Post on 24-Feb-2016
description
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
yearpersonydtdy
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