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  • Level: GCSE
  • Subject: Maths
  • Word count: 2624

Investigate gradients of functions by considering tangents and also by considering chords of the graph and using algebra.

Extracts from this document...

Introduction

Introduction

Aim: to investigate gradients of functions by considering tangents and also by considering chords of the graph and using algebra.

Research: What does “gradient” mean? Generally, the “steepness” of a curve is measured by its gradient.

Firstly in my coursework I will investigate gradients using tangents and find out the bests way to use tangents. I will start of by investigating the gradients of y=x, y=x2, y=x3 because they are likely to be simpler to solve and so I can understand at first then I will move onto more complex equations later.

Then I will look at chords and finally algebra.  I hope I will learn a lot during my time of doing this coursework, as I am sure you will too.

In vertical and horizontal graph lines it is easy to work out the gradient. For a graph of a horizontal line the line has no steepness and so the gradient is zero also for a vertical graph line the graph line is infinite so the gradient is infinite.

Gradients are just as important in other subjects as they are in maths. In physics acceleration is the gradient of a velocity time graph and velocity is the gradient of an instance time graph. Also gradients can be used in radioactivity and decay in physics.

In biology and chemistry population growth is thought of as gradient.

...read more.

Middle

Things are not solved yet; I have to look at the formula for these three times to find any possible relationship. In y=x, g=1=x0; in y=x2, g=2x; in y=x3, g=3x2; as we can see, in the three gradient formulas, the number before x and the indices are both increasing as the indices of the equation get higher, and the difference between the number before x and the index is 1 (the index number is smaller than the number before x). So I predict that the gradient formula for y=x4 is: g=4x3.

Let’s try it out.

The co-ordinates:

x

1

2

3

4

Y

1

16

81

256

The gradients in y=x4:

x

g (The Gradient)

Increment Method

1

4.040601

2

32.240801

3

108.541201

4

256.96101

g1=4=4×1=4x13

g2=32=4×2×2×2=4x23

g3=108=4×3×3×3=4x33

g4=256=4×4×4×4=4x43

So my prediction works this time! Look at the table of the previous steps below:

Equation

y=x

y=x2

y=x3

y=x4

y=xn

Gradient Function

g=x0

g=2x

g=3x2

g=4x3

g=nxn-1

I got the gradient function. It can be written in this way:

When y=xn, g=nxn-1

4

Just to be certain, I will try y=x5:

x

1

2

3

4

y

1

32

243

1024

x

g (The Gradient)

Increment Method

1

5.10100501

2

80.80401001

3

407.70902

4

1286.41602

5×14=3
5×2
4=80
5×3
4=405
5×4
4=1280

My formula does work! And notice that because of the high number, increment method is now not so perfect.

...read more.

Conclusion

Therefore dy/dx =

lim 2x + dx

dx --> 0

When dx becomes zero, dy/dx = 2x.

Therefore the gradient of y = x² is 2x.

For example, at the point (2, 4), the gradient is 2x = 4 .

10

Conclusion

During this coursework I have learnt a great deal about straight lines, curves, algebra and in particular gradients. I have been able to set out my results in a table and I noticed a general pattern. From this pattern I was able to produce a general formula for the gradient of a curve i.e.

if y=xn, G=nxn-1

This result enabled me to find the gradient of more complicated curves, such as the coursework curve i.e.

if y=4x3+2x-5

G=49(3x2)+2-0

=12x2+2

I can now find the gradient at any point on the coursework curve just by substituting the x value at that point into the formula for the gradient i.e.

if at the point P (0, -5), G=2

I have no doubt that the mathematics I have learned will be useful to me at GCSE level and in more advanced work in other subjects in the future.

If I had more time I would like to investigate other curves to see if my result still holds

e.g.  y=1/x2, y=1/xn, y=sin x, y=cos x, y=tan x

I predict that the formula would work for the first two groups but I am not sure if it would work for the trigonometry. This could be investigated in a further coursework project.

11

...read more.

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