Download Application of Differential Equations in different Scenarios

Muhammad Al Fredey
Abdullah Alshaye
1. Radiocarbon dating
 Radiocarbon dating (or simply carbon dating) is a
radiometric dating technique that uses the decay of
carbon-14 (14 C6) to estimate the age of organic
materials, such as wood and leather, up to about
58,000 to 62,000 years Before Present (BP, present
defined) .
 Before Present (BP) years is a time scale used mainly
in geology and other scientific disciplines to specify
when events in the past occurred. Because the
"present" time changes, standard practice is to use …
Radiocarbon dating (Continued)
1st January 1950 as commencement date of the age
scale, reflecting the fact that radiocarbon dating
became practicable in the 1950s.
Carbon dating was presented to the world by Willard
Libby in 1949, for which he was awarded the Nobel
Prize in Chemistry.
Since the introduction of carbon dating, the method
has been used to date many items, including samples
of the Dead Sea Scrolls, the Shroud of Turin, enough
Egyptian artefacts to supply a chronology of Dynastic
Egypt, and Ötzi the Iceman
Radiocarbon dating (Continued)
The Earth's atmosphere contains various isotopes of
carbon, roughly in constant proportions. These
include the main stable isotope (12C6) and an unstable
isotope (14C6). Through photosynthesis, plants absorb
both forms from carbon dioxide in the atmosphere.
When an organism dies, it contains the standard ratio
of 14C6 to 12C6, but as the 14C6 decays with no
possibility of replenishment, the proportion of carbon
14C
6 decreases at a known constant rate. The time
taken for it to reduce by half is known as the half-life
of 14C6 .
Radiocarbon dating (Continued)
The measurement of the remaining proportion of 14C6
in organic matter thus gives an estimate of its age (a
raw radiocarbon age). However, over time there are
small fluctuations in the ratio of 14C6 to 12C6 in the
atmosphere, fluctuations that have been noted in
natural records of the past, such as sequences of tree
rings and cave deposits. These records allow finetuning, or "calibration", of the raw radiocarbon age, to
give a more accurate estimate of the calendar date of
the material. One of the most frequent uses of
radiocarbon dating is to estimate the age of organic
remains from archaeological sites.
Radiocarbon dating (Continued)
 Calculating ages:
While a plant or animal is alive, it is exchanging carbon
with its surroundings, so that the carbon it contains will
have the same proportion of 14C6 as the biosphere (12C6).
Once it dies, it ceases to acquire 14C6 , but the 14C6 that it
contains will continue to decay, and so the proportion of
radiocarbon in its remains will gradually reduce. Because
14C decays at a known rate, the proportion of radiocarbon
6
can be used to determine how long it has been since a
given sample stopped exchanging carbon—the older the
sample, the less 14C6 will be left . The equation governing
the decay of a radioactive isotope is
Radiocarbon dating (Continued)
 N  Noe  t where N0 is the number of atoms of the
isotope in the original sample (at time t = 0), and N is
the number of atoms left after time t. The mean-life,
i.e. the average or expected time a given atom will
survive before undergoing radioactive decay denoted
by τ, of 14C6 is 8,267 years, so the equation above can be
rewritten as
No
t  8267. ln(
)
N
Radiocarbon dating (Continued)
The ratio of 14C6 atoms in the original sample, N0, is
taken to be the same as the ratio in the biosphere 12C6,
so measuring N, the number of 14C6 atoms currently in
the sample, allows the calculation of t, the age of the
sample.
Radiocarbon dating (Continued)
The half-life of a radioactive isotope (the time it takes
for half of the sample to decay, usually denoted by T1/2)
is a more familiar concept than the mean-life, so
although the equations above are expressed in terms of
the mean-life, it is more usual to quote the value of
14C half-life than its mean-life. The currently accepted
6
value for the half-life of radiocarbon is 5,730 years. The
mean-life and half-life are related by the following
equation
T1 2   .ln2
Radiocarbon dating (Continued)
For over a decade after Libby's initial work, the
accepted value of the half-life for 14C6 was 5,568 years;
this was improved in the early 1960s to 5,730 years,
which meant that many calculated dates in published
papers were now incorrect (the error is about 3%).
However, it is possible to incorporate a correction for
the half-life value into the calibration curve, and so it
has become standard practice to quote measured
radiocarbon dates in "radiocarbon years", meaning
that the dates are calculated using Libby's half-life
value and have not been calibrated.
Examples (Radiocarbon dating)
(Continued)
 A fossil bone is found to contain 1/1000 the original
amount of 6C14 Determine the age of the fossil.
We have the equation governing the given phenomenon
e  
as under , which gives the amount
of
y(t) = y0.e  kt
where y0 is the initial amount of the radioactive
substance. For t = 5730 years ( the half – age of 6C14 ),
this equation gives,
y(t) = y0 /2.
Radiocarbon dating (Continued)
Thus from the above equation we can determine the
value of the constant k as under,
5730k
y0/2 = y0.e
or
- 5730 k = In (1/2).
Thus we get,
k= In(1/2)/(-5730) 0.000120
Therefore the above equation becomes,
y(t) = y0. e 0.00012097t
Thus , when y(t) = y0 / 1000, as given , we get,
0.00012097t
y0/1000 = y0. e
Radiocarbon dating (Continued)
On taking log-natural of both sides this equation gives,
-0.00012097 t = In (1/1000) = - In ( 1000 ).
So that ,
t = In ( 1000 ) /0.00012097 = 57136 years.
Note : In the above method radioactive Carbon dating much
of the accuracy of result depends upon the chemical analysis
of the fossil. In order to obtain better estimations of 6C14
present in the fossil , destruction of large samples of the
specimen are required. The same may not always be possible
from archival point of view. In view of the same the age
estimation of the fossil in the above example is not very
accurate.
Radiocarbon dating (Continued)
In the recent years geologists have shown that age
estimation of the fossils by the above mentioned
method may be out by as much as 3500 years in certain
cases. One of the possible reasons for this error is the
fact that 6C14 levels in the air very with time. They have
devised another method for the purpose, based on the
fact that the living organisms ingest traces of Uranium.
By measuring the relative amounts of Uranium and
Thorium (the isotope into which Uranium decays ), and
by knowing
Radiocarbon dating (Continued)
the half-lives of these elements, one can determine the
age of the fossil. By this method one can estimate the
ages of even 5000,000 old fossils. However, this method
is not applicable to marine fossils. Some other
techniques can also be used for the purpose including
the use of Potassium – 40 and Argon 40 (which can
estimate ages up to millions of years), and non-isotopic ,
methods based on the use of amino acids.
2. Newton's Law of Cooling
 Newton's Law of Cooling states that the rate of change of
the temperature of an object is proportional to the
difference between its own temperature and the ambient
temperature (i.e. the temperature of its surroundings).
Newton's Law makes a statement about an instantaneous
rate of change of the temperature. We will see that when
we translate this verbal statement into a differential
equation, we arrive at a differential equation. The solution
to this equation will then be a function that tracks the
complete record of the temperature over time.
Newton's Law of Cooling
(Continued)
 Crime Scene
 A detective is called to the scene of a crime where a dead
body has just been found. She arrives on the scene at 6:00
pm and begins her investigation. Immediately, the
temperature of the body is taken and is found to be 85o F.
The detective checks the programmable thermostat and
finds that the room has been kept at a constant 72o F for
the past 3 days.
Newton's Law of Cooling
(Continued)
After evidence from the crime scene is collected, the
temperature of the body is taken once more and found
to be 78o F. This last temperature reading was taken
exactly three hour after the first one (i.e., at 9:00 pm).
The next day the detective is asked by another
investigator, “What time did our victim die?”
Assuming that the victim’s body temperature was
normal (98.6o F) prior to death, what is her answer to
this question? Newton's Law of Cooling can be used to
determine a victim's time of death.
Newton's Law of Cooling
(Continued)
The governing equation for the temperature, T of the
body is
dT
 k (T  To)
dt
where
T = temperature of the body, o F
o
To = ambient temperature F
t = time in hrs
k = constant based on thermal properties of body and
air
Newton's Law of Cooling
(Continued)
dT
 k (T  To)
dt
dT
 kT  kTo
dt
The characteristic equation of the above differential
equation is
mk  0
m  k
Tc  Ae  kt
Newton's Law of Cooling
(Continued)
Let the particular solution is T p  B
Substituting this particular solution into the ordinary
differential equation 0  kB  kTo
B  To
The complete solution is
T  Tc  Tp
T  Ae  kt  To
Newton's Law of Cooling
(Continued)
Given is
To  72
T (6)  85
T (9)  78
T ( B )  98.6
where B = time of death,
We get
6 k
85  Ae
 72
(1)
78  Ae9 k  72
(2)
98.6  Ae  KB  72
(3)
Newton's Law of Cooling
(Continued)
Use equations. (1) and (2) to find A and K, we get
A  61.028
k  0.25773
Substituting the values of A and K into equation (3), to
find B
98.6  61.028e 0.25773B  72
B  3.221
The time of death is 3.221 hrs, that is 0.3221*60 = 13.326
minutes after 3 pm.
Time of death = 3:13 pm
3. Exponential Population Growth
 Bacterial growth
Suppose the population of bacteria doubles every 3
hours. What exactly does that mean? Imagine you
inoculate a fresh culture with N bacteria at 12:00 pm.
At 3 pm, you will have 2N bacteria, at 6 pm you will
have 4N bacteria, at 9 pm you will have 8N bacteria,
and so on. If these cell divisions occur at EXACTLY
each of these time points the cells are said to be
growing synchronously. If this were the case, the
growth process would be geometric.
Exponential Population
Growth(Continued)
A geometric growth model
predicts that the population
increases at discrete time points
(in this example hours 3, 6, and 9).
In other words, there is not a
continuous increase in the
population.
Exponential Population
Growth(Continued)
Exponential Population
Growth(Continued)
However, this is not what actually happens.
Imagine you take a small sample of the culture
every hour and count the number of bacteria cells
present. If bacterial growth were geometric, you
would expect to have N bacteria between 12 pm
and 3pm, 2N bacteria between 3 pm and 6 pm, etc.
However, if you perform this experiment in the
laboratory, even under the best experimental
conditions, this will not be the case.
Exponential Population
Growth(Continued)
If you go a step further
and make a graph with
the number of bacteria
on the y-axis and time
on the x-axis, you will
get a plot that looks much
more like exponential
growth than geometric growth.
Exponential Population
Growth(Continued)
 Why does bacterial growth look like exponential
growth in practice?
 The answer is because bacterial growth is not
completely synchronized. Some cells divide in fewer
than 3 hours; while others will take a little longer to
divide. Even if you start a culture with a single cell,
synchronicity will be maintained only through a few
cell divisions. A single cell will divide at a discrete
point in time, and the resulting 2 cells will divide at
ABOUT the same time, and the resulting 4 will again
divide at ABOUT the same time.
Exponential Population
Growth(Continued)
As the population grows, the individual nature of cells
will result in a smoothing of the division process. This
smoothing yields an exponential growth curve, and
allows us to use exponential functions to make
calculations that predict bacterial growth. So, while
exponential growth might not be the perfect model of
bacterial growth by binary fission, it is the appropriate
model to use given experimental reality.
Exponential Population
Growth(Continued)
 Problem -Calculate the number of bacteria in a
culture at a given time
How many bacteria are present after 51 hours if a
culture is inoculated with 1 bacterium? Use the model,
N(t) = Noe kt, and assume the population doubles every
3 hours. (N(t) is the population size at time t and k is a
constant.)
4. Radioactive Decay
 In nature, there are a large number of atomic nuclei
that can spontaneously emit elementary particles or
nuclear fragments. Such a phenomenon is called
radioactive decay. This effect was studied at the turn of
19-20 centuries by Antoine Becquerel, Marie and Pierre
Curie, Frederick Soddy, Ernest Rutherford, and other
scientists. As a result of the experiments, F.Soddy and
E.Rutherford derived the radioactive decay law, which
is given by the differential equation
Radioactive Decay(Continued)
dN
 kN
dt
where N is the amount of a radioactive material, λ is a
positive constant depending on the radioactive
substance. The minus sign in the right side means that
the amount of the radioactive material N(t) decreases
over time. The given equation is easy to solve, and the
solution has the form
N (t )  Ce kt
Radioactive Decay(Continued)
 To determine the constant C, it is necessary to indicate
an initial value. If the amount of the material at the
moment t = 0 was N0, then the radioactive decay law is
N  Noe  kt
written as
The half life or half life period T of a radioactive
material is the time required to decay to one-half of
the initial value of the material. Hence, at the moment
No
T:
N (t ) 
 Noe kt
2
e  kt 
1
2
1
k  ln 2
t
Example ( Radioactive Decay )
(continued)
 In a certain radioactive substance the rate of decrease
in mass is proportional to the current mass. If the
mass present is reduced by half, in half , an hour, what
percent of the original mass is expected to remain
present at the end of 0.9 hours?
Radioactive Decay (continued)
Let the mass at time is y(t). Then the given problem can be represented by
dy/dt=ky,
where k is constant of proportionality. On solving at we get,
y= c e-kt
where c is the constant of integration. If y0 is the mass at
t =0, the above equation becomes,
y0 = c e0 =
1⁄2 y0 = c e-k/2 = y0 e-k/2
Thus,
1⁄2 = e-k/2 ,
Radioactive Decay (continued)
Taking log-natural of both sides we get 1⁄2 k = In(1/2)= In (2),
or k = In 4.
Therefore, the desired solution is,
y = y0 e-(In4)t
The above solution can be put as : y = y0 / et(In4) = y0/4t .
Therefore , when t = 0.9 (given ), then the above equation
gives: y = y0 / 40.9 = 0.2871745 , y0 = 29% of y0 ( approx)
Thus at the end of 0.9 hours about 29 percent of the original quantity of
the given radioactive substance would be left.
5. Compound Interest
 Compound Interest: Non-Continuous
r

A  P 1  
 m
mt
• P = principal amount invested
• m = the number of times per year interest
is compounded
• r = the interest rate
• t = the number of years interest is being
compounded
• A = the compound amount, the balance
after t years
Compound Interest(Continued)
Notice that as m increases, so does A. Therefore, the maximum amount of
interest can be acquired when m is being compounded all the time continuously.
Compound Interest(Continued)
 Compound Interest: Continuous
Compound Interest: Continuous
A  Pert
• P = principal amount invested
•r = the interest rate
• t = the number of years interest is
being compounded
• A = the compound amount, the
balance after t years
Compound Interest(Continued)

EXAMPLE
(Continuous Compound) Ten thousand dollars is invested at 6.5% interest
compounded continuously. When will the investment be worth $41,787?
SOLUTION
We must first determine the formula for A(t). Since interest is being
rt
compounded continuously, the basic formula to be used is A  Pe .
Since the interest rate is 6.5%, r = 0.065. Since ten thousand dollars is being
invested, P = 10,000. And since the investment is to grow to become $41,787,
A = 41,787. We will make the appropriate substitutions and then solve for t.
A  Pert
41,787  10,000e
0.065t
This is the formula to use.
P = 10,000, r = 0.065, and
A = 41,787.
Compound Interest(Continued)
4.1787  e 0.065t
ln 4.1787  0.065t
22  t
 Therefore, the $10,000 investment will grow to $41,787, via 6.5% interest
compounded continuously, in 22 years.
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