12MC Advanced Periodic Functions - Weebly · 2018. 9. 4. · bridges, swings ... I could...
Transcript of 12MC Advanced Periodic Functions - Weebly · 2018. 9. 4. · bridges, swings ... I could...
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Chapter8A
WearerevisitingthissowecanuseEulerFormlaterintheChaptertoassistusIntegratedifferentfunctions(amongstotherfunthings).Butfirstletsgooversomeoldground:
Recall
𝑧 = 𝑎 + 𝑏𝑖
where
𝑧 = 𝑟 = 𝑎( + 𝑏(
𝜃 = tan-.𝑏𝑎
DeMoivre:
If
𝑧 = 𝑟 cos 𝜃 + 𝑖𝑟 sin 𝜃 = 𝑟 cis 𝜃
then
𝑧4 = 𝑟4 cos 𝑛𝜃 + 𝑖𝑟4 sin 𝑛𝜃 = 𝑟4 cis 𝑛𝜃
NowletsdosomethingNEW…somegroovyalgebra…
𝑧4 = 𝑧4
𝑟4 cis 𝑛𝜃 = 𝑟 cis 𝜃 4
𝑟4 cis 𝑛𝜃 = 𝑟4 cis 𝜃 4
so
cis 𝑛𝜃 = cis 𝜃 4
therefore,
cos 𝑛𝜃 + 𝑖 sin 𝑛𝜃 = cos 𝜃 + 𝑖 sin 𝜃 4
Soclearly,weget:
cos 𝑛𝜃 = 𝑅𝑒 cos 𝜃 + 𝑖 sin 𝜃 4
sin 𝑛𝜃 = 𝐼𝑚 cos 𝜃 + 𝑖 sin 𝜃 4
Whydowecare…becauseitallowsustogetridofthecoefficientinfrontofTheta!(InadifferentwaytohowwediditlasttermwiththoselovelyTrigrules.)
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AnotherrulethatcanhelpistheMultipleangleformula(pge330)
cos 𝑛𝜃 =𝑧4 + 𝑧-4
2
sin 𝑛𝜃 =𝑧4 − 𝑧-4
2𝑖
yesyoucanjustrememberthisthese…!
BothofthesegetusedinProofsaswell…asyouwillseeoverthepage!
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eg.Q1c Prove,sin 4𝑥 = 4 sin 𝑥 cos> 𝑥 − 4 cos 𝑥 sin> 𝑥
LHS = sin 4𝑥
= 𝐼𝑚 cos 𝑥 + 𝑖 sin 𝑥 ?
(useTi-Nspiretoexpandanythingaboveacubic)
= 𝐼𝑚 cos? 𝑥 + 4𝑖 sin 𝑥 cos> 𝑥 + 6𝑖( sin( 𝑥 cos( 𝑥 + 4𝑖> sin> 𝑥 cos 𝑥 + 𝑖? sin? 𝑥
= 𝐼𝑚 cos? 𝑥 + sin? 𝑥 − 6 sin( 𝑥 cos( 𝑥 + 4 sin 𝑥 cos> 𝑥 − 4 sin> 𝑥 cos 𝑥 𝑖
= 4 sin 𝑥 cos> 𝑥 − 4 sin> 𝑥 cos 𝑥
= 𝑅𝐻𝑆
***Strategy…startwiththesidethathasacoefficientinfrontofTheta.
eg.Q2a. Prove,2 sin 2𝑥 cos 𝑥 = sin 3𝑥 + sin 𝑥
LHS= 2 sin 2𝑥 cos 𝑥
= 2×𝑧( − 𝑧-(
2𝑖 ×𝑧 + 𝑧-.
2
= 2×𝑧( − 𝑧-( 𝑧 + 𝑧-.
4𝑖
=𝑧> + 𝑧. − 𝑧-. − 𝑧->
2𝑖
=𝑧> − 𝑧->
2𝑖 +𝑧. − 𝑧-.
2𝑖
= sin 3𝑥 + sin 𝑥
= 𝑅𝐻𝑆
***Didyougetthat?…Pushitintothedoubleangleform…juggleabit…getitbackintothedoubleangleform…andthenpopitbacktostandardtrigform***
***Strategy…startonthesidethatisFactored…dothatlastquestionagain,butstartontherighthandside.Althoughyoucanpopitinto‘doubleangle’form,becausethereareplusseseverywhere,wedon’thaveanyindexlawsthatwecanuse,sowedon’tgettoofar…startingonthefactoredsidewecanexpandbracketsandapplyindexlawstogetourwantedoutcome!
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eg.Q3C Prove, sin? 𝑥 = .E(cos 4𝑥 − 4 cos 2𝑥 + 3)
Given, sin 𝑥 = HI-HJI
(K
LHSbecomes 𝐿𝐻𝑆 = HI-HJI
(K
?
=𝑧. − 𝑧-. ?
2?𝑖?
=𝑧? − 4𝑧( − 4𝑧-( + 𝑧-? + 6
16
=𝑧? + 𝑧-? − 4𝑧( − 4𝑧-( + 6
16
=𝑧? + 𝑧-?
16 +−4𝑧( − 4𝑧-(
16 +616
=18𝑧? + 𝑧-?
2 +−4𝑧( − 4𝑧-(
2 +62
=18𝑧? + 𝑧-?
2 −4(𝑧( + 𝑧-()
2 + 3
=18 cos 4𝑥 + 4 cos 2𝑥 + 3 = 𝑅𝐻𝑆
So,it’samatterofunderstandingwhentoputininwhatform…andyes,youarecorrectifyouarethinkingthatcomesdowntoPRACTICE…:-)
***ifthereisasingletermwithacoefficientinfrontofTheta…useDeMoivre
***wherethereisafactorfromoftrig,usedoubleangleformula
***AnythingwithaPowerinit…usethedoubleangleformula
Don’tlookatmysolutionfirst…
Prove:
cos 6𝐴 = 32 cosP 𝐴 − 48 cos? 𝐴 + 18 cos( 𝐴 − 1
Hence,usethisresulttosolve32𝑥P − 48𝑥? + 18𝑥( = >(,(giving6answers).
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Solution:
cos 6𝐴 = 𝑅𝑒 cos 𝐴 + 𝑖 sin 𝐴 P
= 𝑅𝑒 cosP 𝐴 + 6 cosQ 𝐴 isin 𝐴 + 15 cos? 𝐴 𝑖 sin 𝐴 ( + 20 cos> 𝐴 𝑖 sin 𝐴 >
+ 15 cos( 𝐴 𝑖 sin 𝐴 ? + 6 cos𝐴 𝐼 sin 𝐴 Q + 𝑖 sin 𝐴 P
= cosP 𝐴 − 15 cos? 𝐴 sin( 𝐴 + 15 cos( 𝐴 sin? 𝐴 − sinP 𝐴
= cosP 𝐴 − 15 cos? 𝐴 (1 − cos( 𝐴) + 15 cos( 𝐴 1 − cos( 𝐴 ( − 1 − cos> 𝐴 >
= cosP 𝐴 − 15 cos? 𝐴 + 15 cosP 𝐴 + 15 cos( 𝐴 1 − 2 cos( 𝐴 + cos? 𝐴− 1 − 3 cos( 𝐴 + 3 cos? 𝐴 − cosP 𝐴
= cosP 𝐴 − 15 cos? 𝐴 + 15 cosP 𝐴 + 15 cos( 𝐴 − 30 cos? 𝐴 + 15 cosP 𝐴 − 1+ 3 cos( 𝐴 − 3 cos? 𝐴 + cosP 𝐴
∴ cos 6𝐴 = 32 cosP 𝐴 − 48 cos? 𝐴 + 18 cos( 𝐴 − 1𝑄𝐸𝐷
Now,set𝑥 = cos𝐴,substituteandsolve;
32𝑥P − 48𝑥? + 18𝑥( =32
32 cosP 𝐴 − 48 cos? 𝐴 + 18 cos( 𝐴 =32
32 cosP 𝐴 − 48 cos? 𝐴 + 18 cos( 𝐴 − 1 =12
cos 6𝐴 =12
6𝐴 =𝜋3,
5𝜋3 ,
7𝜋3 ,
11𝜋3 ,
13𝜋3 ,
17𝜋3
𝐴 =𝜋18,
5𝜋18,
7𝜋18,
11𝜋18 ,
13𝜋18 ,
17𝜋18
andwehave;
𝑥 = cos𝐴 = cos𝜋18, cos
5𝜋18 , cos
7𝜋18 , cos
11𝜋18 , cos
13𝜋18 , cos
17𝜋18
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Chapter8B Euler’sformulaandIntegration
Recall:
General/CartesianForm 𝑧 = 𝑎 + 𝑏𝑖
PolarForm 𝑧 = 𝑟(cos 𝜃 + 𝑖 sin 𝜃) = 𝑟 cis 𝜃
IntroducingEULERwhotellsus(page333):
𝑒K[ = cis 𝜃
since 𝑧 = 𝑟 cis 𝜃 then 𝑧 = 𝑟𝑒K[
(youdon’treallyneedtounderstandthatseriesinthetextbookonpage333!,thisisjustanotherwayofdisplayingacomplexnumber)
sonowwehaveEULERForm 𝑧 = 𝑟𝑒K[
Hence:
𝑟4 = 𝑎 + 𝑏𝑖 4 = r]cos 𝑛𝜃 + 𝑖𝑟4 sin 𝑛𝜃 = 𝑟4 cis 𝑛𝜃 = 𝑟4𝑒K4[
Now,hereiswhereEulercanhelpwithintegration…
𝑓 𝑥 sin 𝑎𝑥 𝑑𝑥 = 𝐼𝑚 𝑓 𝑥 𝑒K`a 𝑑𝑥
𝑓 𝑥 cos 𝑎𝑥 𝑑𝑥 = 𝑅𝑒 𝑓 𝑥 𝑒K`a 𝑑𝑥
Noneedtobeabletoprovewheretheseformulacamefrom…justknowthem!
Howaboutthis…Sinehasan‘i’init,soiftheintegrandhasSineinit,thesolutionisthe’i’maginarypartoftheintegral?
Butfirst,letsjustdosomeconvertingbetweenforms:
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Firstthingisfirst…gettingintoandoutofEulerForm.
FromEulerform…itjust‘pops’out
Eg.Q1a
𝑒(bK
becomes
cis 2𝜋
becomes
cos 2𝜋 + 𝑖 sin 2𝜋
becomes
1 + 0𝑖
so
𝑒(bK = 1
whatafantasticequationthatis…itincorporatessomany“identities”…e,𝜋,𝑖and1…andtheyareallinarelationshiptogether…awesome!
Similarly…
𝑒bK = −1
GofromGeneral,intoPolar,andthenfromPolar,toEuler…!
Eg.Q2a
−1 + 𝑖
becomes
2 cis3𝜋4 𝜃
becomes
2𝑒cde K
so
−1 + 𝑖 = 2𝑒cde K
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ThetrickinIntegratingispracticingsoyougetusedtowhentoswapfromoneformtoanother…Thisisacondensedsettingouttofititononepage…
Eg.Q4a
𝑒a sin 2𝑥 𝑑𝑥
thetriggoestoEulerform = 𝐼𝑚 𝑒a𝑒(Ka 𝑑𝑥
thenjointhebases = 𝐼𝑚 𝑒af(Ka 𝑑𝑥
thenfactorisethatindex = 𝐼𝑚 𝑒(.f(K)a 𝑑𝑥
nowitseasytoIntegrate = 𝐼𝑚 g(Ihij)k
.f(K
rationalisedenominator = 𝐼𝑚 gIhij k(.-(K)
Q
simplifyandexpandtheindexportion = 𝐼𝑚 .Q𝑒af(Ka 1 − 2𝑖
andexpandfurthertoget = 𝐼𝑚 .Q𝑒a𝑒(Ka 1 − 2𝑖
useEulertogetridofoneofthee’s = 𝐼𝑚 .Q𝑒a cos 2𝑥 + 𝑖 sin 2𝑥 1 − 2𝑖
expand = 𝐼𝑚 .Q𝑒a cos 2𝑥 + 2 sin 2𝑥 + sin 2𝑥 − 2cos 2𝑥 𝑖
= 𝐼𝑚 .Q𝑒a cos 2𝑥 + 2 sin 2𝑥 + .
Q𝑒a sin 2𝑥 − 2cos 2𝑥 𝑖
extracteitherReorImasneededtogetyourresult
𝑒a sin 2𝑥 𝑑𝑥 = 15 𝑒
a sin 2𝑥 − 2cos 2𝑥 + 𝐶
ThisgetsaquickeranswerthanwhatwedidlastTerm…?Orelse,wejustlikedoingthesamethingdifferentwaysJ
TASK:Dothisagainandprovetoyourselfthatitdoesn’tmatterwhichwayyouarrangetheindicies‘above’𝑒.Thatisfromline4fromabove,solvefor;
𝑒((Kf.)a 𝑑𝑥
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eg.Q4d
𝑒-a cos 2𝑥 𝑑𝑥
= 𝑅𝑒 𝑒-a𝑒K(a 𝑑𝑥
= 𝑅𝑒 𝑒(aK-a 𝑑𝑥
= 𝑅𝑒 𝑒((K-.)a 𝑑𝑥
= 𝑅𝑒 𝑒((K-.)a
2𝑖 − 1
= 𝑅𝑒 𝑒(Ka𝑒-a
2𝑖 − 1 ×2𝑖 + 12𝑖 + 1
= 𝑅𝑒 𝑒(Ka𝑒-a 2𝑖 + 1
−5
= 𝑅𝑒 −15 𝑒-a 𝑒(Ka2𝑖 + 𝑒(Ka
= 𝑅𝑒 −15 𝑒-a cis 2𝑥 ×2𝑖 + cis 2𝑥
= 𝑅𝑒 −15 𝑒-a cos 2𝑥 + 𝑖 sin 2𝑥 2𝑖 + cos 2𝑥 + 𝑖 sin 2𝑥
= 𝑅𝑒 −15 𝑒-a 2 cos 2𝑥 𝑖 − 2 sin 2𝑥 + cos 2𝑥 + 𝑖 sin 2𝑥
= 𝑅𝑒 −15 𝑒-a −2 sin 2𝑥 + cos 2𝑥 + (2 cos 2𝑥 + sin 2𝑥)𝑖
=−15 𝑒-a cos 2𝑥 − 2 sin 2𝑥 + 𝐶
…or
𝑒-a cos 2𝑥 𝑑𝑥 = 15 𝑒
-a 2 sin 2𝑥 − cos 2𝑥 + 𝐶
HowawesomeisQuestion5…tyingtogetherthemultipleangletriglawsfromlastterm,andlinkingittothesecurrentmultipleanglerules!
EVERYTHINGinmathsagreeswitheachother…J
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Chapter8C
Idon’tthinkso
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Chapter8D DampedFunctions
***Iwon’tbeaskingaboutLimits,buttheyaren’tthathardtoworkoutanyway!
DampedfunctionsareVERYsimilartoalltheperiodicfunctionsyouhavelearntinMathsB…allthesameprinciplesapply.
Thebookhasthentaketheform:
𝑦 = 𝑒`a sin 𝑏𝑥
butmypreferenceisforthisform:
𝑦 = 𝐴𝑒`a sin 𝑏𝑥
***MytaskwillshowyouwhyIpreferthisform!***
Youneedtobeabletofindminimumandmaximumvaluesforanygivendomain,soensureyourecallco-terminalanglesandgeneralminima/maximascenariosthatwehavedonethroughALLofyourlearningsofar(MathsB&C).
Don’tbefooled…thenewintegrationrulesinthischapteraretheSAMEasinChapter8B…youdoNOThavetorememberthesespecificintegrationforms,asyoucanalwaysjustworkfromthegeneralformfromChapter8B…
𝑒`a cos 𝑏𝑥 = 𝑅𝑒 𝑒a `fnK 𝑑𝑥
𝑒`a sin 𝑏𝑥 = 𝐼𝑚 𝑒a `fnK 𝑑𝑥
InfactdoNOTremembertheseas“rules”.Ifaskedinyourexam,pleaseworkfromFirstprinciples,thatis,followtheprocessfromChapter8Bandmanuallyconvertthetrigterm,intoEulerform,manuallycombinethelike𝑒terms,andfactorisetheexponentandthatgetsyoutotheaboverules.
PanelwillbeSoooooimpressedwithhowwellIhavetaughtyou…Imeantheywillbeimpressedwithhowthoroughlyyouknowthecurriculum…J
IhaveNOIdeawhywearenowintroducingNewtitlesforderivatives…???…Ihaveneverseenthesebefore,but‘whatever’…J
𝑦 = 𝑦o =𝑑𝑦𝑑𝑡
𝑦 = 𝑦oo =𝑑(𝑦𝑑𝑡(
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Ex8DQ3ai
𝑒-a sin 𝑥 𝑑𝑥q
r
= 𝐼𝑚 𝑒-a𝑒Ka𝑑𝑥q
r
= 𝐼𝑚 𝑒(-.fK)a𝑑𝑥q
r
= 𝐼𝑚𝑒(-.fK)a
−1 + 𝑖 r
q
= 𝐼𝑚𝑒-a(cos 𝑥 + 𝑖 sin 𝑥)
−1 + 𝑖 ×−1 − 𝑖−1 − 𝑖 r
q
= 𝐼𝑚𝑒-a(− cos 𝑥 − 𝑖 cos 𝑥 − 𝑖 sin 𝑥 + sin 𝑥)
2 r
q
= 𝑒-a(− cos 𝑥 − sin 𝑥)
2 r
q
= −12 𝑒-a(cos 𝑥 + sin 𝑥)
r
q
= −12 𝑒-q(cos 𝐶 + sin 𝐶) −
−12 𝑒r(cos 0 + sin 0)
=−12 𝑒-q cos 𝐶 + sin 𝐶 +
12
=12 1 − 𝑒-q sin 𝐶 + cos 𝐶
TASK:Determinethedampedfunction,where𝐵 = 1,thatgoesthroughpointsb(, 9.620954762 and >b
(, −222.635557 .
Trytogetitintheform𝑦 = 𝑒`a sin 𝑥
Thentry𝑦 = 𝐴𝑒`a sin 𝑥
Howdidyougo…thefunctionwas𝑦 = 2𝑒a sin 𝑥…canyouthinkofwhyitishardwithoutthe“A”?
TASK:Makeupsomeofyourownquestionslikethisforpractice.Ifyouaren’tcareful,youwillmakeupaquestionthatisNotsolvable…J
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Chapter8E RealLifeApplications
Dampedfunctionshaveapplicationsinmanydifferentthings…carsuspensions,bridges,swings...Icould‘contrive’manymodelstofitthis,soitisjustaboutlookingatthecontextofthequestion,andapplyingthesameprocesstoanysituation.
Electricalcircuitsalternate(wehavea240ACsystem),sowehavethisperiodicnatureofelectricitythatcanbemodeledbytrigfunctions,sowewillincludeelectricalcircuitsinthischapter.
Physicsmayhaveyouuptospeed,butifnot,thinkaboutgettingabattery.ThebiggertheChargeofthebattery,thenthebiggertheCurrentitputsout.Sothereisarelationshipbetweenthem.
Charge=𝑞 Current=𝐼
Justasthegradientofadistancegraphmapstovelocity,
Similarly,theGradientofaChargegraph,givesCurrent.
SotheIntegralofaCurrentgraph,worksbackwardtoobtaintheCharge.
Let’ssaytheCURRENTcanbemodeledby 𝐼 = 𝐴𝑒`w cos 𝑏𝑡
IfwewanttofindtheChargeatanytime(t),wecouldsay;
𝐼 = 𝑞o = 𝐴𝑒`w cos 𝑏𝑡
So,
𝑞 = 𝐼 𝑑𝑡 = 𝐴𝑒`w cos 𝑏𝑡 𝑑𝑡
ThatIntegrallooksfamiliar(IhopesoanywayJ).Soinreality,thischapterdoesnotintroduceanythingnew,itismerelyapplying‘current’knowledge.(didyoulikemypunthere?)
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Chapter8F Malthusian
Don’tbefooled…thisisjustafancynameforwhatwehavedoneinMathsB…J
Iftherateofpopulationgrowthisdirectlyproportionaltothepopulation,thenwecansay:
𝑑𝑃𝑑𝑡 ∝ 𝑃
then,
𝑑𝑃𝑑𝑡 = 𝑘𝑃
ButnowweareinMathsC,weneedtoIntegratetoarriveatthePopulationfunction…!
**Wedidthisinthelastchapter,somovethroughquicklyasitgetsalotharder!)
Viaalgebra,
.{𝑑𝑃 = 𝑘𝑑𝑡
integratebothsides
1𝑃 𝑑𝑃 = 𝑘𝑑𝑡
ln 𝑃 = 𝑘𝑡 + 𝐶
𝑃 = 𝑒}wfq
𝑃 = 𝑒}w×𝑒q
andaswhen𝑡 = 0,𝑃 = 𝑥q ,weget,
𝑃 𝑡 = 𝑃r𝑒}w
Thischaptershouldn’tposetoomuchtrouble…!ButyoumustbeabletoshowtheIntegrationtoarriveatthepopulationfunction!!
Somemodelswilldifferslightly,butitisjustaboutfollowingtheprocess,andworkingoutthedifferentIntegrationtocomeupwiththecorrectmodel.
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Eg.Q2
𝑑𝑁𝑑𝑡 = 𝑘(90 − 𝑁)
190 − 𝑁 𝑑𝑁 = 𝑘𝑑𝑡
190 − 𝑁 𝑑𝑁 = 𝑘𝑑𝑡
−−1
90 − 𝑁 𝑑𝑁 = 𝑘𝑑𝑡
− ln(90 − 𝑁) = 𝑘𝑡
− ln 90 − 𝑁 = −𝑘𝑡 + 𝐶
90 − 𝑁 = 𝑒-}wfq
𝑁 = 90 − 𝐾𝑒-}w
at𝑡 = 0, 𝑁 = 20
20 = 90 − 𝐾, 𝑠𝑜𝐾 = 70
at𝑡 = 10,𝑁 = 40
40 = 90 − 70𝑒-.r}
𝑒-.r} =5070
𝑘 = −ln Q
�10
𝑁 𝑡 = 90 − 70𝑒II� �]
��w
a)set𝑡 = 25
𝑁 25 = 90 − 70𝑒II� �]
��(Q = 59.81591947
b)tofind70%potential,mustfindmaxpotential,so,set𝑡 = ∞
𝑁 ∞ = 90 − 70𝑒II� �]
��� = 90
Becauseln Q�isnegative,as𝑡getsbigger,𝑒}wgetssmaller,andapproacheszeroat
thelimitof𝑡 → ∞…thus,70%potentialis63units…thus,solvefortasfollows
63 = 90 − 70𝑒II� �]
��w
𝑒II� �]
��w =
2770
110 ln
57 𝑡 = ln
2770
𝑡 = 28.31313471𝑑𝑎𝑦𝑠
16
Chapter8G Verhulst/LogisticModel
WelearntinMathsBthatpopulationstendtogrowinanexponentialway,butthisisclearlyNOTthecase.(theassignmentquestionaboutwhentherewillonlybestandingroomontheplanetisclearly‘silly’!)
Therearemanydifferentpopulationgrowthmodelsandthischapterinvestigatesjustoneofthem;aLogisticModel.
Commonsensetellsusthattherateofchangeofpopulationisinfluencedbytheactualpopulation.Aswehavemorepotentialtobreed,ourrateofgrowthincreases,however,theresimplyhastobealimittohowbigthepopulationgets(sufficientfood,air,room).So,aswegetcloserandclosertoourpopulationlimit(carryingcapacity),therateofgrowthactuallystartstoslow.Forthismodel,wesay:
𝑑𝑁𝑑𝑡 = 𝑎𝑁 − 𝑏𝑁(
andweIntegrate(referpage370)toget;
𝑁 𝑡 =`n𝑁r
𝑁r +`n− 𝑁r 𝑒-`w
Note:Thelimitingpopulationisdefinedby:`n
Note:Themaximalpopulation(whenpopulationisincreasingatitsmaximumrate),isdefinedby: `
(n
Beclearonthisterminologyandrememberthese‘Notes’!
Thischapterhasarangeofskills…:
Straightrecall,byimplementingtheLogisticequationdirectly,
Verifyingthisequationusingdifferentiation(verificationisinyourrubric…J)
Usingcomplexintegrationtechniquesandlog’stoarriveattheLogisticequation
Howaboutrefiningamathematicalmodel,fromaMalthusiantoaVerhulstmodel.
17
Eg.Q1a
Given ���w= 0.2𝑁 − 0.0004𝑁(
𝑎 = 0.2,𝑏 = 0.0004,𝑁r = 100
𝑁 𝑡 =`n𝑁r
𝑁r +`n− 𝑁r 𝑒-`w
becomes,
𝑁 𝑡 =r.(
r.rrr?100
100 + r.(r.rrr?
− 100 𝑒-r.(w
𝑁 𝑡 =50000
100 + 400𝑒-r.(w
𝑁 𝑡 =50000
100 1 + 4𝑒-r.(w
andwehave,
𝑁 𝑡 =500
1 + 4𝑒-r.(w
Fromherewecanfindthepopulationatanytime𝑡.
Itmaybebeneficialtorevisitthischapterfromthebeginningagainandtryandunderstandwhatishappening.
Rateofchange(judgedagainsttime)isaffectedbythepopulation.Howevertheinitialrelationshipdoesnotgiveusafunctionoftime…!...(it’sjustadifferentialequation)
Weareunabletofindthepopulationortherateofchangeofpopulationgivenatimeperiod…!Thatdoestendtolimitourmodel.Soweneeda‘t’intheresomewhere?
OncewegetthePopulationfunction,weknowtherelationshipbetweenPopulationandTime,wecannotonlyfindthepopulationatanytime𝑡,butwecanalsofindtherateofchangeofpopulation(populationgrowth)atanytime𝑡also!
LetusproceedtoVERIFYourabovesolutionthroughdifferentiation…
18
Eg.2b
Task:Prove ���w= 0.2𝑁 − 0.0004𝑁(
Thisisalittlelikeproofbyinductiontechnique,notintheprocess,butifyouhavedoneitafewtimes,youknowwhichformeachsideshouldtake,thatis,theaimistogetthefirstsectiontoasingledenominator(youwillsee)…
Thatis,asweknow, 𝑁 = Qrr.f?gJ�.i�
wecansubitintothegivenequation
Weneedtoprove;
𝑑𝑁𝑑𝑡 = 0.2
5001 + 4𝑒-r.(w − 0.0004
5001 + 4𝑒-r.(w
(
𝑑𝑁𝑑𝑡 =
1001 + 4𝑒-r.(w −
1001 + 4𝑒-r.(w (
needcommondenominatortoaddfractions,
𝑑𝑁𝑑𝑡 =
1001 + 4𝑒-r.(w ×
1 + 4𝑒-r.(w
1 + 4𝑒-r.(w − 100
1 + 4𝑒-r.(w (
𝑑𝑁𝑑𝑡 =
100 1 + 4𝑒-r.(w − 1001 + 4𝑒-r.(w (
𝑑𝑁𝑑𝑡 =
100 + 400𝑒-r.(w − 1001 + 4𝑒-r.(w (
So,simplified,weneedtoprove;
𝒅𝑵𝒅𝒕 =
𝟒𝟎𝟎𝒆-𝟎.𝟐𝒕
𝟏 + 𝟒𝒆-𝟎.𝟐𝒕 𝟐
**ThisisourTarget**
Now,backtothegivenfunctiontodifferentiateit…
𝑁 𝑡 =500
1 + 4𝑒-r.(w
𝑁 𝑡 = 500 1 + 4𝑒-r.(w -.
ViatheChainrule,Set;
𝑁 = 500𝑢-. → 𝑑𝑁𝑑𝑢 = −
500𝑢(
where,
𝑢 = 1 + 4𝑒-r.(w → 𝑑𝑢𝑑𝑡 =
−4𝑒-r.(w
5
19
Now,
𝑑𝑁𝑑𝑡 =
𝑑𝑁𝑑𝑢 .
𝑑𝑢𝑑𝑡
𝑑𝑁𝑑𝑡 = −
500𝑢( ×−
4𝑒-r.(w
5
𝒅𝑵𝒅𝒕 =
𝟒𝟎𝟎𝒆-𝟎.𝟐𝒕
𝟏 + 𝟒𝒆-𝟎.𝟐𝒕 𝟐
Q.E.D.
So,justlikeinproofbyinduction,youneedtohaveaninitialaimforyourfirstexpression…Thisisbecauseitisverydifficulttoworkbackwardsfromthatlastline,togetallthewaybacktoourinitialequationJ
20
Nexttogetthefunctionof𝑡throughintegration:
Q1c
𝑑𝑁𝑑𝑡 = 0.2𝑁 − 0.0004𝑁(
𝑑𝑁𝑑𝑡 = 0.0004𝑁(500 − 𝑁)
10.0004𝑁(500 − 𝑁) 𝑑𝑁 = 𝑑𝑡
2500𝑁(500 − 𝑁) 𝑑𝑁 = 𝑑𝑡
beforeweIntegratebothsides,wewillneedtousethepartialfractionstechniqueontheLHS.Iwillskipthisprocessandgetyoutoreviseseparately.
5𝑁 +
5500 − 𝑁 𝑑𝑁 = 1𝑑𝑡
5𝑁 +
5500 − 𝑁 𝑑𝑁 = 1 𝑑𝑡
5𝑁 𝑑𝑁 +
5500 − 𝑁 𝑑𝑁 = 1 𝑑𝑡
51𝑁 𝑑𝑁 − 5
−1500 − 𝑁 𝑑𝑁 = 1 𝑑𝑡
5ln𝑁 − 5 ln 500 − 𝑁 = 𝑡 + 𝐶
5 ln𝑁 − ln 500 − 𝑁 = 𝑡 + 𝐶
5 ln𝑁
500 − 𝑁 = 𝑡 + 𝐶
ln𝑁
500 − 𝑁 = 0.2𝑡 + 𝐶
𝑒r.(wfq =𝑁
500 − 𝑁
𝑒q𝑒r.(w =𝑁
500 − 𝑁
weknow,at𝑡 = 0,𝑁 = 100,
𝑒q𝑒r.(×r =100
500 − 100
𝑒q =14
So,
21
14 𝑒
r.(w =𝑁
500 − 𝑁
𝑒r.(w =4𝑁
500 − 𝑁
500𝑒r.(w − 𝑁𝑒r.(w = 4𝑁
4𝑁 + 𝑁𝑒r.(w = 500𝑒r.(w
𝑁(4 + 𝑒r.(w) = 500𝑒r.(w
𝑁 =500𝑒r.(w
4 + 𝑒r.(w
𝑁 =500𝑒r.(w
4 + 𝑒r.(w ×𝑒-r.(w
𝑒-r.(w
𝑁 =500
4𝑒-r.(w + 1
𝑁 𝑡 =500
1 + 4𝑒-r.(w
Q.E.D.
Nicehey…!!!
Don’tforgetourLimittheory.Inthesemodelstheimplementationof‘limits’won’tbedifficult,asthelogisticmodelreachesahorizontalasymptoteas𝑥 → ∞