#### Math Portfolio 1

ST. ROBERT CATHOLIC HIGH SCHOOL PORTFOLIO ASSIGNMENT 1 OCTOBER 2009 LOGARITHM BASES SL TYPE I This task consists of two parts. While both parts consider logarithms with different bases of the same argument, these parts are not necessarily directly related to each other. PART 1 Exploring Determine the numerical values of the following sequences. Explain how you got these values. Justify your answers using technology. Use of Technology log28 = x log48 = x log88 = x log168 = x log328 = x 2x = 8 4x = 8 8x = 8 16x = 8 32x = 8 2x = 23 22x = 23 23x = 23 24x = 23 (2)5x = 23 x = 3 2x = 3 3x = 3 4x = 3 5x = 3 log381 = x log981 = x log2781 = x log8181 = x 3x = 81 9x = 81 27x = 81 81x = 81 3x = 34 32x = 34 33x = 34 34x = 34 x = 4 x = 2 log525 = x log2525 = x log12525 = x log62525 = x 5x = 25 25x = 25 125x = 25 625x = 25 5x = 52 52x = 52 53x = 52 54x = 52 x = 2 x = 1 Write the next two terms in each of these sequences, in both logarithmic and numerical forms. Explain how you got these values. The next two terms for the first sequence is based off this equation: where n represents the term

• Word count: 1123
• Level: International Baccalaureate
• Subject: Maths

#### Stellar Numbers Portfolio

MATHEMATICS SL INTERNAL ASSESSMENT TYPE 1 STELLAR NUMBERS In this Internal Assessment, triangular and stellar number patterns were treated with a thorough investigation. Triangular Numbers To begin with, like square number patterns (such as 1, 4, 9, 16) triangular number patterns work in a similar manner. The first five triangular numbers are: 1 3 6 10 15 Following the same arrangement pattern of the dots within the triangular pattern, the 6th, 7th, and 8th triangular numbers are as follows: 21 28 36 An instantly noticeable aspect of the pattern is that from the triangular numbers 1 to 3, 2 is added; from 3 to 6, 3 is added; 6 to 10, 4 is added; from 10 to 15, 5 is added; from 15 to 21, 6 is added; from 21 to 28, 7 is added; and from 28 to 36, 8 is added. From the above triangular number pattern, it is possible to deduce a general statement that represents the nth triangular number in terms on n. Sn=1+2+3+...+(n-2)+(n-1)+n + Sn=n+(n-1)+(n-2)+...+3+2+1______________________ 2Sn=(n+1)+(n-1+2)+(n-2+3)+...+(n-2+3)+(n-1+2)+(n+1) 2Sn=(n+1)+(n+1)+(n+1)+...+(n+1)+(n+1)+(n+1) 2Sn=n(n+1) Sn= So, the general statement that represents the nth triangular number in terms of n is Sn=. In order to test the validity of this statement, let us insert the number 7 for n in an effort to find the 7th triangular

• Word count: 2481
• Level: International Baccalaureate
• Subject: Maths

#### Stopping Distances

Stopping Distances Portfolio Mathematics SL Yr. 1 Period A Speed (x) Vs. Thinking Distance (y) Points plotted on the graph: (32, 6) (48, 9) (64, 12) (80, 15) (96, 18) (112, 21) As you can see in the graph above all six points seem to line up in a fairly straight line. As speed increases, so does the thinking distance, always by a similar amount. This means that the speed to thinking distance ratio is constantly increasing as speed increases. Functions: Slope: m= (y2 - y1) / (x2 - x1) m= (12 - 6) / (64 - 32) m= 6/32 m= 3/16 (32,6): y - y1 = m(x - x1) y - 6 = (3/16) (x - 32) y - 6 = (3/16)x -6 y = (3/16)x (64, 12): y - y1 = m(x - x1) y - 12 = (3/16) (x - 64) y - 12 = (3/16)x - 12 y = (3/16)x Percentage of error" (64, 12) y = (3/16)64 y = 12 (12 - 12) /12 ˜ 0 % The function, y = (3/16)x, matches the points, plotted on the graph, almost perfectly. You can see this by the calculated percentage of error which equals 0%. The function goes through every single point graphed. In the real life situation this means that the function represents the data recorded, while measuring how long it took for a person to think to apply the brakes, at different speeds, perfectly. The fact that a linear function resembles this situation in a very accurate manner means that the speed to thinking distance ratio is constant. Speed (x) Vs. Braking Distance (y)

• Word count: 1642
• Level: International Baccalaureate
• Subject: Maths

#### Koch Snowflake

THE KOCH SNOWFLAKE A MATHEMATICS PORTFOLIO MADE BY:- AYUSH AGARWAL IB-HL INTRODUCTION The Koch Snowflake was created by the Swedish mathematician Niels Fabian Helge von Koch. In his 1904 paper entitled "Sur une courbe continue sans tangente, obtenue par une construction géométrique élémentaire" he used the Koch Snowflake to show that it is possible to have figures that are continuous everywhere but differentiable nowhere. In order to reach the Koch snowflake, we need to start with a Koch curve. A Koch curve is a straight line which is divided into three parts and an equilateral triangle is constructed keeping the base as the middle part of the divided line segment. Finally, the base is removed which gives us the first iteration of the Koch curve (as shown in the figure above). From the Koch curve, comes the Koch snowflake. Now, instead of one line it starts with an equilateral triangle. The steps in Koch curve are applied now to each side of the equilateral triangle, which ultimately gives the shape of a "snowflake". STAGE 0 STAGE 1 STAGE 2 STAGE 3 PROCEDURE & OBSERVATION ASSUMPTIONS - Let Nn = Number of sides, Ln = the length of a single side, Pn = the length of the perimeter, An = the area of the snowflake; at the nth stage. In this portfolio, the sign '*' indicates multiplication whereas the sign '/' indicates division.

• Word count: 2921
• Level: International Baccalaureate
• Subject: Maths

#### Parallels and Parallelograms

QASMT Mathematics IA Parallels and Parallelograms Jeremiah Joseph 24/05/2009 This internal assessment will investigate the relationship between vertical transversals, horizontal lines and parallelograms. Vertical transversals are lines that intersect horizontal lines. To create parallelograms two or more parallel vertical transversals needs to intersect with two or more horizontal lines. This is shown in figure 1.1 and figure 1.2. Figure 1.1 Figure 1.2 These above figures demonstrate how vertical transversals (red) intersect with horizontal lines (black) to create parallelograms. These parallelograms are demonstrated in figure 1.1, A1, A2, A1 U A2. Furthermore, the parallelograms are illustrated in figure 1.2, A1, A2, A3, A1 U A2, A2 U A3, and A1 U A2 U A3. If parallel transversals are continually added an increasing number of parallelograms would be formed and a general formula can be deduced. Vertical Transversals Parallelograms Notation 2 A1 3 3 A1, A2, A1 U A2 4 6 A1, A2, A3, A1 U A2, A2 U A3, A1 U A2 U A3 5 0 A1, A2, A3, A4, A1 U A2, A2 U A3, A3 U A4, A1 U A2 U A3 , A2 U A2 U A3, A1 U A2 U A3 U A4 6 5 A1, A2, A3, A4, A5, A1 U A2, A2 U A3, A3 U A4, A4 U A5, A1 U A2 U A3, A2 U A3 U A4, A3 U A4 U A5, A1 U A2 U A3 U A4, A2 U A3 U A4 U A5, A1 U A2 U A3 U A4 U A5 7 21 A1, A2, A3, A4, A5, A6, A1 U A2, A2 U A3, A3 U A4, A4 U A5, A5 U A6, A1 U

• Word count: 1904
• Level: International Baccalaureate
• Subject: Maths

#### Infinite Surds

MATH PORTFOLIO Infinite Surds Introduction Surds: a surd is a number that can only be expressed exactly using the root sign "" in other words it can be defined as a positive irrational number. Thus, a number is a surd if and only if: (a) It is an irrational number (b ) It is the root of a positive rational number. The symbol V is called the radical sign. The index n is called the order of the surd and x is called the radicand. For example: V2 = (1.414.......) is a surd. V3= (1.732......) is a surd. V4= (2) is not a surd. V5= (2.236....) is a surd. Note: if n is a positive integer and a be a real number, then if a is irrational, is not a surd. Again if is rational, then also is not a surd. Surds have an infinite number of non-recurring decimal. Hence, surds are irrational numbers and are considered infinite surds. Following expression is the example of an infinite surd: Considering this surd as a sequence of terms where: = = = etc. Q: to find a formula for in terms of Answer: › = = 1.4142... › = or = 1.5537.... › = or = 1.5980.... As it is observed that = ; therefore, it can be understood that = as the trend has been observed as this until now. Hence, = = 1.6118... = = 1.6161... = = 1.6174... = = 1.6178... = = 1.6179... = = 1.6179... = = 1.6179... [NOTE: VALUES OBTAINED USING THE GRAPHICAL DISPLAY CALCULATOR:

• Word count: 1021
• Level: International Baccalaureate
• Subject: Maths

#### Infinite Surds

March 9, 08 Infinite Surds .) (V1+V1+V1+V1+...) A1 V1+V1 .4142 A2 V(1+ A1) .5537 A3 V(1+ A2) .5980 A4 V(1+ A3) .6118 A5 V(1+ A4) .6161 A6 V(1+ A5) .6174 A7 V(1+ A6) .6178 A8 V(1+ A7) .6179 A9 V(1+ A8) .61801 A10 V(1+ A9) .61802 > As n increases, An also increases, however when n gets very large An - An + 1 > An = V(1+ An -1) > An+1 = V(1+ An ) An = V(1+ An -1) An = V(1+ An ) An2 = 1 + An Quadratic formula : a = 1 b = -1 c = -1 An = -b ±V(b2-4ac) / 2a An = 1 ±V(-1)2-4(1)(-1)) / 2(1) An = 1 ±V(5) / 2 a.) 1 +V(5) / 2 b.) 1 -V(5) / 2 < 0 c.) answer: 1 +V(5) / 2 An2 - An - 1 = 0 : As n gets very large An - An-1 = 0 ; An = An-1 2.) V(2+V2+V2+V2+...) A1 V2+V2 .8477 A2 V(2+ A1) .9615 A3 V(2+ A2) .9903 A4 V(2+ A3) .9975 A5 V(2+ A4) .9993 A6 V(2+ A5) .9998 A7 V(2+ A6) .99996 A8 V(2+ A7) .99999 A9 V(2+ A8) .999997 A10 V(2+ A9) .999999 > As n increases, An also increases, however when n gets very large An - An + 2 > An = V(2+ An -1) Therefore: > An+1 = V(2+ An ) An = V(2+ An -1) An = V(2+ An ) An2 = 2+ An Quadratic formula : a = 1 b = -1 c = -2 An = -b ±V(b2-4ac) / 2a An = 1 ±V(-1)2-4(1)(-2)) / 2(1) An = 1 ±V(9) / 2 1 ± 3 / 2 = 4/2 Answer : An = 2 An2 - An - 2 = 0 :

• Word count: 504
• Level: International Baccalaureate
• Subject: Maths

#### Infinite Surds

Infinite Surds An infinite surd is a never ending irrational number. Its exact value would be left in square root form. The following example is the infinite surd of 1. This is the first infinite surd being investigated: ... When we see this, we can generate a pattern. a1 = a2 = a3 = So using this pattern, we can find the next 10 consecutive terms. a1 1.414213562 a2 1.553773974 a3 1.598053182 a4 1.611847754 a5 1.616121206 a6 1.617442798 a7 1.617851290 a8 1.617977531 a9 1.618016542 a10 1.618028597 We start to see a pattern form. We see that for each consecutive surd there is 1+ added to the previous. This can be shown using an+1 in terms of an:. This pattern can also be clarified in a graph: In this graph, one can see the pattern as well. One sees that the graph is already approaching its asymptote, suggesting that even if an is greater than a10, the numbers will still be very close to 1.618. One can already see its horizontal trend. To find what an is, one must perceive an and an+1 as being the same variables. an+1 = <-- Get rid of the subscripts. (a = 2<-- Square it to get rid of the radical sign. a2 = 1 + a<-- Get all numbers on one side. Then set theme equal to 0. a2 - a - 1 = 0<-- One now uses the quadratic formula because this is unable to be factored. --> --> --> The asymptote is approximately 1.618. This makes sense

• Word count: 981
• Level: International Baccalaureate
• Subject: Maths

Maths Internal Assessment: Shady Areas Introduction: In this investigation you will attempt to find a rule to approximate the area under a curve (i.e. between the curve and the x - axis) using trapeziums (trapezoids). Let us consider using the function: . Before finding the area under the curve using trapeziums. I will use integration to help me find the exact area. This will enable me to compare the results I acquire using the trapezium method and to find out whether the use of more trapeziums will provide me with a more accurate estimation of the area under the graph. Using the integration method, I have found the exact area for the function is I will now use the trapezium method to find the approximation of the area under the curve. I have used Autograph to show the graph for the function of g. It can clearly be seen from the graph that the area under the curve is roughly by the sum of the areas of two trapeziums. The area of a trapezium can be found using the formula: Let a = the length of one side of the trapezium b = the length of the other side of the trapezium h = the vertical height between the two lengths In order, Although there are two trapeziums, there are three different values and in order to determine the area of the trapeziums at each value I will work out the values on the graph using the trapezium method. I will now input the

• Word count: 891
• 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