PI009E21 GROUP 1
MAT099 GROUP PROJECT REPORT TASK 1: DECAY PROBLEM OF RADIOACTIVE SUBSTANCE TASK 2: NEWTON-RAPHSON METHOD
NO.
NAME
STUDENT ID
1
MUHAMMAD FAUZAN BIN AMINUDDIN
2018642646
2
HARISA RABIHAH BINTI MOHD BASRI
2018414958
3
MUHAMAD ARIF HAKIMI BIN MUHAMAD ARIFIN
2018257322
4
MUHAMMAD HELMI AMINULLAH BIN JELISUN
2018290286
5
NUR SAMIHAH BINTI MD RUSLI
2018256494
6
NURAMINUDDIN BIN ABDUL RASHID
2018218804
ACKNOWLEDGEMENT
First and foremost, we would like to express our gratitude to Allah for giving us the opportunity to settle this assignment successfully. Thanks to Him because He gave us his blessings since we can complete this assignment peacefully without any distractions and obstacles. Other than that, we also want to portray our loves for our parents at home because they gave us this opportunity to study here with no worries and let us learn more about this project. Without them, we did not have the chance to know about the differential equation and the Newton-Raphson method. Plus, we also want to express our gratitude to Miss Nurashikin Bt Abdullah for teaching us a lot about this assignment. She gave us the best guidance to ensure that we can complete this assignment with minor error. She also prepared a template for us to refer in order to make this report as she wanted us to be more prepared in our future especially when we step in degree life after completing this foundation. Miss Nurashikin is an approachable lecturer which students can see her whenever we need helps from her. Lastly, not to forget our beloved classmates, PI009E21 members for giving us the cooperation in order to complete this project even though there are still another assignment to be done. They also help one another by spreading all the information that comes from the Miss Nurashikin. Thank you ď Š
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TABLE OF CONTENTS ACKNOWLEDGEMENTS
Page 2
TABLE OF CONTENTS
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LIST OF TABLES
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LIST OF FIGURES
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ASSIGNMENTS
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Task 1(Decay problem of radioactive substance) 1.1 Introduction
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1.2 Implementation
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1.3 Result and Discussion 2
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Task 2(Newton-Raphson method)
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2.1 Introduction
11
2.2 Implementation
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2.3 Result and Discussion
CONCLUSION
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BIBLIOGRAPHY
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LIST OF TABLES Page Table 1: Data for Radioactive Substance Decay Process
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LIST OF FIGURES Page Figure 1: Graph plotted by excel
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Figure 2: Graph from the exponential solution
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Figure 3: Calculation data from Excel
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ASSIGNMENT
1 1.1
Task 1 (Decay Problem- Radiation of Substance) Introduction Decay problem is related to radioactivity of a substance. The radioactive elements follow the differential equation. It decomposes at a rate proportional to its mass. This rate called decay rate. đ???đ??Œ đ?œśđ??Œ đ???đ?? đ???đ??Œ = −đ?’Œđ?‘´ đ???đ?? where M is the mass of the substance and k is the constant of proportionally.
Another parameter that is useful for characterizing radioactive decay is the half-life. The half-life of a radioactive substance is the time it takes for half of a given number of radioactive nuclei to decay. The original term, half-life period, dating to Ernest Rutherford’s discovery of the principle in 1907, was shortened to half-life in the early 1950s. Rutherford applied the principle of a radioactive element’s half-life to studies of age determination of rocks by measuring the decay period of radium to lead-206.
This task was done to see the benefits of half-life method to society. We also can see the relation between the solution of differential equation and half-life method to determine the decay process of radioactive substance. Every radioactive substance has different decay rate. This is the reason why every substance carries different benefits to the society in many aspects such as medical field and engineering field.
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The objectives of this project are: 1. To solve the differential equation based on the data that we found for the decay problem. 2. To plot a graph from the exponential solution that we calculated and compare the graph with the graph from the original data. 3. To predict the future event based on the half-life method either near future and distant future. 4. To find the think that might affect the change (decay process) in the future.
1.2 Implementation Data for radioactive substance for a decay process is a secondary data that are retrieved from https://prezi.com/vmxqpi-la-pj/0801-half-life-and-radioactive-decay-half-life-lab/ . Table 1: Data for radioactive substance decay process
Time(sec)
Mass Remained(u)
0
200.0
3
104.5
6
54.0
9
31.0
12
14.0
15
5.5
18
2.5
21
1.5
24
1.0
27
0
30
0
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SOLUTION The decay rate was obtained from the definition of the decay problem. Let M : mass remaining at time(sec) đ?‘‘đ?‘€ âˆ?đ?‘€ đ?‘‘đ?‘Ą đ?‘‘đ?‘€ đ?‘‘đ?‘Ą
= −đ?‘˜đ?‘€, where k is the constant of proportionality.
Separating variables, đ?‘‘đ?‘€ = −đ?‘˜ đ?‘‘đ?‘Ą đ?‘€ Integrating both sides, to get the constant value of k âˆŤ
đ?‘‘đ?‘€ = âˆŤ −đ?‘˜ đ?‘‘đ?‘Ą đ?‘€
ln đ?‘€ = −đ?‘˜đ?‘Ą + đ?‘? đ?‘€ = đ?‘’ −đ?‘˜đ?‘Ą+đ?‘? đ?‘´ = đ?’†âˆ’đ?’Œđ?’• . đ?’†đ?’„
equation of integration
đ?‘™đ?‘’đ?‘Ą đ?‘’ đ?‘? = đ??´ as A is a constant value. đ?‘¤â„Žđ?‘’đ?‘› đ?‘Ą = 0, đ?‘€ = đ?‘€đ?‘œ , substitute this value into the equation of integration above. đ?‘€đ?‘œ = đ??´đ?‘’ −đ?‘˜(0) đ?‘€đ?‘œ = đ??´ đ??´ = đ?‘€đ?‘œ ∴ đ?‘´ = đ?‘´đ?’? đ?’†âˆ’đ?’Œđ?’•
general equation
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Since this decay problem are related with half-life method, so the mass of the radioactive substance 1 are always half of the original mass, 2 đ?‘€đ?‘œ . From the secondary data obtained, the time taken for the initial mass to half is 3 seconds. 1
đ?‘¤â„Žđ?‘’đ?‘› đ?‘€ = 2 đ?‘€đ?‘œ , đ?‘Ą = 3 đ?‘ đ?‘’đ?‘?, substitute this value into the general equation. 1 2
đ?‘€đ?‘œ = đ?‘€đ?‘œ đ?‘’ −3đ?‘˜ , the đ?‘€đ?‘œ are cancel out each other.
đ?‘’ −3đ?‘˜ =
1 2
To solve that equation, taking natural logarithm both sides and solve the equation to get the value of constant k. ln đ?‘’ −3đ?‘˜ = đ?‘™đ?‘›
đ?‘˜=
1 2
1 ln (2) −3
= 0.23104 đ?‘€ = đ?‘€đ?‘œ đ?‘’ −(0.23104)đ?‘Ą Particular solution to find mass of radioactive substance after t time, ∴ đ?‘€ = 200đ?‘’ −(0.23104)đ?‘Ą
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1.3 Results and Discussion Graph plotted by excel (by using the raw data obtained shows at table 1)
Figure 1: Graph plotted by excel
Graph from the exponential solution đ?‘€ = 200đ?‘’ −(0.23104)đ?‘Ą (using desmos)
Figure 2: Graph from the exponential solution
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From the particular solution, the near future and distant future value are predicted. Near future value đ??ľđ?‘Ś đ?‘˘đ?‘ đ?‘–đ?‘›đ?‘” đ?‘’đ?‘žđ?‘˘đ?‘Žđ?‘Ąđ?‘–đ?‘œđ?‘› đ?‘€ = đ?‘€đ?‘œ đ?‘’ −(0.23104)đ?‘Ą đ?‘¤â„Žđ?‘’đ?‘› đ?‘Ą = 21đ?‘ đ?‘’đ?‘? đ?‘€ = 200đ?‘’ −(0.23104)(21) = 1.562 đ?‘˘ Distant future value đ??ľđ?‘Ś đ?‘˘đ?‘ đ?‘–đ?‘›đ?‘” đ?‘’đ?‘žđ?‘˘đ?‘Žđ?‘Ąđ?‘–đ?‘œđ?‘› đ?‘€ = đ?‘€đ?‘œ đ?‘’ −(0.23104)đ?‘Ą đ?‘¤â„Žđ?‘’đ?‘› đ?‘Ą = 120 đ?‘ đ?‘’đ?‘? đ?‘€ = 200đ?‘’ −(0.23104)(120) = 1.80 Ă— 10−10 đ?‘˘
We can see that there is some different between the data that we found and the data that we calculated using the half-life method. At time, 21 seconds, the data from the table 1 is 1.5u while the value after calculated using particular solution is 1. 562u.From the distant future calculation, at time 120 seconds, the mass of the radioactive substance is still can calculate and not varnish. When compared to the data from table 1, at time 27 seconds, the mass of the radioactive substance is already zero. By this, we conclude that the experimental value is not really accurate. This is because the time when an individual radioactive atom decays is completely random. It is impossible to predict when an individual radioactive atom will decay. Hence, to know how much amount of substance will be left after particular amount of time, this half-life method becomes very helpful. This half-life method using particular solution will predict the accurate mass of the substance left. This method is used in many fields like nuclear chemistry, medical filed, etc. It also can be used to determine the ages of very old artefacts.
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2 2.1
Task 2 (Newton-Raphson Method) Introduction Newton–Raphson method, named after Isaac Newton and Joseph Raphson, is an iterative
method that can be use to solve the complicated equation or function especially a non-linear equation. This definition was defined by Thomas Simpson described Newton's method as an iterative method for solving general nonlinear equations using calculus in 1740. This method is used a tangent method in order to find an approximate value of roots of an equation. The application of Newton-Raphson method in real life can be seen in electric power systems engineering where Newton-Raphson method is use to solve power flow such as fault analysis, relays coordination, security, contingency. Plus, it is also use for Logistic Regression by fitting ideal parameters, and maximizing the predictive accuracy in order to maximize log-likehood function. From the task given, by taking the initial approximation đ?‘Ľ1 = 1, use Newton-Raphson method to estimate the root of equation sin 4đ?‘Ľ − 3√2 − đ?‘Ľ + đ?‘Ľđ?‘’ 3đ?‘Ľ = 3đ?‘Ľ 4 correct to six decimal places.
2.2 Implementation Step taken for this method: 1. From the equation given, we arranged the equation equal to zero. sin 4đ?‘Ľ − 3√2 − đ?‘Ľ + đ?‘Ľđ?‘’ 3đ?‘Ľ = 3đ?‘Ľ 4 3đ?‘Ľ 4 − sin 4đ?‘Ľ + 3√2 − đ?‘Ľ − đ?‘Ľđ?‘’ 3đ?‘Ľ = 0 2. Then, we differentiate the equation. đ??żđ?‘’đ?‘Ą đ?‘“ (đ?‘Ľ ) = 3đ?‘Ľ 4 − sin 4đ?‘Ľ + 3√2 − đ?‘Ľ − đ?‘Ľđ?‘’ 3đ?‘Ľ đ?‘‘ đ?‘‘ (3đ?‘Ľ 4 − sin 4đ?‘Ľ + 3√2 − đ?‘Ľ − đ?‘Ľđ?‘’ 3đ?‘Ľ ) = 3 (đ?‘Ľ 4 ) − đ?‘‘đ?‘Ľ đ?‘‘ đ?‘‘đ?‘Ľ
đ?‘‘đ?‘Ľ
đ?‘‘
đ?‘‘
(sin 4đ?‘Ľ) + 3 đ?‘‘đ?‘Ľ (√2 − đ?‘Ľ) − đ?‘‘đ?‘Ľ
(đ?‘Ľđ?‘’ 3đ?‘Ľ )
= 3(4đ?‘Ľ 3 ) − cos 4đ?‘Ľ . đ?‘‘đ?‘Ľ
đ?‘‘
đ?‘‘
(4đ?‘Ľ) + 3 . đ?‘‘đ?‘Ľ
1 2
1
(2 − đ?‘Ľ )2−1 .
đ?‘‘ đ?‘‘đ?‘Ľ
(đ?‘Ľ) . đ?‘’ 3đ?‘Ľ )
= 12đ?‘Ľ 3 − 4 cos(4đ?‘Ľ) −
3 2√2 − đ?‘Ľ
− (3đ?‘Ľđ?‘’ 3đ?‘Ľ + đ?‘’ 3đ?‘Ľ )
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(2 − đ?‘Ľ ) − (đ?‘Ľ .
đ?‘‘ đ?‘‘đ?‘Ľ
(đ?‘’ 3đ?‘Ľ ) +
= 12đ?‘Ľ 3 − 3đ?‘Ľđ?‘’ 3đ?‘Ľ − đ?‘’ 3đ?‘Ľ −
3 2√2 − đ?‘Ľ
đ?‘“′(đ?‘Ľ ) = 12đ?‘Ľ 3 − 3đ?‘Ľđ?‘’ 3đ?‘Ľ − đ?‘’ 3đ?‘Ľ −
− 4 cos(4đ?‘Ľ) 3
2√2 − đ?‘Ľ
− 4 cos(4đ?‘Ľ)
3. To find the approximate root, we use the general formula below: đ?‘“(đ?‘Ľđ?‘› ) đ?‘Ľđ?‘›+1 = đ?‘Ľđ?‘› − đ?‘“′(đ?‘Ľđ?‘› ) Initial approximation, đ?‘Ľ1 = 1 From equation:
đ?‘Ľ2 = đ?‘Ľ1 −
�(�1 ) �′(�1 )
đ?‘“(1) =1− đ?‘“′(1)
Substitute the đ?‘Ľ1 = 1
3(1)4 − sin 4(1) + 3√2 − 1 − (1)đ?‘’ 3(1) 3 12(1)3 − 3(1)đ?‘’ 3(1) − đ?‘’ 3(1) − − 4 cos(4(1)) 2√2 − 1 −13.328734 =1− −67.227573 =1−
Substitute the �1 = 1 in �(�) and �′(�)
= 0.801737 The steps is repeated by using the final answer đ?‘Ľ3 = đ?‘Ľ2 −
�(�2 ) �′(�2 )
= 0.801737 −
from previous step until the answer equal to final answer from previous step achieved.
�(0.801737) �′(0.801737)
= 0.653257
đ?‘Ľ4 = đ?‘Ľ3 −
�(�3 ) �′(�3 )
= 0.653257 −
�(0.653257) �′(0.653257)
=0.581447
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đ?‘Ľ5 = đ?‘Ľ4 −
�(�4 ) �′(�4 )
= 0.581447 −
�(0.581447) �′(0.581447)
=0.569696
đ?‘Ľ6 = đ?‘Ľ5 −
�(�5 ) �′(�5 )
= 0.569696 −
�(0.569696) �′(0.569696)
= 0.569454
đ?‘Ľ7 = đ?‘Ľ6 −
�(�6 ) �′(�6 )
= 0.569454 −
�(0.569454) �′(0.569454
= 0.569453
đ?‘Ľ8 = đ?‘Ľ7 −
�(�7 ) �′(�7 )
= 0.569453 −
�(0.569453) �′(0.569453
= 0.569453 Stop at here because đ?‘Ľ7 = đ?‘Ľ8 have the same answer The approximate root for this equation is 0.569453 (6 decimal places)
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This is the calculation that we done in Microsoft Excel.
đ?‘Ľ7 đ?‘Ľ8 Figure 3: Calculation Data from Excel
For f(x): use formula: =(3*(B2^4))-(SIN(4*B2))+(3*SQRT((2-B2)))-(B2*EXP(3*B2)) For f’(x): use formula: =(12*(B2^3))-(3*B2*EXP(3*B2))-(EXP(3*B2))-(3/(2*SQRT(2-B2)))(4*COS(4*B2))
For Answer: use formula: =B2-(C2/D2)
For đ?‘Ľđ?‘› (Row 3 and so on) except Row 1: use formula: =B2-C2/D2
*Repeat the step until the same value of roots(at least 2) found.
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2.3
Results and Discussion
In this method, we get the equation sin 4đ?‘Ľ − 3√2 − đ?‘Ľ + đ?‘Ľđ?‘’ 3đ?‘Ľ = 3đ?‘Ľ 4 (from question task 2) and we arrange it equal to zero. We get đ?‘“ (đ?‘Ľ) = 3đ?‘Ľ 4 − sin 4đ?‘Ľ + 3√2 − đ?‘Ľ − đ?‘Ľđ?‘’ 3đ?‘Ľ , we differentiate the equation and we get đ?‘“′(đ?‘Ľ ) = 12đ?‘Ľ 3 − 3đ?‘Ľđ?‘’ 3đ?‘Ľ − đ?‘’ 3đ?‘Ľ −
3 2√2−đ?‘Ľ
− 4 cos(4đ?‘Ľ).Then, we create the
function( đ?‘“(đ?‘Ľ ) and đ?‘“′(đ?‘Ľ ) ) in the Microsoft Excel. Next, with initial approximation đ?‘Ľ1 = 1, we find approximation value using formula đ?‘Ľđ?‘›+1 = đ?‘Ľđ?‘› −
�(�� ) �′(�� )
. Then, the step was repeated until we
get the answer is equal to previous final answer. The objective of this method is to find the approximate value of the root for the non-linear equation by using Newton-Raphson method.
From the excel, the approximation of the root we get are đ?‘Ľ2 = 0.801737, đ?‘Ľ3 = 0.653257, đ?‘Ľ4 = 0.581447, đ?‘Ľ5 = 0.569696, đ?‘Ľ6 = 0.569454, đ?‘Ľ7 = 0.569453, đ?‘Ľ8 = 0.569453 .We stop at đ?‘Ľ8 = 0.569453, because the value of đ?‘Ľ8 is equal to đ?‘Ľ7 . That indicate, the number iteration must be stop because the approximate value we get is so closed to the root of the function(the root from the graph when đ?‘“ (đ?‘Ľ ) = 0 ). Increasing the number of iterations, means we increase the approximation value of the root to get the precise value for the root to solve this non-linear equation.If we plotted the tangent for đ?‘Ľ2 we can see that đ?‘Ľ2 is closed to the root compare to the tangent for đ?‘Ľ1 . Then if we plotted the tangent for đ?‘Ľ8 ,we can see that đ?‘Ľ8 is closer to the root compare to đ?‘Ľ2 .Considering that, the number of iterations can increase the approximation value of the root accurately.
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CONCLUSION From this project, we can conclude that the half-life method can help us in many field such as to determine the ages of very old artifacts. This is because the accuracy that the method gives is very high. Newton-Raphson method is very useful to find the approximate value of the root for the nonlinear equation. We also learn how to plot a graph using desmos website while doing this project.
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BIBLIOGRAPHY
[1] Lopez, A. (2014). 08.01 Half-life and Radioactive Decay: Half-life Lab. Retrieved March,16,2019 from https://prezi.com/vmxqpi-la-pj/0801-half-life-and-radioactive decay-half-life-lab/ [2] Growth and decay problem. (n.d) Retrieved March 13, 2019, from https://en.wikipedia.org/wiki/Growth_and_decay_problem [3] Half-life. (n.d) Retrieve March 13, 2019, from https://en.wikipedia.org/wikki/Half-life [4] Serway, R. & Vuille, C. (2018). College Physic. China: Cengage Learning.
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