Application of Differential Equations in different Scenarios - PowerPoint Presentation

Slide1

Application of Differential Equations in different Scenarios

Muhammad Al

Fredey

Abdullah

Alshaye

Slide2

1. Radiocarbon dating

Radiocarbon dating

(or simply

carbon dating

) is a

radiometric dating

technique that uses the decay of

carbon-14

(

14

C

6

) 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 …

Slide3

Radiocarbon dating (Continued)

1

st 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

Slide4

Radiocarbon dating (Continued)

The Earth's atmosphere contains various

isotopes

of

carbon

, roughly in constant proportions. These include the main stable isotope (

12

C

6

) and an unstable isotope (

14

C

6

). Through photosynthesis, plants absorb both forms from

carbon dioxide

in the atmosphere. When an organism dies, it contains the standard ratio of

14

C

6

to

12

C

6

, but as the

14

C

6

decays with no possibility of replenishment, the proportion of carbon

14

C

6

decreases at a known constant rate. The time taken for it to reduce by half is known as the

half-life

of

14

C

6

.

Slide5

Radiocarbon dating (Continued)

The measurement of the remaining proportion of

14

C

6

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

14

C

6

to

12

C

6

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 fine-tuning, 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.

Slide6

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

14

C

6

as the biosphere (

12

C

6

). Once it dies, it ceases to acquire

14

C

6

, but the

14

C

6

that it contains will continue to decay, and so the proportion of radiocarbon in its remains will gradually reduce. Because

14

C

6

decays at a known rate, the proportion of radiocarbon can be used to determine how long it has been since a given sample stopped exchanging carbon—the older the sample, the less

14

C

6

will be left . The equation governing the decay of a radioactive isotope is

Slide7

Radiocarbon dating (Continued)

where

N

0

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

Slide8

Radiocarbon dating (Continued)

The ratio of

14

C

6

atoms in the original sample,

N

0

, is taken to be the same as the ratio in the biosphere

12

C

6

, so measuring

N, the number of 14C6 atoms currently in the sample, allows the calculation of t, the age of the sample.

Slide9

Radiocarbon dating (Continued)

The

half-life

of a radioactive isotope (the time it takes for half of the sample to decay, usually denoted by

T

1/2

) 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

14

C

6

half-life than its mean-life. The currently accepted value for the half-life of radiocarbon is 5,730 years. The mean-life and half-life are related by the following equation

Slide10

Radiocarbon dating (Continued)

For over a decade after Libby's initial work, the accepted value of the half-life for

14

C

6

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.

Slide11

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 as under , which gives the amount of

y(t) =

y0.

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.

Slide12

Radiocarbon dating (Continued)

Thus

from the above equation we can determine the value of the constant k as under,

y0/2 =

y0. or

- 5730 k = In (1/2).

Thus we get,

k= In(1/2)/(-5730) 0.000120

Therefore the above equation becomes,

y(t) =

y0.

Thus , when y(t) = y0 / 1000, as given , we get,

y0/1000 =

y0.

Slide13

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.

Slide14

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

Slide15

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.

Slide16

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.

Slide17

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 85

o

F. The detective checks the programmable thermostat and finds that the room has been kept at a constant 72

o

F for the past 3 days.

Slide18

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 78

o

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 (98.6

o

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.

Slide19

Newton's Law of Cooling (Continued)

The governing equation for the temperature, T of the body is

where

T = temperature of the body,

To = ambient temperature

t = time in hrs

k = constant based on thermal properties of body and air

Slide20

Newton's Law of Cooling (Continued)

The characteristic equation of the above differential equation is

Slide21

Newton's Law of Cooling (Continued)

Let the particular solution is

Substituting this particular solution into the ordinary differential equation

The complete solution is

Slide22

Newton's Law of Cooling (Continued)

Given is

where B = time of death,

We get

Slide23

Use equations. (1) and (2) to find A and K, we get

Substituting the values of A and K into equation (3), to find B

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

Newton's Law of Cooling (Continued)

Slide24

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.

Slide25

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.

Slide26

Exponential Population Growth(Continued)

Slide27

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.

Slide28

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.

Slide29

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.

Slide30

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.

Slide31

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

) = 

N

o

e

 

kt

, and assume the population doubles every 3 hours. (

N(t) is the population size at time t and k

is a constant.)

Slide32

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

Slide33

Radioactive Decay(Continued)

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

Slide34

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

N

0

, then the radioactive decay law is 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 T:

Slide35

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?

Slide36

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 ,

Slide37

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.

Radioactive Decay (continued

)

Slide38

5. Compound Interest

Compound Interest: Non-Continuous

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

Slide39

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.

Slide40

Compound Interest(Continued)

Compound Interest: Continuous

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: Continuous

Slide41

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 compounded continuously, the basic formula to be used is

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

.

This is the formula to use.

P

= 10,000,

r

= 0.065, and

A

= 41,787.

Slide42

Compound Interest(Continued)

Therefore, the $10,000 investment will grow to $41,787, via 6.5% interest compounded continuously, in 22 years.

Slide43

Thanks