• 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)

    xn P (2, 2�) and Q (2.6, 2.6�) (2.6� - 2�) / (2.6 - 2) = 2.76 / 0.6 = 4.6 P (2, 2�) and Q (2.4, 2.4�) (2.4� - 2�) / (2.4-2) = 1.76/0.4 = 4.4 P (2, 2�)

  2. Numerical Method (Maths Investigation)

    Hence it can't find the required roots here using both methods. COMPARISON MADE WITH ALL THREE METHODS After doing all three methods, i.e. Change of Sign Method, Newton-Raphson Method and Rearrangement Method, with each different examples showing how it works.

  1. Sequences and series investigation

    - 2(4) + c = 8 this can then be simplified to, _(2) - 2(2) + c = 4 by dividing the equation by two. We will continue with the calculation using the simplified version of the equation; To find 'c' we will use the following calculations: 1)

  2. Investigate the number of winning lines in the game of connect 4.

    Formula For Connect 4 The formula for any box with a width of 4 is C=7H-18 The formula for any box with a width of 5 is C=11H-27 The formula for any box with a width of 6 is C=15H-36 There is a visible pattern the first number always goes

  1. Mathematical Investigation

    The exact value of a stretched period of one complete cycle (two wave peaks or 2pi radians) can be obtained by the formula 2pi/b. Part III Investigate the family of curves y=sin(x+c) Figure#5: Graphs of different functions of the family curve y=sin(x+c): y=sin(x); y=sin[x+ (pi/4)] and y=sin[x-(pi/4)].

  2. Repeated Differentiation

    = a.11.10.9x8 yIV = a.11.10.9.8x7 yV = a.11.10.9.8.7x6 We can see that if a constant multiplying factor is inserted into the equation then this constant is unaffected by differentiation. From observing what happens to this function we can simply multiply the above rule by a.

  1. Solutions of equations

    ROOT B ANSWER = 0.66667 (5 d.p.) ROOT A As mentioned earlier the method failed for Root A as there was no change of sign between f(0.4) and f(0.5). I know there is a root because of the graph and f(x) = 0 for x = (4/9) or 0.44444 (5.d.p).

  2. Investigating the Quadratic Function

    We already established that adding values to x� will move the vertex of a parabola up and down the y-axis. We also recognized that adding a value directly to x would move the vertex left or right upon the x-axis.

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