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

315

6

36

3)2ln(331

31

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(2ln312ln32

)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

hourbacteriaydtdy

hourbacteriatydtdy

tyhaveWe

hoursTT

yyyTy

ThenquadrupletosizecolonytheforneededtimethebeTLet

t

tt

t

T

T

/3

2ln)10(16)16(3

2ln)10()2(3

2ln)10()12(.4

/)2(3

2ln)10()31(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(2ln312ln32

)10(100)10(200

)10(100)3()10(200

&)10(100)(:

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

)1997()10(100)0(

)0()(

)()(

5353

315

3

33

3)2ln(331

31

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

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yearpersontydtdy

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()31(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

21

6

26

2)4ln(221

21

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(4ln214ln24

)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

ThenreachtocolonytheforneededtimethebeTLet

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(380003

2810008000)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(

byyearsinRiyalsreachwillloanTheT

TyThen

RiyalsreachloanthewhichofendtheatyearsofnumberthebeTLetty

etkt

ke

ey

inythenisrateterstintheSinceinyhaveWe

eeyty

yearsinttimeaatloantheofsizethebetyLet

TT

t

tt

k

k

ktkt

t

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

yandyhaveWeeyty

yearsinttimeamassthebetyLet

t

tkt

t

k

k

kt

kt

t

25.6425)

321(200)

21(200)15(.1

)21(200)(

)21()

21ln(])

21ln([

)21ln(

21ln

31

21ln3

21

200100

200)3(100

&200)(:

100)3(200)0(

)0()(

)()(

315

3

3)

21ln(

331

31

3

)3(

6

3

yeargydtdy

yeargtydtdy

tyhaveWe

yearsTT

yyyTy

Thensizeoriginaklitsofquartertodecaytomassthefor

neededtimethebeTLet

tyhaveWe

t

tt

t

T

T

t

/48

2ln200)21(

32ln200)

21(

32ln200)12(.4

/)21(

32ln200

31)

21ln()

21(200)(

)21(200)(:.3

6232

141)

21(

41)

21(200

41)(

:

)21(200)(:.2

4312

12

33

3

23

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

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tkt

t

k

k

kt

ty

ee

ktk

ke

eysoand

yearyearmonthsandyearthatnote

yandyhaveWe

eyty

yearsinttimeatmassthebetyLet

t

253

253

10099ln

253

253

)325(

)325(

1000999)(

1000999

1000999ln0

100999ln

1000999ln

253

3251000999ln

1000999ln)

325(

1000999

1000)325(999:

325

31848(

999)325(1000)0(

)0()(

)()(

253

yearsT

T

yTyy

Then

sizeoriginaklitsofhalftodecaytomasstheforneededtimethebeTLet

ytyhaveWe

sizeitsofhalftodecayoelementtheofmassanytakesittimetheislifeHalf

T

T

t

5773

9991000ln3

2ln25

1000999ln

253

21ln

21ln

1000999ln

253

21

1000999

1000999)(

21

:

1000999)(:

:.2

21

21

21

21

21

21

253

253

00

253

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

2ln1.2006(10)

2ln

1.2006(10) 999

1000ln53

1000999ln

53,

,

2ln21ln

21ln

21)(

21

:

:

44-

4-

00

21

21

21

21

21

21

21

T

khadWe

itfindingafterformulathistoinksubstiuedhavecouldweproblemlasttheinThus

kkT

kTeeyTyy

Then

sizeoriginalitsofhalftodecaytomasstheforneededtimethebeTLetlifehalftheandknttaconsthebeweenprelaionshitheDeducing

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 decided that a discovered fossilized bone of an animal has 25% of the C-14 that a bone of that live animal has. Knowing that the half-life of C-14 is approximately 5730 years and that its decay follows exponential model, find, the age of the bone.

Solution:

)5730(000

21

0

0

21

21

,573014)(,

14

kkT

kt

eyeyy

thenTCoflifehalftheSinceeytyThen

yboneliveainCofamounttheLet

yearst

ytyy

thenboneliveainyamounttheofhaditfoundwasbonethettimetheatSince

yeyeyty

tkt

kke

eyeyy

t

t

t

t

kt

t

k

kkT

t

11460)5730(25730

221

21

21

41

21)(

10025

.,%2521)(

21ln

21ln

21ln

573021ln

573021ln

21

21

57302

5730

5730

00

0

5730

021ln

00

573057301

57301

5730

)5730(000

5730

21