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

Critically consider one invasive method and one non-invasive method of studying the brain There are a number of invasive methods of studying the brain. One such technique is ablation and lesion production. The ablation technique involves surgically removing or destroying brain tissue. A study into this technique was conducted by Flourens (1820); who showed that the removal of thin slices of brain tissue; resulted in them displaying poor co-ordination and sense of balance but, they experienced no other difficulties. Meanwhile, lesion production involves deliberately injuring a specific area of the brain and observing the behavioural changes that occur as a result. 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). In this experiment, they found that when they stimulated the hypothalamus of rats, they

  • Word count: 574
  • Level: AS and A Level
  • Subject: Maths
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Arctic Research (Maths Coursework)

MATHS COURSEWORK "ARCTIC RESEARCH" Introduction: In my investigation, a group of researchers are to set up a series of observation sites at equal distances from each other in the Arctic on the circumference of a circle of radius 50 km. The aim of my investigation is to figure out how long the researchers should be prepared to travel for on each journey. I am also to find where in the circle that the base camp should be located in order to minimize these journey times as short as possible. Using some calculations I will determine all of this taking into account that there is a strong westerly wind present constantly at all times in the area. I found out, from research, that average wind velocities in the Arctic are from 25 - 35 km/h. I will use this range in my investigation. The bearing at which the planes must depart is affected by this westerly wind therefore, this angle must be calculated. If this investigation was to be physically carried out there would be some factors that needs to be considered. In the arctic there are many mountains and obstacles a plane would have to take into account during flight. To take into account these factors in my calculations would be impractical therefore, I must establish variables that could be controlled and make some assumptions in order to make it practical. These are the variables that I must establish; * Strength of the

  • Word count: 3157
  • Level: AS and A Level
  • Subject: Maths
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Find out whether there is a correlation between the melting points and boiling points of the chemical elements.

Bivariate Data Coursework Aim The aim of this investigation is to find out whether there is a correlation between the melting points and boiling points of the chemical elements. This idea is worth pursuing because it is related to my studies in A level chemistry and once data has been analysed the information would be useful to chemists for making predictions about the melting point or boiling point of an element if the other value is known. It is suspected that there will be positive correlation between the two variables because if a large amount of energy is needed to melt a solid it is likely that a large amount of energy will also be required to cause the liquid to boil and vice versa. Data Collection The population to collect data from will be all the elements in the periodic table of which there are about 118 that are known, and of these 50 will be selected randomly to collect data on. Both of the variables (melting and boiling point) are random, because they have unpredictable values and are free to assume any of a particular set of values in a given range. To select the 50 elements on which to find out the bivariate data, a calculator will be used to generate random numbers, until 50 different atomic numbers up to 118 have been noted. These are the atomic numbers of the elements I will find out the melting points and boiling points of in degrees Kelvin. Data on

  • Word count: 1656
  • Level: AS and A Level
  • Subject: Maths
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Estimating the length of the line and the size of the angle

Guestimate Mathematics Coursework My task is to write a hypothesis and to test how well people estimate and to design and carry out an investigation to test my hypothesis. My hypothesis is that people will estimate the length of the line better than the size of the angle because lines are used more commonly than angles are. My hypothesis is also that neither the boys nor the girls will estimate better than one another. I also predict that people in higher sets will estimate the angles and lines better than people in lower sets. I will need to carry out an investigation to test my hypothesis. The investigation which I will carry out will be a questionnaire. The questionnaire will be conducted with two lines and two angles which I will ask twenty people in year eleven to estimate the length of the two lines and the size of the two angles. I have chosen twenty people because it will give me a set of eighty results, which is enough to prove my hypothesis. I will do this using a stratified sample to find out how many people I will ask in each set. Once I have found out how many people I will ask in each set, I will randomly pick an even number of boys and girls by picking boys and girls' names out of a hat. If the number of people that I will ask is an odd number e.g. five, then I will pick two boys from one hat and two girls from the other, then I will mix the names of the

  • Word count: 4461
  • Level: AS and A Level
  • Subject: Maths
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Decimal search.

Decimal search The equation f (x) = 0, where f (x) = x3-x2+x-2, has only one real root, but there is no simple analytical method of finding it. Therefore, a spreadsheet had been used to solve the equations numerically using decimal search. x f(x) -3 -41 -2 -16 -1 -5 0 -2 -1 2 4 3 9 The table and graph above illustrated the first approximations to the roots of the equation x3-x2+x-2=0. As the curve crosses the x-axis, f (x) changes sign, so provided that f (x) is a continuous function, once you have located an interval in which f(x) changes sign, you know that that interval must contain a root. In the table, you first take increments in x of size 1 within the interval 1<x<2, working out the value of the function x5-5x+3 for each one. You do this until you find a change of sign of f(x) between the value of x = 1 and 2 , so you should know that there is a root lying in the interval 1<x<2 on the graph. Having narrow down the interval, you can now continue 1.3<x<1.4, you could now homing in on such root using decimal search. x f(x) -1 .1 -0.779 .2 -0.512 .3 -0.193 .4 0.184 .5 0.625 .6 .136 .7 .723 .8 2.392 .9 3.149 2 4 * Tabulate f (x) for 1<x<2 with increments in x of 0.1, a change of sign reveals that 1.3<?<1.4. * Tabulate f (x) for 1.3<x<1.4 with increments in x of 0.01, a change of sign gives 1.35<?<1.36 x f(x) .3 -0.193 .31

  • Word count: 1974
  • Level: AS and A Level
  • Subject: Maths
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Growing Squares

Growing Squares I have decided to find a formula to find the nth term. To help me find the nth term I shall compose a table including all the results I know. Pattern Number Number of Squares st Difference 2nd Difference 4 4 2 5 8 3 3 4 2 4 25 The 2nd difference is constant; therefore the equations will be quadratic. The general formula for a quadratic equation is an2 + bn +c. The coefficient of n2 is half that of the second difference Therefore so far my formula is: 2n2 + [extra bit] I will now attempt to find the extra bit for this formula. Pattern Number Extra Bit st Difference 2nd Difference 2 6 4 2 8 0 3 8 4 4 4 32 From my table of results I have found the formula to be 2n2 + 2n + 1 I will now check my formula by substituting a value from the table in to my formula: E.g. n = 2 Un = 2 (2) 2 - 2 (2) + 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)

  • Word count: 1164
  • Level: AS and A Level
  • Subject: Maths
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Guestimate the length and size of the angle.

Guestimate 1. Collect relevant data I asked a sample of people to estimate: [image001.gif] [image001.gif] * [image002.gif] The length of this line [image003.gif] * [image004.gif] The size of this angle *Note: The data collected from this investigation is located in the Appendix of this assignment 2. Extract a sample of 30 items I have chosen to extract two random samples of 30 people wearing spectacles and 30 people with out. The reason for this is one of hypotheses 2 involves comparing peoples estimates with spectacles and without and I wanted to make sure it is Hypotheses 1: Boys are more likely to estimate the length of the line and the given angle correctly as opposed to the girls Hypotheses 2: People who wear spectacles are less likely to guess the length and the angle correctly than people who do not wear spectacles Random sample of 30 people (15 boys and 15 girls) for hypothesis 1 Sample data 1: Estimate of length of line (L) to nearest cm Estimate of size of angle (A) to nearest degree Age Sex M or F Wears spectacles Occupation 7 52 30 M No Student 4 50 21 F Yes Student 4 45 25 F No Student 4.5 45 23 M No Student 4.5 45 40 M No Student 2 45 25 M No Student 3 45 28 M No Student 4 45 45 F Yes Lecturer 2.5 45 12 F No Student 4 45 22 F No Swimming teacher 7 45 40 F No

  • Word count: 3803
  • Level: AS and A Level
  • Subject: Maths
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Telepathy Maths Investigation

Telepathy Maths Investigation Introduction I will take a card from a set of specifically designed cards and will look at it, not letting anyone else see the image on the card. Another person will then state what card they think it is, and I will record whether they are correct or not in their guess. Each person will try to receive the images of 10 randomly chosen cards, hopefully, a large enough number of tests to determine whether the subject has any telepathic abilities. I have assigned each card with a set of numbers. Random numbers will then be generated on a calculator for each test, to ensure that every card is picked at random, making sure that each event is independent, as the distribution model requires. The aim of this investigation is to analyse the results of telepathy tests run on a selection of people to determine whether I, as the sender, possess any telepathic powers. These results could be used as a representative sample of other similar people. The people I am targeting to run these telepathy tests on are young adults aged 16-19, from New College. These people are similar in age and have been assumed to be of the same intelligence, so the results for each every person should remain constant, relative to the level of my telepathic ability. The sample size will be 20 people large, as this should be enough to see if I possess any telepathic powers.

  • Word count: 1000
  • Level: AS and A Level
  • Subject: Maths
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Using Decimal search

Pure 2 Coursework Using Decimal search In this investigation, I have chosen x^5-14x+2=0. Such an equation cannot be solved algebraically. The following graph represents y= f(x): - Graph 1 - The graph shows me that x^5-14x+2=0 has three roots, lying in the intervals (-2,-1) , (0, 1) , (1,2). X -3 -2 -1 0 2 3 Y -199 -2 5 2 -11 6 203 Within the change of sign methods, decimal search method shall be used to find out an approximate value for roots of x^5-14x+2=0. In this method I first take increments of x of size 0.1 within the interval (1,2). I do this until I find a change in sign. X .1 .2 .3 .4 .5 .6 .7 .8 .9 Y -11 -11.78 -12.31 -12.48 -12.22 -11.40 -9.91 -7.6 -4.3 0.16 There is a sign change, and therefore a root, in the interval (1.8,1.9). I can now continue with increments of 0.01 within the interval (1.8,1.9). X .8 .81 .82 .83 .84 .85 .86 .87 .88 .89 .9 Y -4.3 -3.91 -3.51 -3.09 -2.66 -2.23 -1.77 -1.31 -0.83 -0.34 0.16 This shows that the root lies in the interval (1.89,1.90). We can express this information as, the root can be taken as 1.895 with maximum error of + 0.005 This can be shown graphically: - Graph 2 - I have also chosen 10x^3-2.5x+0.2=0 which again cannot be solved algebraically. The following graph represents y=f(x) - Graph 3 - The graph shows me that 10x^3-2.5x+0.2=0 has

  • Word count: 2515
  • Level: AS and A Level
  • Subject: Maths
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numerical solutions-Comparison of the three methods and Newton Raphson

Numerical Solutions of equations ) Newton Raphson method Equation to be solved is x³-5x-1=0 The function f(x)= x³-5x-1is shown below There are 3 roots. I will first find the root in the interval [2, 3] I will do the first few lines of calculation manually. The formula to use is: xn+1= xn -f(xn)/ f'(xn) Therefore I must first differentiate x³-5x-1 which is 3x2-5 Using x1=3 X2: 3 - [(33-5 x 3-1)/(3 x 32 -5)] = 2.5 X3: 2.5 - [(2.53-5 x 2.5-1)/(3 x 2.52 -5)] = 2.3455 I will now work out all 3 roots using autograph until 5 significant figures are guaranteed The 3 boxes above show how I obtained the 3 roots of the equation x³-5x-1=0 Which are -0.20164, -2.1284 and 2.3301 Below is the function f(x)= x³-5x-1 showing where I applied the Newton's Raphsons to find each root, showing the tangent made to the curve and how it converges closer each time. I will now illustrate one of these roots graphically, showing closely the changes in the tangent to the curve. I will use the root in interval [2, 3] Error bounds for interval [2, 3] 2.3301 is the root in this interval to 5 significant figures. Therefore the error is 2.3301 ± 0.00005 I will now perform the change of sign test to confirm it is within these limits. Lower limit is 2.33005 then f (2.33005) = -0.000098648 Upper limit is 2.33015 then f (2.33015) = 0.0010302 There is a change of sign

  • Word count: 1020
  • Level: AS and A Level
  • Subject: Maths
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