Surface Area: Volume Ratio Investigation
Surface Area: Volume Ratio Investigation I am going to do an experiment to find out how heat loss is affected by surface area to volume ratio. I will make this investigation as safe as possible by wearing goggles, and using a kettle rather than a Bunsen burner to avoid unnecessary danger. Also I will make the experiment as fair as possible by: - * Making my measurements as accurate as possible * I will repeat each part of the experiment 3 times and work out an average, using a mean average as I think this is most suitable for this experiment. * I will keep everything apart from the surface area to volume ratio the same. Although I will make the experiment as fair as I possibly can there may be other factors (which I cannot control) that may effect the results such as: - * The surface area may not be exact because the variables are not exact cylinder shapes. * Heat may escape from the water while I am pouring it into the glassware. * Human error e.g. misreading measurements I predict that the smaller the surface area compared to the volume the longer it will take for heat loss. I have decided on this prediction after researching homeostasis. Homeostasis is the maintenance of a constant internal environment, by balancing bodily inputs and outputs and removing waste products by an animal. In my research looked at a polar bear. A polar bear lives in a cold, harsh
To investigate the areas of different shapes when they are joined together on square dotted paper
William Alston. G.C.S.E Maths Coursework. 1st October 2001 Introduction: - We are to investigate the areas of different shapes when they are joined together on square dotted paper. To start off with, in this investigation, I will be looking at regular shapes (those with 45° and 90° angles). To break it up into simple parts I will test 1 thing at a time using 5 different shapes. I will start by making regular shapes with 1 dot inside and then 2 dots and then 3 dots on the inside. Then I will make regular shapes and change the area each time e.g. 4, 5 and 6cm's². Also I will try different numbers of dots on the outside if I feel it is necessary to my investigation. That is something that I will be deciding as I go along. Then I will look at irregular shapes (those with angles of say 30º or 75º). After I have drawn the 5 shapes I will put them in a table, so they are easy, to Asses. I will try to draw up a formula for those shapes then I will test the formula by drawing another shape and working out the formula before physically counting the area. Throughout the investigation I will be carrying out an on-going evaluation, as this will help me in finding formulas and noticing patterns. I will also be predicting shape areas and explaining patterns and formulas and trying to justify why the formulas work. I will also be explaining the decisions that I
Assignment 2 Pond area
AS Use of Maths Working with Calculus Assignment 2: Pond Area Introduction: As part of the 'Mathematical Garden' a lake has been built, whose edges can be modelled by y =10COS(0.08x)+15 and y =10SIN(0.08x)+25 between x = 0 and x = 60, which gives the picture below: Finding the area By numerical methods To find an area between 2 lines, you find the area under one line and then subtract the area under the other. As there are 2 methods of finding the approximate area numerically, one will be used to find the area under one curve and the other method will be used for the 2nd curve. The calculator will then be used to find the exact answer to each and hence it will be possible to asses which mothod produces the best results. Finding the area under y =10SIN(0.08x)+25 by the trapezium rule A fairly good approximation could be gained by dividing the line into 3 sections each 20m wide. The formula Area =1/2 width(h1+2h2+ .. +2hn-1+ hn) = 1/2 x 20 (25 + 2 x 35 + 2 x25 + 15) = 1/2 x 20 x 160 = 1600m2 As this curve is mostly concave, the area will be an underestimate, but it would have been more accurate if it had been divided up into more strips But by the calculator the integral is 1614m2, so the error% is (1614-1600)/1614 x 100 =0.87% which is very small and in part due to the last 3rd of the curve being convex Finding the area under y =10Scos(0.08x)+15 by the
fencing problem part 2/8
This is a graph of the areas of isosceles triangles. It contains more details than the table above and there are more results shown. My results show that the triangle with the largest area for a perimeter is the equilateral triangle. This also follows my previous observation but with a slight change. This triangle has no difference between all three sides, not just two sides. As there is only one form of equilateral triangle for each perimeter, I will now work out the area of an equilateral triangle with a perimeter of a thousand metres. To find out the area of this equilateral triangle, I will use Heron's Formula. The formula is Where S stands for semiperimeter. (Half of the total perimeter). 48112.52m2 Now that I know that equilateral triangles are the best shape to use for three sided shapes, I will now investigate what is the best shape to use for four sided shapes (quadrilaterals). To work out the areas of quadrilaterals, I will use Bretschneider's formula. The formula is . In this formula a, b, c and d are the lengths of the four sides of the quadrilateral. The s stands for the semiperimeter and ? is half the sum of two opposite angles. As we want the maximum area for a quadrilateral, the number which has to be square rooted should be as high as it can be. This can only be achieved if the 'abcdCos2? ' part of the formula is a low as possible because this
Proving a2 + b2 = c2 Using Odd Numbers
Proving a2 + b2 = c2 Using Odd Numbers a2 + b2 = c2 (2n + 1) (2n + 1) + (2n2 + 2n) (2n2 + 2n) = (2n2 + 2n + 1) (2n2 + 2n + 1) 4n2 + 8n3 + 4n2 + 4n + 1 = 4n4 + 4n3 + 2n2 + 4n3 + 4n2 +2n + 2n2 +2n +1 4n4 + 8n3 + 8n2 + 4n + 1 = 4n4 +8n3 + 8n2 +4n + 1 The above proves that a2 + b2 = c2 is correct using odd numbers and is a overall formula for Pythagorean triples. Perimeter Odds Perimeter = a + b + c a= 2n + 1 b= 2n2 + 2n c= 2n2 + 2n + 1 = 4n2 + 6n + 2 Proving that this formula is correct: If I take n=3 = 4x9 + 18 + 2 = 36 + 18 + 2 = 56 So if you look along the odds table on n=3 the perimeter is 56 this proves my formulae is correct. Perimeter Evens Perimeter = a + b + c a = 2n + 4 b = n2 + 4n + 3 c = n2 + 4n + 5 = 2n2 + 10n + 12 Proving my formulae is correct: If I take n=3 = 18 + 30 + 12 = 60 Follow n=3 across on the evens table the perimeter says 60 my formulae is correct. Odds Side c c = b + 1 c = 2n2 +2n + 1 So from looking at the table and extending it, I have managed to come up with the conclusion that: ODDS a = 2n + 1 b = 2n2 + 2n c = 2n2 + 2n + 1 Evens Table Length of (a) Shortest Side Length of (b) Middle Side Length of (c) Shortest Side Perimeter Area 4 3 5 6 8 0 3x6 24 24 8 5 7 4x8 40 60 0 24 26 5x10 60 20 2 35 37 6x12 84 210 4 48 50 7x14 12 336 6 63 65 8x16 44 504 From
Using Tai Po as a case study, is clustering move evident in an older of newer shopping area?
Li Po Chun Geography Field Investigation Using Tai Po as a case study, is clustering move evident in an older of newer shopping area? Introduction: Shopping is the prime leisure activity for residents here in Hong Kong, a city of life where virtually every type of commodity can be purchased in you know where to look. So why are some retail outlets of the same category very clustered while others appear dispersed? Does this pattern really exist and is it more likely to occur in old or new shopping areas? The aim of this geography field investigation is to compare the distribution of selected retail types in two contrasting commercial areas in Hong Kong. This information is useful in finding patterns of distribution within the two areas and will give us an idea of the degree of clustering through the use of the NNI (Nearest Neighbor Index). Research Question The aim of this investigation will be to answer the following question: Using Tai Po as a case study, is clustering move evident in an older of newer shopping area? Background The research area for this investigation will be Tai Po (see figure 1) of the many residential and industrial new towns in Hong Kong. Covering an area of 147.43 square kilometers and a population of over 320,000, Tai Po is one of the least densely populated settlements in Hong Kong. This makes Tai Po one of the least crowded areas for
To investigate the ratio of Area:Perimeter for triangles (2) To investigate the suggestion made by students that a 40,60,80 triangle would maximise the above mentioned ratio.
Natalie Pinfield Williams Aims (1) To investigate the ratio of Area:Perimeter for triangles (2) To investigate the suggestion made by students that a 40,60,80 triangle would maximise the above mentioned ratio. (3) Should the students hypothesis be un-sustainable then find out what triangle would maximise the ratio Important facts about triangles Simple formula for the area Area = 1/2 base *height General Formula Area = 1/2 ab sin C Interior Angles A+B+C = 180 Sin rule a = b = c Sin A Sin B Sin c Cosine Rule a2 = b2 + c2 - 2bc Cos A Student Suggestion We will Investigate the claim made by the students that a 40,60,80 triangle would maximise the ratio of area:perimeter Angles The students suggested that a 40 60 80 would maximise the area : perimeter ratio. We will make an initial assumption that the 40 6080 refers to the size of the angle in degrees. We accept that they could have been referring to lengths. Is our assumption possible? Yes as 40 + 60 + 80 =180. The interior angles do sum to 180 which they must for any triangle so the students could have been referring to the angles. Calculation As no side lengths have been give then we could assume any one side is any length we want. By defining the length of one side the others are obviously defined. We can calculate the other length by use of the sine rule. We will make a
The rain forest of Amazonia.
The rain forest of Amazonia "The biggest rain forest on earth is now disappearing!", "Trees have been cut down in an enormous speed in Amazonia."......etc. these headlines have already been around us for a very long time, but have anyone actually done anything to it? I don't think so, in fact, people just not care about it, they scarcely think that it is important! A local rubber tree cutter in the area, who works in 30, hectors big land, but in this big piece of land, with 30 thousand trees in the area and he has got only 60 rubber trees, so how can he get enough rubber to survive? He actually visits every rubber tree in the land and take very little bit of rubber liquid from each tree. The amount of rubber liquid that he gets might be very little, but he does virtually no damage to the forest. One of the local rubber tree cutters tells "It has been really hard to compete with those big rubber tree companies who just cut down all the trees in the area and only plant rubber trees in the area which gives them such a big amount of rubber liquid, but it also damages the forest so much!" Also, the land in the area are free from the government which destroys even faster, another fact is that the big companies can just call a contract killer to kill the owner of the land so that they can take the land for their own use! Another big problem in the forest is the mahogany trees
Cameron balloons does not need to be close to its customers because the balloons can be easily transported by methods of transport such as rail, plane and sea. This means that the whole world is an available market
Danielle Scholz 10R Cameron balloons Customers Cameron balloons does not need to be close to its customers because the balloons can be easily transported by methods of transport such as rail, plane and sea. This means that the whole world is an available market. Cameron balloons customers are the U.K and U.S because Cameron balloons customer base in is both the U.K and U.S. Cameron balloons offer e-mail addresses and phone numbers of their sales teams to its customers so they can contact the business whenever they need to. Raw materials Cameron balloons does not need to be buy its raw materials because it is located near major roads such as the A38, M5, M32 and the M4 which lead to and away from Bristol. It is situated by an international airport which connects Europe and the U.S. It is also on the channel, so deliveries can easily be done by sea. The closest station is under 10miles away so deliveries by rail are also easy. Skilled Labour There is skilled labour available in the area because Cameron balloons employs more than 120 people. You can tell this because there are to universities near Bristol University and the University of West England. Employees could easily be trained, as Bristol is a traditional manufacturing area and has two universities training should be very easy to come by. The cost of Labour The company employs more than 120 people.
Fencing Problem - Math's Coursework
... 50 425 418.330 31374.75 50 475 474.342 1858.55 Graph Conclusion The regular triangle seems to have the largest area out of all the areas but to make sure I am going to find out the area for values just around 333. Base (m) Side (m) Height (m) Area (m2) 333 333.5 288.964 48112.450 333.25 333.4 288.747 48112.518 333.3 333.4 288.704 48112.522 333.5 333.3 288.531 48112.504 333.75 333.1 288.314 48112.410 334 333.0 288.097 48112.233 This has proved that once again, the regular shape has the largest area. Because the last 2 shapes have had the largest areas when they are regular, I am going to use regular shapes from now on. Regular Pentagon Because there are 5 sides, I can divide it into 5 segments. Each segment is an isosceles triangle, with a top angle of 720. This is because it is a fifth of 360. This means I can work out both the other angles by subtracting 72 from 180 and dividing the answer by 2. This gives 540 each, because every isosceles triangle can be split into 2 equal right-angled triangles, I can work out the area of the triangle, using trigonometry. I also know that each side is 200m long (1000 divided by 5), so the base of the triangle is 100m. 88/8588/38588/image017.gif"> 88/8588/38588/image018.gif"> Using Trigonometry I can work out the unlabeled side. SOHCAHTOA Tangent =