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The Gradient Function Maths Investigation

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The Gradient Function I am going to be investigating the function of the gradient. A function is a variable that depends on the value of other independent variables and the gradient is the steepness of a line or curve. I am going to try to work out a formula that will calculate the gradient of any given line or curve; this will be the gradient function. I already know of some methods that can be used to calculate the gradient, these are: 1. The formula: increase in y increase in x This formula represents the vertical value on the graph divided by the horizontal value. It can also be written as: dy dx This notation demonstrates the rate of change at y with respect to x. Which means that as x changes so does y. When using this formula to work out the gradient at of a curve a tangent must first be drawn. 2. 'Omnigraph' is a computer program which will create a graph for a given a formula. It will then draw tangents on the graph and work out the gradient. 3. The small increment method can be used as a more accurate way of calculating the gradient of a graph. To use it you must zoom in on a section of the graph, for example the coordinates (3,9) ...read more.


going to put the formulas I have discovered so far into a table to make them easier to interpret: Equation of graph Gradient Function y=x� 2x1 y=x� 3x� y=x4 4x� n nx(n-1) From the formulas I have found using my graphs I can see the general rule and from them I have worked out this formula: Gradient= nx(n-1) This formula is correct for all the graphs I have done so far but I am now going on other graphs. Firstly I am going to see if it works for negative numbers on the graph y=x -1 (see graph 5). These are the results I have found using this graph. I have also tested them to see if the formula Gradient= nx(n-1) works: x Gradient=nx(n-1) Actual Gradient 1 = -1x1(-1-1) = -1 -1 2 = -1x2(-1-1) = -0.25 -0.25 3 = -1x3(-1-1) = -0.11 -0.11 4 = -1x4(-1-1) = -0.0625 -0.063 This graph is evidence of my formula for graphs with equations with negative integers but not with all numbers. I am now going to try it with a graph with a fraction integer, y=x1/2 (see graph 6). These are the results I have found using this graph. I have also tested them to see if the formula Gradient= nx(n-1) ...read more.


After again consulting the A level textbook I have found that as H is worth such a tiny amount that it does not matter. This is because the gradient of Q is on the line of PQ and is therefore equal to the gradient of P. This means H is worth virtually nothing. Therefore: Gradient of tangent at P =2x dy =2x dx These workings proves that my formula, Gradient= nx (n-1) is correct for the graph y= x�. I am now going to try to prove my formula for the graph y=x� using this method. I have drawn a sketch graph (graph 8) that is very similar to the graph previously used. The symbols all represent the same values as before. (y + K) = (x + H)� (y + K) = x� + 3Hx� + 3H�x + H� K = 3Hx� + 3H�x + H� Gradient of PQ = K H = 3hx� + 3H�x + H� H = 3x� + 3Hx + H� The Hs can now be ignored leaving: dy =3x� dx As I predicted this method shows that the gradient of the graph y=x� is 3x�. I have discovered that the gradient of all the graphs I have used can be found using the formula nx (n-1). This formula works for positive, negative and fractional integers. I have also gone some way to proving my formula algebraically by using differentiation. ?? ?? ?? ?? Sarah Shea 11A Maths Investigation ...read more.

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