#### Stopping distances portfolio. In this task, we may develop individual functions that model the relationship between speed and thinking distance, as well as speed and braking distance. We could also develop a model for the relationship between speed and ov

IB Diploma Standard Level Mathematics Portfolio: Type II Stopping Distances Rami Meziad section 11/8 American college of Sofia STOPPING DISTANCES Introduction Stopping a moving car requires to apply the brakes which must actually stop the vehicle. There must be a special mathematical relationship between the speed of a car and the thinking distance, braking distance and overall stopping distance. In this task, we may develop individual functions that model the relationship between speed and thinking distance, as well as speed and braking distance. We could also develop a model for the relationship between speed and overall distance. This will be achieved by using Excel Graphing Software to create two data plots: speed versus thinking distance and speed versus braking distance, for which the results will be evaluated and described. Subsequently we can develop a function that will model the relationship and behavior of the graphs based on basic knowledge of functions. After all that we can add the thinking and breaking distance to obtain an overall stopping distance, from which we will once again graph the data and describe it's results, whilst evaluating the correlation between the graph obtained from these results and the graphs obtained from the comparison between the speed vs. thinking/braking distance. Finally, we will evaluate how the additional set of data

#### Moss's Egg. Task -1- Find the area of the shaded region inside the two circles shown below. The two large circles have a radius of 6cm.

Moss's Egg The following formulas were used in the solving of the questions of this assessment piece: Area of a Circle Circumference of a Circle Area of a Sector Arc Length Working Out and Explanation Task -1- Find the area of the shaded region inside the two circles shown below. The two large circles have a radius of 6cm. Their centres are A and B. From the information given above, we know that the radii of the two larger circles are 6 cm in length. We define the radius of the circle as a straight line extending from the centre of a circle to its circumference. Since we know that points A and B are the centres of the two large circles, we can conclude that this is the length of the two points from A to B is 6 cm also, since point B is along the circumference of the top larger circle, and vice versa. From lengths A to B is the therefore the diameter of the smaller circle between the two larger ones, and thus we can conclude that the radius of the smaller circle is 3 cm. The area of the small circle can therefore be calculated using the formula indicated: , where A equals the area and r is the radius. Thus: A = (32) = 9 ˜ 28.3 cm2 Task -2- The same circles are shown below. Find the area and perimeter of the triangle ABC. a) In order to determine the area of triangle ABC, we must adopt the formula: , where b is the base and h is the

#### Mathematics Higher Level Internal Assessment Investigating the Sin Curve

Investigating the Sine Curve This report investigates the sine curve in the form o f, and how that relates to the graph of the sine curve. In particular, it would be investigated how the different variables ( effect the way that the graph is drawn and then seeing if the rule can be generalized to apply to any form of the equation. The first thing to do would be to allow and to be 0, which would mean that the equation takes the form of: . It can be seen from Graph 1.1 that when is 1 what graph you get (the red graph) and when is allowed to be 2 what the graph looks like (the blue graph). From the graph below it can be seemn that increasing stretches the graph by the factor of change in . In simple words, the graph of would be twice the height of as is clearly seen from Graph 1.1. When I change the value for , all the value of the sine curve get multiplied by that value of , which is 2 in this case. By multiplying all the values of the curve by the height of each point in the curve increases while there is no change in the position of the graph in the x-axis. For the original sine graph () it is common knowledge that the relative minima and maxima are -1 and 1 respectively, however when the is changed the original minima and maxima are also multiplied by and therefore the new minima and maxima would become and respectively. The rest of the graph also gets

#### Infinite summation SL Portfolio type I. Concerning the portfolio, the evaluation is composed on the sum of the series

Infinite summation - SL Portfolio type I Name: Abdelrahman Lotfy Mosalam Presented to: Mr. Hamed Mokhtar Date: 11/10/2011 Class: DP2 The infinite series is almost considered as the main tool in calculus, it has different utilizes. It guesses the behavior of functions, investigates differential equations and also it's used in numerical analysis. Beside these uses in math, the infinite series may be used in physics and economics as well. Concerning the portfolio, the evaluation is composed on the sum of the series below, where: t0 = 1, t1 = , t2 = , t3 =, ... , tn =.... for the value in the series above, the task gave a notation for the factorial , described as : (Given) To find out the general statement that represents the infinite sum of this sequence, it's always required to determine the sum of the first terms of the infinite sequence for , for where Using the notations given mentioned and the sequence values, we concluded this table containing the variation of and it's effect on the Note for ALL tables and graphs: The only controlled variable in the table is the All of the answers below will be corrected to 6 decimal places Domain of n is Graphs are all done using Microsoft Excel Relation between and ( 0 2 .000000 2 .693147 2 2 .933374 3 2 .988878 4 2 .998496 5 2 .999829 6 2 .999983 7 2 .999999 8 2 2.000000 9 2

#### IB SL Math Portfolio- Body Mass Index

SL PORTFOLIO TYPE II "Body Mass Index" BODY MASS INDEX Throughout this portfolio, various functions will be evaluated, applying the given data. A model function will be determined and extrapolated as it relates to the following real-world example: Body mass index (BMI) is a measure of one's body fat. It is calculated by taking one's weight (kg) and dividing it by the square of one's height (m). The table below provides the median BMI for females of varying ages in the US in the year 2000. Age (yrs) BMI 2 6.40 3 5.70 4 5.30 5 5.20 6 5.21 7 5.40 8 5.80 9 6.30 0 6.80 1 7.50 2 8.18 3 8.70 4 9.36 5 9.88 6 20.40 7 20.85 8 21.22 9 21.60 20 21.65 (Source: http://www.cdc.gov) When graphed: (The independent variable being the age of the women studied <x>, and the dependent being the BMI of these women <y>. Both values must always be greater than zero. ) This graph's behavior is most nearly modeled by the cosine function,, because it is undulating and periodic: it repeats a pattern as it rises and falls. However, due to the limitations presented by the nature of the given information itself, only the portion of the graph in the first quadrant that is positive applies, as both the age of the women and their respective body mass index values are real world examples and could never be negative. Other functions

#### Math SL Circle Portfolio. The aim of this task is to investigate positions of points in intersecting circles.

Circles Jennifer Aim: The aim of this task is to investigate positions of points in intersecting circles. Introduction The following diagram shows a circle with centre O and radius r, and any point P. In this case, the r is the distance between any point (such as A) and the centre point O of the circle . Since the radius is 1 unit, OP will also be 1 unit away from the point O. The circle has centre P and radius . A is one of the points of intersection of and . Circle has centre A, and radius r. The point P' is the intersection of with (OP). The r=1, =2, and P'=0.5. This is shown in the diagram below. This investigation will explore in depth of the relationship between the r value and values, when r is held constant and values modified. It will also investigate the reverse, the relationship when the values are held constant and the r values are modified. In the first example, the r value given is 1. An analytic approach will be taken to find the length of ' when =2. Firstly, one can note that 2 isosceles triangle can be drawn by using the points A, O, P', and P. It is shown in the diagram below. In ?AOP', lines and have the same length, because both points, O and P' are within the circumference of the circle , which means that and are its radius. Similarly, ?AOP forms another isosceles triangle, because the lines and are both radii of the circle .

#### Maths Portfolio - Population trends in China

Mathematics Portfolio Topic: Populations trend in China Date: 28.02.2012 In this portfolio, we should investigate and use mathematical functions and equations that would portray the model in the best way. The relevant variables in this investigation are the population in millions in different years. The parameter is the initial population growth. I have plotted the points given in the table above in the graph shown below. *the year points of the picture above are replaced from 1950 to 50, 1960 to 60, etc. [plotted with Microsoft Excel 2007] We can clearly see from the graph above that the obvious trend occurring is the rise of the population over a period of time and we can see it is increasing gradually, so we could possibly present the model through a linear equation like: Y = aX + b where a are the variables whilst b are the parameters. Now I will try to develop the model using a linear function. In order to start, we will firstly need to find the slope. And to find the slope, we divide the difference between the y-values with the x-values. After the slope, we find the y-intercept, y1. After we find this formula, we plot it into the graph, and we get the line that approximates the original points we got from the data. Below is the graph: *graph above shows the linear function we found, plotted with the actual points. [plotted with GeoGebra4] We can clearly see

#### Infinite summation portfolio. A series is a sum of terms of a sequence. A finite series, has its first and the last term defined, and the infinite series, or in other words infinite summation

International Baccalaureate Math Standard Level Internal Assessment Portfolio Type: I Portfolio Title: Infinite Summation Due date: 9th of December, 2011 Teacher: Mr. Peter Vassilev School: The American College of Sofia Candidate Name: Rami Meziad Candidate number: 002368-008 Examination Session: May 2012 Introduction: A series is a sum of terms of a sequence. A finite series, has its first and the last term defined, and the infinite series, or in other words infinite summation [3] is a series which continues indefinitely. The Taylor's theorem [1] and the Euler-Maclaurin's formula [2] will help us solve our given infinite summation, which is: ,,,, And by adding different values for x and a, we will be able to find a general pattern in which the sequences tends to move with. And this is mainly what this portfolio will ask us to do. Method: For our sequence, which is: , we have to substitute in the case where x = 1 and a = 2. After that, we have to calculate the first n terms which happen to be eleven to fulfill the given condition. So after substitution we get Now let's calculate for n, when : In fact, t9 and t10 are not equal to 0, but since we have to take our answers correct to six decimal places, we can't see the real values. However, the numbers become so small, that they become insignificant, or in other words they are equal to 0. Now, we need to find

#### Infinite Surds portfolio

Internal Assessment number 1 Nazha AlFaraj Ms. Leana Ackerman IB Mathematics SL (year 2) Sunday, February 19, 2012 Infinite Surds This following expression is known as an infinite surd. V1+V1+V1+V1+... The previous infinite surd can be changed into the following sequence: a1= V1+V1= 1,414213 a2= V1+V1+V1= 1,553773 a3= V1+V1+V1+V1= 1,598053 a4= V1+V1+V1+V1+V1= 1,611847 a5= V1+V1+V1+V1+V1+V1= 1,616121 a6= V1+V1+V1+V1+V1+V1+V1= 1,617442 a7= V1+V1+V1+V1+V1+V1+V1+V1= 1,617851 a8= V1+V1+V1+V1+V1+V1+V1+V1+V1= 1,617977 a9=V1+V1+V1+V1+V1+V1+V1+V1+V1+V1= 1,618016 a10= V1+V1+V1+V1+V1+V1+V1+V1+V1+V1= 1,618028 The first 10 terms can be represented by: an+1= V1 + an If we graph the first 10 terms of this sequence we can show that the relationship between n and L can be represented by L= an The data begins to increase by a smaller amount about each consecutive n, suggesting that the data may be approaching as asymptote. As these values get very large, they will probably not get much higher than the value of a10, because there already appears to be almost horizontal trend. The data also suggests that the asymptote is between the value of 6 and seven, although to find the exact value requires a different approach The graph clearly shows that the value of L gradually moves approximately towards 1,618, but it will never reach that number. Furthermore, the relationship

#### Investigating Logarithms

Investigating Logarithms log2 + log3 0.7782 log6 0.7782 log3 + log7 .322 log21 .322 log4 + log 20 .903 log80 .903 log0.2 + log11 0.3424 log2.2 0.3424 log0.3 + log 0.4 -0.9208 log0.12 -0.9208 This table to the left clearly shows that the log of 2 numbers added together will equal the log of the number multiplied. The table below clearly shows that log (?) + log (y) will equal log (?y). Let log x = a, let log y = b. Therefore 10a = x and 10b = y, these two equations can then be simplified to 10(a+b) =x*y. it is then possible to convert this back to log (xy) = a + b. log5 + log4 log20 .301 log3 + log2 log6 0.7782 log4 + log8 log32 .505 log6 + log3 log18 .255 log3 + log26 log78 .892 log7 + log4 log28 .447 log12 - log3 0.6021 log4 0.6021 log50 - log2 .398 log25 .398 log7 - log5 0.1461 log1.4 0.1461 log3 - log4 -0.1249 log0.75 -0.1249 log20 - log40 -0.3010 log0.5 -0.3010 This table to the left clearly shows that the log of 2 numbers subtracted from each other will equal the log of the numbers divided by each other.The table below clearly shows the log () - log () will equal log (). Let log x = a, let log y = b. Therefore 10a = x and 10b = y, these two equations can be converted into 10(a - b) = x/y. Finally, this equation can then be converted back into log x - log y = log (x/y). log6 - log2 log3 0.4771 log18 -