Investigation – Von Koch’s Snowflake Curve
Adrian Zwierzchowski
2 IB
Investigation – Von Koch’s snowflake curve
In this investigation I am going to consider a limit curve named after the Swedish mathematician Niels Fabian Helge von Koch. I will try to investigate the perimeter and area of Von Koch’s curve.
[pic]
The Koch’s curve has an infinite length because each time the steps above are performed on each line segment of the figure there are four times as many line segments, the length of each being one-third the length of the segments in the previous stage.
First of all I am going to suppose c1 has a perimeter of 3 units. I will try to find the perimeter of c2, c3, c4 and c5. c1 s1 = 1 (s – side length) c2 s2 = 1/3 c3 s3 = 1/9 c4 s4 = 1/27 c5 s5 = 1/81
If the original line segment had length s, then after the first step each line segment has a length s · ⅓. For the second step, each segment has a length s ·(⅓)2, and so on.
Assuming a unit length for the starting straight line segment, we obtain the following figures:
|iteration |segment |segment |curve |
|number |length |number |length |
|1 |1 |1 |1.00 |
|2 |⅓ |4 |1.33 |
|3 |1/9 |16 |1.77 |
|4 |1/27 |64 |2.37 |
|5 |1/81 |256 |3.16 |
|6 |1/243 |1024 |4.21 |
|... |... |... |... |
|10 |1/19683 |262144 |13.31 |
Thus c1 has a perimeter of 3 units. c2 is divided into 12 sides. If one side is equal to ⅓ hence c2 has a perimeter of 4 units (12 · ⅓ = 4)
The common ratio of this geometric sequence is equal to
(cn+1) / cn thus I can
2 IB
Investigation – Von Koch’s snowflake curve
In this investigation I am going to consider a limit curve named after the Swedish mathematician Niels Fabian Helge von Koch. I will try to investigate the perimeter and area of Von Koch’s curve.
[pic]
The Koch’s curve has an infinite length because each time the steps above are performed on each line segment of the figure there are four times as many line segments, the length of each being one-third the length of the segments in the previous stage.
First of all I am going to suppose c1 has a perimeter of 3 units. I will try to find the perimeter of c2, c3, c4 and c5. c1 s1 = 1 (s – side length) c2 s2 = 1/3 c3 s3 = 1/9 c4 s4 = 1/27 c5 s5 = 1/81
If the original line segment had length s, then after the first step each line segment has a length s · ⅓. For the second step, each segment has a length s ·(⅓)2, and so on.
Assuming a unit length for the starting straight line segment, we obtain the following figures:
|iteration |segment |segment |curve |
|number |length |number |length |
|1 |1 |1 |1.00 |
|2 |⅓ |4 |1.33 |
|3 |1/9 |16 |1.77 |
|4 |1/27 |64 |2.37 |
|5 |1/81 |256 |3.16 |
|6 |1/243 |1024 |4.21 |
|... |... |... |... |
|10 |1/19683 |262144 |13.31 |
Thus c1 has a perimeter of 3 units. c2 is divided into 12 sides. If one side is equal to ⅓ hence c2 has a perimeter of 4 units (12 · ⅓ = 4)
The common ratio of this geometric sequence is equal to
(cn+1) / cn thus I can