Rates of Growth & Decay

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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.

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

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etysoand

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t

tkt

t

k

k

kt

kt

t

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dy

tyhaveWe

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t

tt

t

T

T

/3

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2ln)10()2(

3

2ln)10()12(.4

/)2(3

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3

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

)2()10()(:.3

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004

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4

4

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

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yhaveWe

eyty

yearsinttimeamassthebetyLet

t

tkt

t

k

k

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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

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t

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t

T

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/48

2ln200)

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1(

3

2ln200)

2

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3

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/)2

1(

3

2ln200

3

1)

2

1ln()

2

1(200)(

)2

1(200)(:.3

6232

1

4

1)

2

1(

4

1)

2

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4

1)(

:

)2

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43

12

12

33

3

2

3

03

00

3

41

414

1

41

41

41

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.

Solution

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tkt

t

k

k

kt

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ee

ktk

ke

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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

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lifeHalf

T

T

t

5773

9991000

ln3

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ln253

21

ln

2

1ln

1000

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25

3

2

1

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999

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999)(

2

1

:

1000

999)(:

:.2

21

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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

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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 (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

2

1

0

0

2

1

2

1

,573014

)(,

14

kkT

kt

eyeyy

thenTCoflifehalftheSince

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yearst

ytyy

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yeyeyty

tkt

kke

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t

t

t

t

kt

t

k

kkT

t

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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 (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

Solution

16

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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

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tkt

kke

eyeyy

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etysoand

yand

eyty

hoursinttimeabodytheindrugthebetyLet

t

kt

t

k

kkT

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kt

t

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hadWe

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t

482

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2

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2

1

8

1

2

1)(

100

5.12

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2

1)(

:

.

163

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16

02

1ln

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