#### Lacsap's fraction math portfolio

Lacsap's fraction Ryohei Kimura IB Math SL 1 Internal Assessment Type 1 Lacsap’ fraction Lacsap is backward word of Pascal. Thus, the Pascal’s triangle can be applied in this fraction. How to find numerator In this project, the relationship between the row number, n, the numerator, and the denominator of the pattern shown below. 1 1 1 1 1 1 1 1 1 1 Figure 1: The given symmetrical pattern (Biwako) Figure 2: The Pascal’s triangle shows the pattern of .It is clear that the numerator of the pattern in Figure 1 is equal to the 3rd element of Pascal’ triangle which is when r = 2. Thus, the numerator in Figure 1 can be shown as, (n+1)C2 [Eq.1] where n represents row numbers. Sample Calculation - When n=1 (1+1)C2 (2)C2 -When n=2 (2+1)C2 (3)C2 -When n=5 (5+1)C2 (6)C2 15 Caption: The row numbers above are randomly selected within a range of 0≤x≤5. Therefore, the numerator of 6th row can be found by, (6+1)C2 (7)C2 x = 21 [Eq. 2] and the numerator of 7th row also can be found by, (7+1)C2 (8)C2 x = 28 [Eq. 3] How to find denominator 1 )+0 1 )+0 Figure 3: The pattern showing the difference of denominator and numerator for each fraction. The first element and the last

• Word count: 937
• Level: International Baccalaureate
• Subject: Maths

#### Math Portfolio Type 1

Math Portfolio Type 1 Task Investigating areas and volumes Miras International School Amrebayeva Nurbala Gr. 12IB Introduction In this portfolio, I'm going to investigate ratios of areas and volumes of power functions, which can be generalized as the following: Y=xn, where n-is the power, n?R This function will be graphed between two arbitrary parameters x=a and x=b such that a<b (in other words boundaries or limits). In this investigation, I'm going to use method of math induction, integration using power rule, application of integration (areas under the curve and solids of revolution), also some knowledge about power functions. Investigation Process Given the power function y=x2, graph of which is parabola. I need to consider the region formed by this function from x=0 to x=1 and the x-axis, let's label this area B; and the region from y=0 to y=1 and the y-axis area A. This can be shown on the illustration (Fig.1) below: Figure 1 To plot this, I used Graph 4-3 Software, which I found very convenient. So the areas formed and illustrated above (Figure 1), need to be considered in order to find a ratio between them, hence area A: area B. To find that, I need to use integration using power rule. Integration, using power rule is the general integration rule, which is: where n?-1 It consists of the following steps: . Calculate area B, which is -0 2. Calculate

• Word count: 3311
• Level: International Baccalaureate
• Subject: Maths

#### Math Portfolio Type II

Modeling the amount of a Drug in the Bloodstream The goal of the following investigation will be to determine the amount of a drug which is present in the human bloodstream for different durations of specified time. Foremost, appropriate variable must be established to deal with the investigation. Let "t" represent the time in hours, and let "a" represent the amount of drug in micrograms (µg). The controlled variable will be the time, because we can establish at what hour the monitoring will start and stop. The uncontrolled variable thus is the amount of drug in the bloodstream; we have no control over that and it is ultimately what we hope to discover. Foremost, a data table is created according to the information provided in the "Amount of Drug in the Blood Stream" graph. This is done by taking all the t values and a values and organizing them into a data point chart. Figure 1 t a 0.5 9.0 .0 8.3 .5 7.8 2.0 7.2 2.5 6.7 3.0 6.0 3.5 5.3 4.0 5.0 4.5 4.6 5.0 4.4 5.5 4.0 6.0 3.7 6.5 3.0 7.0 2.8 7.5 2.5 8.0 2.5 8.5 2.1 9.0 .9 9.5 .7 0.0 .5 Figure 1 shows a data point chart for the Amount of drug vs. the time. After the data chart table has been created, a graph to represent the information should be simple to establish using graphing software on a computer. The software chosen for this specific investigation is Graphmatica. To

• Word count: 2727
• Level: International Baccalaureate
• Subject: Maths

#### Logan's logo

Logan's Logo Abstract This internal assessment focuses on functions and areas under curves. The task at hand was to develop models of functions to best fit the characteristics/behaviours of the curve of Logan's logo. Using an appropriate set of axes data points of the curve were measured using a ruler and then identified and recorded. 11 data points were recorded for each curve. The main objective of this IA was to determine which model, cubic or sin, would be the best fit. It was observed that the cubic function was superior. Another task was to refine the model function to fit on a t-shirt and a business card. The area between the two curves had to be calculated. In doing this assignment many pieces of technology were incorporated when doing this assignment, ranging from a pencil and a ruler to calculators and even spreadsheets and other programs. Introduction: A diagram of a 10cm by 10cm square is divided into three regions by two curves. The logo is the shaded region between the two curves. This investigation is aimed at answering several questions but mainly to develop mathematical functions to model the two curves represented by the two curves. Some key terms that should be understood are: sin function, cubic function, MAE (Mean Absolute Error). These terms will be elaborated on later on this assignment. A few data points were taken from the curves, with the base

• Word count: 4567
• Level: International Baccalaureate
• Subject: Maths

#### Logarithm Bases

Logarithm Bases In the beginning of this problem, we are given multiple sequences of logarithms, and are told to write down the next two terms of each sequence. Here is a table of the first sequence, including the next two terms and the numerical equivalence of each term: # of Term 2 3 4 5 6 7 Term Numerical Equivalence I created the last two terms in this sequence, terms 6 and 7, simply by doubling the base of the logarithm for each term. In more proper mathematical terms, I used the formula , where n is the number of the term. I noticed that each numerical equivalence seems to have 3 as the numerator, and the number of the term as the denominator. In this manner, I created a formula to find the numerical equivalence for the nth term of the sequence in the form , where both p and q are integers: . I used this method of investigation for the other two sequences as well. Here are tables for each of the two sequences: # of Term 2 3 4 5 6 Term Numerical Equivalence # of Term 2 3 4 5 6 Term Numerical Equivalence I noticed that in both of these sequences as well, the numerical equivalences seem to have a constant integer as the numerator, and the number of the term as the denominator. Therefore, the formula for the first sequence is , and the formula for the second sequence is . Now, we are asked to calculate a set of many logarithms, in the form ,

• Word count: 988
• Level: International Baccalaureate
• Subject: Maths

#### Investigating Ratios of Areas and Volumes

Bob Devan Calculus AB/BC 7th & 8th December 6, 2008 Investigating Ratios of Areas and Volumes .Given the function y=, consider the region formed by this function from =0 to =1 and the -axis. Label this area B. Label the region from y=0 to y=1 and the y-axis area A. A. Find the ratio of area A: area B. The area of A for the given function of is. Area A The area of B for the given function of is Area B: Thus the ratio of area A : area B can be given as 2:1. B. Calculate the ratio of the areas for other functions of the type between 0 and . Make a conjecture and test your conjecture for other subsets of the real numbers. Area A: 1- Area B: Conjecture: Given the function the area of A in ratio of the area of B can be given as n:1. Equation Area of A Area of B Ratio 3:1 4:1 5:1 6:1 2. Does your conjecture hold only for areas between x=0 and x=1? Examine for =0 and =2; =1 and =2 etc. Ex 1) The area of A for the function y=x2 [0,2] is 8/3 Area A: The area of B for the function y=x2 [0,2] is 16/3 Area B: Ex 2) The area of for the function is Area A: The area of B for the function is Area B: Ex 3) Area A: Area B: 3. Is your conjecture true for the general case from to such that and for the regions defined below? If so prove it; if not explain why not. Area A: and the y-axis Area B: and the x-axis Area A: = =

• Word count: 617
• Level: International Baccalaureate
• Subject: Maths

#### Investigating ratios of areas and volumes

INVESTIGATING RATIOS OF AREAS AND VOLUMES Introduction The objective of this portfolio assignment is to investigate the ratio of the areas formed when is graphed between arbitrary parameters and such that . This investigation may lead to a conjecture which ends up in a general formula. Given the function , we can consider a region formed by this function from and . The area between the function and the x-axis will be labeled B. The area from to and the y-axis will be labeled A. The formula used to find the area under a curve to the x-axis is . To find the area to the y-axis the formula used is . The area between the function and the x-axis is given by: Therefore area of B is equal to . So as the unit area is 1, the area of A is given by: Another method which may be used is by getting the inverse of the function such that if , the inverse would be . Therefore, the ratio of the areas A and B in the function is simplified into But will this ratio remain as we increase the power of the function? This will be proven by elevating the exponent of the formula by 1 each time so this can lead to a conjecture. Testing variables for n n Function Area Area A Area B Ratio :1 2 2:1 3 3:1 4 4:1 5 5:1 N n:1 Procedure Therefore, the ratio is given by the formula. Graphing From the graphs we can observe that as the value of n keeps increasing the area

• Word count: 1013
• Level: International Baccalaureate
• Subject: Maths

#### Matrix Binomials

Matrix Binomials A matrix is a rectangular array of numbers arranged in rows and columns. Numbers or letters inside the brackets in matrices are called entries. Matrices can be added, subtracted, multiplied, divided, and also raised to a power. A common matrix can look like this where a, b, c, and d are the entries. Given the matrices X = and Y = I calculated X2, X3, X4; Y2, Y3, Y4. Using my GDC (graphic display calculator) I evaluated these matrices. X2 = Y2 = X3 = Y3 = X4 = Y4 = After calculating the powers of X and Y, I observed my solutions. Looking at this I saw a trend that emerged as I increased the power of the matrix. The trend was increasing the power of the matrix by one, caused the product to double its previous solution. When X is to the second power, the entries of the solution are all 2's; when X is to the third power the entries are all 4's; when X is to the forth power the entries are all 8's, and so on. As for Y, the pattern is similar but some entries are negative (-). Using this information I produced an expression to solve for X to a certain power, and it is Xn = letting "n" represent the power. For Y the expression is the same for the numbers in position of a and d. The expression for numbers of b and c is also the same but it is negative. As a matrix the expression is Yn = . I also calculated integer powers of (X

• Word count: 1182
• Level: International Baccalaureate
• Subject: Maths

#### Matrix Binomials IA

Matrix Binomials Type 1 Internal Assessment Matrices are rectangular arrays of numbers that are arranged in rows and columns, however the regular rules of algebra do not apply. Let X= and Y= and calculate X2, X3, X4 ; Y2, Y3, Y4. X2= × = X3= × × = X4= × × × = Because matrices do not follow the algebraic rules of exponents, one can not simply distribute the exponent for each matrix value. Instead the matrix must be multiplied by itself however many times the exponent says. So for example, for X2, the matrix X must by multiplied with itself two times. The pattern that has emerged is that with every increasing power the matrix value increases with an exponential power of 2. 21 is equal to 2 showed by the matrix . 22 is equal to 4 showed by the matrix and similarly 23 is equal to 8 represented in the matrix . Y2= × = Y3= ×× = Y4= ××× = The pattern is very similar to the one above except that all the negatives in the original matrix will also become negatives. Based on the results above we can conclude that for each exponent value Xn the matrix value will result in Xn-1 and the same goes for Yn resulting in Yn-1. Combining these terms would result in (X+Y)n-1. (X+Y)n-1: (X+Y)3-1 = (X+Y)2 = (X+Y)2 = X2+Y2 (X+Y)2 = which is the same as Xn-1+Yn-1 ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Now let A=aX and B=bY where a and b are

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

#### Matrix Powers

Matrix Powers . Consider the matrix M = a. Calculate Mn for = 2, 3, 4, 5, 10, 20, 50. M2 = M3 = M4 = M5 = M10 = M20 = M50 = b. Describe in words any pattern you observe. Using a TI-83 Plus to calculate each of these power matrices, we are able to find the th matrix. The matrix provided is an identity matrix, which is uniquely defined by the property In M = M thus the value of Mn is equated to two to the power of as well; the value zero stays constant for other values of . The matrix Mn can also be calculated using geometric sequence: Un = U1rn-1;as the common ratio between each matrix is a factor of two. c. Use this pattern to find a general expression for the matrix Mn in terms of . Mn = I × 2n 2. Consider the matrices P = and S = P2 = 2 = = 2 S2 = 2 = = 2 a. Calculate Pn and Sn for other values of and describe any pattern(s) you observe. P3 = = 4 S3 = = 4 P4 = = 8 S4 = = 8 P5 = = 16 S5 = = 16 P7 = = 64 S7 = = 64 P10 = = 512 S10 = = 512 From a standard matrix , we can see that the difference between and , and is two. Therefore we could infer that it follows the pattern , and that in matrix P is two whereas in matrix S is three. However, a coefficient had to be factorised in order to have that difference of two. A pattern for the coefficient was found as 2n-1. 3. Now consider matrices of the form . Steps

• Word count: 565
• Level: International Baccalaureate
• Subject: Maths