• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month   # Borders Investigation

Extracts from this document...

Introduction

Joseph Smith

BORDERS INVESTIGATION

## Introduction

The starting point for this investigation is the following sequence of cross shapes: The sequence begins with a single white square, which is then surrounded by black squares to form the second shape. Each new cross is then formed by completely surrounding the previous cross with a border of black squares. In each new cross, the previous cross can be seen as the area of white squares in the centre.

The aim of this investigation is to derive algebraic formulae from the sequence, each expressing one property in terms of another (e.g. defining the area as the diameter squared). These formulae can then be examined and, hopefully, proven, using a wide variety of mathematical tools.

The next stage will then be to apply this same process to the three-dimensional counterpart of this sequence of shapes. This sequence would begin with a single white cube, which would then be surrounded on all six sides by black cubes, and the sequence would continue in the same way as the two-dimensional version.

## 2D

Defining n

Throughout this investigation, the variable n (the value in the phrase “nth term”) is used to refer to terms in sequences. Terms are compared to, and expressed as functions of, n. However, in order to be able to identify the value of n

Middle

i.e. remove the discrete square units) and then rotate it by 45°, we end up with a square, and what was previously the diameter across the middle of the cross is now the diagonal across the square. We know that the diameter is equal to , and therefore this is the length of the diagonal across the square. Using this, we can derive a formula for the area of the square in terms of n.  Pythagoras’ theorem tells us that is equal to (because l forms a right-angled triangle with the square, and l is the hypotenuse while s is the length of both other sides), so we know that . As the area is also , we can say that the area equals . Now l, which is the diagonal across the square, is also the diameter of the cross, which as we have said is , so by substituting this into , multiplying out and simplifying, we get . This final result is, of course, less than the actual formula, because when we round off the edges of the cross to remove the discrete units we also lose some accuracy.  Here we have our smooth approximation of the shape drawn over the top of the

Conclusion

r does range from one to n.

So, we begin with a formula which we gives the volume of the nth 3D cross. It contains a series which is the sum of all those 2D cross layers above the middle. This uses the formula we have already obtained and proven for the area of a 2D cross, and, as we have said, substitutes in values ranging from zero to (because the number of these layers must be n, and the smallest 2D layer, because it is a single cube, must use 0 as the value of r). The sum of this series is multiplied by two (to include both above the middle and below it), and then the volume of the central layer (simply ) is added. Written out, with sigma notation, it is: What we must do next is convert the series so that r can range from one to n. Since the range 0 to is just the same as 1 to n shifted down by one, we can do this by replacing each r in the formula by . We can then substitute in standard results. There are three we must use now, one which we have already come across: What follows, then, is our final proof:   The result is identical to our formula for volume. Both the 2D and 3D formulae have now been proven.

-  -

This student written piece of work is one of many that can be found in our GCSE Fencing Problem 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

# Related GCSE Fencing Problem essays

1. ## Fencing investigation.

S = half the perimeter (in this case 500) V (S (S-a)(S-b)(S-c)) V500 (500-350)(500-200)(500-450) V500 x 150 x 300 x 50 V1125000000 33541.02m2 Review In the previous section: I have noticed that the more regular the triangle is, the larger its area will be.

2. ## Perimeter Investigation

a regular hexagon each equal side would be =1000 / 6 = 166.7m Using TAN = h / 166.7 h = 166.7 / 2 * tan60 Area = 1/2 * 166.7 * 144.35 = 12031.6 * 6 Area of six = 72189.6 = 72 190m In a regular heptagon each

1. ## Geography Investigation: Residential Areas

When the inner city was first being built it was not planned, houses and factories were just built on available land and pushed together so that more and more could be fit into a small piece of land, thus, there will be a lack of green space, more industry and more pollution in the area.

2. ## Biological Individual Investigation What Effects Have Management Had On Grasses In Rushey Plain, Epping ...

This is evident from the soil sample, as the soil was moister the lower, and more clay-like the soil became. Prediction Grass height will be greater in areas of high light intensity, where clearing has occurred, than in shaded areas, where natural succession has occurred, and trees stop light reaching the ground.

1. ## Geography As Environmental Investigation

If I am the only one counting, it is possible that I may miss some vehicles. Questionnaire: (Please circle your preferred answer) 1. How would you rate the environmental quality of the area. Poor 1 2 3 4 5 6 7 8 9 10 Excellent 2.

2. ## Regeneration has had a positive impact on the Sutton Harbour area - its environment, ...

amounts of people that use the area, at different times of the year and on different days. When deciding on how to do this, I had to make a choice of some things that I would differ from the first counts, and some things that I would keep the same in order to analyse the data efficiently.

1. ## Maths Investigation on Trays.

we have the highest volume of the turquoise line. x Area Base Area 4 sides Volume 2.2 184.96 119.68 406.91 2.4 174.24 126.72 418.18 2.6 163.84 133.12 425.98 2.8 153.76 138.88 430.53 3 144 144 432 3.2 134.56 148.48 430.59 3.4 125.44 152.32 426.5 3.6 116.64 155.52 419.9 3.8 108.16 158.08 411.01 4 100 160 400 Here we can

2. ## &amp;quot;Evaluate the success of the economic, social and physical regeneration of The London Docklands.&amp;quot;

However, all this has changed as the derelict land has become regenerated. It has turned out to be the largest urban regeneration scheme in Europe. So can we say that the Docklands redevelopment scheme has been an all round success? • Over 160,000 pieces
of student written work
• Annotated by
experienced teachers
• Ideas and feedback to 