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AS and A Level: Core & Pure Mathematics

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Get help from 80+ teachers and hundreds of thousands of student written documents Differentiation and intergration

1. 1 It is easy to get differentiation and integration the wrong way round. Remember that the power gets smaller when differentiating.
2. 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. 3 If a function is increasing then the first derivative is positive, if a function is decreasing, then the first derivative is negative.
4. 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. 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.

1. 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. 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.

Straight lines

1. 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. 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.

• Marked by Teachers essays 3
1. Exponential decay:The objective of this experiment is to get the trend line of heart rate and to determine the decay constant of this decay.

2.1 Apparatus 17 people, stop watch, playground. 2.2 Procedures 1. Everyone put their hands on carotid artery. 2. Timekeeper calculate the time use. 3. Once the timekeeper call "start!", students can begin to count the heartbeat. 4. Repeat 3 times and get 3 different groups of data. 5. All the students run 5 minutes continuously. 6. Repeat the procedure 1-3 every half a minutes for 7 times. 2.3 Data To get a obvious and simple curve, for each item, I just calculate the average value.

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2. Investigation of circumference ratio - finding the value of pi.

Known: Segment AC=1 Segment BC=1 Angle ACD=60 We know two sides and one included angle. Then we could use the formula to find the area of triangle. As we known, the area of triangle ACD is one sixth of area of hexagon. Let us try 12-sided, 24-sided, and 48-sided polygons. 12-sided: The triangle ACD in 12-sided is one twelfth of the . Known: Segment AC=1 Segment BC=1 Angle ACD=30 Area of triangle ACD: Area of : 24-sided: The triangle ACD in 24-sided is one twenty-fourth of the . Known: Segment AC=1 Segment BC=1 Angle ACD=15 Area of triangle ACD: Area of : 48-sided: The triangle ACD in 48-sided is one forty-eighth of the .

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3. Logarithms (type 1)

Find a general pattern for log x + log y Log x + log y = log (xy) (e) Can you suggest why this is true? This rule is true because when you multiply both of the abscissas they equal another abscissa and adding the two logs will equal the third log. Therefore the general pattern is true. 2. (a) Copy and complete the following table using your calculator. Give your answers to the correct four decimal places. Log 12 - log 3 0.6020 Log 4 0.6020 Log 50 - log 2 1.3979 Log 25 1.3979 Log 7 - log

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4. Maths change of sign coursework

+ 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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5. In this coursework I am going to investigate the relationship between the orbital period and the distance of the planet from the sun. Assume that this relationship is a power law of the form: T = KR^n

After separating the logs we must use the log rule number 3, which is Log x+ k log x. Keep in mind that the transposition should end up to the linear function Y=c + mx. * Log k as C, which is the intercept of the line * N as the gradient of the line and an exponent in the formula * Log R as the as the base of the exponent n Log T = log K + n log R is the same as Y = c + mx I've calculated the log of R

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6. Solving the equation of 0 = 3x^5 - 3x + 1 using different methods

0.333 0.013284074 0.334 0.010469631 0.335 0.007657423 0.336 0.004847471 0.337 0.002039795 0.338 -0.000765585 0.339 -0.003568648 0.34 -0.006369373 x y 0.3377 7.57868E-05 0.33771 4.77377E-05 0.33772 1.96889E-05 0.33773 -8.35975E-06 0.33774 -3.64081E-05 0.33775 -6.44563E-05 0.33776 -9.25042E-05 0.33777 -0.000120552 0.33778 -0.000148599 0.33779 -0.000176647 0.3378 -0.000204694 x y 0.337 0.002039795 0.3371 0.001759153 0.3372 0.001478535 0.3373 0.001197939 0.3374 0.000917366 0.3375 0.000636817 0.3376 0.00035629 0.3377 7.57868E-05 0.3378 -0.000204694 0.3379 -0.000485151 0.338 -0.000765585 Using the Decimal Search/Change of Sign method, it took 32 calculations to reach a conclusive answer to 5 decimal places of 0.337735 �0.000005.

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7. Graph Original of Equation y=3x^5-3x+1

When it reaches the second curve, another line is drawn horizontally. This carries on until a value of the function is calculated to the required accuracy.

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8. Design, make and test a Sundial.

or South Pole (in the southern hemisphere). The gnomon must tilt at the angle of the latitude at the location. Although a sundial seems like a simple device for measuring time, it is not. One cannot simply look at the shadow and find the time. Firstly the sundial will show local time, not GMT. Secondly we must allow for the 'equation of time.' Because the Earth's orbit is not circular, its velocity changes at the perihelion. To allow for this would be awkward, so we use a 'mean sun.'

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9. Investigating the Quadratic Function.

(Refer to attached graphs) 3. The vertex of the equation y = x2 is (0,0). Therefore, the vertex of y = (x - 4)2 + 5 is (4,5) based on previous conclusions above. Using the information gathered above, the translation values (h,k) are (4,5). Thus, the whole function would be moved horizontally +4 units and vertically +5 units. Therefore proving, the original vertex of (0,0) would become (4,5). 4. (a) In order to articulate the expression x2 - 10x + 25 in the form (x - h)2 , it is required to complete the square.

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10. My job is to investigate how many squares would be needed to make any cross shape like this build up in the same way.

the first term here is 4, to find the 0th terms you need to go back one step (take the difference from the first terms to find the 0th term) and n is the term you want to find, e.g. if you want to find the 1st term n=1 if you want to find the second term n=2. a=4, b=0. Now I have the equation, 4n+0, now let's test it! We know the first terms is 4, so 4xn(1)+0, 4x1=4 so the answer is 4 and that is the 1st term.

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11. Solving Equations using Numerical Methods.

2 -1 10 f(x) is found by substituting the values of x in the equation. We look for roots in [0, 1] x 0 0.1 0.2 0.3 0.4 0.5 0.6 f(x) 2 1.6001 1.2016 0.8081 0.4256 0.0625 -0.2704 We look for roots in [0.5, 0.6] x 0.5 0.51 0.52 f(x) 0.0625 0.02765 -0.00688 We look for roots in [0.51, 0.52] x 0.51 0.511 0.512 0.513 0.514 0.515 0.516 0.517 0.518 f(x) 0.02765 0.02418 0.02071 0.01725 0.01380 0.01034 0.00689 0.00344 -0.000002 We look for roots in [0.517, 0.518] x 0.517 0.5171 0.5172 0.5173 0.5174 f(x)

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12. Design and construction of circuits to solve problems.

1 0 0 0 1 1 0 0 0 1 1 0 1 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 1 1 0 0 1 1 0 1 1 0 1 0 0 0 0 1 1 1 1 1 0 0 0 1 0 From the Truth table above I can now create a Boolean equation for each time the red, amber and green light is on. Traffic Light 1 _ _ _ _ _ _ _ _ _ R = X�Y�Z + X�Y�Z + X�Y�Z + X�Y�Z + X�Y�Z +

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13. The Change of Sign method locates the root of an equation by where it crosses the x-axis.

x y -3 -82 -2 -14 -1 2 0 -4 1 -2 2 38 3 146 The change of signs are in the intervals [-2, -1], [-1,0] and [1,2]. To demonstrate the method, I am going to find the root in the interval [1,2].

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14. Scale of Production of signs.

Another reason why this sign is mass produced is that it would not take very long to produce each individual sign and not very long to modify them for the job in hand. This sign is further helped by the fact that a very large quantity of signs use the same shaped sign for their purpose, for example if there were no longer a demand for men at work signs, Give Way or caution signs could be created instead.

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15. Solve the equation: X3- 5X-5 = 0. This equation cannot be factorised nor solved algebraically therefore it must be solved through interval bisection.

Interval Bisection n Xn Y(Xn) Sign X(Endpoint) 0 2.6 -0.424 -1 2.7 1 2.65 0.359625 1 2.6 2 2.625 -0.037109 -1 2.65 3 2.6375 0.1600215 1 2.625 4 2.63125 0.0611477 1 2.625 5 2.628125 0.0119422 1 2.625 6 2.6265625 -0.012603 -1 2.628125 7 2.6273438 -0.000335 -1 2.628125 8 2.6277344 0.0058023 1 2.62734375 9 2.6275391 0.0027333 1 2.62734375 10 2.6274414 0.001199 1 2.62734375 11 2.6273926 0.0004319 1 2.62734375 12 2.6273682 4.837E-05 1 2.62734375 13 2.627356 -0.000143 -1 2.62736816 14 2.6273621 -4.75E-05 -1 2.62736816 15 2.6273651 4.335E-07 1 2.62736206 16 2.6273636 -2.35E-05 -1 2.62736511 After 16 literations it gives the answer of 2.627 to 3 d.p or 2.627 +/- 0.0005.

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16. Fixed-Point Iteration

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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17. Observation of people obeying temporary signs.

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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18. Collect data and investigate whether it can be model by a exponential function

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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19. Maths Coursework: Curve Fitting

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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20. Critically consider one invasive method and one non-invasive method of studying the brain

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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21. Information Technology Development Cycle Diary

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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22. Investigation on Boyles law

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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23. Maths Assignment - trigonometry, trend line, probability and calculus questions.

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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24. Fixed point iteration: Rearrangement method explained

g(x) = 3? (9x+11-2x2). The steps I followed to derive this were: f(x) = x3 +2x2 ? 9x ? 11 0 = x3 +2x2 ? 9x ? 11 9x + 11 ? 2x2 = x3 So: 3? (9x+11-2x2) Now I will need to create a table to see if my ?x? values and g(x) values converge, where they do is the value of a root. Hence the ?x? values should be the same as the g(x) values or extremely close. In this case I will quote my answer to 5 s.f. I need to repeat my iterations until x and g(x)

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25. Solving Equations Using Numerical Methods

This allowed me to work out that the three roots were between the integers of -4 and 1. These three roots are: -4 < ? > -3 -1 < ? > 0 0 < ? > 1 For one root, I worked out where the change of sign took place to 5 sig.fig. The first step to complete this process was to start with entering -4 to -3 (in 0.1?s) into Excel and the formula that I was using into the column next to it to show me where the change of sign was ? between -3.5 and -3.4.

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