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.

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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

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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.

Thus c1 has a perimeter of 4 units. c2 is divided into 20 sides. If one side is equal to ⅓ hence c2 has a perimeter of 6 ⅔ units (20 · ⅔ = 6 ⅔)

The common ratio of this geometric sequence is equal to

(cn+1) / cn thus I can suppose that r = 1 ⅔ (r – common ratio)

c1 = 4 hence c1 has a perimeter of 4 units c2 = 6 ⅔ hence c2 has a perimeter of 6 ⅔ units c3 = c1 · r 3-1 c3 = 4 · (1 ⅔)2 c3 = 11 1/9 hence c3 has a perimeter of 11 1/9 units c4 = c1 · r 4-1 c4 = 4 · (1 ⅔)3 c4 = 18 14/27 hence c4 has a perimeter of 18 14/27 units c5 = c1 · r 5-1 c5 = 4 · (1 ⅔)4 c5 = 30 70/81 hence c5 has a perimeter of 30 70/81 units

cn = c1 · r n-1 cn = 4 · (1 ⅔) n-1

I am going to suppose the area of c1 is 1 unit2. I will try to investigate the areas of c2, and c3.

A2 = 1 + 4/9 units2

A3 = 1 + 4/9 [1 + 13/27] units2

Similarly, its area converges rapidly to a finite limit while the total length of the segments that composed that curve have still no limit.

Koch constructed his curve in 1904 as an example of a non-differentiable curve, that is,

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.

Thus c1 has a perimeter of 4 units. c2 is divided into 20 sides. If one side is equal to ⅓ hence c2 has a perimeter of 6 ⅔ units (20 · ⅔ = 6 ⅔)

The common ratio of this geometric sequence is equal to

(cn+1) / cn thus I can suppose that r = 1 ⅔ (r – common ratio)

c1 = 4 hence c1 has a perimeter of 4 units c2 = 6 ⅔ hence c2 has a perimeter of 6 ⅔ units c3 = c1 · r 3-1 c3 = 4 · (1 ⅔)2 c3 = 11 1/9 hence c3 has a perimeter of 11 1/9 units c4 = c1 · r 4-1 c4 = 4 · (1 ⅔)3 c4 = 18 14/27 hence c4 has a perimeter of 18 14/27 units c5 = c1 · r 5-1 c5 = 4 · (1 ⅔)4 c5 = 30 70/81 hence c5 has a perimeter of 30 70/81 units

cn = c1 · r n-1 cn = 4 · (1 ⅔) n-1

I am going to suppose the area of c1 is 1 unit2. I will try to investigate the areas of c2, and c3.

A2 = 1 + 4/9 units2

A3 = 1 + 4/9 [1 + 13/27] units2

Similarly, its area converges rapidly to a finite limit while the total length of the segments that composed that curve have still no limit.

Koch constructed his curve in 1904 as an example of a non-differentiable curve, that is,