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• Level: GCSE
• Subject: Maths
• Word count: 3331

Extracts from this document...

Introduction

In this coursework I am going to try and find out what the gradient function is for various curved graphs. I will do this by drawing graphs and find the gradient of the tangent of a point.

1.        I am going to draw a graph of y=x². 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=x².

 X 0 1 2 3 4 5 Y=x² 0 1 4 9 16 25

This is the results of the gradient of the tangent:

 At (1,1) 1.81 1.8 = 2(to 1sf) At (2.4) 20.5 4 At (3,9) 4.70.8 5.875 = 6(to 1sf) At (4,16) 30.3 10

I have noticed that the gradient of the tangent doubles each time. Except for at (4,16) so I fink that this is an anomalious result, which would mean that the tangent must be drawn incorrectly.

I am now going to draw a graph of y=x³. 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=x³.

 X 0 1 2 3 4 5 Y=x³ 0 1 8 27 64 125

This is the results of the gradient of the tangent:

 At (1,1) 30.9 3.33333 = 3(to 1sf) At (2.8) 151.3 11.538462 = 12(to 1sf) At (3,27) 200.8 25 At (4,64) 220.5 44

Middle

=

0.40527009

0.001

=405.27009

At (4,1024)

1025.28064016 – 1024

4.001 – 4

=

1.28064016

0.001

=1280.64016

At (5,3125)

3128.12625025 – 3125

5.001 - 5

=

3.12625025

0.001

= 3126.2525

These are quite accurate but when it gets to at (4,1024) and (5,3125) it starts to get inaccurate again so I will go to 0.0001 instead of 0.001.

I will do (4,1034) and (5,3125) again, going to 0.0001.

 At (4,1024) 1024.1280064 – 1024         4.0001 – 4 = 0.1280064    0.0001 =1280.064 At (5,3125) 3125.3125125 – 3125         5.0001 - 5 = 0.3125125   0.0001 = 3125.125

These results are more accurate.

I will also use this method for y=x1, though I know that the answer is one all the time, just to prove that this method is more accurate.

 At (1,1) 1.001 – 1 1.001 – 1 = 0.0010.001 1 At (2,2) 2.001 – 2 2.001 – 2 = 0.0010.001 1 At (3,3) 3.001 – 3 3.001 – 3 = 0.0010.001 1 At (4,4) 4.001 – 4 4.001 – 4 = 0.0010.001 1 At (5,5) 5.001 – 5 5.001 – 5 = 0.0010.001 1

These are very accurate results.

Overall results.

These are the results of the small increments method. The results will be to 1 the nearest whole integer.

 Y= x1 Y= x2 Y= x3 Y= x4 Y=x5 X=1 1 2 3 4 5 X=2 1 4 12 32 80 X=3 1 6 27 108 405 X=4 1 8 48 256 1280 X=5 1 10 75 500 3126

I will now compare the results of the tangent of the gradient method and the small increments method. (TG= gradient method, S.I small increments method)

 Y= x1 Y= x2 Y= x3 Y= x4 Y=x5 TG S.I TG S.I TG S.I TG S.I TG S.I X=1 1 1 2 2 3 3 10 4 5 X=2 1 1 4 4 12 12 30 32 80 X=3 1 1 6 6 25 27 65 108 405 X=4 1 1 10 8 44 48 286 256 1280 X=5 1 10 75 500 3126

The highlighted ones are the anomalous results. This shows me that the small increments method is more accurate. I did not use the gradient method for x=5 or y=x5.

There must be an easier way to find out what the results should be. I will now try and find a formula for each one using the small increment results.

Y = x1.

1        1        1        1        1                n= x0

1               1               1              1

Y=x2

2        4        6        8        10                n=2x1

2              2              2              2

Y=x3

3        12        27        48        75

9              15              21              27                        n=3x2

6        6        6

Y=x4

4        32        108        256        500

32               76        148       244                        n=4x3

48        72          96

24                24

Y=x5

5        80        405        1280          3125

75                325      875         1845

250      550       970                                n=5x4

1. 420

120

This is the table of results for the formulae.

 Y= x1 Y= x2 Y= x3 Y= x4 Y=x5 X=1 1 2 3 4 5 X=2 1 4 12 32 80 X=3 1 6 27 108 405 X=4 1 8 48 256 1280 X=5 1 10 75 500 3126 X X0 2x1 3x2 4x3 5x4

Conclusion

Small Increments Method – this method proves to be more accurate than the tangent method, but it still has a level of inaccuracy when the numbers start to get bigger. So from this method I could find formulas and use the…Formula Method – this method is very accurate and once you know the overall formula you can work out any equation of a curved graph. The overall formula is Y=nxn-1. This formula works for positive, negative and fractional indices.

Throughout the investigation I have used two of the three methods for each equation. This is so that I can check the results of both methods and compare them to double-check it and to find a more accurate method. My overall conclusion of the methods is that the tangent method is very inaccurate as there are many factors that can make it inaccurate (the curve, the tangent drawn etc), and the formula method is very accurate as it makes it easier to find the correct answer and does not cause inaccuracies unless the formula is worked out the wrong way.

I have found out from the positive indices the numbers increase, whereas the negative indices decrease, as well as the fractional indices.

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

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