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By the term mathematical modeling, we
mean the process whereby the behavior of areal-life system or phenomenon, whetherphysical, sociological, or even economics isdescribed by a set of Mathematical relations,after approximation and idealizations.Construction of Mathematical model startswith
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i) Identification of the system we are
modeling. This requires selection of somevariables as important to understanding ordescribing the behavior of the system andignoring other as marginal or irrelevant to
this understanding.
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ii)Making some realistic assumption, orhypothesis, about the system we aredescribing. These assumptions will alsoinclude any empirical laws that may beapplicable to the system.
We translate these assumptions intomathematical language which usuallyresults in a set of equations and /or
inequalities. This gives us a mathematicalmodel which will be an approximation of theactual real-life problem.
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Once we have constructed the mathematicalmodel, we are faced with the problem ofsolving it so as to obtain the unknown variableor variables in terms of the known quantities.
So Modeling steps are the steps thatlead from the physical situation to amathematical formulation and solution, thenphysical interpretation of the result.
In this section we solve some model thatare described by linear first order differentialequations.
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Let x(t) denote the population at any time t.We construct a mathematical model ofpopulation x(t) which grows or decays at a
rate proportional tox.
Setting up a mathematical model
kxdt
dx
Growth and Decay
Step 1
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Step 2
Solving the differential equation obtained.
Step 3
Interpretation of the result.
kxdt
dx
. (1)
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Integrating (1)
kt
cktxkt
(2)
for decay
for growth
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c is any arbitrary constant and can be
determined by using initial condition andvalue of kchanges from problem to problem.
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Example
Bacterial Growth
A culture initially has N0 number of bacteria.At t= 1hr the number of bacteria is measuredto be (3/2)N
0
. If the rate of growth is
proportional to the number of bacteriapresent, determine the time necessary for thenumber of bacteria to triple.
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.
Solution
We first solve the differential equation
kdt
dN. (1)
Subject to N(0)=N0. Then we use theempirical condition to determine theconstant of proportionality k . Now eq (1) is
both separable and linear. When it is put intothe form
0dN
kNdt
...(2)
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We can see by inspection that the integratingfactor is . Multiplying (2) with we getkte
ktd e Ndt
Integrating (3)
.(3)
kt .. .(4)
k te
it follows that, at
so equation (4) takes the form
tcceN
0
0
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kteNtN0
.(5)
so equation (4) takes the form
at t= 1 , we have
kteNN 00
2
3
kte2
3
4055.0)2
3ln(k
Thus0.4055
0( ) t
N t N e
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To find the time at which the bacteria havetripled we solve
t4055.03ln
hrt 71.24055.0
3ln
teNN
4055.0
00
te 4055.0
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Example
Bacteria in a certain culture increases at a rateproportional to the number present. If thenumber of bacteria is 2000 after one hour and4000 after two hours. What was their initial
numbers?
Solution
Let x be the number of bacteria at time t, then
dxkx
dt
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ktcetx )(
Solution of this equation is same as in theabove example i.e.
. (1)
Here we have two conditions
(1) 2000(2) 4000
xx
Using these two conditions in eq (1), we get
values of c & k.
ln 2tx t e
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Now we have to find the initial number when t
= 0. x . (2)
Equation (2) is the required initial number
Radioactive Decay
Radioactive decay is the process in whichan unstable atomic nucleus loses energy by
emitting radiation in the form of particles orelectromagnetic waves
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i) Radium is a radioactive chemical elementwhich has symbol Ra.
ii) Radium is an alkaline earth metal. It isextremely radioactive.
iii) Radium is used in medicines to produceradon gas which in turn is use as a cancertreatment.
iv) Radium is over one million times moreradioactive than the same mass of Uranium.
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Example
Radium decomposes at a rate proportional tothe quantity of radium present. If 100 mgreduces to 90 mg in 20 years. How many milligrams will remain at the end of 1000 years?
How long will it take for one-half of the originalamount to decompose?
Solution
Letx mg denote the amount of Radium presentat time t.
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Here we the differential equation
kxdtdx
Solution of this equation is
ktcetx )( (1)
Here we have two conditions
(0) 100
(20) 90
x
x
.....(2a)
.(2b)
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Appling (2a) in (1) we get
cUsing value of c in equation (1) we get
ktx t e ...(3)
Appling (2b) in (1) we get2090 100 ke
k
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So equation (3) becomes
t
etx0005.0
100)(
Now at t = 1000 gives
txHence after 1000 years nearly 61 mg ofradium will remain.
t = ? Whenx = 50
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Example
Initially there were 100mg of a radioactivesubstance present. After 6hrs the mass decreased
by 3%. If the rate of decay is proportional to the
amount of the substance present at any time, find
the amount remaining after 24 hrs?
Solution
Let be the initial amount, and A(t) be
amount at any time.
0A
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( ) 1 0 0A t m g(1)
( ) 97A t mg (2)
at t = 0
at t = 6
24 , ( ) ?t hr A t
Governing equation is
dAkA
dt
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Its solution isk tA t c e
Using condition (1) in eq (3) we get
(3)
c (4)
Using the value of c in eq (3), we getktA t e
Using condition (2) in eq (4) we get
697 100 ke
0.00507k
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Using the value of k in eq (4) for t = 24 hrs,
we get)24(00507.0100)( etA
mgtA 5.88)(
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Half Life
The half life of a radioactive substance is
the time in which half of the given amountwill disappear. Hence it measures theradioactive decay process .
OrThe time taken for one half of a
radioactive substance to decay is calledhalf life of the substance.
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Example
Determine the half-life of the radioactivesubstance described in last example.
Solution0( )
2
AA t
0.00507( )0 1002
tAe
Eq (1) 0.00507( )100 1002
te
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1ln 0 .0 0 5 0 7
2t
Example
The breeder reactor converts stable uranium238 into the isotope platinum 239. After 15years it is determined that 0.043% of the initial
amount of the platinum has disintegrated. Findthe Half Life of this isotope
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Let A denote the amount of platinum remainingat any time t. Governing equation is
d Ak A
d twith
0A A .. ..(1)
Its solution is
ktA t ce ..(2)
Solution
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Using above condition (1) in eq (2) we get
0( ) kt
A t A e
0AIf 0.043% of have disintegrated, thenremaining amount is 99.957% Or 0.99957
00.99957A A
Now at t = 15
Using this condition in eq (3) we get
(3)
15
0 00.99957 kA A e
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ln 0.99957
15k
0 . 0 0 0 0 2 8 6 7k
0.00002867
0( ) tA t A e
150.99957 ke
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24180t yrs
1ln
2
0.0002867t
1ln 0.00002867
2t
0.0000286700
2
tAA e
0( )2
AA t
Now the half life in the corresponding value oftime for which
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Carbon Dating
About 1950, the chemist Willard deviseda method of using radio active carbon as ameans of determining the approximate ageof fossils, such as remains of woods, bones,
etc.
The half life of a radio active carbon isabout 5600 years. The ratio of Carbon 14 to
ordinary Carbon in the atmosphere isconstant, and this ratio also applies toanimals and plants tissue as long as they arealive.
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However, when the animals or plants die,
the Carbon-14 starts to decay throughradioactivity, while the ordinary carbon doesnot disappear.
Thus knowing the ratio of C-14 to ordinarycarbon we can determine an approximation tothe year of death.
For his work Libby won the Nobel Prize forchemistry in 1960.
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A wooden archeological specimen contains15% of the original Carbon-14. Using a half lifeof 5730 years. Determine the approximate ageof the specimen.
Example
Solution
Governing equation is
dA
kAdtwith
0)0( AA ..(1)
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Its solution is
( ) k tt ce . ..(2)
Using condition (1) in eq (2) we get.
0( ) kt
t A e
5730, ( )2
oA
t A tat .....(3)
Substituting condition (3) in eq (2) we get.
57300
02
kA
A e
1ln 5730
2k
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k
0( ) 0.15A t AWhen
0.000120968
0 0(0.15) tA A e
ln(0.15) 0.000120968t
15682t years