Add Math Project Work 1 2010

June 22, 2018 | Author: Hon Loon Seum | Category: Integral, Calculus, Derivative, Mathematical Concepts, Mathematics
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Additional Mathematics Project Work 12010 Important of Calculus in our Daily Life Name IC Number Class School Teacher : : : : : Seum Hon Loon 931116 - 08 - 6471 5 Sains Iktisas SMK Dato Bendahara C.M. Yusuf Pn. Rohidah Binti Ngah @ Salleh Contents Title Appreciation Objectives Introduction History Procedure and Findings Further exploration Conclusion Reflection Page 2 3 4 6 9 12 23 24 Appreciation First of all, I would like to thank my teacher, Pn. Rohidah Binti Ngah @ Salleh for guiding me and my friends throughout this project. We had some difficulties in doing this task, but she taught us patiently until we knew what to do. She tried and tried to teach us until we understand what we supposed to do with the project work. Not forgotten my parents for providing everything, such as money, to buy anything that are related to this project work and their advise, which is the most needed for this project. Internet, books, computers and all that. They also supported me and encouraged me to complete this task so that I will not procrastinate in doing it. Last but not least, my friends who were doing this project with me and sharing our ideas. They were helpful that when we combined and discussed together, we had this task done. Objectives The aims of carrying out this project are : • To learn the way to apply the formulas of mathematics in our daily life accurately. • To widen our prospective view on mathematics. • To improve our mastering skills. • To use the correct language to express our mathematical ideas properly. • To develop positive attitude towards mathematics • to improve our way of thinking Introduction Integration is an important concept in mathematics and, together with differentiation, is one of the two main operations in calculus. Given a function ƒ of a real variable x and an interval [a, b] of the real line, the definite integral is defined informally to be the net signed area of the region in the xy-plane bounded by the graph of ƒ, the x-axis, and the vertical lines x = a and x = b. The term integral may also refer to the notion of antiderivative, a function F whose derivative is the given function ƒ. In this case it is called an indefinite integral, while the integrals discussed in this article are termed definite integrals. Some authors maintain a distinction between antiderivatives and indefinite integrals. The principles of integration were formulated independently by Isaac Newton and Gottfried Leibniz in the late 17th century. Through the fundamental theorem of calculus, which they independently developed, integration is connected with differentiation: if ƒ is a continuous real-valued function defined on a closed interval [a, b], then, once an antiderivative F of ƒ is known, the definite integral of ƒ over that interval is given by Integrals and derivatives became the basic tools of calculus, with numerous applications in science and engineering. A rigorous mathematical definition of the integral was given by Bernhard Riemann. It is based on a limiting procedure which approximates the area of a curvilinear region by breaking the region into thin vertical slabs. Beginning in the nineteenth century, more sophisticated notions of integrals began to appear, where the type of the function as well as the domain over which the integration is performed has been generalised. A line integral is defined for functions of two or three variables, and the interval of integration [a, b] is replaced by a certain curve connecting two points on the plane or in the space. In a surface integral, the curve is replaced by a piece of a surface in the three-dimensional space. Integrals of differential forms play a fundamental role in modern differential geometry. These generalizations of integral first arose from the needs of physics, and they play an important role in the formulation of many physical laws, notably those of electrodynamics. There are many modern concepts of integration, among these, the most common is based on the abstract mathematical theory known as Lebesgue integration, developed by Henri Lebesgue. History The product rule and chain rule, the notion of higher derivatives,Taylor series, and analytical functions were introduced by Isaac Newton in an idiosyncratic notation which he used to solve problems of mathematical physics. In his publications, Newton rephrased his ideas to suit the mathematical idiom of the time, replacing calculations with infinitesimals by equivalent geometrical arguments which were considered beyond reproach. He used the methods of calculus to solve the problem of planetary motion, the shape of the surface of a rotating fluid, the oblateness of the earth, the motion of a weight sliding on a cycloid, and many other problems discussed in his Principia Mathematica. In other work, he developed series expansions for functions, including fractional and irrational powers, and it was clear that he understood the principles of the Taylor series. These ideas were systematized into a true calculus of infinitesimals by Gottfried Wilhelm Leibniz, who was originally accused of plagiarism by Newton. He is now regarded as an independent inventor of and contributor to calculus. His contribution was to provide a clear set of rules for manipulating infinitesimal quantities, allowing the computation of second and higher derivatives, and providing the product rule and chain rule, in their differential and integral forms. Unlike Newton, Leibniz paid a lot of attention to the formalism – he often spent days determining appropriate symbols for concepts. Leibniz and Newton are usually both credited with the invention of calculus. Newton was the first to apply calculus to general physics and Leibniz developed much of the notation used in calculus today. The basic insights that both Newton and Leibniz provided were the laws of differentiation and integration, second and higher derivatives, and the notion of an approximating polynomial series. By Newton's time, the fundamental theorem of calculus was known. INTEGRATION BY PARTS One of very common mistake students usually do is To convince yourself that it is a wrong formula, take f(x) = x and g(x)=1. Therefore, one may wonder what to do in this case. A partial answer is given by what is called Integration by Parts. In order to understand this technique, recall the formula which implies Therefore if one of the two integrals and is easy to evaluate, we can use it to get the other one. This is the main idea behind Integration by Parts. Let us give the practical steps how to perform this technique: 1 Write the given integral where you identify the two functions f(x) and g(x). Note that if you are given only one function, then set the second one to be the constant function g(x)=1. 2 Introduce the intermediary functions u(x) and v(x) as: Then you need to make one derivative (of f(x)) and one integration (of g(x)) to get Note that at this step, you have the choice whether to differentiate f(x) or g(x). We will discuss this in little more details later. 3 Use the formula 4 Take care of the new integral . The first problem one faces when dealing with this technique is the choice that we encountered in Step 2. There is no general rule to follow. It is truly a matter of experience. But we do suggest not to waste time thinking about the best choice, just go for any choice and do the calculations. In order to appreciate whether your choice was the best one, go to Step 3: if the new integral (you will be handling) is easier than the initial one, then your choice was a good one, otherwise go back to Step 2 and make the switch. It is after many integrals that you will start to have a feeling for the right choice. In the above discussion, we only considered indefinite integrals. For the definite integral to go: , we have two ways 1 Evaluate the indefinite integral which gives 2 Use the above steps describing Integration by Parts directly on the given definite integral. This is how it goes: (i) Write down the given definite integral where you identify the two functions f(x) and g(x). (ii) Introduce the intermediary functions u(x) and v(x) as: Then you need to make one derivative (of f(x)) and one integration (of g(x)) to get (iii) Use the formula (iv) Take care of the new integral . The following examples illustrate the most common cases in which you will be required to use Integration by Parts: EXAMPLE Evaluate First let us point out that we have a definite integral. Therefore the final answer will be a number not a function of x! Since the derivative or the integral of lead to the same function, it will not matter whether we do one operation or the other. Therefore, we concentrate on the other function . Clearly, if we integrate we will increase the power. This suggests that we should differentiate and integrate . Hence After integration and differentiation, we get The integration by parts formula gives It is clear that the new integral is not easily obtainable. Due to its similarity with the initial integral, we will use integration by parts for a second time. The same discussion as before leads to which implies The integration by parts formula gives Since , we get which finally implies Easy calculations give From this example, try to remember that most of the time the integration by parts will not be enough to give you the answer after one shot. You may need to do some extra work: another integration by parts or use other techniques,.... Procedure and Findings Solution: (a)Function 1 Maximum point (0,4.5) and pass through point (2,4) y=a(x-b)²+ c b=0, c=4.5 y=a(x-0)²+ 4.5 y=ax²+ 4.5---(1) Substitute (2,4) into (1) 4=a(2)²+ 4.5 4a= -0.5 a= -0.125 ∴y= -0.125x²+ 4.5 Function 2 Maximum point (0, 0.5) and pass through point (2, 0) y=a(x-b)²+ c b=0, c=0.5 y=a(x-0)²+ 0.5 y=ax²+ 0.5---(2) (2,0) into (2) Substitute 0=a(2)²+ 0.5 4a= -0.5 a= -0.125 ∴y= -0.125x²+ 0.5 Function 3 Maximum point (2, 4.5) and pass through point (0, 4) y=a(x-b)²+ c b=2, c=4.5 y=a(x-2)²+ 4.5---(3) (0,4) into (3) Substitute 4=a(0-2)²+ 4.5 4a= -0.5 a= -0.125 ∴y= -0.125(x-2)²+ 4.5 (b) Area to be painted = Area of rectangle - Area under the curve = ?? Further Exploration Solution: (a)(i) Structure 1 Area = ??? (223m²) Volume = Area x Thickness =223m² x 0.4 m = 1615m³ Cost = 1615m³ x RM840 = RM896 Structure 2 Area=Area of Rectangle-Area of Triangle =1m x 4m-12 x 4m x 0.5m =4m²- 1m² =3m² Volume =Area x Thickness =3m² x 0.4m =1.2m³7 Cost =1.2m³ x RM840 =RM1008 Structure 3 Area=Area of Rectangle-Area of Trapezium =1m x 4m-(4m+1m)² x 0.5m =4m²- 54m² =2.75m² Volume =Area x Thickness =2.75m² x 0.4m =1.1m³ Cost =1.1m³ x RM840 =RM924 Structure 4 Area=Area of Rectangle-Area of Trapezium =1m x 4m-(2m+4m)2 x 0.5m =4m2- 1.5m2 =2.5m2 Volume =Area x Thickness =2.5m2 x 0.4m =1m3 Cost=1m3 x RM840 =RM840 ∴Structure 4 will cost the minimum to construct, that is RM 840. (ii) As the president of the Arts Club, I will decide Structure 4 as the shape of the gate to be constructed. It is because Structure 4 will cost the minimum and it is easier to be constructed compared to Structure 1 which is a curve. (b) (i) k (m) 0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00 Area to be painted(m²) 4 x 1- 0+42 x 0.5= 3 4 x 1- 0.25+42 x 0.5= 2.9375 4 x 1- 0.5+42 x 0.5= 2.875 4 x 1- 0.75+42 x 0.5= 2.8125 4 x 1- 1+42 x 0.5= 2.75 4 x 1- 1.25+42 x 0.5= 2.6875 4 x 1- 1.5+42 x 0.5= 2.625 4 x 1- 1.75+42 x 0.5= 2.5625 4 x 1- 2+42 x 0.5= 2.5 (ii) There is a pattern in the area to be painted. The area to be painted decreases as the k increases 0.25m and form a series of numbers: 3, 2.9375, 2.875, 2.8125, 2.75, 2.6875, 2.625, 2.5625, 2.5 We can see that the difference between each term and the next term is the same. 2.9375-3 2.875- 2.9375 2.8125- 2.875 2.75- 2.8125 2.6875- 2.75 2.625- 2.6875 2.5625- 2.625 2.5- 2.5625 = -0.0625 = -0.0625 = -0.0625 = -0.0625 = -0.0625 = -0.0625 = -0.0625 = -0.06257 ∴ We can deduce that this series of numbers is an Arithmetic Progression (AP), with a common difference, d= -0.0625 By using K = 0 m, the structure may be like this: When the value of K gradually increases, the area will gradually decreases. By using K = 0.25 m, the structure may be like this: When K approaches more to 4, the shape of concrete structure will be something like: In conclusion, when k increases 0.25m, the area to be painted decreases by -0.0625m² (c) The area of the concrete structure to be painted Area of concrete structure to be painted →3-1 →2m² The shape of the concrete structure will be a rectangle with length 4m and breadth 0.5m, which may look like this: Conclusion As I doing this project, I notice that quadratic function and integration can be so close in our daily life. There are many shape of gate outside there. Different shapes of the gate have different cost. From quadratic function and integration, we can know area of the gate. From the area we can get volume og the gate. As the result, we can know cost of the gate by times volume with RM840 (price for 1m³) After doing research, answering questions, drawing graphs and some problem solving, I saw that the usage of calculus is important in daily life.It is not just widely used in science, economics but also in engineering. In conclusion, calculus is a daily life nessecities. Without it, marvelous buildings can’t be built, human beings will not lead a luxurious life and many more. So, we should be thankful of the people who contribute in the idea of calculus. Reflection After spending countless hours,days and night to finish this project and also sacrificing my time for chatting and movies in this mid year holiday,there are several things that I can say... Additional Mathematics... From the day I born... From the day I was able to holding pencil... From the day I start learning... And... From the day I heard your name... I always thought that you will be my greatest obstacle and rival in excelling in my life... But after countless of hours... Countless of days... Countless of nights... After sacrificing my precious time just for you... Sacrificing my play Time.. Sacrificing my Chatting... Sacrificing my Facebook... Sacrificing my internet... Sacrifing my Anime... Sacrificing my Movies... I realized something really important in you... I really love you... You are my real friend You my partner... You are my soulmate... I LOVE U ADDITIONAL MATHEMATIC


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