Infinite surds Maths Portifolio

Mathematics SL Portifolie Infinite Surds An infinite surd is a never-ending positive irrational number. It is a number that can only be expressed exactly using the root sign . This sequence above is known as an infinite surd and can be expressed in the terms of an: a1 = = 1.414213 a2 = = 1.553773 a3 = = 1.598053 a4 = = 1.611847 a5 = = 1.616121 a6 = = 1.617442 a7 = = 1.617851 a8 = = 1.617977 a9 = = 1.618016 etc. This is the first ten terms and the formula for these sequences is: because if we use the term a2 as an example, this could be proven as: a1+1 = a2 = a2 = 1.553773 By plotting a graph of the ten first term of this sequence the relationship between n and an could be shown: As can be seen from this graph is that the values increase, but then flattens out. The values of an moves towards the value of 1.618 approximately, but will actually never reach it. This can be understood by: an - an+1 as n gets very large. lim(an - an+1) ? 0 When n gets very large and approaches infinity, the value approaches 0 because the difference between these two values become very small. This is a way to find the exact value of this infinite surd: an+1 can also be written as an so therefore an = an2 = 1+an an2 - an - 1 = 0 abc- formula: a = a1 = 1.618 a2 = -0.618 Since the value has to be a positive number, the exact value of this infinite surd is

  • Word count: 911
  • Level: International Baccalaureate
  • Subject: Maths
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Gold Metal Heights in High Jump.

Hannah Krohn Gold Metal Heights Aim: The aim of this task is to consider the winning height for the men’s high jump in the Olympic games. The graph below gives the Winning height (in centimeters) by the gold medalists at various Olympic games form 1932 to 1980 with the exception of 1940 and 1944. Constraints to this graph: Since there were no Olympic games in 1940 and 1944 the slope is from 1936 to 1948 is not defined on the same domain as the other data points. This is limiting because the amount of data points vs. time is not consistent throughout the whole graph so it will be more difficult to create an accurate model. The function that is modeled after the data points and graph is a linear graph seen below: y The following variables on the next page are the variables that are used in the linear equation. To achieve these numbers plug in the data in the form of linear regression, where the height (y-value) is the dependent variable while the year (x-value) an explanatory value or independent. Also, the value (coefficient of determination) is used in statistics whose main purpose is to predict the future outcomes on the basis of related information. In the case of linear regression, between the outcomes and the values of the single regressor being used for prediction. In such cases, the coefficient of determination ranges from 0 to 1. Contextually, the

  • Word count: 898
  • Level: International Baccalaureate
  • Subject: Maths
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Shady Areas

Maths Internal Assessment: Shady Areas Introduction: In this investigation you will attempt to find a rule to approximate the area under a curve (i.e. between the curve and the x - axis) using trapeziums (trapezoids). Let us consider using the function: . Before finding the area under the curve using trapeziums. I will use integration to help me find the exact area. This will enable me to compare the results I acquire using the trapezium method and to find out whether the use of more trapeziums will provide me with a more accurate estimation of the area under the graph. Using the integration method, I have found the exact area for the function is I will now use the trapezium method to find the approximation of the area under the curve. I have used Autograph to show the graph for the function of g. It can clearly be seen from the graph that the area under the curve is roughly by the sum of the areas of two trapeziums. The area of a trapezium can be found using the formula: Let a = the length of one side of the trapezium b = the length of the other side of the trapezium h = the vertical height between the two lengths In order, Although there are two trapeziums, there are three different values and in order to determine the area of the trapeziums at each value I will work out the values on the graph using the trapezium method. I will now input the

  • Word count: 891
  • Level: International Baccalaureate
  • Subject: Maths
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Quadratic Polynomials. Real and Imaginary components

MATHS IA Alex Chen ________________ PART A (Quadratic Polynomials) The investigation is to find out if the zeros and to determine the real and imaginary components of the complex zeros of . From the function given, The coordinates of the vertex is by using the Quadratic equation: where Hence, has zeros , and By subbing in different numbers of into the equation: For: , it is given that , which is equal to : : Value of a value of b value of y1 1 1 2 2 3 3 4 4 Value of a value of b value of y1 1 1 2 2 3 3 4 4 For: : : Value of a value of b value of y1 1 1 2 2 3 3 4 4 Value of a value of b value of y1 1 1 2 2 2 3 3 4 4 After subbing different values for and From the above results, by comparing with , it can be seen that their values are opposite, have negative results, ’s results are always a positive number or bigger than 0. A graph of y1 and y2 is shown below when a= 3 and b=5, We know that has zeros , while has opposite concavity to ,which is in the form . From the graph, it can be seen that, is a reflection of , Therefore, the equation of the quadratic is : . When

  • Word count: 881
  • Level: International Baccalaureate
  • Subject: Maths
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MATRIX BINOMIALS. In this investigation, we will identify a general statement by examining the patterns of the matrices.

MATHEMATICS STANDARD LEVEL INTERNAL ASSESSMENT MATRIX BINOMIALS Summary of Investigation: A matrix can be defined as a rectangular array of numbers of information or data that is arranged in rows and columns. There are a number of operations in which these matrices can perform (i.e., addition, multiplication, etc). In this investigation, we will identify a general statement by examining the patterns of the matrices. Investigation: To begin with, we are given the two matrices, X and Y. We let X=and Y=, and calculated = == ==== ==== === ==== === = And therefore, = , = , = = , = , = Here, it seems reasonable to suggest a pattern for the X and Y values. And so, by considering integer powers of X and Y, we can find the expressions for,: = , =, With the aforementioned expressions for the value of and, we will now determine the value for. This can be done through substituting the value of n to find a pattern for the matrices, as done so when determining the value of and. Thus, with these patterns, the following expression can be suggested: The matrices X and Y can now be used to form two new matrices A and B. Here, we will use a and b as constants for the matrices A and B, respectively. And hence the following: , Now, the different values of a and b can be used to calculate the values of And therefore, , , , , With the patterns from these

  • Word count: 880
  • Level: International Baccalaureate
  • Subject: Maths
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Shady Areas

Shady Areas I shall investigate different ways of finding a working rule to approximate the area under a curve using trapeziums. The area under the curve represents area between the f(x) and x values under the curve in the specified area. Therefore, through integration to find the area, integration of the integrated area will find the volume. Consider the function g(x) = x² + 3 g(x) = x² + 3 (see Graph 1) The area under this curve from x=0 to x=1 is approximated by the sum of the area of two trapeziums. The approximation can be discovered by working out the area of the square in the trapezium and then by working out the triangle in the trapezium. The sum of this will give the area of the two trapeziums, which can be summed together. The area of a trapezium can therefore be worked out using the formula (b x c) + 1/2(a x d) = area of trapezium bc + 1/2(e - b)c c [ b + 1/2(e - b) ] c [ 2b/2 + (e-b)/2 ] = c[ (e + b)/2 ] x = g(x) for x² + 3= g(x) 0.2 3.04 0.4 3.16 0.6 3.36 0.8 3.64 .0 4 0.0 3 Graph 1 area = 0.5 [ (3.25 + 3)/2 ] = 1.5625 = area of first trapezium 0.5 [ (4 + 3.25)/2 ] = 1.8125 = area of second trapezium .5625 + 1.8125 = 3.375 For Graph 1 the sum of the area of the two trapeziums gives a result of 3.375 By increasing the number of trapeziums we can gain a more accurate estimate of the area under the curve. Using the

  • Word count: 880
  • Level: International Baccalaureate
  • Subject: Maths
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Integral Portfolio

Trapezium Portfolio Raghav "Ron" Nair Scot Millhollen IB/AP Calculus SL January 21, 2009 Table Of Contents Introduction Page.................................................................................... 1 Table of Contents..................................................................................... 2 Goal........................................................................................................... 3 First Function and Five Trapeziums...................................................... 4 Seven Trapeziums................................................................................... 5 Eight Trapeziums..................................................................................... 6 Nine Trapeziums...................................................................................... 7 General Expression with "n" Trapeziums............................................. 8 The General Statement........................................................................... 9 Three New Curves.................................................................................... 10 Curve # 1.................................................................................................... 11 Curve # 2.................................................................................................... 12 Curve #

  • Word count: 860
  • Level: International Baccalaureate
  • Subject: Maths
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Portfolio Type II: Stopping Distances

Maths Portfolio Type II- Stopping Distances The table below shows the average stopping and thinking distances when a person is driving a car and needs to apply the brakes at various speeds. Speed (kmh-1) Thinking Distance (m) Braking Distance (m) 32 6 6 48 9 4 64 2 24 80 5 39 96 8 55 12 21 75 From this data we can graph two data plots: one showing the relationship between speed versus and the other between speed versus braking distance. Figure 1 Figure 2 Figure 1 shows a straight line, and therefore it can be said that it is a linear graph. It shows that the correlation between speed and thinking distance is directly proportional and shows the trend that as the speed increases the thinking distance also increases. The equation of this graph is in the form, where m is the gradient, and c is the y intercept. The equation can be worked by following these steps: * Find m (the gradient) using the following equation: * b is the y intercept but as the graph passes through the origin this value is 0 because while the car is moving at 0 kmh-1 the braking distance is 0 m. * The final linear equation is for the domain 32<x<112 because we do not know how long the thinking distance will be for any higher speeds. We could guess from the graph but in a real life situation this would be putting the drivers' life at risk. Although finding the equation of

  • Word count: 856
  • Level: International Baccalaureate
  • Subject: Maths
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IB Design Tech Design Cycle

Majid Jamialahamdi MJ Tallest Straw Design Identify Needs In December 10, 2009 Mr.Youmans told us to create the tallest straw building by just using straws. The point of this project was to be creative and use our engineering skills to create a straw building that will have a strong support. Researching I thought of a way to create the tallest straw building by creating support inside the building and thickening the base. I opened Netscape internet browser and started researching the best straw building. I used the picture on http://www.mysciencebox.org/files/images/Earthquake%20tower.img_assist_custom.jpg to create my building and the reason why I chose it is because it resembles the buildings that are used today. Generating Ideas For the building I used 55 straws and 55 straw papers to tie the straws together so they won't fall apart. I've realized in my research that in order to have a strong building the base has to be well-built to hold the straws and also the joints should be thickened to hold the tower straight. For the bridge, I thought that tying 3 straws to form a bundle and using 4 bundles to form a square and 2 other bundles to cross inside the square will create a strong base. Every 5 inches I will place a base so that the building will hold up straight. My goal for the tower was to be 2 feet tall or even taller. For the joints there should be a 40

  • Word count: 851
  • Level: International Baccalaureate
  • Subject: Maths
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IB Math HL Portfolio Type I Series and Induction

Mathematics Higher Level Portfolio Type I Series and Induction Acknowledgement Question sheet should directly be given from your IB Mathematics HL teachers. This is due to the fact that IB students are not allowed to hold any question paper; every candidate must finish this coursework and return the question sheet in five days. Therefore I was not able to include any questions in this portfolio. Introduction This Investigation of the Series and Induction Portfolio for Math HL brings out that the sum of terms of series following a certain pattern can be predicted as expressions by studying these patterns. With the resourceful use of a calculator, studying the graphs and undertaking regression of the data we can easily deduce the general term and by further considering similar series we can predict they're expressions which emerge rational and true when induced in terms of a proposition. Recalling that 1+2+3+4+5+........+. 1. Considering where is the general term and 'n' is the number of terms, Let us take into consideration the first term, The first term implies that . Since = 1, We can also say that, Let us also take into consideration, The second term implies that and so on. Question 1 a1 = 1•2 = 2 a2 = 2•3 = 6 a3 = 3•4 = 12 a4 = 4•5 = 20 a5 = 5•6 = 30 a6 = 6•7 = 42 an = n(n+1) = n2+n Question 2 A) S1 = a1 = 2 S2 = a1+ a2 =8 S3 = a1+...+ a3 =

  • Word count: 847
  • Level: International Baccalaureate
  • Subject: Maths
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