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AS and A Level: Core & Pure Mathematics
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Differentiation and intergration
- 1 It is easy to get differentiation and integration the wrong way round. Remember that the power gets smaller when differentiating.
- 2 Differentiation allows you to find the gradient of a tangent at any point on a curve. The first derivative describes the rate of change.
- 3 If a function is increasing then the first derivative is positive, if a function is decreasing, then the first derivative is negative.
- 4 When asked to find the area under a curve, it is asking you to integrate that curve between two points. Even if you don’t know the points, pick two numbers. You’ll get marks for methods.
- 5 When referring to a min/max/stationary point, the gradient equals 0. Differentiate the curve and set this to equal 0. The second derivative tells you whether it is a maximum or minimum. If the second derivative is positive, the point is a minimum, if the second derivative is negative, then the point is a maximum.
Quadratics and circles
- 1 When solving a quadratic inequality, always draw a picture. The inequality is less than 0, where the curve is below the x-axis and bigger than 0 when the curve is above the x-axis.
- 2 Sometimes in part (a) of a question you are asked to find something, for example a radius. In part (b) you might be then asked to use the radius that you found. If you couldn’t do part (a), don’t give up, choose a random radius.
- 1 To find the distance of a straight line, draw the straight line with the co-ordinates. Then make a right angle triangle, find the lengths of the horizontal and vertical lines, then use Pythagoras.
- 2 When a question asks you for a straight line. The first thing to do is to write down the equation of a straight line. Then find out what information you know, and what information you need. Even if you don’t understand the whole question, it is important to start.
Although everyone who gambles at all probably tries to make a quick mental marginal analysis of the game, in depth analysis of the figures shall reveal how a rational player reacts to better odds, or a lower entry price, or a higher potential payout.
Case 3 can be represented by a draw, where E is set, and both O and P are known. A process of linear regression will be used to determine the actual relationship between P and O for each game. Analysis Case 1 The lottery chosen is from Ontario, lotto 649. The player pays $1 to enter the game, and then chooses 6 non-repeating numbers from 1 to 49. Thus the number of possible combinations a player can choose equals. 6 numbers are drawn and the payout scheme is shown below in Table 1: (This data is from the January 12th 2002 drawing; although it is only one sample of data, it is an accurate representation of the average value of prizes)
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Nevertheless, we would need to convert most of the data into logs, Experiment Log Volume of KI (dm3) Time taken in seconds Log (1/time) A -2.30 596 -2.77 B -2.00 336 -2.53 C -1.82 247 -2.39 D -1.70 207 -2.31 E -1.60 160 -2.20 Thus, we could derive the gradient of the graph, by drawing it, as seen upon the nest page. We could also use the statistical method of finding the equation of the regression line, which is: Y = a +bx (where y and x are the axis's) B = ((xI - (x)(yI - (y) ((xI - (x)2 a = (y - B(x Where (y = average of all y values (x = average of all
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Iterations of: f(x) = 1.25x - 2 from the left side of graph: First "guess": 5 f(5) = 1.25(5) - 2 = 4.25 Second iteration: f(4.25) = 1.25(4.25) - 2 = 3.3125 Third iteration: f(3.3125) = 1.25(3.3125) -2 = 2.140625 Fourth iteration: f(2.140625) = 1.25(2.140625) -2 = 0.67578125 - Copied axes located on page C. Iterations of f(x) = 1.25x - 2 from the right side: First "guess": 10 F(10) = 1.25(10) -2 = 10.5 Second iteration: F(10.5) = 1.25(10.5)
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However, if we cannot use the computer, we have to use one of the numerical methods - interval estimation. A root must be lying on the interval a and b where an interval have been marked in which F(x) change signs. We thus know that there must be a root in that interval. The equation F(x) = x�-9x+3 has 3 roots which are in the interval of (-3,-4), (0,1) and (2,3). We can find the roots through decimal search, interval bisection and linear interpolation. Decimal search It is one of the methods of interval estimation through dividing each interval into 10 parts and looking the sign change.
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Null Hypothesis: There will be no difference in obedience between the signs written by the authority figure and school pupil Variables: The independent variable was whom the signs were written by. The dependent variable was whether the pupil obeyed the signs or not. Population: Sixth Form pupils (16 and over), in UK. Sample: An opportunity sample of 40 Sixth Form pupils from a secondary school in Letchworth per lunch hour. They included male and female subjects, ranging from 16 to 18 years old.
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+ 1 = 8. For Diagrams 1 - 4 I can see a pattern with square numbers. The diagrams numbers squared added to one less than the diagram number squared gives the correct number of squares. For diagram n it should be: Un = (n - 1) 2 + n2 (n - 1)(n - 1) + n2 n2 - n - n - n + 1 + n Un = 2n2 - 2n + 1 This is correct. Growing Hexagons I will now repeat my investigation, and change the original shape of the square to hexagons, and try to find the formula as before. I shall start by finding the width of each hexagon. Pattern Number (n)
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f(x) = x� - 12x + 5 I am going to use the root in the interval [0,1] f(0) > 0 f(1) < 0 f(0+1) � f(0.5) = -0.875 2 Therefore f(0.5) < 0 f(0) > 0 f(0.5) < 0 f(1) < 0 Root is in interval [0, 0.5] I now test the mid-point of [0, 0.5] f(0+0.5) � f(0.25) = 2.015625 2 Therefore f(0.25) > 0 f(0) > 0 f(0.25) > 0 f(0.5) < 0 Root is in interval [0.25, 0.5] I will now test mid-point of [0.25, 0.5] f(0.25+0.5) � f(0.375) = 0.552734375 2 Therefore f(0.375) > 0 f(0.25)
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I interval that I measured in was 2.5 cm, which gave me 10points for my graph. To get my y points I had to pour water up to the cylindrical section and made a hole at the bottom of the bottle. As the water passes a point I record the time. 2) A mathematical representation of the data Graphically Algebraically The equation for this data is 3.429801*1.126417^x. To get the equation I used the Graphics calculator and find the equation in the data matrix editor. Although when I was trying to validate it the computer gave me a different equation (this is using excel).
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and touches the x-axis at (3,0). I used the same method as before, of drawing what I thought the graph would look like: (0,9) (3,0) I then put the numbers into brackets again (as below), because I worked out that when Y=0, X=3, and no other number. Then once again expanded the brackets to find the formula: Y=(x-3)(x-3) = 0 I worked out the formula to be: Y=x2-6x+9 I could be sure that this was the correct equation because the co-ordinate was (0,9)
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The diagram shows a house built with dominoes. This house has four stories and uses 24 dominoes.Simon broke the World Record by building a domino house with 73 stories. How many dominoes did he use?
I will try to work systematically to make the project simple to understand. I will then draw a results table where I will comment on why I am using the results and what purpose it has in my project. I will also explain if I can spot any patterns. After I spot the patterns I will work out the differences. Next, using the differences I will draw a differences table, which will give us the rules (either being linear and quadratic). Using these rules I will draw a graph, and if the graph has a straight line then this determines the rule is linear if the line is a smooth curve this determines the rule is quadratic.
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this is a drawback when there are two or more roots in between two integer values. Method 2: Newton-Raphson Method Using Autograph again, the roots for y=x5 -3x2 + 1 were found using the Newton-Raphson Method: This illustration demonstrates how we acquire the first root of the equation. The tangent is found at a point and then the point where this tangent crosses x=0 is found. The process is then repeated from this point. The equation for Newton-Raphson iterations is: If we find the root closest to -0.5, starting at -1, the values we find during the Newton-Raphson method are:
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The aim of lesion studies is to tell us something about how different areas of the brain are connected. However, there are several flaws with this invasive method of studying the brain. One of these is that, since experiments of this type are carried out on animals, the results cannot really be generalised to humans. Also, there are ethical issues involved in the use of animals in experiments that could cause distress. Other invasive methods of investigating the brain include; chemical stimulation of the brain and Electrical stimulation of the brain (ESB). One such investigation was conducted by Olds + Milner (54).
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I did not set-up any validation checks, macros or cell protection. I simply wanted to see if this section of the system would function correctly. Input Process Problem On the process sheet I had difficulty setting up the 'Discount on spending' cell. I used the formulae: =IF(G21>300,40,G21>200,20,G21>100,10,0)
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- b F (a)]/[F (b)- F (a)]. Newton-Raphson Method: Xr+1 = Xr - f (Xr)/ f' (Xr), X0 is given. Fix-point iteration: The reason why I don't use Fixed-point method is because to rearrange the formula to x= g (x) is very difficult and I tried this method but found the duration didn't go to convergence. (Appendix 1) So I didn't use this method any more. As the spreadsheet (Appendix 2) shown Newton-Raphson method is most efficient method to find a root.
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Sorting the Data You notice that the teams are not listed in alphabetical order and you decide that you want to put this right. 1. You must first select columns B and C It is very important that you do this. If you just sort the list of team names, the teams will be rearranged but the goals scored won't. That would make your data incorrect. 2. From the Data menu, select Sort. 3. Sort by Column B (the teams list)
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We used a foot pump and pumped the air into the column, which increased the pressure. We then took readings of pressure and volume every five seconds and noted down the results. Then after we had taken about thirty results we re-tested them again to insure that the results were totally accurate and no error had occurred. From our results we had taken we plotted a graph with pressure (lb/in2) on the vertical axis and volume (cm3) on the horizontal axis. From all these points we drew a line of best fit. Analysis There are a few strange points, which may have been from human error of reading.
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Simplifying the expression we get the value observed. c) Finally graph, with a suitable window . What are the roots of ? We find that the roots are: Therefore, the smallest positive root, expressed as a fraction is This values correspond also to those of the graph in part (b) and we find that this equation has the same coefficients as in the previous one, only that multiplied by 3. Even though the second and fourth terms have changed, the other two roots are kept the same. Given this aspect, we may assume that the relative proportions of the coefficients and not their actual values define the position of the roots.
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EQUATION USED: = 0 has only 1 roots in [0,-1] From Graphmatica: Results obtained from Microsoft Excel: X-value F(X) -1 -1.5 0 0.5 Observe the change of y-value from negative value to positive value in the table. That shows that the root of the equation where the y-value is 0, is between x = 0 and x = -1. Now this is step 1, we're moving on to step 2. X-value F(X) -0.9 -0.7882969 -0.8 -0.3497152 -0.7 -0.0723543 -0.6 0.1120064 Now the search is narrowing in and it comes to one decimal place in between x = -1 and x = 0.
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The only exception would be where nothing is being added to the x� (A) in which case the parabola would have its vertex's turning point on the origin. The first graph (A) shows us that y = x� gives us a parabola with the vertex on the origin, the positive leading coefficient is shown by the parabola opening up meaning the parabola is positive. This graph is showing neither dilations, translations nor reflections, because no other coefficients are available. Nevertheless graph B (y = x� + 3) and graph C (y = x� - 2)
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- x3/2 = 0 This equation can not be solved numerically because x appears inside the function ln(x) as well as in a power term. Therefore we can not find the inverse function and isolate x. This comparatively innocuous looking function crosses the x-axis at some point around the interval [1, 2] and again at some point around the interval [8, 9]. For small values of x the term 12ln(x) dominates and is large and negative. For larger values of x the term x3/2 dominates and is large and positive, making the f(x) value large and negative. My centre of interest lies within the range 0<=x<=4.
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We know from the rules of differentiation for such a function that for the (n+1) th differential you multiply the nth function by its power and lower its power by one. However, if we don't know the nth differential and we know only the original function this simple method is obviously not possible. Looking at the above example suggests that a solution would have to have something to do with factorials. The way the power and the differential number add together to equal b is of particular interest.
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After 4hrs distance between the cyclist and man 2. Question: If experimental data doesn?t produce a straight trend line it doesn?t mean there isn?t a relationship. The use of software packages can allow quick and accurate testing of data to different trend lines so that a suitable relationship can be found if it exists. Atmospheric pressure P is measured at varying altitudes h as shown below. The data is thought to be of the exponential form p = aekh Altitude 500 1500 3000 5000 8000 Metres Pressure 73 68 62 54 43 cm a)
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X = 1.1155 Error Bounds = ±0.00005 Solution bounds = (1.11545≤ x ≤1.11555). n xn f(xn) f `(xn) xn+1 0 x0 = 1 f(x0)= -3 f '(x0)= 23 x1 = 1.1304 1 x1 = 1.1304 f(x1) = 0.4376 f '(x1) = 29.826 x2 = 1.1157 2 x2 = 1.1157 f(x2) = -0.000748 f '(x2) = 29.015 x3 = 1.1155 Solving 6x3+7x2-9x-7=0 for second root ________________ The root lies between -0.63488 and -0.634875 f(x) is negative -0.634885 -0.63488 -0.634875 f(-0.634885)= 0.000065017 f(-0.63488)=0.000011852 f(-0.634875)= -0.000041312 X= -0.63488 (5 sig fig) Error bounds = ±0.000005 Solution bounds = (-0.634875≤ x ≤-0.634885).
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x y=x³+3x²–3 0.8793 -0.00065 0.87931 -0.00057 0.87932 -0.0005 0.87933 -0.00042 0.87934 -0.00034 0.87935 -0.00027 0.87936 -0.00019 0.87937 -0.00012 0.87938 -4E-05 Change of sign 0.87939 3.61E-05 0.8794 0.000112 The change of sign here tells us that there is a root in the interval [0.87938, 0.87939]. So to 4 d.p. the root has value 0.8794. x x³+3x²–3 0.87938 -4E-05 0.87939 3.61E-05 The solution bounds are 0.87938and 0.87939. x x³+3x²–3 -4 -19 -3 -3 -2 1 -1 -1 0 -3 1 1 2 17 Error bounds: 0.879385± 0.000005.
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Use of technology I used Microsoft Excel throughout my method along with many algorithms, these algorithms allowed me to reach a more accurate approximation thanks to the fact that there were no rounding errors usually gained through using a calculator that gives answers to fewer decimal places. Overall then Excel is more accurate than the use of a calculator due to being able to work to more decimal places, it is also easier to use and saves a lot of time.
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