• 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
  18. 18
    18
  19. 19
    19
  20. 20
    20
  21. 21
    21
  22. 22
    22
  23. 23
    23
  24. 24
    24
  25. 25
    25
  26. 26
    26
  27. 27
    27
  28. 28
    28
  29. 29
    29
  30. 30
    30
  31. 31
    31
  32. 32
    32
  33. 33
    33
  34. 34
    34
  35. 35
    35
  36. 36
    36

The Gradient Fraction

Extracts from this document...

Introduction

Mathematics Investigation:

The Gradient Function

My aim in this investigation is to find the different gradients for all curves, including parabolas and straight line curves.

The Gradient is the “steepness” of a curve. The gradient is usually founded by a method.

To find a gradient of a line, there is a strict method;

GRADIENT = VERTICAL DIFFERENCES

                HORIZONTAL DIFFERENCES

In order to use this method you must divide the vertical difference of the point by the horizontal difference of the point.

y= mx + c is the general equation for a straight line graph.

This equation consists of two parts that you need to remember:

“m” is the gradient of the graph

“c” is the value where it crosses the y-axis and is called the intercept.

In my investigation I will try to use a number of methods .The methods I will be using are methods like, the triangle method, the Increment Method, Tan θ Method, Number of Differences and Differentiation. I will also show examples of Maximum and Minimum curves. I will also explain the concept of a ‘limit’.

The first method to use is the “triangle method”. The triangle method is where you have to draw a gradient line which best fits the “x” value point on the graph. From there you have draw down a line to the “x” value and draw a line across the value where the gradient line finishes.

image17.png

Here is an example:image00.png

image01.png

image18.png

The second step is to find the differences from the points on the graph. First, you have to look at the “x” value, and then find its point at the “y” value. (In my case that was x=3 and the “y” value was 9). Then you have to find the differences from the points. (y= 9-0 = 9, x= 3-1.5 = 1.5).

...read more.

Middle

-32/1.8]

Above are the differences in the ‘x’ and ‘y’ scales.

As you can see, I have investigated all the positive points on the ‘x’ scale. Some results are not all to 1 decimal place. The pattern in this results table is that the ‘x’ value has been multiplied by 4. If you round the values up and down to their lowest terms and highest terms, they will equal to the following:

x=2: 7.82 (round up to 8) the 2 has been multiplied by 4.

x=3: 12.41 (round down to 12) the 3 has been multiplied by 4.

x=4: 16.8 (round down to 16) the 4 has also been multiplied by 4.

x=-2.5: -9.23 (round to -10) the -2.5 has been multiplied by 4.

x=-4: -17.77 (round down to -16) the -4 has been multiplied by +4.

These results are not exactly accurate due to the graph scales, but they show an approximate of how close they are. The gradients obtained are all multiples of 4.

The Increment Method

The increment method is another method which calculates the gradient of a line. This method is much more accurate than the ‘triangle method’.  

Here is an example of how to use the Increment Method:

image35.png

The first thing you need to do is to draw the tangent across the point you want to calculate. (In my case, x=4). After you have done that, then draw down the line of differences.

After you have done that, on the ‘x’ axis you have to move across to a value which is very close to your previous ‘x’ value. (In my case, x=4; move to 4.01). After you have done that, then square that number and then the value you get from that draw a line from that point on the ‘y’ axis. [4.012 =16.0801]

...read more.

Conclusion

x=1 = 2 x (multiplied) by 2 x (multiplied) by the x value (1)   = 4.

x=2 = 2 x (multiplied) by 2 x (multiplied) by the x value (2) =8

So it seems like my method is working for both equations.

I will now test it for y=x3

y=x3

y=3x2 (this would be the new formula). The power has been subtracted by 1 and the previous power has been placed in front of the ‘x’.

So now that I have tested and worked with my method, I will need to simplify it a little.

Step 1:

For example: y=x3

Subtract the power by 1. So then that becomes:

y=x3-1

Step 2:

You will need to bring the previous power in front of the equation. So that becomes y=3x3-1

So the final equation becomes:

y=3x2

And if we want to test this method for a different equation, for example:

y=4x2

If we want to find the gradient for the x value of 2, the method would work like this:

Y=4x2

= y=2 x 4x(x=in our case 2)

So that is y=2 x 4 x 2

               = 16

The gradient for the ‘x’ value 2 is = 16.

This is where ‘Differentiation’ has taken place. I have differentiated the equations into a simple formula = nx n-1

 ‘n’ being the power and ‘x’ the value of the equation. Differentiation is the co-efficient placement of the power in the equation.

In the beginning of my investigation I was achieving the gradient function by the tangent method. However, when progressing throughout the investigation, it became very clear that the tangent was not necessary in finding the gradient function of a graph, as the increment was enough to prove that alone.

Progressing further along using the concept of a ‘limit’ showed that there was a formula which gave the gradient of a curve.

At the last stages, I have used number differences to obtain different formulae for different equations. Each stage of investigating got closer and closer to the algebraic formula, which is:

 nx n-1

                

...read more.

This student written piece of work is one of many that can be found in our AS and A Level Core & Pure Mathematics 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 AS and A Level Core & Pure Mathematics essays

  1. Marked by a teacher

    The Gradient Function

    5 star(s)

    Here I have taken the point x = 2. I first get the co-ordinates for both points, which are 2 and 22, and Q, 3 and 32. After this I used the formula: y - x 2 2 = 3^2 - 2^2 = 9-4 = 5 y - x 3

  2. Sequences and series investigation

    Sequence 5 : N = 5 _(5�) - 2(52) + 2Y(5) - 1 = 166Y - 50 + 135 - 1 = 129 The formula on this sequence seems to be successful. I will now apply it to another sequence to be 100% correct: N = 4 _(4�) - 2(42)

  1. Solving Equations. Three numerical methods are discussed in this investigation. There are advantages and ...

    g'(xn) 1 0.500000 0.812500 -0.625 2 0.812500 0.608032 -0.63477 3 0.608032 0.742693 -0.66151 4 0.742693 0.653240 -0.658 5 0.653240 0.712654 -0.6664 6 0.712654 0.673094 -0.66349 7 0.673094 0.699419 -0.6666 8 0.699419 0.681886 -0.66506 9 0.681886 0.693559 -0.66632 10 0.693559 0.685785 -0.66558 11 0.685785 0.690961 -0.66612 12 0.690961 0.687514 -0.66578 13 0.687514

  2. maths pure

    = (x-1)(x+2)2 but since there is no change of sign, it will not be detected by the decimal search. 2. Newton-Raphson Method The Newton-Raphson Method states that if a is a first approximation to a root of f(x) = 0, a better approximation is, in general,???

  1. Numerical Differentiation

    bounds is that assuming the root lies exactly in-between the two bounds (midpoint). 1. Firstly finding the midpoint of the two gives: 2.3921495 2. Then taking the lower bound away from the midpoint gives 0.0000005 3. Therefore you can write ?= 2.3921495 � 0.0000005 Therefore you can write ?= 2.392145 (6s.f.)

  2. Methods for Advanced Mathematic

    This would indicate that it has only one root within the interval and the second root is over looked. I will use the equation 0 = x4+5x3-7x+2 to demonstrate how this method would fail. Here is a graph of the function f(x)

  1. Solutions of equations

    = 0 for x = (4/9) or 0.44444 (5.d.p). The table is shown again below. x F(x) 0 -32 0.1 -16.33 0.2 -6.776 0.3 -1.859 0.4 -0.128 0.5 -0.125 0.6 -0.392 0.7 0.529 0.8 4.096 0.9 11.767 1 25 I will investigate this further: A table is produced of values between 0.4 and 0.5.

  2. Fractals. In order to create a fractal, you will need to be acquainted ...

    + z0 ever lands outside of the circle of radius 2 centered at the origin, then this orbit definitely tends to infinity. Therefore, 2i does not exist in the Mandelbrot since it quickly leaves the boundary after only one iteration: z0 = 0 z1 = 0 + 2i Enough of

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