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Snowflake island fractal The fractal called the snowflake
Chapter 11, Problem 71E(choose chapter or problem)
Snowflake island fractal The fractal called the snowflake island (or Koch island) is constructed as follows: Let \(I_{0}\) be an equilateral triangle with sides of length 1. The figure \(I_{1}\) is obtained by replacing the middle third of each side of \(I_{0}\) by a new outward equilateral triangle with sides of length 1/3 (see figure). The process is repeated where \(I_{n+1}\) is obtained by replacing the middle third of each side of \(I_{n}\) by a new outward equilateral triangle with sides of length \(1 / 3^{n+1}\). The limiting figure as \(n\ \rightarrow\ \infty\) is called the snowflake island.
a. Let \(L_{n}\) be the perimeter of \(I_{n}\). Show that \(\lim_{n\rightarrow\infty}\ L_n=\infty\).
b. Let \(A_{n}\) be the area of \(I_{n}\). Find \(\lim_{n\rightarrow\infty}\ A_n\). It exists!
Questions & Answers
QUESTION:
Snowflake island fractal The fractal called the snowflake island (or Koch island) is constructed as follows: Let \(I_{0}\) be an equilateral triangle with sides of length 1. The figure \(I_{1}\) is obtained by replacing the middle third of each side of \(I_{0}\) by a new outward equilateral triangle with sides of length 1/3 (see figure). The process is repeated where \(I_{n+1}\) is obtained by replacing the middle third of each side of \(I_{n}\) by a new outward equilateral triangle with sides of length \(1 / 3^{n+1}\). The limiting figure as \(n\ \rightarrow\ \infty\) is called the snowflake island.
a. Let \(L_{n}\) be the perimeter of \(I_{n}\). Show that \(\lim_{n\rightarrow\infty}\ L_n=\infty\).
b. Let \(A_{n}\) be the area of \(I_{n}\). Find \(\lim_{n\rightarrow\infty}\ A_n\). It exists!
ANSWER:
Problem 71ESnowflake island fractal The fractal called the snowflake island (or Koch island)is constructed as follows: Let I0 be an equilateral triangle with sides of length 1. The figure I1 is obtained by replacing the middle third of each side of I0 by a new outward equilateral triangle with sales of length 1/3 (see figure). The process is In+1 is obtained by replacing the middle third of each side of In by a new outward equilateral triangle with sides of length 1/3n+1. The limiting figure as n is called the snowflake island.a. Let Ln be the perimeter of In. Show that .b. Let An be the area of In. Find. It exists! Solution:Step 1The perimeter of the .The perimeter of the The perimeter of the The perimeter of the The limit of the perimeter as ( )