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Rates of Growth & Decay

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Rates of Growth & Decay. Example (1 ). - PowerPoint PPT Presentation
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Rates of Growth & Decay
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Page 1: Rates of Growth & Decay

Rates of Growth & Decay

Page 2: Rates of Growth & Decay

.;)0()(

:,.,,,

.,,

)0()(

)0(:

,

)()(ln

,)()(

);'()()(

),'(,)()(:)(

0

1

1

12

1

11

positiveiskwhereeyty

notationtextbooktheadoptwillwecasethisFordecayofratethecalledischangeofrateitsandfunctioningdecreasahavewenegativeiskwhenWhile

growthofratethecalledischangeofrateitsandfunctioningincreasanhavewepositiveiskWhen

eyty

CCeyhaveWe

CasentconstathengrenamiCe

eeetycktty

cktydy

ntconstaaiscwherecktdxtyty

antconstisyofchangeofraterelativethesThatktyty

timethatatyofvaluethetoalpropertiondirectlyisttimeanyatyofchangeofratethesThatkntconstasomefortyktysatisfyingfunctionabetfyLet

kt

kt

ckt

cktckt

Page 3: Rates of Growth & Decay

Example (1)

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.

Page 4: Rates of Growth & Decay

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

Page 5: Rates of Growth & Decay

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

Page 6: Rates of Growth & Decay

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.

Page 7: Rates of Growth & Decay

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

Page 8: Rates of Growth & Decay

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

Page 9: Rates of Growth & Decay

Example (2) – b

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.

Page 10: Rates of Growth & Decay

Solution

t

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

Page 11: Rates of Growth & Decay

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

Page 12: Rates of Growth & Decay

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.

Page 13: Rates of Growth & Decay

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

itfindingafterformulathistoinksubstiuedhavecouldweproblemlasttheinThuskk

T

kTeeyTyy

Then

sizeoriginalitsofhalftodecaytomasstheforneededtimethebeTLetlifehalftheandknttaconsthebeweenprelaionshitheDeducing

kTkT

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

Page 14: Rates of Growth & Decay

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.

Page 15: Rates of Growth & Decay

Example (3)

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

21

0

0

21

21

,573014)(,

14

kkT

kt

eyeyy

thenTCoflifehalftheSinceeytyThen

yboneliveainCofamounttheLet

Page 16: Rates of Growth & Decay

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

Page 17: Rates of Growth & Decay

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.

Page 18: Rates of Growth & Decay

Example (4)

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

Page 19: Rates of Growth & Decay

Solution

162ln2ln:

21)(

21ln

21ln

21ln

1621ln

1621ln

21

21

,1632)(:

32)(

)()(

21

16

021ln

00

16161

161

)16(

)16(000

21

0

0

16

21

TkformulathengusibykfoundcouldWeNote

yeyeyty

tkt

kke

eyeyy

thenTdrugtheoflifehalftheSinceetysoand

yandeyty

hoursinttimeabodytheindrugthebetyLet

t

kt

t

k

kkT

kt

kt

t

Page 20: Rates of Growth & Decay

hoursTyTyy

amountoriginalitsofreachtodosefirstthetakesittimethebeTLet

yeyeyty

hadWea

TTT

t

kt

t

4821

21

21

81

21)(

1005.12

%5.1221)(

:.

163

1616

00

16

021ln

00

16

Page 21: Rates of Growth & Decay

dosefirstthetakingfromhoursaftersThatdosecondsethetakingfromhoursaftermgreachesdrugtheofamounttheThus

tt

tZ

thenmgreachtomomentthisfromdrugthetakesittimethebetLet

ZtZ

isthatafterttimeanyatpresentZdrugtheofamountTheZamountthisandttoheredingcorrresponmomentthisfromStarting

givenwasdosecondsetheafterpresentamountThe

mgy

hoursafterremainingstilldosefirsttheofamountThe

ty

haveWeb

ttt

t

tt

t

963264'32:5.8,

3216

2

21

21

21

41

21

345.8

2134)(5.8

,5.8)(2134

21)(

:)()34()0(

34322

22132

2132)64(

642132)(

:.

11

162

1616

16

1

1

1616

0

0

41664

16

111

1


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