# Investigation – Von Koch’s Snowflake Curve

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. c1s1 = 1 (s – side length) c2s2 = 1/3

c3s3 = 1/9

c4s4 = 1/27

c5s5 = 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 suppose that r = 1 ⅓ (r – common ratio)

c1 = 3hence c1 has a perimeter of 3 units

c2 = 4hence c2 has a perimeter of 4 units

c3 = c1 · r 3-1c3 = 3 · (1 ⅓)2c3 = 5 ⅓hence c3 has a perimeter of 5 ⅓ units c4 = c1 · r 4-1c4 = 3 · (1 ⅓)3c4 = 7 1/9hence c4 has a perimeter of 7 1/9 units c5 = c1 · r 5-1c5 = 3 · (1 ⅓)4c5 = 9 39/81hence c5 has a perimeter of 9 39/81 units

Remembering that Von Koch’s curve is cn, where n is infinitely large, I am going to find the perimeter of Von Koch’s curve.

cn = c1· r n-1cn = 3 · (1 ⅓) n-1hence the total length increases by one third and thus the length at step n will be (4/3)n of the original triangle perimeter.

I am going to suppose the area of c1 is 1 unit2. I will try to explain why the areas of c2, c3, c4 and c5 are A2 = 1 + ⅓ units2

A3 = 1 + ⅓ [1 + 4/9] units2

A4 = 1 + ⅓ [1 + 4/9 + ( 4/9) 2] units2

A5 = 1 + ⅓ [1 + 4/9 + ( 4/9) 2 + ( 4/9) 3] units2

analogously

A6 = 1 + ⅓ [1 + 4/9 + ( 4/9) 2 + ( 4/9) 3 + ( 4/9) 4] units2 A7 = 1 + ⅓ [1 + 4/9 + ( 4/9) 2 + ( 4/9) 3 + ( 4/9) 4 + ( 4/9) 5] units2

The height of one of these two triangles is ½(a · √3) and the base is ½(a). So, the area of the equilateral triangle is A0 = ¼(a2 · √3). In each step k, we add the area of nk little equilateral triangles with sides sk. The area at step k can be written: Ak+1 = Ak + nk · ¼(√3) · (sk) 2

The sides of the little triangles are scaled down by a factor of 3 in each step, i.e., sk = (⅓)na. The area recurrence formula can now be written: Ak+1 = Ak + 1/12 · (√3) · (4/9) k-1 · a2

The total area can be computed by evaluating the geometric series formed by the above recurrence formula. The total area after an infinite number of steps is: An = 2/5 · (√3) · a2

The Von Koch’s snowflake has an infinite perimeter, but a finite area.

As said above, the Von Koch’s curve is enclosed in a finite area. Putting aside the very first step of curve drawing which is a simple straigth line, we...

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