Volumes of Cones

IB Mathematics HL Portfolio (Type III) Volumes of Cones By Diana Herwono IB Candidate No: D 0861 006 May 2003 In this assignment, I will use WinPlot, a graphing display program. . Find an expression for the volume of the cone in terms of r and ?. The formula for the volume of a cone is: V = 1/3 x height x base area To find the base area, we must find the radius of the base. The circumference of the base of the cone = the length of arc ABC Therefore: 2? ? rbase = r? rbase = r? / (2?). The area of the base can therefore be calculated: Abase = ? ? rbase 2 = ? ? ( r? / (2?))2 Next, we must find the height of the cone, h. Notice that for the cone, h rbase r is a right angled triangle, with r as the hypotenuse. Therefore, using the Pythagorean Theorem, we can find h. r2 = h2 + R2 r2 = h2 + (r?/2?)2 h2 = r2 - (r?/2?)2 h = ? [r2 - (r2?2/4?2)] Therefore, substituting the values for rbase and h, we can find the volume of the cone. V = 1/3 x height x base area V = 1/3 ? ? [r2 - (r2?2/4?2)] ? ? (r2?2/4?2) 2. By using the substitution x = ?/2? , express the volume as a function of x. x = ?/2? ? = x2? V = 1/3 ? ? [r2 - (r2?2/4?2)] ? ? (r2?2/4?2) V = 1/3 ? ? [r2 - (r2(x2?)2/4?2)] ? ? (r2(x2?)2/4?2) = 1/3 ? ? [r2 - (r24x2?2/4?2)] ? ? (r24x2?2/4?2) = 1/3 ? ? [r2 - (r2x2)] ? ? (r2x2) 3. Draw the graph of this function using the calculator. Hence find the

• Word count: 823
• Level: International Baccalaureate
• Subject: Maths

Investigation of Area Under a Curve and Over a Curve

In this investigation I will be examining the relationships of the area under a curve and over a curve between two specified limits. Here I will call A the area surrounded by and the x-axis between the limits x=a and x= b. B on the other hand is the area defined by And the y-axis between the limits and . The vales of a and b will be whole real numbers. The investigation will test the ratio between A and B. Initially, I will begin by finding the ratio of A to B using a simple function, . The area of B is surrounded by this curve and the x-axis from x = 0 to x = 1 (a=0 and b=1). The area of A will be surrounded by the curve , and the y-axis from To which are y = 0 and y = 1. The graph below shows the graph of With the areas of A and B shaded respectively Graph of First I will find the area of B by using the formula of . This formula however translates into . In here a=0 and b=1 and . Now that I have calculated the area of B I will calculate the area of A, but to do so there are 3 possible methods that can be used. Method 1: The first method consists of taking the area of the entire triangle formed by the boundaries and subtracting from it B, the area under the curve which we already found to be , hence leaving only the area of A. Method 2: The second method to find A is by subtracting the area underneath lower curve from the area of the top curve. Here I can use

• Word count: 1669
• Level: International Baccalaureate
• Subject: Maths

Height limitations enforced by bone strength

An Analysis of Height Limitations Enforced by Bone Strength Wouldn't it just be amazing if there was no limit to anything? What if we could eat all we want none stop and nothing would happen? Or what if we could write a paragraph and call it an essay? Yes, but unfortunately our world doesn't work like that. There are limits on everything and anything even our growth. There is a certain height limit that we cannot exceed or else our bones won't be able to support our weight and just tumble down. But what is this incredulous height of which we cannot exceed? Well that is one answer I am ready to find out. We know that that no human can exceed 200 Mpa. Mammals like us have a ratio of 3 to five times the safety factors of the working stress to the breaking stress. It would be ridiculous to use 5 times because a human cannot, as I said before, exceed 200 Mpa, and five times the safety factor would exceed it. So I have chosen to work with 3. Since I'm using 3, I have to create a table that shows when our bone stress reaches 200. I am using an average male which would be 189lbs. and 5'9. The bone stress is measured because half of your weight, when walking, goes to one leg and the other half to the other. Multiplication Factor Height (cm) Weight(lbs) Bone Stress Initial X 75 89 50 Double(2X) 350 378 00 Triple (3X) 700 756 200 In this table we see the

• Word count: 600
• Level: International Baccalaureate
• Subject: Maths

Math Portfolio

IB Math Standard Level Portfolio Assignment Type 1- Mathematical Investigation Logarithm Bases Jaclyn Markowitz IB Candidate ID: 0000602-013 I have completed this assignment in accordance with the Newark Academy honor code. X___________________________. This investigation will determine the relationship between different sets of sequences including logarithms. This investigation will be tested using technology and general mathematic relationships. There are a few rules, which govern all the concepts of logarithms: , where a>0, a=1, b>0 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 number in the logarithm remains constant. The following two terms of each sequence were determined: log28, log48, log88, log168, log328, log648, log1288 log381, log981, log2781, log8181, log24381, log72981 log525, log2525, log12525, log62525, , log1562525 : : : Find an expression for the nth term of each sequence. Write down your expression in the form , where p, q . Justify your answers using technology. Begin with the first sequence (log28, log48, log88, log168, log328, log648, log1288) and determine an expression for the nth

• Word count: 1253
• Level: International Baccalaureate
• Subject: Maths

Math Portfolio - Weather Analysis

Introduction The purpose of this mathematics assignment is to explore the cosine function transformations with the data collected from the average temperature of a city in Ontario. Using www.weatherbase.com, we found average temperatures from the city of our choice. In my case, I selected Bloomfield, Ontario. The data points collected form a cosine function by nature that is already transformed and translated. Cosine Equation y=AcosB(x - C)+D Breakdown In the function, A represents the amplitude. The amplitude is a vertical dilation of either compression or stretch. It directly affects the size of the wavelengths in the function. In this situation, the amplitude is the range of the temperature over a twelve-month period. By calculating subtracting the maximum and minimum temperatures, you will fall upon the middle value. This value is 13.5. Considering the first values of the graph are in the negatives - obviously because of the fact that January and February are winter months - the A value is going to be negative. A period is defined as the time for one full cycle to take place in the graph. The period of this graph is 12 since the data was collected over the span of a year, or 12 months. In this function, the B-value represents the horizontal dilation. The horizontal dilation is the horizontal stretch or compression. This affects how long it takes for the data to be

• Word count: 593
• Level: International Baccalaureate
• Subject: Maths

Tarea Tipo 1 (nm)

Alfonso González Ortiz 2H Matemáticas NM Paralelas y Paralelogramos Prof. Antonio Jané Colegio Americano de Puebla Paralelas y Paralelogramos NM TIPO 1 Figura 1 . Cuando tenemos una figura como la que se muestra arriba y añadimos a la misma una cuarta transversal, podemos notar y comprobar que se forman seis paralelogramos. Para comprobar lo dicho anteriormente utilizaré la notación de conjuntos. ,, 2. Al repetir el proceso anterior pero ahora con cinco, seis y siete rectas transversales obtenemos lo siguiente: Cuando tenemos cinco rectas transversales se forman diez paralelogramos y para esto podemos utilizar la notación de conjuntos para comprobar lo dicho anteriormente. , Asimismo podemos notar que al aumentar otra recta transversal se forman quince paralelogramos y utilizaré la notación de conjuntos para comprobar esto. ,,,,,,,,, Por último añadiré una recta transversal más a la figura teniendo así siete rectas transversales. Luego de contar los paralelogramos puedo decir que son de un total de veintiún paralelogramos en la figura y esto lo puedo comprobarlo utilizando el método de conjuntos. ,,,,,,,,,, Al final de estos ejercicios y comprobar con el método de paralelogramos y estar más familiarizado con este tema y notación me di cuenta que existe cierta relación entre las rectas paralelas que cortan a dos rectas horizontales y los

• Word count: 1242
• Level: International Baccalaureate
• Subject: Maths

Math 20 Portfolio: Matrix

MATH 20IB SL PORTFOLIO TYPE 1 STELLAR NUMBERS By: Bob Bao March 10, 2010 TABLE OF CONTENTS Introduction Triangular Pattern Stellar Shape Pattern Conclusion Introduction: In this portfolio, I will investigate a pattern of numbers that can be represented by a regular geometric arrangement of equally spaced points. The simplest examples of these are square numbers, 1, 4, 9, 16, which can be represented by squares of side 1, 2, 3 and 4. Other geometric shapes which can lead to special numbers are the triangular shapes and the stellar (star) shapes. Our calculations will be based on the sets of triangular and stellar diagrams that are already provided and those that will be constructed. The aim of this investigation is to examine and determine the general statement for geometric patterns that lead to special numbers as well as to demonstrate a good and clear understanding of patterns and the operations that can be done with them. At the end of the project, we should be able to generate expressions and recognise patterns of other various geometric arrangements. Triangular Pattern: In the following, I will examine how we can derive general statements of patterns within triangular shapes. To begin, we can use a sequence of diagrams to better illustrate the pattern: Firstly, looking at the pattern, we can see that the number of dots is the sum of a series of

• Word count: 3011
• Level: International Baccalaureate
• Subject: Maths

Stellar Numbers Portfolio. In this task I will consider geometric shapes, which lead to special numbers

The English School IB Mathematics SL Math Portfolio (Type1) Stellar Numbers Vladislav Tajc Leal 2th of September 2011 Bogotá, Colombia Stellar Numbers Aim: In this task I will consider geometric shapes, which lead to special numbers Task 1 Complete the triangular sequence with more than 3 terms. Following the triangle sequence: 0+1=1 +2=3 3+3=6 6+4=10 0+5=15 Therefore, 5+6=21 21+7=28 28+8=36 After looking at the sequence I could realize that these triangular numbers are simply the sum of numbers from 1 to the term number. As the pattern continues the adding number increases arithmetically. Example: Triangular number 4 is 10 so, 6+4=10; or the Triangular number 7 is 28 so, 21+7= 28 Hence, the general statement that represents the triangular number in terms of is the equation: , Were is the triangular number we want to find For example, if we want to find the 10th triangular number we replace with 10 = = 55 This means that the 10th term will have 55 dots in its triangular shape. Task 2 Find the number of dots (the stellar number) in each stage up to S6 At S1 the number of dots is 1, at S2 the number of dots is 13, at S3 the number of dots is 37 and in S4 the number of dots is 73. The complete pattern in this 6-stellar is at follows: 0+1= 1 +12=13 3+24=37 37+36=73 Therefore, S5 = 73+ 48=121 S6 = 121 + 60= 181 Similarly to the

• Word count: 1319
• Level: International Baccalaureate
• Subject: Maths

The purpose of this paper is to investigate an infinite summation patter where Ln(a) is a constant and the coefficient of x is an increasing factor to Ln(a).

Infinite Summation Math IA The purpose of this paper is to investigate an infinite summation patter where Ln(a) is a constant and the coefficient of x is an increasing factor to Ln(a). Consider the following sequence of terms where x=1 and a=2 under the terms that 0?n?10: tn= n t(n) S(n) 0.000000 .000000 0.69314718 .693147 2.000000 0.24022651 .933373 3.000000 0.05550411 .988877 4.000000 0.00961813 .998495 5.000000 0.00133336 .999829 6.000000 0.00015404 .999983 7.000000 .5253E-05 .999998 8.000000 .3215E-06 .999999 9.000000 .0178E-07 .999999 0.000000 7.0549E-09 2 As n › +?, Sn › +2 Consider the following sequence of terms where x=1 and a=3: tn= n t(n) S(n) 0.000000 .000000 .000000 .000000 .098612 2.098612 2.000000 0.603474 2.702087 3.000000 0.220995 2.923082 4.000000 0.060697 2.983779 5.000000 0.013336 2.997115 6.000000 0.002442 2.999557 7.000000 0.000383 2.999940 8.000000 0.000053 2.999993 9.000000 0.000006 2.999999 0.000000 0.000001 3.000000 As n › +?, Sn › +3 There is a horizontal asymptote as n approaches positive infinite (?). As n approaches positive infinite then Sn will approach positive three. Sn approaches a horizontal asymptote when y=3. There is a y-intercept at (0,1). As n › +?, Sn › +4 There is a horizontal asymptote as n approaches positive infinite (?). As n approaches

• Word count: 2111
• Level: International Baccalaureate
• Subject: Maths

Solution for finding the sum of an infinite sequence

Internal Assessment 1 Solution for finding the sum of a infinite sequence The objective of this assignment is to find out the sum of infinite sequences , where In this equation, is defined by the term number. For example, is the first term, whereas is the nth term. are various variables. In this equation, which is defined by . Likewise 3! is defined by 3, but there is an exception First I will break down the equation so that it will be easier for me to find out he formula. I will examine the defined as the sum of the first . For example, , . I am first going to use this equation , where , , and where is . So it should look like this: I will find out the for . In order to find , I used my TI-84 Plus to figure this out. I will plug in the equation , where the is 1, is 2 and value is from 0 to 10. The method is shown in the appendix. I came up with: In order to check if I got it right, I used Microsoft Excel 2010. The method is shown in the appendix. After seeing that my result matches the results in Excel, I decided to then find out the sum. I found out the sum using my TI-84 Plus. The method is shown in the appendix. This is what I did in order to get for . <-- I got 1 because of . <-- Here is equal to plus because is the sum of the pervious term. So if I add which is term number 1, it will give me the . I will use

• Word count: 1272
• Level: International Baccalaureate
• Subject: Maths