Add Math Essay 2 Essays

Add Math Essay 2 Essays Add Math Essay 2 Essay Add Math Essay 2 Essay Additional Mathematics Project Work 2 Written By : Nurul Hazira Syaza Abas I/C : 940602-01-6676 Angka Giliran : School : SMK Kangkar Pulai Copyright 2011  ©. Hazira Syaza, All Right Reserve Numb| Title| Page| 1| Acknowledge| 1| 2| Objective| 2| 3| Introduction Part I| 3| 4| Mathematics In Cake Baking And Cake Decorating| 4 5| 5| Part II| 6 14| 6| Part III| 15 17| 7| Further Exploration| 18 21| 8| Reflection| 22 23| 9| Conclusion| 24| 10| Reference| 25| Table. of. Content Copyright 2011  ©. Hazira Syaza, All Right Reserve Acknowledge First of all, I would like to say Alhamdulillah, for giving me the strength and health to do this project work. Not forgotten my parents for providing everything, such as money, to buy anything that are related to this project work and their advise, which is the most needed for this project. Internet, books, computers and all that. They also supported me and encouraged me to complete this task so that I will not procrastinate in doing it. Then I would like to thank my teacher, Puan Andek for guiding me and my friends throughout this project. We had some difficulties in doing this task, but he taught us patiently until we knew what to do. He tried and tried to teach us until we understand what we supposed to do with the project work. Last but not least, my friends who were doing this project with me and sharing our ideas. They were helpful that when we combined and discussed together, we had this task done. Copyright 2011  ©. Hazira Syaza, All Right Reserve 1 Objective The aims of carrying out this project work are: * To apply and adapt a variety of problem-solving strategies to solve problems * To improve thinking skills * To promote effective mathematical communication To develop mathematical knowledge through problem solving in a way that increases students’ interest and confidence * To use the language of mathematics to express mathematical ideas precisely * To provide learning environment that stimulates and enhances effective learning * To develop positive attitude towards mathematics Copyright 2011  ©. H azira Syaza, All Right Reserve 2 Introduction Part 1 Cakes come in a variety of forms and flavours and are among favourite desserts served during special occasions such as birthday parties, Hari Raya, weddings and others. Cakes are treasured not only because of their onderful taste but also in the art of cake baking and cake decorating Baking a cake offers a tasty way to practice math skills, such as fractions and ratios, in a real-world context. Many steps of baking a cake, such as counting ingredients and setting the oven timer, provide basic math practice for young children. Older children and teenagers can use more sophisticated math to solve baking dilemmas, such as how to make a cake recipe larger or smaller or how to determine what size slices you should cut. Practicing math while baking not only improves your math skills, it helps you become a more flexible and resourceful baker. Copyright 2011  ©. Hazira Syaza, All Right Reserve 3 MATHEMATICS IN CAKE BAKING AND CAKE DECORATING GEOMETRY To determine suitable dimensions for the cake, to assist in designing and decorating cakes that comes in many attractive shapes and designs, to estimate volume of cake to be produced When making a batch of cake batter, you end up with a certain volume, determined by the recipe. The baker must then choose the appropriate size and shape of pan to achieve the desired result. If the pan is too big, the cake becomes too short. If the pan is too small, the cake becomes too tall. This leads into the next situation. The ratio of the surface area to the volume determines how much crust a baked good will have. The more surface area there is, compared to the volume, the faster the item will bake, and the less inside there will be. For a very large, thick item, it will take a long time for the heat to penetrate to the center. To avoid having a rock-hard outside in this case, the baker will have to lower the temperature a little bit and bake for a longer time. We mix ingredients in round bowls because cubes would have corners where unmixed ingredients would accumulate, and we would have a hard time scraping them into the batter. Calculus (DIFFERENTIATION) To determine minimum or maximum amount of ingredients for cake-baking, to estimate min. or max. amount of cream needed for decorating, to estimate min. or max. Size of cake produced. Copyright 2011  ©. Hazira Syaza, All Right Reserve 4 PROGRESSION To determine total weight/volume of multi-storey cakes with proportional dimensions, to estimate total ingredients needed for cake-baking, to estimate total amount of cream for decoration. For example when we make a cake with many layers, we must fix the difference of diameter of the two layers. So we can say that it used arithmetic progression. When the diameter of the first layer of the cake is 8? nd the diameter of second layer of the cake is 6? , then the diameter of the third layer should be 4?. In this case, we use arithmetic progression where the difference of the diameter is constant that is 2. When the diameter decreases, the weight also decreases. That is the way how the cake is balance to prevent it from smooch. We can also use ratio, because when we prepare the ingredient for each layer of the cake, we need to decrease its ratio from lower layer to upper layer. When we cut the cake, we can use fraction to devide the cake according to the total people that will eat the cake. Copyright 2011  ©. Hazira Syaza, All Right Reserve 5 Part 11 Best Bakery shop received an order from your school to bake a 5 kg of round cake as shown in Diagram 1 for the Teachers’ Day celebration. 1) If a kilogram of cake has a volume of 38000cm3, and the height of the cake is to be 7. 0 cm, the diameter of the baking tray to be used to fit the 5 kg cake ordered by your school 3800 is Volume of 5kg cake = Base area of cake x Height of cake 3800 x 5 = (3. 142)( d/2)? x 7 1900/7 (3. 142) = ( d/2)? 863. 872 = (d/2 )? d/2 = 29. 392 d d = 58. 784 cm Copyright 2011  ©. Hazira Syaza, All Right Reserve 2) The inner dimensions of oven: 80cm length, 60cm width, 45cm height a) The formula that formed for d in terms of h by using the formula for volume of cake, V = 19000 is: 19000 = (3. 142)(d/2)(d/2)   ? h 1900/(3. 142)2 = d   ? /4 24188. 415/h = d   ? d = 155. 53/ h Copyright 2011  ©. Hazira Syaza, All Right Reserve 7 Height,h| Diameter,d| 1. 0| 155. 53| 2. 0| 109. 98| 3. 0| 89. 79| 4. 0| 77. 76| 5. 0| 69. 55| 6. 0| 63. 49| 7. 0| 58. 78| 8. 0| 54. 99| 9. 0| 51. 84| 10. 0| 49. 18| Table 1 b) i) h 7cm is NOT suitable, because the resulting diameter produced is too large to fit into the oven. Furthermore, the cake would be too short and too wide, making it less attractive. b) ii) The most suitable dimensions (h and d) for the cake is h = 8cm, d = 54. 99cm, because it can fit into the oven, and the size is suitable for easy handling. c) i) The same formula in 2(a) is used, that is 19000 = (3. 142)( )? h. The same process is also used, that is, make d the subject. An equation which is suitable and relevant for the graph: Copyright 2011  ©. Hazira Syaza, All Right Reserve 8 1900= (3. 1420(d/2)   ? h 119000/(3. 142)h = d   ? /4 24188. 415/h = d   ? d = 155. 53/ vh d = 155. 53h(1-/? ) log d = log 155. 3h(-1/? ) log d = (-1/? )log h + log 155. 53 Table of log d = (-1/? )log h + log 155. 53 Height,h| Diameter,d| Log h| Log d| 1. 0| 155. 53| 0. 00| 2. 19| 2. 0| 109. 98| 0. 30| 2. 04| 3. 0| 89. 79| 0. 48| 1. 95| 4. 0| 77. 76| 0. 60| 1. 89| 5. 0| 69. 55| 0. 70| 1. 84| 6. 0| 63. 49| 0. 78| 1. 80| 7. 0| 58. 78| 0. 85| 1. 77| 8. 0| 54. 99| 0. 90| 1. 74| 9. 0| 51. 84| 0. 95| 1. 71| 10. 0| 49. 18| 1. 0| 1. 69| Table 2 Copyright 2011  ©. Hazira Syaza, All Right Reserve 9 Graph of log d against log h Copyright 2011  ©. Hazira Syaza, All Right Reserve 10 ii) Based on the graph: a) d when h = 10. 5cm h = 10. 5cm, log h = 1. 21, log d = 1. 680, d = 47. 86cm b) h when d = 42cm d = 42cm, log d = 1. 623, log h = 1. 140, h = 13. 80cm 3) The cake with fresh cream, with uniform thickness 1cm is decorated a) The amount of fresh cream needed to decorate the cake, using the dimensions Ive suggested in 2(b)(ii) My answer in 2(b)(ii) ==; h = 8cm, d = 54. 99cm Amount of fresh cream = volume of fresh cream needed (area x height) Amount of fresh cream = volume of cream at the top surface + volume of cream at the side surface The bottom surface area of cake is not counted, because were decorating the visible part of the cake only (top and sides). Obviously, we dont decorate the bottom part of the cake Volume of cream at the top surface = Area of top surface x Height of cream = (3. 142)(54. 99/2) ? )x 1 = 2375 cm? Volume of cream at the side surface = Area of side surface x Height of cream = (Circumference of cake x Height of cake) x Height of cream = 2(3. 142)( 54. 99/2)(8) x 1 = 1382. 23 cm? Therefore, amount of fresh cream = 2375 + 1382. 23 = 3757. 23 cm? Copyright 2011  ©. Hazira Syaza, All Right Reserve 11 c) Three other shapes (the shape of the base of the cake) for the cake with same height which is depends on the 2(b)(ii) and volume 19000cm? The volume of top surface is always the same for all shapes (since height is same), My answer (with h = 8cm, and volume of cream on top surface =1900/8 = 2375 cm? ) 19000 = base area x height base area = 1900/8 length x width = 2375 By trial and improvement, 2375 = 50 x 47. 5 (length = 50, width = 47. 5, height = 8) Therefore, volume of cream = 2(Area of left and right side surface)(Height of cream) + 2(Area of front and back side surface)(Height of cream) + volume of top surface = 2(50 x 8)(1) + 2(47. 5 x 8)(1) + 2375 = 3935 cm? Copyright 2011  ©. Hazira Syaza, All Right Reserve 12 Triangle-shaped base 19000 = base area x height base area = 1900/8 base area = 2375 x length x width = 2375 length x width = 4750 By trial and improvement, 4750 = 95 x 50 (length = 95, width = 50) Slant length of triangle = v (95? + 25? )= 98. 23 Therefore, amount of cream = Area of rectangular front side surface(Height of cream) + 2(Area of slant rectangular left/right side surface)(Height of cream) + Volume of top surface = (50 x 8)(1) + 2(98. 23 x 8)(1) + 2375 = 4346. 68 cm? Copyright 2011  ©. Hazira Syaza, All Right Reserve 13 3 – Pentagon Shaped Base 19000 = base area x height ase area = 2375 = area of 5 similar isosceles triangles in a pentagon therefore: 2375 = 5(length x width) 475 = length x width By trial and improvement, 475 = 25 x 19 (length = 25, width = 19) Therefore, amount of cream = 5(area of one rectangular side surface)(height of cream) + vol. of top surface = 5(19 x 8) + 2375 = 3135 cm? c) Based on the values above, the shape that require the least amount of fresh cream to be used is: Pentagon-shaped cake, since it requires only 3135 cm? of cream to be used. Copyright 2011  ©. Hazira Syaza, All Right Reserve 14 Part III When theres minimum? or maximum? well, theres differentiation and quadratic functions. The minimum height, h and its corresponding minimum diameter, d is calculated by using the differentiation and function. Method 1: Differentiation Two equations for this method: the formula for volume of cake (as in 2(a)), and the formula for amount (volume) of cream to be used for the round cake (as in 3(a)). 19000 = (3. 142)r? h (1) V = (3. 142)r? + 2(3. 142)rh (2) From (1): h =19000/(3. 142)r? (3) Sub. (3)into (2): V=(3. 142)r? +2(3. 142)r(19000/(3. 142)r? ) V=(3. 142)r? +(38000/r) V=(3. 142)r? +38000r-1 (dV/dr)=2(3. 142)r-(38000/r? ) 0=2(3. 142)r-(38000/r? minimun value therefore dv/dr=0 38000/r? =2(3. 142)r 38000/2(3. 142)=r^3 6047. 104=r^3 R=18. 22 Sub. r = 18. 22 into (3) H=19000/(3. 142)(18. 22) Therefore,h = 18. 22cm,d=2r=2(18. 22)=36. 44cm Copyright 2011  ©. Hazira Syaza, All Right Reserve 15 Method 2 : Quadratic Functions Two same equations as in Method 1,but only the formula for amount of cream is the main equation used as the quadratic functions. Let f(r)=volume of cream,r = radius of round cake: 19000 = (3. 142)r ? h (1) F(r)=(3. 142)r ? +2(3. 142)hr (2) From (2): F(r) = (3. 142)(r ? +2hr) factorize (3. 142) = (3. 142)[(r+2h/2) ? (2h/2) ? ] completing square,with a =(3. 142),b=2h and c=0 = (3. 142)[(r+h) ? -h ? ] = (3. 142)(r+h) ? -(3. 142)h ? (a=(3. 142)(positive indicates min. value),min. value = f(r)=-(3. 142)h ? ,corresponding value of x = r = -h) Sub. r =-h into (1): 19000=(3. 142)(h) ? h h^3=6047. 104 h=18. 22 Sub. h=188. 22 into (1) 19000=(3. 142)r ? (18. 22) r ? =331. 894 r=18. 22) therefore,h=18. 22cm,d=2r=2(18. 22)=36. 44cm Copyright 2011  ©. Hazira Syaza, All Right Reserve 16 I would choose not to bake a cake with such dimensions because its dimensions are not suitable (the height is too high) and therefore less attractive. Furthermore, such cakes are difficult to handle easily. Copyright 2011  ©. Hazira Syaza, All Right Reserve 17 Futher Exploration Copyright 2011  ©. Hazira Syaza, All Right Reserve 18 Best Bakery received an order to bake a multi-storey cake for Merdeka Day celebration, as shown in Diagram 2. The height of each cake is 6. 0 cm and the radius of the largest cake is 31. 0 cm. The radius of the second cake is 10% less than the radius of the first cake, the radius of the third cake is 10% less than the radius of the second cake and so on. Given: height, h of each cake = 6cm radius of largest cake = 31cm adius of 2nd cake = 10% smaller than 1st cake radius of 3rd cake = 10% smaller than 2nd cake 31, 27. 9, 25. 11, 22. 599,†¦ a = 31, r = 9/10 V = (3. 142)r? h, a) By using the formula for volume V = (3. 142)r? h, with h = 6 to get the volume of cakes. Volume of 1st, 2nd, 3rd, and 4th cakes: Copyright 2011  ©. Hazira Syaza, All Right Reserve 19 Radius of 1st cake = 31, volume of 1s t cake = (3. 142)(31)? (6) = 18116. 772 Radius of 2nd cake = 27. 9, 9, volume of 2nd cake = (3. 142)(27. 9)? (6) 14674. 585 Radius of 3rd cake = 25. 11, Radius of 4th cake = 22. 59 volume of 3rd cake = (3. 42)(25. 11)? (6) 11886. 414 , volume of 4th cake = (3. 142)(22. 599)? (6) 9627. 995 The volumes form number pattern: 18116. 772, 14674. 585, 11886. 414, 9627. 995,†¦ (it is a geometric progression with first term, a = 18116. 772 and ratio, r = T2/T1 = T3 /T2 = †¦ = 0. 81) b) The total mass of all the cakes should not exceed 15 kg ( total mass ; 15 kg, change to volume: total volume ; 57000 cm? ), so the maximum number of cakes that needs to be baked is Copyright 2011  ©. Hazira Syaza, All Right Reserve 20 Sn =(a(1-rn))/(1-r) Sn = 57000, a = 18116. 772 and r = 0. 81 57000 =(18116. 22(1-(0. 81)n))/(1-0. 81) 1 0. 81n = 0. 59779 0. 40221 = 0. 81n og0. 81 0. 40221 = n n=log0. 40221/log0. 81 n = 4. 322 therefore, n ~ 4 Verifying the answer: When n = 5: S5 = (18116. 772(1 ( 0. 81)5)) / (1 0. 81) = 62104. 443 ; 57000 (Sn ; 57000, n = 5 is not suitable) When n = 4: S4 = (18116. 772(1- (0. 81)4)) / (1 -0. 81) = 54305. 767 ; 57000 (Sn ; 57000, n = 4 is suitable) Copyright 2011  ©. Hazira Syaza, All Right Reserve 21 Reflection Copyright 2011  ©. Hazira Syaza, All Right Reserve 22 TEAM IS IMPORTANT! BE HELPFUL ALWAYS READY TO LEARN NEW THINGS BE A HARDWORKING STUDENT Copyright 2011  ©. Hazira Syaza, All Right Reserve 23 CONCLUSION * Geometry is the study of angles and triangles, perimeter, area and volume. It differs from algebra in that one develops a logical structure where mathematical relationships are proved and applied. * An arithmetic progression (AP) or arithmetic sequence is a sequence of numbers such that the difference of any two successive members of the sequence is a constant * A geometric progression, also known as a geometric sequence, is a sequence of numbers where each term after the first is found by multiplying he previous one by a fixed non-zero number called the common ratio * Differentiation is essentially the process of finding an equation which will give you the gradient (slope, rise over run, etc. ) at any point along the curve. Say you have y = x^2. The equation y = 2x will give you the gradient of y at any point along that curve. Copyright 2011  ©. Hazira Syaza, All Right Reserve 24 REFERENC E * Wikipedia * one-school. net/ * Additional Mathematics text book form 4 and form 5 Copyright 2011  ©. Hazira Syaza, All Right Reserve 25