#### infinite surds

• Word count: 800
• Level: International Baccalaureate
• Subject: Maths

#### Type I - Logarithm Bases

IB Standard Portfolio Assignment Type I - Mathematical Investigation Logarithm Bases This investigation will determine the relation between different sets of sequences. The sequences include logarithms. This investigation will be tested using technology. The sets of sequences are as follows: Log28 , Log48 , Log88 , Log168 , Log328 , ... Log381 , Log981 , Log2781 , Log8181 , ... Log525 , Log2525 , Log12525 , Log62525 , ... : : : , ,,, ... By following these sequences a pattern can be shown. The base of each term in the sequences changes but the exponents are constant. The following two terms of each sequence were determined: Log28 , Log48 , Log88 , Log168 , Log328 , , Log381 , Log981 , Log2781 , Log8181 , , Log525 , Log2525 , Log12525 , Log62525 , , , , ,,,, Let's start with the first sequence (Log28 , Log48 , Log88 , Log168 , Log328 , , ) and determine an expression for the nth term: 2 3 4 5 6 7 2 2 2 2 2 2 2 2 4 8 6 32 64 28 The value 2 was used to determine the nth term. is worked out from the table above in the form of by applying the change of base rule which states that: An application of this rule is as follows: The equations shown below will be used to determine and solve the sequences shown in the portfolio: = = Terms in the form of : The numerator and denominator both have the value log2 in them which

• Word count: 777
• Level: International Baccalaureate
• Subject: Maths

#### Verify Newtons second law.

AIM To verify Newton's second law and to determine the friction force acting on the wooden block. BACKGROUND THEORY When a wooden block (cart) slides over a surface, the net force acting on the block is given by, In the following set up, the applied is given by, ('m' is the mass suspended) The net force provides acceleration 'a' to the wooden block. Hence, ('M' is the mass of the wooden block) APPARATUS : Vernier lab pro , photo gates , wooden block , string meter scale , standard weights , pulley, laser source. RAW DATA : Least count of Vernier Lab Pro : 0.000001 s Uncertainty of Vernier Lab Pro : ± 0.000001 s Least count of meter scale : 0.1 cm Uncertainty of meter scale : ± 0.05 cm Sr. No. Weight suspended ( Kgs) Length of the Wooden block (± 0.05 cm) Distance between the photo gates (± 0.05 cm) Initial time at gate1 (± 0.000001 s) ( T1) Final time at gate 1 (± 0.000001 s) ( t1 ) Initial time at gate 2 (± 0.000001 s) ( T2 ) Final time at gate 2 (± 0.000001 s) ( t2 ) 0.10 .90 49.5 0.791711 0.807784 .139284 .151821 2 0.15 .90 49.5 0.733984 0.745485 0.982284 0.991184 3 0.20 .90 49.5 .749819 .759811 .982129 .990108 4 0.25 .90 49.5 .429185 .437991 .625008 .632085 5 0.30 .90 49.5 .356000 .364286 .539210 .545795 DATA PROCESSING Sr. No. Time taken for wooden block to pass GATE 1 ( ± 0.000002

• Word count: 772
• Level: International Baccalaureate
• Subject: Maths

#### Investigating the graphs of sine function

IB - SL Math Portfolio Locci Jonathan Investigating the graphs of sine function In this investigation, we are going to look at the different graph of y=sin x. We are going to compare and investigate the graphs of y=A sin x, Y=sin Bx and y= sin(x+C). We are going to look what the different value and what are the effects on the sin graph. Part 1: Investigation of the graphs y = sin x In the graphic 5 sin x, as you can see is much stretched because the amplitude is greater in 5 sin x than 2 sin x. But you can also see that in 2 sin x, the graph is much wider the 5 sin x. So we can conclude that more the amplitude is low and more the graph is wide. I put -2 sin x to show that when the value of the amplitude is negative, the graph will flip up sat down. Part 3: Investigation of the graphs y = sin (x+C) I choose sin (x+1) and sin x to show that when the number is positive, the period will move to the left by one because it will start at 1 for the amplitude. Here you can see that when the value in the parenthesis is negative, the period will shift to the right by one and the amplitude will start at -1. Part 4: Prediction of the shape and position of the graphs So, based on what I did,I saw that the value of the amplitude was 3 so a positive but I didn't use well the (x + 2) because I thought it would

• Word count: 738
• Level: International Baccalaureate
• Subject: Maths

#### Portfolio: Body Mass Index

Mathematical Portfolio Task II: Body Mass Index Vesela Germanova Germanova The data given in the task is plotted in the following graph: The values on the x-axis express the age and the values of the y-axis - the Body Mass Index. As far the age of a person must be always a positive number, than the BMI must also be a number bigger than 0. The graph of the function behaves like a Gaussian function. The Gaussian function has a general expression of this look: , where a, b, c are bigger than 0, and e ˜ 2.718281828 (Euler's number). I chose this function, because the curve line, that the Gaussian function forms fits the best to the graph I got from the data for the age of females and their BMI. To be more sure, I drew a simple Gaussian function over the graph, which I already have had. The new graph looks as follows (the red line is the function that the given data form and the black line is the Gaussian function): The exact equation of the Gaussian function, which fits the data graph, is: y =, where A=-6,933 B= 5,510 C=8.846 D=22.15 It must be noted that it is highly possible the Gaussian function to be inappropriate if the data is for elder people. The biological development of the human body leads in most of the cases to changes in the weight, height and the mass amount, that if put in a data set and represented in graph those quantities would not look

• Word count: 733
• Level: International Baccalaureate
• Subject: Maths

#### Math Portfolio Type 2 - PATTERNS FROM COMPLEX NUMBERS

Erzurum İhsan Doğramacı Vakfı Özel Bilkent Laboratuvar Lisesi Mathematics Portfolio HL Type I Topic PATTERNS FROM COMPLEX NUMBERS Name – Surname Selim TEPELER IB No: DPR160 (006040-047) Introduction In this portfolio we are gonna use complex number patterns which are related to roots of complex number , and we connect them our knowledge about analytic geometry ( which is distance formula) . Finally we will try to formulate a conjecture. This conjecture will help us to finding distance between roots. PART A . Use de Moivre’s theorem to obtain solutions to the equation Questions asks ; and firstly we should turn it into complex form; De Moivre’s theorem claims that ; where k = 0, 1, 2 we found ; - Cis(0) -cis(120) -cis(240) (by the way cis(α) means ‘’ cos(α)+isin(α)) 1. Use graphing software to plot these roots on an Argand diagram as well as a unit circle with centre origin. 1. Choose a root and draw line segments from this root to the other two roots. 1. Measure these line segments and comment on your results. As we see that the length of line segments are equal to each other. (one segment is 0.01 unit bigger than the others because of accuracy of graphing software) 1. Repeat the above for the equations and Comment on your results and try to formulate a conjecture. For 0 For Our conjecture is finding distance with using

• Word count: 720
• Level: International Baccalaureate
• Subject: Maths

#### Math SL Portfolio: Matricies

Mathematics SL Portfolio Assignment 1 Title: Matrix Powers Type 1 . Consider the matrix M = Calculate Mn for n = 2, 3, 4, 5, 10, 20, 50. M²= * = = M³=*== M4=*== M5=*== M¹°=*== M²°=*== M5°=*== * To obtain the matrices above I multiplied (a*b+b*g), (a*f+b*h), (c*e+d*g), (c*f+d*h)as shown in the general formula seen below Describe in words any pattern you observe. In the above matrices the pattern I observed is shown in relationship to the exponents and the numbers within the matrix. As the exponent increases consecutively the matrix is in turn multiplied by two. Use this pattern to find a general expression for the matrix Mn in terms of n. As a result of the pattern expressed, the General Formula for these matrices is Mn=M*2n-¹ ......................................................................................................................................................................... 2. Consider the matrices P = and S = Calculate Pn and Sn for other values of n and describe any pattern(s) you observe. P²=*== Determinant: 100-36=64 P³=*= = Determinant: 1296-784=512 P4=*== Determinant: 18496-14400=4096 P5=*= Determinant: 278784-246016=32768 S²=*== Determinant: 400-256=144 S³=*== Determinant:

• Word count: 720
• Level: International Baccalaureate
• Subject: Maths

#### Properties of quartics

Ecolint - DP 2009/10 Properties of Quartics Math HL - Portfolio Assignment Kathia Zimba 6/23/2009 Introduction Quartic functions are functions that the highest exponent is 4. These types of functions are of the form The graphs of a Quartic functions usually exert two shapes; "W" shape or "M" shape. For this investigation, an analysis of a "W" shaped function is to be carried out to explore the properties of the function. The points of inflection of the Quartic function, will be looked at very closely so that the ratio between the distances of the points if intersection when the Quartic graph is cut by a straight line is found. Analysis Let's take . The second derivative of this function will gives the points inflection at ; provided . FIND 2ND DERIVATIVE OF when and At and is where the points of inflection are located at the original function i.e. Substitute the values in the R (3, 15) Since for both points both points of inflection are non-horizontal points of inflection. Once the two points of inflection are found, a straight line is drawn so that the two points of inflection (Q and R) meet the quartic function i.e. . When this straight line is drawn, the line meets the Quartic again at another two points P and S creating three identical segments. It is important the coordinates of these two points, so that the ratio PQ:QR:RS is determined. The

• Word count: 711
• Level: International Baccalaureate
• Subject: Maths

#### Triangular and Stellar Numbers

Math I.B Internal Assessment: SL Type 1 Stellar Numbers 6/26/2011 St. Dominics International School Raj Devraj TRIANGULAR NUMBERS TRIANGULAR NUMBERS WITH THREE MORE TERMS GENERAL STATEMENT: NTH TRIANGULAR NUMBERS IN TERMS OF N. The differences between the sequences of terms: X Y Y= number of dots on triangle X= number of dots on 1 side of triangle 0 0 2 3 3 5 According to Finite Differences if the 3rd difference of a pattern is 1, then the general term is a quadratic equation: Thus to find the general term we must first find the value of 'c': In order to find the values of 'a' and 'b' we must solve a quadratic using simultaneous equations, thus: Substitute the value of 'a' of one equation: To find the General statement that represents the nth triangular number in terms of 'n', we substitute the value of 'y' by Un and the value of 'x' by n, thus: STELLAR NUMBERS NUMBER OF DOTS TO S6 STAGE S1 S2 S3 S4 S5 S6 3 37 73 21 81 Thus, using finite difference: The most obvious pattern is that the 1st row all numbers and odd and the second row all are even. Also all these numbers are some multiples of 12 + 1, for example: 12 also turns out to be the half of 6. 6 STELLAR NUMBER AT STAGE S7 + 1 (12) + 2 (12) + 3 (12) + 4(12) + 5(12) + 6(12) = 253 GENERAL STATEMENT FOR 6 STELLAR NUMBER AT STAGE SN IN TERMS OF N If you notice the multiples are

• Word count: 705
• Level: International Baccalaureate
• Subject: Maths

#### IB Math Portfolio Investigating Ratios

Investigating Ratios of Areas and Volumes The aim of this portfolio is to investigate the ratio of areas form when is graphed between two arbitrary parameters and such that . . Given the function, consider the region formed by this function from to and the x-axis. This area is labeled B. The region from and and the y-axis is labeled A. Finding the ratio of area A: area B: The ratio of area A: area B is 2:1. Calculate the ratio of the areas for other functions of the type , between and . Let , Finding the ratio of area A: area B: The ratio of area A: area B is 3:1. Let , Finding the ratio of area A: area B: The ratio of area A: area B is 4:1. Let , Finding the ratio of area A: area B: The ratio of area A: area B is 10:1. From the findings above, it is evident that when , the ratio of area A: area B is always n : 1 when n is a positive integer more than 0. To further investigate my conjecture, test the conjecture for other subsets of the real numbers: Fractions: Let , Finding the ratio of area A: area B: The ratio of area A: area B is :1, which means that the conjecture is true to fraction values of n. Negative Integers: Let , Finding the ratio of area A: area B: The conjecture is not true to negative integers. Irrational Numbers: Let , Finding the ratio of area A: area B: The ratio of area A: area B is :1. This means the conjecture is

• Word count: 700
• Level: International Baccalaureate
• Subject: Maths