5k 4 1 = k = ln 4. 5. . . . . . 51. Finding k 38 = T(5) = 80e5k + 18 20 = 80e5k1= e5k ( )41ln= 5k41= k = ln 4.5Nowweneedtosolve t20 = T(t) = 80e 5 ln 4 + 18 for t.. . . . . . 52. Finding tt 20 = 80e 5 ln 4 + 18 t 2 = 80e 5 ln 41 t = e 5 ln 4 40 t ln 40 = ln 4 5 ln 405 ln 40= t =1 = 13 min5 ln 4 ln 4. . . . . . 53. Example A murdervictimis discoveredatmidnightand thetemperatureofthebody isrecordedas 31 C. One hourlater, thetemperatureof thebodyis 29 C. Assume thatthesurroundingair temperatureremains constantat 21 C. Calculate thevictimstimeofdeath. (Thenormaltemperatureof alivinghumanbeingis approximately 37 C.). . . . . . 54. Solution Lettime 0 bemidnight. Weknow T0 = 31, Ts = 21, and T(1) = 29. Wewanttoknowthe t forwhich T(t) = 37.. . . .. . 55. Solution Lettime 0 bemidnight. Weknow T0 = 31, Ts = 21, and T(1) = 29. Wewanttoknowthe t forwhich T(t) = 37. Tond k:29 = 10ek1 + 21 = k = ln 0.8. . . .. . 56. Solution Lettime 0 bemidnight. Weknow T0 = 31, Ts = 21, and T(1) = 29. Wewanttoknowthe t forwhich T(t) = 37. Tond k: 29 = 10ek1 + 21 = k = ln 0.8 Tond t: 37 = 10etln(0.8) + 211.6 = etln(0.8)ln(1.6)t= 2.10 hrln(0.8)Sothetimeofdeathwasjustbefore10:00pm... . . . . 57. OutlineRecallTheequation y = kyModelingsimplepopulationgrowthModelingradioactivedecay Carbon-14DatingNewtonsLawofCoolingContinuouslyCompoundedInterest. . . . . . 58. Interest Ifanaccounthasancompoundinterestrateof r peryearcompounded n times, thenaninitialdepositof A0 dollarsbecomes( r )nt A0 1 +nafter t years. . .. . . . 59. Interest Ifanaccounthasancompoundinterestrateof r peryearcompounded n times, thenaninitialdepositof A0 dollarsbecomes( r )nt A0 1 +nafter t years.Fordifferentamountsofcompounding, thiswillchange. Asn , weget continouslycompoundedinterest(r )nt A(t) = lim A0 1 + = A0 ert . n n . .. . . . 60. Interest Ifanaccounthasancompoundinterestrateof r peryearcompounded n times, thenaninitialdepositof A0 dollarsbecomes( r )nt A0 1 +nafter t years.Fordifferentamountsofcompounding, thiswillchange. Asn , weget continouslycompoundedinterest(r )nt A(t) = lim A0 1 + = A0 ert . n nThusdollarsarelikebacteria. . .. . . . 61. Example Howlongdoesittakeaninitialdepositof$100, compounded continuously, todouble? . .. . . . 62. Example Howlongdoesittakeaninitialdepositof$100, compounded continuously, todouble?Solution Weneed t suchthat A(t) = 200. Inotherwords ln 2 200 = 100ert = 2 = ert = ln 2 = rt = t = .r Forinstance, if r = 6% = 0.06, wehaveln 2 0.69 69t=== 11.5 years. 0.06 0.06 6 .. . .. . 63. I-bankinginterviewtipofthedayln 2Thefractioncan ralsobeapproximatedaseither70or72dividedbythepercentagerate(asanumberbetween0and100, notafractionbetween0and1.)Thisissometimescalledthe ruleof70 or ruleof72.72haslotsoffactorssoitsusedmoreoften.. . . . . . 64. Whathavewelearnedtoday? Whensomethinggrowsordecaysataconstant relative rate, thegrowthordecayisexponential. Equationswithunknownsinanexponentcanbesolvedwith logarithms. Yourfriendlistislikecultureofbacteria(nooffense).... . . .