The Koch Snowflake
Math Mock Exploration
Math SL 1
The Koch Snowflake
The Koch Snowflake is a fractal identified by Helge Von Koch, that looks similar to a snowflake. Here are the diagrams of the first four stages of the fractal -
1. At any stage (n) the values are denoted by the following – Nn - number of sides
Ln - length of each side
Pn - length of perimeter
An - Area of snowflake
Mentioned below are the values of these above variables, for the first 4 stages of the fractal.
Number of Sides
As the stages of the snowflake progress, each side is divided into thirds, with two equal line protruding from the middle third to form an edge. I.e – Each straight line -
Becomes this -
Hence now for everyone 1 line, 4 new ones are formed. Hence we can say that there is geometric progression, by the factor 4.
Hence, the formula for the number of sides is
Nn = 3(4)n
Length of Side
The length of the next side is one-third the previous length. This is once again geometric progression.
Therefore, the equation for the nthterm is:
The perimeter of any shape = Length of each side x Number of sides Considering this, the formula for the perimeter can be obtained by multiplying the formulae of the length and number of sides of the fractal. Hence
Pn = 3(4)n x
The area of every new snowflake would be = area of the earlier stage of snowflake + area of new triangles.
The area of the the first snowflake, or stage zero, is √¾(side)2 . And since the length of the side is 1 unit, the area will be √¾ too. So to obtain the area for stage two, we add the area of the original triangle, to the area of the three new triangles. The new triangles will have a length scale factor of 1/3, and hence an area scale factor of 1/9.
For the second stage, again the same logic is used. However this time, the length of the new triangles added will be 1/32 the size of S, the original side in stage 0. Hence, their area scale factor = 1/33
For the third triangle, we will add the current area, to the area of the 48 new triangles. Their length scale factor compared to S will now be 1/33, making their area scale factor 1/34. We add their area to the area obtained in the earlier stage, to get –
Hence we can see a pattern emerging here. . At the kth iteration we add 3*4^(k-1) additional triangles of area . This means we add a total area of .
This graphs seams to exhibit an exponential increase curve. There is a very high rate of growth in the number of sides, which grows exponentially against the progression of stages of the Koch snowflake.
This graphs seams to exhibit an exponential decrease curve. There is a very high rate of reduction in the length of sides, which decreases exponentially against the progression of stages of the Koch snowflake.
This graph represents a gentle increase. There is a slow rise in the values of the perimeter as the stages of the fractal progress.
This graph displays two main characteristics, through the values of the area of the snowflake as the stages progress, and those are increase, and convergence.
4. Values for n = 4
As we have found out earlier, each of the above graphs follow a geometric progression, and have a general formula. Hence, it is possible to find the values of for n= 4, as well as n=6 using them. 1) Nn = 3(4)n
N4 = 3(4)4 = 768
N6 = 3(4)6 = 12288
2) Ln = (1/3)n
L4 = (1/3)4 = 1/81
L6 = (1/3)6 = 1/729
3) If Pn = 3 ×, then-
P4 = 3 ×= 3 ×= =
P6 = 3 × = 3 × =
Hence, knowing the these 4 values for n=4, we can deduce what the shape of one side of this Koch Snowflake would look like.
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