1

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.)

2

AnotherrulethatcanhelpistheMultipleangleformula(pge330)

cos 𝑛𝜃 =𝑧4 + 𝑧-4

2

sin 𝑛𝜃 =𝑧4 − 𝑧-4

2𝑖

yesyoucanjustrememberthisthese…!

BothofthesegetusedinProofsaswell…asyouwillseeoverthepage!

3

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!

4

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).

5

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

6

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:

7

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

8

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

9

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

10

Chapter8C

Idon’tthinkso

11

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(

12

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

13

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?)

14

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.

15

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 → ∞