# Koch Snowflake Math Portfolio

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Introduction

Urvi Mittal

Maths SL

The Koch Snowflake

Stage (n) | Number of sides(N) | Length of a side(ln)(unit) | Perimeter of the snowflake(Pn)(unit) | Area of the snowflake(An)(unit2) |

0 | 3 | 1 | 3 | /4 |

1 | 12 | 1/3 | 4 | /3 |

2 | 48 | 1/9 | 16/3 | 10/27 |

3 | 192 | 1/27 | 64/9 | 94/243 |

- Nn – The no. of sides:

The relationship between the successive values of the number of sides is that the values increase by 4 times with each stage. The number of sides is directly proportional to the stage number.

Nn ∞ n

- ln – The length of each side:

The relationship between the successive values of the length of each side is that the value decreases by (1/3)n times with each stage. The length of each side is inversely proportional to the stage.

ln ∞ 1

n

- Pn – The perimeter:

The relationship between the successive values of the perimeter is that the value increases to Nn × ln times with each stage. The perimeter is directly proportional to the stage.

Pn ∞ n

- An – The area:

The relationship between the successive values of the area is that the values increase by 3 × 4n-1 times with each stage. The area is

9n

Middle

= 1 (1/3)0

= 1

n = 1

a r n

= 1 (1/3)1

= 1/3

n = 2

a r n

= 1 (1/3)2

= 1/9

Hence Proved.

n (stage) | Pn (perimeter) |

0 | 3 |

1 | 12 |

2 | 16/3 |

3 | 64/9 |

The trend seen in the above data and the graph is that with each successive increase in the stage, the perimeter is equal to the number of sides × the length of 1 side; for that particular stage.

Therefore, it is seen that the formula that generalizes the behaviour of the given set of values is:

Pn = Nn × ln

Verification:

n = 0

Nn × ln

= 3 × 1

= 3

n = 1

Nn × ln

= 12 × 1/3

= 4

n = 2

Nn × ln

= 48 × 1/9

= 48/9

= 5 3/9

Hence proved.

n (stage) | An (area) |

0 | √3/4 |

1 | √3/3 |

2 | 10√3/27 |

3 | 94√3/243 |

The trend seen in the above data and the graph is that with each successive increase in the stage, the area increases by (3 × 4n-1)/9n.

Therefore, it is seen that the formula that generalizes the behaviour of the given set of values is:

An =

Where ‘n’ is the stage, k is the nth iteration of the Koch snowflake. The ‘s’ is taken as 1(i.e. from above), and is therefore not taken into consideration.

Verification:

n = 0

= /4

n = 1

=

=/3

n = 2

=

= 10/27

Hence Proved.

4.

Conclusion

As n tends to, the area of the fractal geometrically progresses and the area becomes a constant, to six decimal places, when n = 17. Therefore we can conclude by saying that the snowflake has a finite area but an infinite perimeter.

Comment on the results obtained:

The results obtained while doing this seem to be accurate as there is no room for error in such an investigation. If any of the results obtained were not correct then there would have not been a common ratio for geometric progression in either of the cases, for the perimeter or the area. If there was no common ratio, there would have been no geometric progression, in which case, there would have been no relationship between the successive terms and the whole investigation would not make any sense.

Bibliography:

Sources from the net:

- www.ecademy.agnesscott.edu/
- www.google.com

Use of Technological Resources/ Software Packages:

- Graphic Display Calculator
- MathType
- Microsoft Word
- Microsoft Excel

In question 4, the drawing of 1 side of the 4th iteration of the Koch snowflake was drawn using Web Turtle:

http://www.sonic.net/~nbs/webturtle/webturtle.cgi

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

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