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• Level: GCSE
• Subject: Maths
• Word count: 2487

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Introduction

Introduction

In this project, I intend to investigate the Gradient Function and then prove how the formula that I get works.  The Gradient Function is an important part of calculus and mathematics as a whole.  It allows us to calculate accurately the gradient of any point on any graph without using the lengthy procedure of drawing a graph and then a tangent which is still extremely inaccurate – the error can be 20 units.

To investigate the Gradient Function, there are two aspects of mathematics that must be made clear from the beginning: what is a gradient? and what is a tangent?

A gradient can be described as the steepness of a curve or line and is found by the formula:

Gradient of a         change in y

straight line           change in x

This is shown in the graph on the left.

A tangent  is defined as a straight line that just ‘touches’ the circumference of a circle (as shown in the picture on the right).  It is often used in mathematics to find an estimate of the gradient of a curve as it is a straight line, using the formula above.

Investigating a simple curved graph

I decided to start with the simplest curved graph, x2 to begin my investigation.

 x 0 1 2 3 4 y 0 1 4 9 16

Middle

Therefore the gradient of the graph y = x2 is 2x + d.

At first, I thought that this must be wrong because it was not simply 2x.  However, I realised that d was so small that in the context of finding the gradient, it was almost negligible.  Hence, to calculate the gradient normally it is simply Gradient  = 2x.

*        *        *

I was not able to find a formula for the graph x3, so I decided to use the above method to find out what it was.

Gradient = (x + d)3 –x3

x + d – x

I expanded the brackets using Pascal’s Triangle and the Binomial Theorem which will be mentioned after I have found the formula for y = x3.

Gradient = x3 + 3x2d + 3xd2 + d3 – x3

x + d – x

As with the previous example I can cancel out the x3s and the xs. Therefore:

Gradient = 3x2d + 3xd2 + d3

d

All of the expressions on the top have d in, so I can factorise the top half of the equation:

Gradient = d(3x2 + 3xd + d2)

d

Thus, having cancelled out the ds (as shown):

the formula for the gradient of y = x3 is 3x2 + 3xd + d2

Conclusion

δ    cos 0.5 = cos (0.5 +0.00001) – cos 0.5

δx                        0.00001

= cos(0.50001) – cos 0.5

0.00001

= - 0.000004794

0.00001

= -0.4794298

Conclusion

Having investigated the gradient function for graphs of the type xn, I have come to the conclusion that the formula nx(n-1) is true for all types that I have investigated except for the trigonometrical graphs.  I have proved that this is true graphically, numerically and algebraically.  I have looked at tables of values, found formulas for each type of graph and then proved that these formulas are true, using algebra.

I have furthered my investigation by looking at the gradient of trigonometrical graphs and shown that the theory that I found in a textbook is true.  Naturally, this investigation could be expanded further and further, looking at all types of curves such as negative curves and those to the power of different numbers.

I have completed the objectives that I set myself at the beginning of the investiagtion.  I am now able to calculate any gradient accurately of any point on almost any graph without drawing a tangent.  I have also completed the objective of proving my formulas and generalising the rule.  Thus to find the gradient the formula is

nx(n-1)

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

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