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GCSE: Gradient Function
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- Peer Reviewed essays 3
- Develop your confidence and skills in GCSE Maths using our free interactive questions with teacher feedback to guide you at every stage.
- Level: GCSE
- Questions: 75
The next range of graphs I am going to investigate will be parabolas. These have the function y=ax�. Here the gradient is different at different points of the graph. So I will have to use a different method. I will draw a tangent as near as possible to a point at which I would like to know the gradient, as the gradient of a point on this graph is defined as the gradient of the tangent that touches this point. I will draw two graphs on the next two pages. From each of these two graphs I will choose three points at which I will use this method to determine the gradient and hope to find a pattern.
- Word count: 3186
It must not touch more than one particular point or intersect the curve, as this will distort the outcome. As a result, we can say that the curve's steepness at a particular point is identical to the tangent formed in that specific curve. I will draw four graphs: y=x2, y=x3 y=2x2 and y=x-1. From each of these graphs I will infer different gradients and compare my results with the theoretical result that is found by using the 'Small Increments Method'. By comparing the two results, I can distinguish if my practical method and theoretical method give similar results that is the steepness of the curve at a given point.
- Word count: 1048
It is the relationship between these gradients that will be explored in the first part of this investigation. The first task to be carried out was to find several perfect tangents of the graph y=x2. This presented a challenge, as there is no easy way to discover lines that are perfect tangents of a curve. In my first attempt, I created a line in autograph and gradually changed the y intercept parameters in an attempt to "move" the line into position as a tangent of the graph y=x2. This proved time consuming and ineffective however, as the following screenshots show; Here the blue line appears to cross the red line at y=0.25, x=0.45, but this is unclear because of the distance from the point.
- Word count: 520
metres be the height of drop and (n) be the number of drops. The graph below is of n against h From the graph above, the number of drops to crack open the nuts depends on the height of the drop. Therefore the independent variable is the height of the drop (h) while the dependent variable is the number of drops (n). The control variable is the size of the nuts which is large for the above specific graph.
- Word count: 604
cut out which gives the largest volume is between 3 and 4 but more towards 3 since 3 gives a larger volume. To obtain a more accurate result I will zoom in the shaded region. Cut Out x Width 20-2x Length 20-2x Volume 3.1 13.8 13.8 590.364 3.2 13.6 13.6 591.872 3.3 13.4 13.4 592.548 3.4 13.2 13.2 592.416 3.5 13 13 591.5 3.6 12.8 12.8 589.824 3.7 12.6 12.6 587.412 3.8 12.4 12.4 584.288 3.9 12.2 12.2 580.476 4 12 12 576 Cut Out x Width 20-2x Length 20-2x Volume 3.31 13.38 13.38 592.571 3.32 13.36 13.36 592.585 3.33
- Word count: 2528
= 360/365 = 0.986 This gives a first model of y = 6.35snin (0.986t) + 10.15 The initial estimate for the phase shift was 110 to the right, so by translation t --> (t - 110). This result was not on the original data, but further modifications gave rise to t --> ( t -113) which appeared closest. On checking the values for the average absolute error for the points given by the data, this also gave the smallest error. So y = 6.35snin (0.986(t - 113)) + 10.15 is the model for the data Checking the model works The data give (197, 16.5)
- Word count: 585
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 1 1 3 1 3 1.1 1.21 2.79 0.9 3.1 1.2 1.44 2.56 0.8 3.2 1.3 1.69 2.31 0.7 3.3 1.4 1.96 2.04 0.6 3.4 1.5 2.25 1.75 0.5 3.5 1.6 2.56 1.44 0.4 3.6 1.7 2.89 1.11 0.3 3.7 1.8 3.24 0.76 0.2
- Word count: 3172
Theoretically this should make it a more accurate approximation. To find out to what degree this is true,I a computer program was designed which carried out the Trapezium Rule and Simpson's Rule, for a given integral, for a specified number of strips. For each number of strips i will calculate the error value between the numerical answer and the exact answer. The formula I will be integrating will be y = 1/x, with limits 1 and 10. The exact value is found by integrating this function. dy/dx = lnx + c.
- Word count: 2721
I think it is that the second difference goes up in 6. Although I have noticed that for y=x� the first gradient was 2 and for y=x� the first gradient was 3. So I predict that the first gradient will the number that is the power of x, so in y=x4 the first gradint should be 4. I am now going to draw a graph of y=x4. 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=x4.
- Word count: 3331
Aim: To find out where the tangent lines at the average of any two roots intersect the curve again in cubic functions. Functions with Three Roots
Before finding the function for the tangent line, we will need to first find its "y" value, simply by substituting the value "-2.25" into the original function, f(x).The slope of the tangent line is also required to find out its linear function, and we can do this by substituting the value "-2.25" into the first derivative function, f'(x). Calculations for both of these are shown as below.) (Now that we know the values for x, y, and m, we can simply substitute them into the linear function, and solve for "c", then we would have the function of the tangent line.)
- Word count: 1282
0.2 3.8 1.9 3.61 0.39 0.1 3.9 1.99 3.9601 0.0399 0.01 3.99 1.999 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 -0.1 4.1 2.2 4.84 -0.84 -0.2 4.2 2.3 5.29 -1.29 -0.3 4.3 2.4 5.76 -1.76 -0.4 4.4 2.5 6.25 -2.25 -0.5 4.5 2.6 6.76 -2.76 -0.6 4.6 2.7 7.29 -3.29 -0.7 4.7 2.8 7.84 -3.84 -0.8 4.8 2.9 8.41 -4.41 -0.9 4.9 3 9 -5 -1 5 Power: 2 Coefficient: 1 Fixed point: 2 My second fixed point is 5, 25 x y change in y change in
- Word count: 3665
This gradient function will apply to all graphs of y=axn. I also predict that there will be different ways to work the gradients or gradient function. I will be drawing a tangent on the curve y=x2, at a particular point to find the gradient of that point. You cannot calculate the gradient of the tangent directly. To find the gradient of the tangent at point P on a curve with a given equation (in this case x2) two points need to be known.
- Word count: 2783
So even if graphs are drawn, even on a huge scale, there is a great possibility of errors in the gradient. In my graphs I have drawn a normal line, which is supposed to be perpendicular to the tangent. I have drawn the normal line using a capillary tube. My aim in using the normal line is to make sure that inaccuracies in the tangent are kept minimal. Increment Method: Apart from the tangent method there is another way to find the gradient.
- Word count: 7068
I am going find out the gradient of each line and then look at all the different gradients. With these different gradients I am going to comment on the difference of the value of the different gradients. Observations: I can see that the gradients have got bigger, the bigger the line.
- Word count: 205
Using this method, when adding on 0.1, the gradient would be like this: G = 4.41 - 4 2.1 - 2 G = 0.41 0.1 G = 4.1 I am expecting for the gradient given to become more like the gradient of its tangent if I decrease the amount added onto x for the second point. The most accurate value would be zero as at that point it would be a tangent, but it is impossible as it would involve dividing by zero.
- Word count: 2700
By comparing the two tables I can see that the height is what Y equals (AXN) and the width is the 1 over the power (X/2 for X2 and X/3 for X3). So if the formula for the gradient is Height/Width then, by replacing the height with AXN and the width with X/N we get XN/(X/N). We can simplify this by multiplying both sides by N to get ANXN/X and we can simplify this by dividing both sides by X to get ANXN-1. I shall now you this formula in the graph Y=X4 to test it.
- Word count: 1964
I will now go on to construct the graph of y=x2. This is because I know that the graph of x2 will be a curve and it is curves that I am investigating X 1 2 3 4 Y 1 4 9 16 To find the gradient of this line, I will use the tangent method. If I have a point on the curve, I will draw a tangent so only the point on the curve is touching the tangent. I will then draw a right-angled triangle with the tangent. I will then take the values of the opposite line y and the adjacent line, x.
- Word count: 2459
The first being drawing a tangent at the point, working out the distances on the tangent using the scale on the graph and then using this formula: dg/dc However there is another way called small increment method. This method gives a more accurate approximation on the gradient. What you do is zoom in on the graph and take part of the curve you take a co-ordinate e.g. (3,9) and (3.01,9.0601). Now you connect the two points together with a straight line.
- Word count: 2064
Gradient Function - To discover different curves and their relations, and the formula for the gradient.
The increment method, which we use as it is more accurate than the tangent one. You zoom into the graph and take the x point and y point just after a whole figure. For instance with y-x2 , using point (1,1) we would do 1.0012 - 1 for the y value and 1.001 - 1 for the x value. So therefore - Method = 0.002001 / 0.001 = 2.001, which we can say is roughly 2. Y=X Please see graph y=x As the co-ordinates are the same, the distances between the parts of the tangent (excluding the hypotenuse)
- Word count: 1038
This instrument enabled me to obtain an accurate curve hence enabling me to draw accurate tangents to obtain accurate gradients. I then checked the points again once I had plotted the curve to make sure that I had not misplaced any points because this could affect my tangents. I then drew the tangents to do this I first took a ruler and a sharp pencil, I then aligned the ruler along the curve making sure that the tangent that was about to be drawn was equally balanced along the line and that the two closest points were the same distance away from the tangent.
- Word count: 971
Investigate the elastic properties of a strip of metal (hacksaw blade) and use the results to determine the value of Young's Modulus of the metal.
The gradient (m) will equal the part of the equation above. This will be re-arranged to give the Young's Modulus as follows: E = 6Mgm bd3 Appearance of the graph - from the gradient we can say that the graph will be straight line through the origin. This is because the gradient consists only of constants: E, b, d, M and g. therefore, the overall gradient will remain the same throughout the graph. The graph will pass through the origin because when Cos?
- Word count: 1351
Investigate gradients of functions by considering tangents and also by considering chords of the graph and using algebra.
In biology and chemistry population growth is thought of as gradient. So moving on to curves now I need to find out the best way to find the gradient of any curve. 1 Tangents A curve does not have a constant gradient. The point A has been marked and the tangent XY drawn. At any point on a curve, the gradient is equal to the gradient of the tangent at that point (a tangent to a curve is a line touching the curve at one point only). For example, the gradient of the below curve at A is equal to the gradient of the tangent at A, which is XY.
- Word count: 2624
These results however are not accurate to the tangents I drew on the graph. There is always going to be an inaccuracy in a graph, even if that inaccuracy is 0.25 of a millimetre. Therefore, I can only accept these results as estimates. As you can see, they are all twice their x value. Therefore, for the graph of y=x2, the formula for the gradient of a tangent is g=2x. I am left with the question of accuracy. I cannot get total accuracy, but there are ways I can get very close to an accurate answer. One such way is to use the method of using a line inside the curve. (4,16) (3.5,12.25) (3.01,9.0601)
- Word count: 882
We decided that in our experiment we would measure the speed of the trolley in m/s. We also decided that we would take 4 readings at each gradient so we could later work out an average. We chose to use the computer to measure the speed as this was a more accurate method than timing it using a stop clock. We chose to use gradient intervals of five books as this would produce results that were more dramatic so would show a more obvious pattern. Fair test To ensure that our results were valid the following measures had to be maintained: * We have to use the same trolley and slope for each experiment - If the trolley was changed, the change in mass could affect the speed of the trolley.
- Word count: 972
I will then fix constants a and c, and focus on constant b. I will examine the path of the turning point, and also the gradient of the curve - using both tangents and calculus. I will then fix constant c, and examine what happens when I change constants a and b simultaneously. Finally, I will plot graphs using all three constants. I will make a step by step plotting of an equation, plotting each constant in turn to emphasise their effects. I will prove the location of the turning point, and the points at which the curve crosses the x and y axis.
- Word count: 1745