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The Koch Snowflake Portfolio

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Introduction

The Koch Snowflake

image03.png

Nn = the number of sides

ln = the length of a single side

Pn = the length of the perimeter

An = the area of the snowflake

1. Using an initial side length of 1, create a table that shows the values of Nn , ln , Pn and An for n = 0, 1, 2 and 3. Use exact values in your results. Explain the relationship between successive terms in the table for each quantity Nn , ln , Pn and An.

image00.png

image02.pngimage01.pngimage01.png

ll

l→     hh                       →    h

Equilateral triangle        →               Two halves                 →     Rectangle

If l is the length of an equilateral triangle and h is the height then the rectangle above has the area of:

(1 / 2) * l * h

image01.png

hl

l / 2

Quoting the Pythagorean theorem,

c2 = a2 + b2

l2 = h2 + (l / 2)2

Thus,

h = √[(3 / 4)l2] = √(3) / 2 l

Hence the area of the equilateral triangle is

(1/2) * l * h

= √(3) / 4 l2

Stage 0:

Number of sides:

Is determined by counting the sides of the triangle.

N0 = 3

Length of a single side

Is stated through the question.

L0 = 1

Area of the snowflake

Is calculated through the formula established for the equilateral triangle (as stated on page 2).

A0 = √(3) / 4

...read more.

Middle

1 = 0.5773502692

A2 = [(3 x 4) (1 / 9)2 √(3) / 4] + A1

A2 = [(12) (1 / 32)2 √(3) / 4] + A1

A2 = 0.0641500299 + A1

A2 = 0.0641500299 + 0.5773502692

A2 = 0.6415002991

Perimeter of the snowflake

Is calculated through the perimeter of an equilateral triangle formula:

l2 * N2 = P2

(1 / 9) * 48 = 5.333333333

P2 = 5.333333333

Stage 3:

Number of sides:

Is determined by multiplying N2 * 4 or proven through counting the sides.

N3 = 192

Length of a single side

Is determined through dividing l2 by 3.

l3 = 1 / 27

Area of the snowflake

Is calculated through the formula established for the equilateral triangle (as stated on page 2). As well as adding the A2 onto the answer.

A2 = 0.6415002991

A3 = [(3 * 4 * 4) (1 / 27)2 √(3) / 4] + A2

A3 = [(48) (1 / 33)2 √(3) / 4] + A2

A3 = 0.0285111244 + A2

A3 = 0.0285111244 + 0.6415002991

A3 = 0.6700114235

Perimeter of the snowflake

is calculated through the following formula:

l3 * N3 = P3

(1 / 27) * 192 =

P3 = 7.111111111

Stage 0

Stage 1

Stage 2

Stage 3

Number of sides

3

12

48

192

Length of a single side

1

1/3

or

0.333333333

1/32

or

0.111111111

1/33

or

0.037037037

Length of the perimeter

3

4

5.333333333

7.111111111

Area of the snowflake

0.4330127019

0.5773502692

0.6415002991

0.6700114235

3. For each of the three graphs above develop a statement in terms if the four sets of values shown in this graph. Explain how you arrived at your generalizations. Verify that your generalizations apply consistently to the sets of values produced above.

...read more.

Conclusion

7. Explain what happens to the perimeter and area, as n gets very large. What conclusions can you make about the area as n → infinity? Comment on your results.

As n gets very large the perimeter continues growing exponentially. There is not limit to the number that n can reach due to the calculation formula having no boundary. Eventually the snowflake would be made up of only sharp corners with no smooth lines connecting them, however this would not limit the infinite series because as long as the perimeter of the snowflake is connected the fractal will maintain its shape.

The area of the snowflake would also follow this trend. Although as n reached infinity the increase in area would become microscopic and possibly unmeasurable. However theoretically speaking the area would still increase.

...read more.

This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.

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