- Level: International Baccalaureate
- Subject: Maths
- Word count: 1412
The Koch Snowflake Portfolio
Extracts from this document...
Introduction
The Koch Snowflake
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.
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
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
Middle
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.
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.
This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.
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