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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.
Then the final step, how to find the gradient using the differences. To find the gradient, you have to use the method: GRADIENT = VERTICAL DIFFERENCE HORIZONTAL DIFFERENCE GRADIENT= 9 1.5 = 6 So the gradient at x=3 is 6. This shows an example of how to obtain the gradient at different "x" values. In my investigation, I will first start off by investigating simple graphs, such as straight line graphs. Then I will investigate further onto "Parabola" curves. Straight line graphs are simple and comprehensible. Parabola curves consist of a power which is larger than 1.
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We can see that there is a sign change, and therefore a root, in the interval?? [-1.1, -1.0] since the function is continuous. Having narrowed down the interval, we can now continue with increments of 0.01 within the interval [-1.1, -1.0]. x -1.1 -1.09 -1.08 -1.07 -1.06 -1.05 -1.04 -1.03 -1.02 -1.01 -1.0 f(x) -0.3 -0.2 -0.04 0.1 This has further narrowed down the interval within which the root lies. The interval has been narrowed down to [-1.08, -1.07]. Continuing to find the third decimal value of the root: x -1.08 -1.079 -1.078 -1.077 -1.076 -1.075 -1.074 -1.073 -1.072 -1.071 -1.07 f(x)
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curve crosses the x-axis changing its sign from positive ( + ) to negative ( - ). An initial interval of where a root lies can be obtained from a sketch. By taking the values of the initial interval we can increase the value of x by increments of 0.1 within the initial interval. Then if each of the increments of x is substituted into function of x a value can be obtained, and where there is change in sign from the values of f(x); it will state a closer interval (second interval)
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In this investigation I am going to investigate three methods of finding the roots to equations and then compare them. I will be using those techniques fully. The three methods that I am going to examine are:
The most ingenuous way to do this is to search for where the value of f(x) changes. This, of course, will only occur when the line crosses the x-axis locating the roots. Because the number line is infinite we cannot just check all the integer values from -? to ? so it is always a good idea to plot a graph of the equation to find out what region its roots lie in. The equation that I have chosen to demonstrate how Decimal Search is successful is f(x)=3x3-7x2-11x+17. In order to find the root of the function f(x)=3x3-7x2-11x+17, the decimal search method was used.
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x f(x) -5 -111 -4 -53 -3 -19 -2 -3 -1 1 0 -1 1 -3 2 1 3 17 4 51 5 109 One can see that there is a change of sign, and therefore a root of f(x)=0, between x=1 and x=2. If I wanted to find the root of the equation between x=1 and x=2 to five decimal places, I would do the following: The change of sign method will be applied for values of x between 1 and 2, taking 0.1 to be the interval in x values.
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Here is the table of results. From this table, I can see that the change in sign appears in between -1.53 and -1.52. To be certain that the root is between -1.53 and -1.52, I will calculate f(x) in each case to make sure that a change of sign occurs. F(x) = y=x�-2x+0.5 F(-1. 53) = -0.0216 F(-1. 52) = 0.0282 As a change of sign has been found and confirmed here, the root does lie between these points. I will now use the same method and use increment of 0.001 to find the root between -1.53 and -1.52.
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Then test this value to see if it's positive or negative, dependent on the result the interval then becomes smaller, hence the answer gets closer to the real answer (converges). This method can be used to solve this equation as we have been given the interval estimate, 0 < x < ?/2. Hence to obtain a better estimate we test, (0 + ?/2) / 2, which is ?/4. We now do f(?/4) in order to see if its negative or positive hence making the interval smaller.
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x < 2.40 x y 2.390 0.015369 2.391 0.008216971 2.392 0.001068288 2.393 -0.006077043 2.392 < x < 2.393 x y 2.3920 0.001068288 2.3921 0.000353604 2.3922 -0.000361047 2.3921 < x < 2.3922 x y 2.39210 0.000353604 2.39211 0.000282137 2.39212 0.000210671 2.39213 0.000139205 2.39214 6.77397E-05 2.39215 -3.72549E-06 2.39214 < x < 2.39215 x y 2.392140 6.77397E-05 2.392141 6.05932E-05 2.392142 5.34467E-05 2.392143 4.63001E-05 2.392144 3.91536E-05 2.392145 3.20071E-05 2.392146 2.48606E-05 2.392147 1.7714E-05 2.392148 1.05675E-05 2.392149 3.42102E-06 2.392150 -3.72549E-06 2.392149 < x < 2.392150 x= 2.39215 (6s.f)
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I will also be studying the symmetry, maximum/minimum points and asymptotes of each graph. The first transformation combination I will be looking at is ab using the trigonometric function of sine. Sine is shown by a blue line in Figure 1.0 A sine graph has no asymptotes but has a rotational symmetry or 180o about the origin(0,0), therefore the only remaining thing to look at is the maximum and minimum values. y=asin(bx) I am going to change a and b to the following values Graph 1 Graph 2 Graph 3 Graph 4 Graph 5 a 1 2 2 -2 -2 b 1 2 -2 2 -2 Predictions Graph 1 - Has a maximum of 1 by point (0.5?,1)
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lies between [-0.411,-0.410], these will be the error bounds, so I will find the midpoint of these to determine which bound is closer to the root. x y -0.411 -0.0002117 -0.4105 +0.001643 -0.410 +0.003495 There is a change of sign between -0.4105 (the midpoint) and -0.411; so the root to 3 d.p will be -0.411. Failure of the Change of Sign Decimal Search Method I will now demonstrate a case where decimal search will fail to find the root. I will use the equation Here is a graph of the function The graph clearly shows that there is a root
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To apply various numerical methods to find roots of equations and to appreciate the limitations of these methods
This shows that the root I am trying to find lies between these two values. X= Ans 0 -1 0.1 -0.848 0.2 -0.584 0.3 -0.196 0.4 0.328 0.5 1 0.6 1.832 0.7 2.836 0.8 4.024 0.9 5.408 1 7 I can get a more accurate answer as to what the root may be if I search to more significant figures. X= Ans 0.3 -0.196 0.31 -0.14992 0.32 -0.10246 0.33 -0.05363 0.34 -0.00339 0.35 0.04825 0.36 0.101312 0.37 0.155806 0.38 0.211744 0.39 0.269138 0.4 0.328 The graph above shows the values when tested to two decimal places.
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I then chose to look at the x values between these two values. I decreased the x value by 0.001 between x = -1.07 and -1.08. The change of sign between these two values occurred between x = -1.073 and x = -1.074. The root needs to be given to 3 decimal places and therefore I need to check the values between these two values to see whether the root is closer to the lower or higher value of x. To 3 decimal places, one root of the equation y=0.5x5-3x3-3 is x = -1.074 as the change of sign occurs between the interval x= -1.0737 and x = -1.0738 therefore I have taken
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I intend to use the trapezium and Mid-point rule as these are on of the basis in which the approximation can be found. Note: X is in radians. 1. Mid - point rule: The mid point rule was adopted, because it is used to approximate the area underneath the graph using rectangles. Below is the formula which is involved in the calculation. 2. Trapezium rule; This is also similar to the mid - point rule. It was adopted, because it would help me to approximate the region under the graph using strips of trapezium and calculating their area.
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On the other hand, if it was considered the flight of the plane would be affected. * The flight velocity of the plane is constant and the speed at which the plane travels will be achieved instantaneously. This will prevent complications concerned with takeoff and landing times. * Wind velocity is also constant and always blowing in the exact, same direction of the west, as the flight times for a constantly changing wind velocity would be almost impossible to calculate.
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is a. b. 5. Find the sums of the following series a. 5+9+13+...+101 b. -17-12-7-...+33 c. 1+ 1 1/4 + 1 1/2 + ... + 9 3/4 The sum of an A.P. with n terms, where (this is a rearrangement of the formula for the last term) a. It can be seen that a=5, l=101 and d=4, therefore and so. b. Here a=-17, l=33 and d=5. So and . c. a=1, l=9 1/4 and d= 1/4 and so . The sum is then given by . 6. Evaluate a. b. a. b. 7. Find the sum of the A.P.
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In that case, my formula is as follows: - (F + L) x 2 = T By using this formula I can calculate any first, last and top number. Test my formula (F + L) x 2 = T The number pyramid I am going to test has base numbers of 10, 11 and 12. Using this I will work out the top number. Formula: (F + L) x 2 = T Substituting: (10 + 12) x 2 = 44 By substituting the formula into the number pyramid, you will notice that: - 44 21 23 10 11 12 By
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Part 1 a) i The value of ?(3) is 2. This is because 3 is a prime number, so its prime factor is naturally 3 and it cannot share a factor with any of the numbers smaller than it. In this case, 1 and 2 are the only numbers smaller than 3 and they are both co-prime with 3, so the value of ?(3) is 2, because there are two numbers (1 and 2) that are co-prime with 3. ii The value of ?(8) is 4, because the prime factor of 8 is 2 and1, 3, 5 and 7 are the only four positive integers less than 8 which do not have 2 as a factor.
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X (3dp) f(x) -1 1 -0.91 0.028387 -0.99 0.890203 -0.909 0.017823 -0.98 0.780824 -0.908 0.007264 -0.97 0.671881 -0.907 -0.00329 -0.96 0.563392 -0.95 0.455375 -0.94 0.347848 -0.93 0.240829 -0.92 0.134336 -0.91 0.028387 -0.9 -0.077 x=-0.9075 +/- 0.0005 Error Bounds The calculations above show that the root lies in the interval 0.908 to 0.907. I can now conclude that the root to the equation 0 = 3x3+11x2+2x-5 has a root of -0.9075 +/- 0.0005 (3 d.p) being half the interval between the error bounds.
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June 1 152 4.60h July 1 182 4.62h August 1 213 5.07h September 1 244 5.65h October 1 274 6.22h November 1 305 6.87h December 1 335 7.50h All the values of the sunrise times for Toronto are converted to a decimal. The values of the sunrise times (y-value on the graph) and the number of days (x-value on the graph) are listed into Microsoft Excel. A graph is created from the values and everything on the graph is labelled.
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I tabulated this one more time. x y -1.9238 0.002248669 -1.92379 0.002003871 -1.92378 0.001759078 -1.92377 0.001514289 -1.92376 0.001269505 -1.92375 0.001024725 -1.92374 0.00077995 -1.92373 0.000535179 -1.92372 0.000290412 -1.92371 4.56502E-05 -1.9237 -0.000199108 The other root of the equation can be found using this method. It is, to 5 significant figures, 1.1144. Failure of the Change of Sign Method The Change of Sign method only works for certain equations, e.g. where there is a change of sign or where a change of sign is shown in the first table of values.
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Steps taken: 1. First we find the gradient m by taking any two points of coordinates and calculating m using the following equation The two points taken: P1 (48,0.009) P2 (112,0.021) x1 = 0.009 x2 = 0.021 y1 = 48 y2 = 112 2. The value of b is the y-intercept but since the car is at rest the speed is 0 and therefore the breaking distance is also 0. 3. Equation obtained: Graph 2. Model of Speed versus Thinking distance This model is a very good and accurate fit for the graph because the correlation coefficient, r, is 1 which means that it is a very strong correlation.
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This process can now be continued using intervals of width 0.01 then by using intervals of width 0.001 etc. Below I will show those intervals and I will bold and highlight those where change of sign occurs in. x f(x) x f(x) x f(x) x f(x) 0.6 -0,48224 0,68 -0,00501 0,68 -0,00501 0,680700 -0,000447 0.61 -0,42714 0,681 0,00151 0,6801 -0,00436 0,680710 -0,000382 0.62 -0,37079 0,682 0,00804 0,6802 -0,0037 0,680720 -0,000316 0.63 -0,31316 0,683 0,014585 0,6803 -0,00305 0,680730 -0,000251 0.64 -0,25423 0,684 0,021144 0,6804 -0,0024 0,680740 -0,000186 0.65 -0,19397 0,685 0,027718 0,6805 -0,00175 0,680750 -0,000121 0.66 -0,13237 0,686 0,034306 0,6806 -0,0011
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Reciprocal graphs equation would be f(x) a/x In this part I will have to divide x by a, a would be the same number and x will change. Out of all these formula I am looking at graph to see what is difference if I change a number or symbol e.g. plus I change it to a minus etc. - What am I going to do - Aim: investigate the functions in the graph also talk about shapes and the positions that has coefficients in the function.
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I now know that a root lies between (-4,-3). To investigate further, I need to find just where between these points the root is. I can do this by the same method as before accept I will use numbers between the points I already know the root lies. This table shows the values of Y when X is between -4 and -3: X -4 -3.9 -3.8 -3.7 -3.6 -3.5 -3.4 -3.3 Y -21 -17.22 -13.67 -10.35 -7.26 -4.38 -1.7 3.03 This shows that the root lies between (-3.4,-3.3). As the sign changed at -3.3, it was needless to continue to 3.
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+ 1 = 1 X2 = 0.1 ?f(0.1) = 3(0.1) 3 - 6(0.1) + 1 = 0.403 X f(X) 0 1 0.1 0.403 0.2 -0.176 Root [0.1 , 0.2] X f(X) 0.1 0.403 0.11 0.34399 0.12 0.28518 0.13 0.22659 0.14 0.16823 0.15 0.11013 0.16 0.052288 0.17 -0.005261 Root [0.16 , 0.17] X f(X) 0.16 0.052298 0.161 0.046520 0.162 0.040755 0.163 0.034992 0.164 0.029233 0.165 0.023476 0.166 0.017723 0.167 0.011972 0.168 0.0062249 0.169 0.00048043 0.170 -0.005261 Root [0.169 , 0.170] X f(X)
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