#### 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 =

#### trigonometric functions

Portfolio Type 1 investigation Transformation of Trigonometric Functions Investigate the function: y = a sin b(x-c) + d in respect to the transformation of the base curve of y = sin x, depending on the values of a, b, c, and d. Be sure to consider all possible values for a, b, c, and d. Describe the base curve Start with a (try some values) Then try values in b (sin2x, sin -3x, sin1/2x) Does your hypothesis hold true for y = cos(x) and y = a cos b (x-c) +d? How about tan(x)? Transformation of Trigonometric Functions Introduction The purpose of this study is to examine the transformation of trigonometric functions of y=A sin B (x-C) + D and determine the effect on the base curve y=sin(x). I am going to be systematically changing the values of A, B, C and D in the equation y= A sin B (x-C) +D. First I am going to examine different numbers for the value of A. I am going to use whole numbers, negative whole numbers, positive rational numbers and negative rational numbers for the value of A and see how this affects the Sine curve. Then I will examine different numbers for B, then C then D. After examining the Sine function I test to see if changing the values of A, B, C and D will have the same effect on the Cosine function. Sine Curve: Figure 1 Figure 2 I will use y-sin (x) as my base curve. This has a maximum value of 1, minimum value of -1

#### Mathematics SL Parellels and Parallelograms. This task will consider the number of parallelograms formed by intersecting m horizontal parallel lines with n parallel transversals; we are to deduce a formula that will satisfy the above.

Parallels and Parallelograms Mathematics Coursework This task will consider the number of parallelograms formed by intersecting m horizontal parallel lines with n parallel transversals; we are to deduce a formula that will satisfy the above. Methodology . We started out the investigation with a pair of horizontal parallel lines and a pair of parallel transversals. One parallelogram (A1) is formed (shown in Figure 1) 2. A third parallel transversal is added to the diagram as shown in Figure 2. Three parallelograms are formed: A1, A2, and A1?A2 3. When a fourth transversal is added to Figure 2 (Figure 3), six parallelograms are formed. A1, A2, A3, A1?A2, A2?A3, A1?A3 4. Figure 4 has 5 transversals cutting the pair of horizontal parallels, forming ten parallelograms. A1, A2, A3, A4, A1?A2, A2?A3, A3?A4, A1?A3, A2?A4, A1?A4 5. A sixth transversal was added to Figure 5, forming 15 parallelograms shown in Figure 6. A1, A2, A3, A4, A5, A1?A2, A2?A3, A3?A4, A4?A5, A1?A3, A2?A4, A3?A5, A1?A4, A2?A5, A1?A5 6. When a seventh transversal is added, twenty-one parallelograms are formed (Figure 7). A1, A2, A3, A4, A5, A6, A1?A2, A2?A3, A3?A4, A4?A5, A5?A6, A1?A3, A2?A4, A3?A5, A4?A6, A1?A4, A2?A5, A3?A6, A1?A5, A2?A6, A1?A6. 7. I then used technology (table 1.0) to record the above and calculate the differences between the parallelograms formed with each addition of a

#### Type I - Parallels and Parallelograms

Parallels and Parallelograms Math Portfolio Introduction: This investigation aims at finding a relationship between the numbers of horizontal parallel lines and the transversals. When these lines intersect they form parallelograms. The aim of this investigation is examine and determine a general statement for transversals and horizontal lines and how they affect the number of parallelograms formed within the figure. A diagram of a parallelogram and a transversal is shown below. These lines represent two transversals; however they are supposed to intersect with a horizontal parallel line. These lines represent two horizontal parallel lines. They are intersected with a number of transversals. When there is one horizontal line and two transversals, this makes one parallelogram, A1, as shown below: This diagram shows that when there are two horizontal lines and two transversals, one parallelogram is formed. This parallelogram is called A1. However, when another transversal is added to the same diagram and same pair of horizontal lines, it looks like this: This figure shows that when another transversal is added to a pair of horizontal parallel lines, three parallelograms are formed. Although the third parallelogram is not labeled, it is clear that the total of A1 A2 makes a big parallelogram, A3. So therefore, there is A1, A2, and A3. Another transversal added to this

#### portfolio Braking distance of cars

. Use a GDC or graphing software to create data plot of speed versus thinking distance and data plot of speed versus braking distance. Describe your results. The data plot of speed versus thinking distance. The speed is relative to the thinking distance. So, this graph shows the direct proportion. There is no way that these values are minus because it doesn't make sense that thinking distance is minus. In this report, let the time of stepping on the brake is equal. The data plot of speed versus braking distance. The graph should show exponential function. It is conceivable that it causes friction when the car breaks. When the car starts breaking, the speed of car is on the table below. During the car keep breaking, speed is decreasing due to friction. Speed (km/h) Speed(m/s) 0 0 32 8.888 48 3.33 64 7.77 80 22.22 96 26.66 12 31.11 The relationship between the speed and advanced distance is shown the graph below. x = speed (m/s) y=advanced distance First, the speed of car is constant. Speed will be decreasing according to braking cause friction. Thus the advanced distance will be decreasing as well. At least, the car will stop which indicate point 0. Therefore, area of triangle represents braking distance (BD). Speed and distance are not constant because friction affects these aspects. BD = = = That's why the relationship between speed and

#### Math IB SL Shady Areas Portfolio

This portfolio is an attempt at deriving and examining the scope and limitations of a general statement that can approximate the area under a curve using trapezoids. Generally, calculus - specifically the method of integration, is used to find the exact area under a curve. Although this method will be explored in comparison later in the portfolio, this investigation deals mainly with investigating a method to approximate the area using high school level math. First, this portfolio will attempt to derive a general statement that will give an approximation of the area under the curve of any function in any closed interval using trapezoids. Then, by applying the formula to sample functions, the answers given can be compared to the integral answers, allowing an examination into the accuracy of the trapezoid method of approximation. Lastly, by examining different behaviors of a graph, this portfolio will investigate the cause of any inaccuracies in this method. The graph: is given: Figure 1 - Graph of Using two trapezoids mapped onto the curve in the domain , the area under the curve in that domain can be approximated as the sum of the areas of the two trapezoids. In order to map the trapezoids onto the graph, one must first divide the domain by the number of trapezoids being used, in order to find the height of the trapezoids (which is equivalent to each other). It should

#### Shady areas; math portfolio type 1

Shady areas This investigation is carried out in order to find a rule to approximate the area under a curve using trapeziums (trapezoids). From calculus it is known that by the help of integration, the definite area under a curve could be calculated. In this case a geometrical method will be used in order to approximate the area under a curve. The function g(x) = x2+3 is considered. The graph of g is shown below. The area under the curve from x = 0 to x = 1 is approximated by the sum of the area of two trapeziums. In order to approximate the area under this curve, the composite trapezoidal rule is used. The composite trapezoidal rule is a numerical approach for approximating a definite integral. g(x) dx = Area x2 + 3 dx = Area According to composite trapezoidal rule: Area = 1/2 (width of strip) [first height + 2(sum of all middle heights) + last height] Since there are two trapeziums from 0 to 1 in fig 1.1 hence the width of each strip: 1/2 = 0.5 x g(x) or height 0 02 + 3 = 3 /2 (1/2)2 + 3 = 3.25 2 + 3 = 4 So: Area = 1/2 (0.5) [3 + 2( 3.25) + 4] Area = 3.37 area units (a.u.) Increasing the number of trapeziums to five: Approximating the area in this case with five trapeziums: Width of every strip: 1/5 = 0.2 x g(x) or height 0 3 /5 3.04 2/5 3.16 3/5 3.36 4/5 3.64 4 Area = 1/2 (1/5) [3 + 2(3.04 + 3.16 + 3.36 + 3.64) + 4] Area = 3.34 a.u.

#### 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

#### The segments of a polygon

Portfolio mathematics HL Assignment 1 The segments of a polygon Author: Luka Dremelj Candidate number: Subject: Mathematics HL Teacher: Barbara Pecanac Date written: 25/5/2009 Introduction First Mathematics HL Portfolio is about investigating the segments of a polygon, using graphical methods and analytical proofs. The task is to find conjecture between ratio of the sides and the ratio of the areas, first of triangles developing it to general conjecture of polygons. . In an equilateral triangle ABC, a line segment is drawn from each vertex to a point on the opposite side so that the segment divides the side in the ratio 1:2, crating another equilateral triangle DEF. (a) What is the ratio of the areas of the two equilateral triangles? To answer this question, (i) create the above diagram with your geometry package. (ii) measure the lengths of the sides of the two equilateral triangles == = 3 units = = 1.13 units (iii) find the areas of the two equilateral triangles and the ratio between them. Formula for area of triangle: (u=unit) I got the same results for the areas, as before calculated with program. (see the picture above) (b) Repeat the procedure above for at least two other side ratios, 1:n. I decided to draw triangle with ratio: 1:3 == = 3 units = = 1.67 units Third ratio: 1:4 == = 3 units = = 1.96 units (c) For: n=2 we

#### matrix binomilas-portfolio

IB PORTFOLIO- MATRIX BINOMIALS MATH SL JAIME CASTRO A. 0-2 PRESENTED TO: CLAUDIA GOMEZ ANGLO COLOMBIAN SCHOOL MATH DEPARTMENT BOGOTA, MARCH 2009 INTRODUCTION This project is about 2x2 matrices or matrix binomials; the aim of this work is to demonstrate a good and clear understanding of matrices and the operations that can be done with them. At the end of the project we should have been able to generate different expressions for some different matrices powered by an n integer as well as a general statement to calculate the addition of specific matrices powered to an n integer. To have a correct development of this piece of work it is essential that each question is solved in order since the result of a question will be necessary for the development of the following one. Since there are many basic calculations that aren't really important we will use a GDC (graphic display calculator) in order to speed up the development of the project. QUESTIONS ) Let and. Calcúlate , ; , . With the help of a GDC (Graphic display calculator) the calculations for the matrices were done and registered below. 2) By considering different integer powers of X and Y find an expression for, and. We have already calculated the resulting matrices for three different integer powers of X and Y in the first question, however to find an expression for, and and be sure that it is