• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month
Page
  1. 1
    1
  2. 2
    2
  3. 3
    3
  4. 4
    4
  5. 5
    5
  6. 6
    6
  7. 7
    7
  8. 8
    8
  9. 9
    9
  10. 10
    10
  11. 11
    11
  12. 12
    12
  13. 13
    13
  14. 14
    14
  15. 15
    15
  16. 16
    16
  17. 17
    17
  • Level: GCSE
  • Subject: Maths
  • Word count: 2700

The Gradient Function

Extracts from this document...

Introduction

The Gradient Function Introduction The gradient of any line is the steepness at which it slopes; on straight lines it can be worked out by drawing a right angled triangle using the line itself as the hypotenuse to find out the ?y, and ?x. The gradient of a line can then be worked out by dividing ?y by ?x. The following graphic shows an example: However, with a curved graph, the gradient is different at different points. To work out the gradient at a point of a curved graph, a tangent would have to be drawn, and the gradient of it measured. The longer the tangent is, the more accurate the result if done by eye. The following graphic is an example: Because this method is inherently inaccurate, to improve the accuracy we could either use a computer program to draw an accurate tangent, or use the small increment method. The small increment method is where the gradient of a chord, from the point, to another point of the line a short distance away, is worked out to find the gradient between the two points. So, for example, if it was a curve of y=x2, the gradient at x=3 would be measured using a chord from [3, 9] to [3.0001, 9.00060001] and so the gradient would be 0.00060001 divided by 0.0001 which is 6.0001 or 6 as an integer. The Small Increment Method I given a curve, y=x2, and a point, x=2, I can calculate an approximate gradient by using a chord with a second point a small distance away from x=2. ...read more.

Middle

I will use this data to work out the nth term of the gradient in the graph of y=x4. 4 32 108 256 500 +28 +76 +148 +244 +48 +72 + 96 +24 +24 As the differences are equal at the third level, the pattern for the nth term is an3 + b, if the value of a is the difference divided by 3, then the result for n=1 would be 8 which is twice the real value, so the value must be the difference over 6, this applies for all the values I have. This means that the nth term is 4n3; because it already fits, the value of b must be zero. To test this pattern, I can use it to predict the gradient at x=6. Using the nth term pattern it should be 4 x 63 which is 864. Using the small increment method, the answer is (6.0014 -64) divided by 0.001 which rounds to 864. Therefore the nth term is correct. Patterns for graphs of form y=xn In the graphs of this form I have worked out patterns for, there seems to be a global pattern in the nth term of each sequence: This means that for graphs of form y=xn the gradient at any point could be worked out using G=nxn-1. I will test this on the graph y=x5: PREDICTIONS X=1 5 x 14 = 5 X=2 5 x 24 = 80 X=3 5 x 34 = 405 X=4 5 x 34 = 1280 X=5 5 x 34 = 3125 Graphs of form y=ax2 I will extend my work to see what effect the value of a has on my results so far, initially I kept a =1. ...read more.

Conclusion

Using calculus to calculate gradients If I was to simplify (x+f)n it would be the following: xn + nxn-1f + ?xn-2f2... The value of ? is not relevant, but would in fact be a triangular number, because the line in the calculation that relates to this is as follows: G = (x+f)n - xn f This means that if it is simplified it would be: G = xn + nxn-1f + ?xn-2f2... - xn f But as it is subtracted by xn and divided by f it cancels out to: G = nxn-1 + ?xn-2f ... Therefore as f was zero (which it should be) the value of the ? makes no difference as all that is left is: G = nxn-1 Which is in fact the formulae I originally came up with. I have now demonstrated that calculus can be used to derive the gradient function for all graphs of the type y=axn. Conclusion To conclude, I have worked out that a gradient at any point on a graph of form y=axn can be worked out using the following formulae: G = anxn-1 I have tried four different methods to work this out: * Drawing tangents by hand * Using the small increment method * Using a computer program * Calculus From using these, I know: * That hand drawn tangents are inaccurate and so no use for working out a pattern * That the small increment method is not entirely accurate * That computer programs can generate perfectly accurate gradients * That calculus can work out the formulae in a single equation. I also know that in graphs of form y=axn + bxm can be treated as two gradients from graphs of form y=axn added together. ...read more.

The above preview is unformatted text

This student written piece of work is one of many that can be found in our GCSE Gradient Function section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related GCSE Gradient Function essays

  1. Peer reviewed

    The Gradient Function Coursework

    5 star(s)

    and that the first point is called P and has the coordinates (x, y). The small y-difference of these two points would thus be ?y and the tiny x-difference would be ?x. ?x or ?y are both a single symbol and are often called the increment in x or y.

  2. Gradient function

    y= x4 My first fixed point: 3, 81 x y increase in y increase in x gradient 2 16 -65 -1 65 2.1 19.4481 -61.5519 -0.9 68.391 2.2 23.4256 -57.5744 -0.8 71.968 2.3 27.9841 -53.0159 -0.7 75.737 2.4 33.1776 -47.8224 -0.6 79.704 2.5 39.0625 -41.9375 -0.5 83.875 2.6 45.6976 -35.3024

  1. Gradient Function

    0.3 70.59 4.8 110.592 14.408 0.2 72.04 4.9 117.649 7.351 0.1 73.51 4.99 124.2515 0.748501 0.01 74.8501 4.999 124.925 0.074985001 0.001 74.985 5 125 5.001 125.075 -0.075015 -0.001 75.015 5.01 125.7515 -0.751501 -0.01 75.1501 5.1 132.651 -7.651 -0.1 76.51 5.2 140.608 -15.608 -0.2 78.04 5.3 148.877 -23.877 -0.3 79.59 5.4

  2. The Gradient Function

    If this gradient function, 3ax2 is correct, then the gradient function of 10x3 will be 30x2. To prove that it is 30x2 I am going to draw up a table of values for certain x values. For the curve y=10x3, using the gradient function 30x2 I think the gradients will

  1. The Gradient Function

    = -0.37 rounded to 2.d.p -3.25-(-2.5) L2: Gradient at X = -2 Y2-Y1 X2-X1 = -1.375-(-1.625) = -0.71 -2.15-(-1.8) L3: Gradient at X =2 Y2-Y1 X2-X1 = 1.675-1.3 =-0.75 1.75-2.25 Graph: -2/x The table below shows the points I have used to plot the graph: X Y -4 0.5 -2

  2. Investigate the gradients of the graphs Y=AXN

    16 0.5 32 3 4*34-1 = 108 3,3.001 81,81.108 0.108/0.001 108 81 0.75 108 4 4*44-1 = 256 4,4.001 256,256.256 0.256/0.001 256 256 1 256 The above table shows that both methods agree for my formula for cases in which A=1 and N= a positive integer, so I shall now

  1. The Gradient Function

    This shows that the gradient is actually 8. Therefore the small increment method is an accurate estimate I will now use the small increment method to work out the gradients for other curves in the hope of spotting a pattern.

  2. The Gradient Function.

    and the point (3.01,9.0601). Now you use the formula: dg/dc Which gives you 0.0601�0.01= 6.01 I have discovered that the formula for the equation g=c� is 2c. As you can see from the results table the accuracy using the tangent method hasn't been perfect, but using the small increment method it is possible to get much more accurate set of results.

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work