Rates of Growth & Decay

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Example (1) - a

The size of a colony of bacteria was 100 million at 12 am and 200 million at 3am. Assuming that the relative rate of increase of the colony at any moment is directly proportional to its size ( the rate of the growth of the bacteria population is constant), find:1. The size of the colony at 3pm.2. The time it takes the colony to quadruple in size.3. Find the (absolute) growth rate function4. How fast the size of the colony was growing at 12 noon.

Solution

8656

3

156

36

3)2ln(33

1

3

1

6

63

)3(66

6

6

6

)10(32)10(3200)2()10(100

)15:3()2()10(100)15(.1

)2()10(100)(

)2()2ln()2ln(

)2ln(2ln3

12ln32

)10(100

)10(200

)10(100)3()10(200

&)10(100)(:

)3:3()10(200)3(

)12()10(100)0(

)0()(

)()(

3

hoursafterpmaty

ty

eetkt

kke

ey

etysoand

hoursafteramatyand

amatyhaveWe

eyty

hoursinttimeaatbacteriaofnumberthebetyLet

t

tkt

t

k

k

kt

kt

t

hourbacteriaydt

dy

hourbacteriatydt

dy

tyhaveWe

hoursTT

yyyTy

Then

quadrupletosizecolonytheforneededtimethebeTLet

t

tt

t

T

T

/3

2ln)10(16)16(

3

2ln)10()2(

3

2ln)10()12(.4

/)2(3

2ln)10()

3

1(2ln)2()10()(

)2()10()(:.3

623

24)2(

4)2(4)(

:

.2

883

128

12

38

38

38

4423

03

004

4

4

4

Example (1) - b

The population of a town was 100 thousands in the year 1997 and 200 thousands in the year 2000. Assuming that the rate of increase of the population at any moment is directly proportional to its size ( the relative rate of the growth of the population is constant), find:1. The population of the town in 2012.2. The number of years it takes for the population to quadruple. 3. Find the (absolute) growth rate function4. How fast the population was growing in 2012.

Solution

3200000)10(32)10(3200)2()10(100

)15:2012()2()10(100)15(.1

)2()10(100)(

)2()2ln()2ln(

)2ln(2ln3

12ln32

)10(100

)10(200

)10(100)3()10(200

&)10(100)(:

)3:2000()10(200)3(

)1997()10(100)0(

)0()(

)()(

5353

3

153

33

3)2ln(33

1

3

1

3

33

)3(33

3

3

3

3

yearsafteriny

ty

eetkt

kke

ey

etysoand

yearsafterinyand

inyhaveWe

eyty

yearsinttimeaattowntheofnumberthebetyLet

t

tkt

t

k

k

kt

kt

t

Solution

yearpersonydt

dy

yearpersontydt

dy

tyhaveWe

yearsTT

yyyTy

ThenquadrupletopopulationtheforneededtimethebeTLet

t

tt

t

T

T

/3

2ln)10(16)16(

3

2ln)10()2(

3

2ln)10()12(.4

/)2(3

2ln)10()

3

1(2ln)2()10()(

)2()10()(:.3

623

24)2(

4)2(4)(

:..2

553

125

12

35

35

35

4423

03

004

4

4

4

Example (1) – a*

The size of a colony of bacteria was 100 million at 12 am and 400 million at 2am. Assuming that the rate of increase of the colony at any moment is directly proportional to its size, find:1. The size of the colony at 1am.2. The time it takes the colony to reach 800 million.

Solution

866

2

16

26

2)4ln(22

1

2

1

6

62

)2(66

6

6

6

)10(2)10(200)2()10(100

)1:1()4()10(100)1(.1

)4()10(100)(

)4()4()4ln(

)4ln(4ln2

14ln24

)10(100

)10(400

)10(100)2()10(400

&)10(100)(:

)2:2()10(400)2(

)12()10(100)0(

)0()(

)()(

2

hourafteramaty

ty

eetkt

kke

ey

etysoand

hoursafteramatyand

amatyhaveWe

eyty

hoursinttimeaatbacteriaofnumberthebetyLet

t

tkt

t

k

k

kt

kt

t

hoursT

Ty

Then

reachtocolonytheforneededtimethebeTLet

TT

T

322)4(8

)4()10(100)()10(800

:

)10(800.2

32

266

6

Example (1) - c

A loan shark lends money at an annual compound interest rate of 100%. An unfortunate man borrowed 1000 Riyals at the beginning of 2014. 1. How much will the loan reach in 2015 ( the end of 2014)?2. When will the loan reach reach 8000 Riyals?

Solution

)2017(38000

3

281000

8000)2()2(1000)(8000

8000

)2(1000)(

)2()2ln(2ln

2ln2)10(100

)10(200

&1000)1(2000.2

)2015(200010001000)1(%,1001

)2014(1000)0(

1000)0()(

)()(

3

)2ln(

6

6

)1(

byyearsinRiyalsreachwillloanThe

T

TyThen

RiyalsreachloanthewhichofendtheatyearsofnumberthebeTLet

ty

etkt

ke

ey

inythenisrateterstintheSince

inyhaveWe

eeyty

yearsinttimeaatloantheofsizethebetyLet

TT

t

tt

k

k

ktkt

t

Example (1) - d

The rate of consumption of energy for an industrial Gulf zone at the beginning of 2000 was 1000 megawatt. The rate was expected to increase at annual growth rate of 100%. Find:1. The function that gives he the rate of energy consumption.2. The total energy used during 2004.3. The function that gives the total cumulative amount of energy used between 2000 and any time t in years)

Solution

)(0;10001000)(

)2(0;)2(1000)(

)2()2ln(2ln

2ln2)10(100

)10(200

&1000)1(2000

)2001(200010001000)1(

%,100)(1

)2000(1000)0(

1000)0()(

)(

,)(

693.02ln

)2ln(

6

6

)1(

ebasewithteetyalsoOr

basewithtty

eetkt

ke

ey

inythen

ofrateannualtheatincreasestpSince

yearinyhaveWe

eeyty

yearsinttimeaat

usedenergyofratethedt

dEgivesthatfunctionthebetyLet

tt

t

tktt

k

k

ktkt

t

)1(693.0

1000

693.0

10001000

:

1)2(2ln

1000

2ln

1000

2ln

)2(1000

2ln

)2(1000)2(1000

:years)in(ttimeanyand2000beweenusedenergyeacumulativThe3.

)(migawatts14436 1442.

)1(693.0

1000

693.0

10001000

)10(2000

e)bsase(withformothertheusingOr,

migawatts14437 1442.7 1442.

2ln

1000)12(

2ln

1000

2ln

)2(1000)2(1000

)10(2000

)1000)(:()2(1000)(:.2

693.0

0

693.0

0

693.0

00

693.01

0

693.01

0

693.0

1

0

1

0

693.0

tttt

t

tt

ttt

t

tt

tt

tt

ee

dte

alsoOr

dt

answersame

ee

dte

ttotfromduringusedenergyThe

dt

ttotfromduringusedenergyThe

etydt

dEAlsoty

dt

dEhadWe

Example (2) - a

A mass of a radioactive element has decreased from 200 to 100 grams in 3 years. Assuming that the rate of decay at any moment is directly proportional to the mass( the relative rate of the decay of the element is constant), find:1. The mass remaining after another15 years.2. The time it takes the element to decay to a quarter of its original mass.3. The half-life of the element3. Find the (absolute) growth decay function4. How fast the mass was decaying at the twelfth year.

Solution

gy

ty

eetkt

kke

ey

etysoand

yand

yhaveWe

eyty

yearsinttimeamassthebetyLet

t

tkt

t

k

k

kt

kt

t

25.64

25)

32

1(200)

2

1(200)15(.1

)2

1(200)(

)2

1()

2

1ln(])

2

1ln([

)2

1ln(

2

1ln

3

1

2

1ln3

2

1

200

100

200)3(100

&200)(:

100)3(

200)0(

)0()(

)()(

3

15

3

3)

2

1ln(

33

1

3

13

)3(

6

3

yeargydt

dy

yeargtydt

dy

tyhaveWe

yearsTT

yyyTy

Then

sizeoriginaklitsofquartertodecaytomassthefor

neededtimethebeTLet

tyhaveWe

t

tt

t

T

T

t

/48

2ln200)

2

1(

3

2ln200)

2

1(

3

2ln200)12(.4

/)2

1(

3

2ln200

3

1)

2

1ln()

2

1(200)(

)2

1(200)(:.3

6232

1

4

1)

2

1(

4

1)

2

1(200

4

1)(

:

)2

1(200)(:.2

43

12

12

33

3

2

3

03

00

3

41

414

1

41

41

41

Example (2) – a*

A mass of a radioactive element has decreased from 1000 g to 999 grams in 8 years and 4 months. Assuming that the rate of decay at any moment is directly proportional to the mass( the relative rate of the decay of the element is constant), find the half-life of he element.

Solution

t

tkt

t

k

k

kt

ty

ee

ktk

ke

eysoand

yearyearmonthsandyearthatnote

yandyhaveWe

eyty

yearsinttimeatmassthebetyLet

t

25

3

25

3

100

99ln

25

3

25

3

)3

25(

)3

25(

1000

999)(

1000

999

1000

999ln0

100

999ln

1000

999ln

25

3

3

251000999

ln

1000

999ln)

3

25(

1000

999

1000)3

25(999:

3

25

3

1848(

999)3

25(1000)0(

)0()(

)()(

25

3

yearsT

T

yTyy

Then

sizeoriginaklitsofhalftodecaytomasstheforneededtimethebeTLet

ytyhaveWe

sizeitsofhalftodecayoelementtheofmassanytakesittimetheis

lifeHalf

T

T

t

5773

9991000

ln3

2ln25

1000999

ln253

21

ln

2

1ln

1000

999ln

25

3

2

1

1000

999

1000

999)(

2

1

:

1000

999)(:

:.2

21

21

21

21

21

21

25

3

25

3

00

25

3

0

Note

We can find the half-life T1/2 in terms of the constant k or the latter in terms of the former ( T1/2 = ln2/k Or k = ln2 / T1/2) and use that to find k when T1/2 is known or find T1/2 when k is known.

57733.5773)10(2006.1

2ln

1.2006(10)

2ln

1.2006(10) 999

1000ln

5

3

1000

999ln

5

3,

,

2ln21

ln

2

1ln

2

1)(

2

1

:

:

44-

4-

00

21

21

21

21

21

21

21

T

khadWe

itfindingafterformulathistoinksubstiuedhavecouldwe

problemlasttheinThuskk

T

kTeeyTyy

Then

sizeoriginalitsofhalftodecaytomasstheforneededtimethebeTLet

lifehalftheandknttaconsthebeweenprelaionshitheDeducing

kTkT

Deducing the Relationship Between half-life T1/2 and the constant k

Carbon Dating

• Carbon (radiocarbon) dating is a radiometric dating technique that uses the decay of carbon-14 (C-14 or 14C) to estimate the age of organic materials or fossilized organic materials, such as bones or wood.

• The decay of C-14 follows an exponential (decay) model.

• The time an amount of C-14 takes to decay by half is called the half-life of C-14 and it is equal about 5730 years. Measuring the the remaining proportion of C-14, in a fossilized bone, for example to the amount known to be in a live bone gives an estimation of its age.

Example (2) - b

It was determine that a discovered fossilized bone has 25% of the C-14 of a live bone. Knowing that the half-life of C-14 is approximately 5730 years and that its decay follows exponential model, estimate the age of the bone.

Solution:

)5730(000

2

1

0

0

2

1

2

1

,573014

)(,

14

kkT

kt

eyeyy

thenTCoflifehalftheSince

eytyThen

yboneliveainCofamounttheLet

yearst

ytyy

thenboneliveainyamounttheofhaditfoundwasbonethettimetheatSince

yeyeyty

tkt

kke

eyeyy

t

t

t

t

kt

t

k

kkT

t

11460)5730(25730

22

1

2

1

2

1

4

1

2

1)(

100

25

.,%25

2

1)(

2

1ln

2

1ln

2

1ln

573021

ln5730

2

1ln

2

1

2

1

57302

5730

5730

00

0

5730

02

1ln

00

57305730

1

5730

1

5730

)5730(000

5730

2

1

Pharmacokinetics

• Pharmacokinetics (PK) is a branch of pharmacology concerned with knowing what happens to substances ( such as drugs, food or toxins) administered to a living body.

• This includes understanding the process by which such substance is assimilated, eliminated or affected by the body.

• With some exceptions ( such as in the case of liquor) the absorption of drugs follows an exponential (decay) model.

Example (2) - c

Two doses of 32 mg of a drug with a half-life of 16 hours were administered to a patient. The second was administered 64 hours after the first.a. How long would it take the drug to reach 12.5% of its first dose.B. How long would it take the drug after the second dose to reach 8.5 mg

Solution

16

2ln2ln:

2

1)(

2

1ln

2

1ln

2

1ln

1621

ln16

2

1ln

2

1

2

1

,16

32)(:

32

)(

)()(

2

1

16

02

1ln

00

1616

1

16

1

)16(

)16(000

2

1

0

0

16

2

1

TkformulathengusibykfoundcouldWeNote

yeyeyty

tkt

kke

eyeyy

thenTdrugtheoflifehalftheSince

etysoand

yand

eyty

hoursinttimeabodytheindrugthebetyLet

t

kt

t

k

kkT

kt

kt

t

hoursTyTyy

amountoriginalitsofreachtodosefirstthetakesittimethebeTLet

yeyeyty

hadWe

a

TTT

t

kt

t

482

1

2

1

2

1

8

1

2

1)(

100

5.12

%5.12

2

1)(

:

.

163

1616

00

16

02

1ln

00

16

dosefirstthetakingfromhoursaftersThat

dosecondsethetakingfromhoursaftermgreachesdrugtheofamounttheThus

tt

tZ

thenmgreachtomomentthisfromdrugthetakesittimethebetLet

ZtZ

isthatafterttimeanyatpresentZdrugtheofamountThe

ZamountthisandttoheredingcorrresponmomentthisfromStarting

givenwasdosecondsetheafterpresentamountThe

mgy

hoursafterremainingstilldosefirsttheofamountThe

ty

haveWeb

ttt

t

tt

t

963264'

32:5.8,

3216

2

2

1

2

1

2

1

4

1

2

1

34

5.8

2

134)(5.8

,5.8)(

2

134

2

1)(

:)(

)34()0(

34322

22

132

2

132)64(

64

2

132)(

:.

11

162

1616

16

1

1

1616

0

0

416

64

16

111

1