I am going to investigate the gradients of different curves and try to work out a pattern that I could use to find the gradient of any curve.

Maths Coursework Introduction I am going to investigate the gradients of different curves and try to work out a pattern that I could use to find the gradient of any curve. I will draw graphs of a selection of curves, some by hand, some using Autograph and some using Excel. I will use three methods to investigate the graphs. Firstly, I will draw tangents to the curves at 4 or 5 points and measure the gradients. Secondly, I will draw chords between x = 1 and 4 or 5 points and measure the gradients. Thirdly, I will use algebra to work out a formula for the gradient and see how this matches the first two methods. At first I split up the coursework into 3 main families (for each family there are additional equations to investigate): Part One: Curves involving x2 . y = x2 2. y = 2x2 3. y = 3x2 4. y = 4x2 5. y = x2 + 1 6. y = 7x2 + 6 Part Two: Curves involving x2 + x . y = x2 + x 2. y = x2 + 2x 3. y = 7x2 + 4x + 5 Part Three: Curves involving x3 + x . y = x3 2. y = 2x3 3. y = 4x3 + 2x - 5 Finally, I will summarise my results in a series of tables and work out an overall formula that I could use to predict the gradient of any curve. PART ONE: CURVES CONTAINING X2 (1) y = x2 I am investigating the changes in gradient for the curve y = x2. To plot the curve, I will use the table of values given below. x 0 2 3 4 5 6 y 0 4 9 6 25 36 I

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  • Level: GCSE
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The Gradient Function

The Gradient Function Aim: To investigate the gradient of different curves I will investigate the gradient of different equations for which the general formula is Y = a?n In this equation I will investigate the gradient by varying the values of 'a' and 'n'. Gradient: The property possessed by a line or surface that departs from the horizontal is called the gradient of the line. In mathematical terms, the gradient of the line simply tells us how steep a line is. The gradient for all lines parallel to the X-axis is 0. Gradient Function: Gradient function is the name of a rule, specific to a graph ex: Y = X3 which can be used to find the gradient at any point of the graph. The 'X' value is substituted in the equation and this gives the exact gradient for that specific graph. I plan to find the gradient using these two methods: Tangent Method: The tangent method involves making a tangent at a point on the graph on which the gradient is to be found. The figure below shows a tangent: To find the gradient of a straight line, we use the equation: Y2-Y1 X2-X1 This equation would give us the gradient. [X1, Y1], [X2, Y2] can be any two points on the graph. This equation can be used only on straight-line graphs. To find the gradient of curves is much more difficult than the straight-line graphs. For the gradient of curves, a tangent is drawn at a point. A tangent is a line,

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The Gradient Function

The Gradient Function Part 1 We have been told to investigate the Gradient Function for the set of graphs: y=axn Where a and n are constants. One way to find out the gradient, is to do what I have done for the curve of y=x2. After drawing the tangent of a specific point of x, you draw a triangle, with the x value in the middle of a line on the tangent. You then find the change in y, and the change in x. You then have to apply them to the following formula: The Change in Y The Change in X You simply have to divide the change in y, by the change in x. On all my graphs, there will be small triangles that have been measured out on the tangent. They will be marked with the letters A, B, and C. The line of the triangle running in parallel with the 'y axis' is The Change in Y. The line of the triangle running in parallel with the 'x axis' is The Change in X. This will give you the gradient. However, this is quite an inaccurate method of working out the gradient. This is because if you draw the co-ordinates for the triangle incorrectly, then it will alter the answer that you will discover. There is also the small increment method that you can apply to the curve. I will go into this in a later graph, as well as using this method to work out the gradient. Firstly, I thought that the best thing to do, would be to investigate the graph of y = x2. Below I have drawn a table of

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I have been given the equation y = axn to investigate the gradient function for varied values of "a" and "n".

I have been given the equation y = axn to investigate the gradient function for varied values of "a" and "n" Firstly I will choose n = 1 for varied values of a When n = 1 then the graph will be a straight line The general equation of a straight line is y = ax I am going to keep n constant and use varied values for "a".The gradient for a straight line remains the same, therefore I am going to calculate the gradient by taking the difference in the y co-ordinates divided by the difference in x co-ordinates. In order to calculate the gradient for a straight line we will have to go through the process of calculating the gradient by taking the difference in y co-ordinates divided by the difference in the x co-ordinates. I would then choose n = 2 for varying values of a where a = 1,2,3,4 The equations would be ; y = x2 , y = 2x2 , y = 3x2 , y = 4x2 These equations would get us curved graphs and the gradient function at each point in a curve is different therefore I will draw tangents using a capillary tube and calculate the gradient by taking the change in the value of the y co ordinates divided by the change in t value of the x co ordinates. I would then choose n = 3 for varying values of "a". The equations would be ; y = x3, y = 2x3,y = 3x3. Again these equations would get me curved graphs and I will establish a relationship Between the curve and the

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Gradient Function

Gradient Function Mr. Bailey Adnan Sarayqum CA9 Plan I am aiming to find the relationship between the X values and the slopes/gradients from given points on a curve. The first graph I will use is for the equation Y=X2. I will choose a fixed point e.g. (2, 4), (4, 16). I can choose the values of X which can be negative numbers or positive numbers. I will find gradient from different points on the curve using my fixed points. I will choose a higher value and a lower value than my fixed point and move towards my fixed point. I will see how the gradient (change in y/change in x) changes as I move closer to my fixed point. I will see the relationship between my fixed point and the gradient and try to look for a formula. I will also use other equations, for example, Y=3X2, Y=X2 - 4 and Y=X3. Then I will look at my results tables and graphs and try to formulate a general formula for gradients. Y=X2 My first fixed point is 2, 4 x y change in y change in x gradient 3 3 .1 .21 2.79 0.9 3.1 .2 .44 2.56 0.8 3.2 .3 .69 2.31 0.7 3.3 .4 .96 2.04 0.6 3.4 .5 2.25 .75 0.5 3.5 .6 2.56 .44 0.4 3.6 .7 2.89 .11 0.3 3.7 .8 3.24 0.76 0.2 3.8 .9 3.61 0.39 0.1 3.9 .99 3.9601 0.0399 0.01 3.99 .999 3.996001 0.003999 0.001 3.999 2 4 2.001 4.004001 -0.004001 -0.001 4.001 2.01 4.0401 -0.0401 -0.01 4.01 2.1 4.41 -0.41

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The Gradient Function Investigation

The Gradient Function Introduction Curves on a graph can be of varying steepnesses. This steepness also varies from point to point on many graphs. The steepness of a curve at a point is called its gradient. There are several methods for calculating the gradient at a certain point on a curve including the 'Tangent Method' and the 'Small Increments Method'. The Tangent Method Calculating the gradient of a straight line is simple. The formula is: Gradient = Change In Y Change In X This formula is demonstrated on Graph A. A curve proves more of a problem as the gradient is constantly changing. To calculate the gradient at a certain point, we must somehow be able to create a straight line from which to calculate this gradient. This can be achieved by drawing a tangent to the curve at the point in question. A tangent is a straight line which touches the curve at one point and one point only. Calculating the gradient of this tangent will give the gradient of the curve at this point. This is illustrated on Graphs B1, C1 and D1. This method is not very accurate though and large discrepancies in gradient can occur. This is because this method involves manually drawing a tangent which will often be drawn incorrectly. Errors in the measurement of changes in Y and X can also occur, even on a very large scale graph. The Small Increments Method This method is more accurate than the

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Gradient Fucntion

The Gradient Function In this coursework I am going to try and find out what the gradient function is for various curved graphs. I will do this by drawing graphs and find the gradient of the tangent of a point. . I am going to draw a graph of y=x². I will use values from 0-5, I will obtain the gradient of the tangent at different points. This is the table of the results of y=x². X 0 2 3 4 5 Y=x² 0 4 9 6 25 This is the results of the gradient of the tangent: Gradient of y=x². At (1,1) .8 .8 = 2(to 1sf) At (2.4) 2 0.5 4 At (3,9) 4.7 0.8 5.875 = 6(to 1sf) At (4,16) 3 0.3 0 I have noticed that the gradient of the tangent doubles each time. Except for at (4,16) so I fink that this is an anomalious result, which would mean that the tangent must be drawn incorrectly. I am now going to draw a graph of y=x³. I will use values from 0-5, I will obtain the gradient of the tangent at different points. This is the table of the results of y=x³. X 0 2 3 4 5 Y=x³ 0 8 27 64 25 This is the results of the gradient of the tangent: Gradient of y=x³. At (1,1) 3 0.9 3.33333 = 3(to 1sf) At (2.8) 5 .3 1.538462 = 12(to 1sf) At (3,27) 20 0.8 25 At (4,64) 22 0.5 44 I haven't noticed a pattern in this set of results, but I think that at (3,27) and (4,64), my gradients are a bit wrong. I think it is that the second difference goes

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The Gradient Function Coursework

The Gradient Function Coursework In this piece of coursework I am going to do research on the gradient of various graphs at various points, in order to find a function, which will determine the gradient of these points without drawing or using approximations. I will only need to know the coordinates of the point as well as the type of graph I am considering, to submit them into the gradient function and determine the gradient at this point. The formulae I will use and produce will have particular parameters. Now I am going to explain them. a: this letter will stand for the coefficient of x in the function y=ax^n and determines how steep the graph will be. n: this letter will be the power to which x is raised in the function y=ax^n and determines the shape of the graph. m: this letter will stand for the gradient at any point of any graph. I can say for example the gradient at the point P(1;1) of the graph y=x is 1. Therefore here m=1. The first range of graphs I am going to investigate will have the function y=ax. I will draw three graphs on the next pages and hope to see a pattern between the gradient and the function of the graphs. I do not need to consider the coordinates of the points at which I will determine the gradient, as the gradient is the same at any point on the graph y=ax. From these three graphs I clearly recognise a pattern. I will show how I noticed

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Gradient function

My coursework on the gradient function is to investigate the gradient on different points on the line and curves. I will start of my investigation with y = x2. This will be a parabolic curve and the gradient will move from point to point. I will need to start of with a fixed point. I have chosen (2, 4). I will use a table to get close to the point. My table will have five columns. The first column will be x, which will have any numbers between 1and 3. The second column will be y, which will be the result of squaring an x numeral. The third column will be the increase in y, where squared value gets subtracted from 4. The fourth column will be the increase in x, this is where the x values get subtracted from 2. The fifth and last column will be the gradient, where the change in y divided by the change in x, gives me the results. I will do different fixed points so the numbers will vary. I will then do other functions such as Y=4x2 Next I will move to a function y = x4 and investigate the gradient at different points. I will use the same method used in the equation y = x2 but instead of squaring the numbers I will cube them. After my investigating finishes I will come up with a conclusion which will summarize my investigation. y=x2 My first fixed point is 2, 4 x y change in y change in x gradient 3 3 .1 .21 2.79 0.9 3.1 .2 .44 2.56 0.8 3.2 .3 .69

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In the following coursework, will investigate the gradient functions using the formula y=ax^n, where a is a constant and n is a number.

Gradient functions In the following coursework, will investigate the gradient functions using the formula y=ax^n, where a is a constant and n is a number. a n Y=ax^n x 2 2x 3 3x 4 4x 5 5x a n Y=ax^n 2 x 2 2 4x 3 2 6x 4 2 8x 5 2 0x a n Y=ax^n 3 3x^2 2 3 6x^2 3 3 9x^2 4 3 2x^2 5 3 5x^2 a n Y=ax^n 4 4x^3 2 4 8x^3 3 4 2x^3 4 4 6x^3 5 4 20x^3 I will plot the graphs of the functions above and I will find their gradient using the formula gradient=increase in y-axis /increase in x-axis. Straight line graphs Straight line graphs are graphs with the equation y=mx+c or y=ax^1,where is stand for the gradient and c is the y- intercept. Gradient calculations . y=x graph Gradient of A= increase in y -axis/increase in x-axis = 2/2 =1 Gradient of B= increase in y-axis/increase in x-axis = 2/2 =1 2. y=2x graph Gradient of D= increase in y-axis/increase in x-axis = 4/2 =2 Gradient of E= increase in y-axis/increase in x-axis = 4/2 =2 Gradient of F= increase in y-axis/increase in x-axis = 4/2 =2 3. y=-2x graph Gradient of G= increase in y-axis/increase in x-axis = -4/2 =-2 Gradient of H= increase in y-axis/increase in x-axis = -4/2 =-2 Gradient of H = increase in y-axis/increase in x-axis = -4/2 =-2 Gradient of I = increase in y-axis/increase in x-axis = -4/2 =-2 4. y=3x graph Gradient of J=

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  • Level: GCSE
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