I am a passionate, persistent individual who wills to become a bachelor in British Law, in order to, further on, exert the profession of a lawyer. I am and have always been deeply interested in the principles of justice, which I try to put into action in my everyday life. My intention on entering a Law School is acquiring a deeper knowledge on the subject which has always been my main interest, both in my social and in my personal life. One might perhaps marvel that a Brazilian would rather join an British law school rather than one in his home country; the main reason why I am convinced that studying law in the United Kingdom, more precisely in England, is the absolute passion and interest I have always had for this country throughout my whole life, also extending to the field of law. My interest in Law and Rights first blossomed at the age of seven, when problems like alimony and share of goods showed up after my parents’ divorce, raising my interest to the mutual dissatisfaction of each with different matters concerning law, which made me wonder to what extent Divorce Law in Brazil was fair to both sides of the divorce. Moreover, I also have been told the basics of the profession by uncles  who are bachelors in Brazil – and about how timedemanding and complex it can be. My uncle also told me, though, that as long as it is exerted for passion, and not for other reasons, the high charge of work and responsibility are not a burden, but a pleasure. So high was my interest in analyzing processes and things alike that I got to spend hours with these uncles, asking questions about their procedures and their opinion about processes. I believe that my differential contribution to the university is based on the cultural interchange interchanged I have been through between the years 2011 and 2013, during which I have been living in Germany in a boarding school, understanding the functionality of a community in...
...MATHS PORTFORLIO SL TYPE I
CIRCLES
In this portfolio I am investigating the positions of points in intersecting circles. (These are shown on the following page.
The following diagram shows a circle C1 with centre O and radius r, and any point P.
The circle C2 has centre P and radius OP. Let A be one of the points of intersection of C1 and C2. Circle C3 has the centre A, and radius r. The point Pʹ is the intersection of C3 with (OP). This is shown in the diagram below.
As shown on the assignment sheet, r=OA. We therefore need to find the values of OPʹ when r=OA=1 for the following of the values of OP: OP=2, OP=3 and OP=4.
We first of all extract the triangle OPA from the above diagram and since we have the values for all the three sides we can finds the angle AOP which will later on help to get the value of length OPʹ. The circle C2 and triangle OPA are shown below with all side of OPA indicated.
OP=AP since they are the radii of the same circle, C2. Having all the three sides, we can now calubulate the angle AOP using the cosine rule. Angle AOP is calculated below:
Cos...
...OPA Circle Style
The three circles; O, P and A intersect to create an interesting investigation regarding circles. Since this is a Calculus course, the investigation does have to deal with Derivatives. The most important and the focus of this portfolio is the line segment, OP’. Using the given diagram above, this investigation consists of finding the general equation for discovering OP’.
The values that are given are that r, is the radius of C1 and C2. OP and AP are the radii of C3. This information allows for the information to be manipulated too create two isosceles triangles. The first triangle and the one that is given, ∆OPA is an isosceles triangle therefore it can be concluded, thanks to the Isosceles Triangle Theorem that angle O and A are congruent to each other in this triangle. ∆OPA is not the only triangle that can be created, ∆OP’A is the second triangle created with a radius from C2. Therefore ∆OP’A is also an isosceles triangle. Now in both the triangles stated above, they share a common angle, O. With the help of this information we can analyze the preliminary relationship between OP and OP’.
We can find OP’, we have the tools. The two tools that we will be using mainly are the sine law and cosine law (respectively);
Sin Aa = Sin Bb = Sin Cc a2= b2+ c22bc cosA
and the two triangles that are going to be used;
For the first calculations, r=1 and OP =2. By finding the ∠O in one...
...Luna
Math IA (SL TYPE1)
CirclesCircles
Introduction
The objective of this task is to explore the relationship between the positions of points within circles that intersect.
The first figure illustrates circle C1 with radius r, centre O, and any point P. r is the distance between the centre O and any point (such as A) of circle C1.
Figure 1
The second diagram shows circle C2 with radius OP and centre P, as well as circle C3 with radius r and centre A. An intersection between C1 and C2 is marked by point A. The intersection of C3 with OP is marked by point P’.
Figure 2
Through this investigation I will examine how the r values correlate with the values of OP in determining the length of OP’ when r is held as a constant variable and the value of OP is the variable that is subject to change. I will then venture on to study the inverse, the relationship when the r values becomes the variable that is changed and the OP value is held constant.
r as a Constant
If we let the value of r be equal to 1, we can use that information to find the length of OP’ when OP=2, 3, and 4. The first thing one can deduce is that by using the points A, O, P’, and P two isosceles triangles can be formed; ∆AOP and ∆AOP’. To rationalize this assertion...
...Introduction
In this task, I will develop model functions representing the tolerance of human beings to Gforce over time.
In general, humans have a greater tolerance to forward acceleration than backward acceleration, since blood vessels in the retina appear more sensitive in the latter direction.
As we all know, the large acceleration is, the shorter time people can bear. Using the data shown in the task and Mat lab analysis, we can get several model functions to represent the tolerance of human beings to Gforce over time. Then the model functions are improved through further investigation. The validation result shows that the models represented are efficient for the tolerance of human beings to Gforce over time.
Problem formulation
To define appropriate variables and parameters, and identify any constraints for the data and use technology plot the data points on a graph. Comment on any apparent trends shown in the graph. Find function to model the behavior of the graph and explain the reason to choose the function. Create an equation to fit the graph. On new set axes, draw the model function and the function of the original data points. Comment on any differences and revise the model if necessary. Discuss the implications of the model in terms of Gforce acting on human beings. Use technology to find another function to model the data. On a new set of axes, draw the model function and comment on the differences. Then, check whether the first model fits...
...Math SL Portfolio – Tips and Reminders Checklist
Notation and Terminology
Check for the following:
• I did not use calculator notation. (I didn’t include things like ‘x^2’ for or Sn for Sn)
• I used appropriate mathematical vocabulary.
Communication
Check for the following:
• The reader will not need to refer to the list of questions in order to understand my work.
• My responses are not numbered.
• I have an introduction, conclusion, title page, and table of contents.
• All graphs are labeled – Each graph has a title, labeled axes, and appropriate scale.
• My graphs and tables are within the body of my work. They are not separate or in an appendix.
• I have explained why I made the choices I did when going through the task.
• I did not include key stroke sequences, e.g. “I pressed the 2nd key, then TRACE…”
• My tables do not straddle pages.
• My tables are labeled well, including my variable definitions in each column.
Use of Technology
Check for the following:
• I used technology to illustrate my points and ideas. I didn’t just “stick in” a graph.
• Each graph or table (or other piece of tech.) is accompanied by explanations and my ideas.
• I did not include too many graphs on the same axes – my graphs are easy to read.
Mathematical Process (Type 1)
Check for the following:
• I explicitly defined variables and parameters the first time I used them, even if they were
already defined in...
...Taipei European SchoolMath Portfolio

VINCENT CHEN 
Gold Medal Heights
Aim: To consider the winning height for the men’s high jump in the Olympic games
Years  1932  1936  1948  1952  1956  1960  1964  1968  1972  1976  1980 
Height (cm)  197  203  198  204  212  216  218  224  223  225  236 
Height (cm)
Height (cm)
As shown from the table above, showing the height achieved by the gold medalists at various Olympic games, the Olympic games were not held in 1940 and 1944 due to World War II.
Year (1=1932, 2 = 1936 and so on)
Year (1=1932, 2 = 1936 and so on)
Using autograph, the graph above is a scatter graph showing the high jump results from the table.
The plot suggests that the high jump heights start off with a steep positive slope then coming to a decreasing negative slope, however without the 1940 and 1944 high jump competitions, it may not be certain. Then finally, it starts increasing again with a fairly steep positive slope.
However, it would not be realistic if the function has an infinitely increasing range, such as quadratic, exponential and linear because of the limitations that humans have due to natural forces like gravity. Therefore, narrowing down the options that may fit this graph to natural logarithm and logistics
Since the statistics given starts from year 1896, in order to make sure that calculations can be as simplified as possible, I have decided to rearrange the table with the...
...Logarithm Base
IBMath SL
Type I Portfolio
Lisa Phommaseng
Logarithm Base
Consider the given sequences:
Log28, log48, log88, log 168, log328,…
Log381, log981, log 2781, log8181,…
Log525, log2525, log12525, log62525,…
:
:
:
,…
For the first sequence Log28, log48, log88, log 168, log328,… you are to find the next two terms of each of the sequences you would have to determine the pattern. As we can see, the value of the bases, 2,4,8,16,32 are increasing by squares of 2 (ex: 22= 4, 23=8, 24=16 etc.,). The two terms following log328 will be log648 and log1288. To find the expression for the nth term of each sequence, you would have to write the expression in the form of . First you would find log28 to begin to look for a pattern. A formula that could be useful to solve this would be logbx = . The variable b would be the base, x would be the number of the logarithm. Saying that log28 = x, we are able to determine that x is 3 because 23 equals 8. The same would follow for the next terms : log48= x, 8 is equivalent to 23 and the base of 4 is equivalent to 22 (Log28 =). So, to that term would be . You would continue on with the next and you would notice that the pattern is 3 over whatever the number that follows. Now we see that the nth term of this sequence is an=.
The next sequence Log381, log981, log 2781, log8181,…you would determine the following two terms as the one above. A we can see the terms are...