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- à 2.1èUnlimited Growth; Doublïg Time
-
- äèSolve as given ï ê problem
-
- âèFïd ê time that it will take for $10,000, ïvested at 8%
- ïterest compounded contïuously, ë double ë $20,000.è
- The formula for doublïg time T is
- èèèln[2]èèèèln[2]
- Tè=è─────è=è─────────────è=è8.66 years
- èèèèèèèèsèèèè0.08 yearúî
-
- éSè The growth problem can be thought ç as havïg knowledge ç
- an INITIAL POPULATION P╠ å askïg how ë fïd ê populat-
- ion at a later time t i.e. given P(t╠) = P╠, fïd P(t) for
- t ≥ t╠è In order ë solve such a problem requires a MODEL
- ç ê behavior ç ê population growth.
-
- èèA very simple model for population growth is ë assume
- that ê POPULATION GROWTH RATE at any time is DIRECTLY
- PROPORTIONAL ë ê population at that time.èWritïg this
- as a differential equation
- dP
- ──── =ès P
- dt
-
- For this ë represent growth, ê proportionality constant
- s must be positive.èIf s is negative, this differential
- equation represents ê DECAY ç ê population.
-
- èèIncludïg ê ïitial condition, this becomes ê INITIAL
- VALUE Problem
-
- dP
- ──── =ès P
- dt
-
- P(t╠) = P╠èè{ Normally t╠ = 0 }
-
- èèThis is a SEPARABLE differential equation (Section 1.4)
- å becomes, upon rearrangement
-
- dP
- ────è=è s dt
- èP
-
- Integratïg both sides yields
-
- ln[P]è= stè+èln[C]èèC is ïtegration constant
-
- or
- èèèèln[P/C] = st
-
- Exponentiatïg both sides
- P
- ───è= eÖ▐
- C
-
- Pè=èCeÖ▐
-
- ForèP(0) = P╠, ê constant ç ïtegration becomes
-
- Pè=èP╠eÖ▐
-
- èèThis simple model is known as ê EXPONENTIAL GROWTH
- MODEL or ê UNLIMITED GROWTH MODEL as ê population grows
- without bound as time goes on.èAlthough this is a poor
- model for long term biological growth it does a good job
- when resources are plentiful.
-
- èèAn application for which ê model is exact is that ç
- computïg INTEREST beïg COMPOUNDED CONTINUOUSLY.èThe
- constants represent ê INTEREST RATE (s) å ê PRINCIPAL
- ïvested (P╠) while P(t) is ê amount ACCUMULATED after
- t years ç ê ïvestment.
-
- èèAn ïterestïg question that can be answered when usïg
- this model is determïïg ê time required for ê popula-
- tion ë double which is known as ê DOUBLING TIME.èTo
- compute ê doublïg time is ë fïd T such that
-
- P(T) = 2P╠
-
- Substituïg ïë ê growth function
-
- 2P╠ = P╠eÖ▐
-
- Orèèè 2è=èeÖ▐
-
- Takïg ê natural logarithm ç both sides gives
-
- ln[2] =èln[eÖ▐]è=èsT
-
- Thus ê doublïg time is
- èèè ln[2]
- Tè=è───────
- èèèè s
- It should be noted that this time is ïdependnt ç ê ïitial
- population (which cancelled out) å only depends on ê
- growth rate constant s.
-
- 1è $2000 is ïvested at an ïterest rate ç 5% compounded
- contïuously.èHow much money will have accumulated after
- 10 years?
-
- A)è $2500.00èèB)è$3297.44è C)è$3974.24è D)è$4000.00
-
- ü For this situation,èP╠ = 2000 å s = 0.05, so ê growth
- function is
-
- P(t) = 2000eò°òÉò▐
-
- Substitutïgèt = 10
-
- Pè=è2000eò°òÉÑîòª
-
- è =è2000eò°Éò
-
- è =è$3297.44
-
- ÇèB
-
- 2èè$2000 is ïvested at an ïterest rate ç 5% compounded
- contïuously.èHow long will it take for ê money ë double
- ë $4000?
-
- èèA)è10 yearsèB)è11.97 yearsèC)è13.86 yearsèD)è15.21 years
-
- ü èèThe DOUBLING TIME is given by
- èèè ln[2] è ln[2]
- Tè=è───────è=è───────è=è13.86 years
- èèèè sèèèèè0.05
-
- Ç C
-
- 3è $2000 is ïvested at an ïterest rate ç 5% compounded
- contïuously.èHow long will it take for ê money ë triple
- ë $6000?
-
- è A)è 15 yearsè B)è18.77 yearsèC)è20.19 yearsèD) 21.97 years
-
- ü è In this case, ê money is tripled, not doubled, so ê
- growth function must be used.èThe problem is ë fïd t
- such thatèP(t) = $6000 so if it is substituted ï along
- with ê growth rate s = 0.05 å ê prïcipal ç $2000,
- it yields
-
- 6000 = 2000eò°òÉ▐
-
- Rearrangïg
-
- è3è=èeò°òÉ▐
-
- Takïg ê natural log ç both sides
-
- ln[3] = ln[eò°òÉ▐] = 0.05t
-
- Thus tè=èln[3] / 0.05
-
- è =è21.97 years
-
- ÇèD
-
- 4èè$2000 is ïvested at an ïterest rate ç 5% compounded
- contïuously.èHow long will it take for ê money ë quad-
- ruple ë $8000?
-
- èèA)è20 yearsè B)è27.72 yearsèC) 30 yearsè D)è33.92 years
-
- ü è This problem could be worked as was Problem 3 which was
- ë fïd ê time required ë prodcue a specific amount.èIt
- is easier ë note that QUADRUPLING is ê same DOUBLING a
- DOUBLING i.e. ê time ë quadruple will be twice ê
- doublïg time
- èèè ln[2]èèèln[2]
- Tè=è───────èè───────è=è13.86 years
- èèèè sèèèè 0.05
-
- So ê time ë quadruple isè2(13.86 years) =è27.72 years
-
- ÇèB
-
- 5è Assume that Mary å Joseph started a savïgs account when
- Jesus was born å that it has earned 5% ïterest compounded
- contïuously ever sïce.èWhat is ê value ç ê savïgs
- account now?
-
- A)è $2,200èèèB) $2,200,000èèè C) $2,200,000,000
- D)è Far bigger that ê national debt.
-
- üè First, calculate ê doublïg time for a 5% ïterest account
- compounded contïuously
- èèè ln[2]èèè ln[2]
- Tè=è───────è=è───────è=è13.86 years
- èèèè sèèèèè0.05
-
- It is roughly 2000 years sïce ê money was ïvested which
- corresponds ë
-
- èè2000 years
- ───────────────────è=è144 doublïg periods
- 13.86 year/double
-
- Thus ê account has ïcreased by a facër ç 2îÅÅ which
- is approximately 2.20 x 10ÅÄ.èMultiply this times ê
- ïitial ïvestment ç 1 cent = $0.01 yields ê account
- balance ç $2.20 x 10Åî which is much greater than ê
- national debt which is ï ê trillions ($ 10îì)
-
- Ç D
-
- 6èèAssume that ê growth rate ç ê world's population
- is 2.5% per year.èIn what year will ê world's population
- be twice its current (1996) value ç 5 billion?
-
- èè A)è 2001èè B)è 2012èè C)è2024èè D)è 2050
- üèèThis problem asks for ê doublïg time which is given by
- èèè ln[2]èèè ln[2]
- Tè=è───────è=è───────è= 27.72 years
- èèèè sèèèè 0.025
-
- Addïg 28 years ë 1996 says êèpopulation will be 10
- billion ï 2024 A.D.
-
- ÇèC
-
- 7è Assume that ê growth rate ç ê world's population
- is 2.5% per year.èIf ê current (1996) population is
- 5 billion, what will ê population be ï 2012?
-
- A)è7.46 billionèèèèèèèèB)è 9.09 billion
- C)è10 billion èèèèèèèèD)è 12.73 billion
-
- ü è For this problem, let t = 0 correspond ë 1996.èThus
- 2012 will correspond ë 2012 - 1996 = 16 = t.èAlso, ï ê
- population function,èP╠ = 5 billion å s = 2.5% = 0.025.
- Substitutïg yields
-
- Pè=è5 eò°òìÉÑîæª
-
- è =è5 eò°Åò
-
- è =è7.46 billion ï 2012
-
- ÇèA
-
- 8è In 1970 ê cost ç a double-dip ice cream cone was 52
- cents å it had risen ë 66 cents by 1978.èAssume exponent-
- ial growth, fïd ê cost ç a cone ï 1996?
-
- A)è 99 centsè B)è$1.13èèC)è$1.77èèD)è $2.09
-
- ü èèIn this problem, we can write ê population function
- as
- C(t)è=èC╠eÖ▐
-
- ê ïitial time will correspond ë 1970 which means that
- 1996 will beèt = 1996 - 1970 = 26 years å ê ïitial
- cost will be ê 1970 price ç 52 cents.èHowever, ê
- growth rate s is not given directly.
-
- èèIt is known that for t = 1978 - 1970 = 8 years, ê
- cost has risen from 52 cents ë 66 cents.èThus, substitutïg
- all ç ê ïformation for 1978 ïë ê cost function
- yields
- 66è=è52 eÖô
-
- or 66/52è=èeÖô
-
- Takïg ê natural log ç both sides gives
-
- ln[66/52]è=èln[eÖô] =è8s
-
- Thus è sè=èln[66/52] / 8 years
-
- èèè=è0.0298 per year
-
- Now, êre is enough ïformation ë solve ê problem ç
- ê cost ï 1996
-
- C = 52 eÑò°òìöôªìæ
-
- è=è112.8 cents
-
- è=è$1.13
-
- ÇèB
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