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Chapter 12: Problem 62

These Applet Bytes* are exercises built around the applet Geometric Fractals (available in MyMathLab in the Multimedia Library or Tools for Success.) Exercises 61 through 64 deal with the Sierpinski carpet, a geometric fractal that is a square version of the Sierpinski gasket. The applet allows you to see, step-bystep, how the Sierpinski carpet is generated. Suppose that the area of the seed square (Step 0 ) of the Sierpinski carpet is \(A\). (a) Find the area (in terms of \(A\) ) of the figures at Steps 1,2 , and \(3 .\) (b) Find the area of the figure at Step 6. (c) Give a general formula (in terms of \(A\) and \(N\) ) for the area of the figure at Step \(N\) of the construction.

Short Answer

Expert verified
The areas at Steps 1,2 and 3 are \(8A/9\), \(64A/81\), and \(512A/729\) respectively. The area of the figure at Step 6 is approximately \(0.262 A\). And the general formula for the area at any step \(N\) of the construction is \((8/9)^N \times A\).

Step by step solution

01

Part (a) Calculation of area for Steps 1, 2 and 3

For Step 1, note that a single square of length \(1/3\) of the total length is removed from the center of the seed square. Hence, \(1 - 1/3^2 = 8/9\) of the initial area remains. This is equivalent to \(8A/9\). For Step 2, every remaining square has a smaller square of area \((1/3)^2\) of the larger square removed. Again \(8/9\) of the total area from the previous step remains. This area is \((8/9)^2 \times A\) or \(64A/81\). The pattern continues for Step 3 with an area of \((8/9)^3\times A\) or \(512A/729\).
02

Part (b) Calculation of area for Step 6

Following the pattern identified in part (a), the area at Step 6 can be found using the formula \((8/9)^6 \times A\) which results in approximately \(0.262 A\).
03

Part (c) Derivation of a generic formula for any Step N

From the understanding in parts (a) and (b), one can observe a pattern - at each step, the remaining area is \((8/9)\) times the previous area. Thus, for any Step \(N\), the formula becomes \((8/9)^N \times A\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sierpinski Carpet
The Sierpinski Carpet is a fascinating example of a geometric fractal. It originated as a variation of the Sierpinski triangle, but instead of triangles, it applies the pattern to squares. To create a Sierpinski Carpet, you begin with a solid square - known as the seed square. The process involves repeatedly removing the middle one-ninth portion of each of the smaller squares, creating an intricate, infinitely repeating pattern.

This fractal showcases a recursive process, where each iteration of removal creates more complexity and beauty. With each step into the infinite, more of the middle squares are removed, and the pattern gains more holes, while the side squares are preserved. This recursive nature is central to understanding fractals like the Sierpinski Carpet.
  • Starts with a square
  • Middle section removed repeatedly
  • Infinite pattern of holes develops
Understanding this fractal gives insight into how complex patterns can emerge from simple, repetitive actions.
Area Calculation
Calculating the area of figures at each step in the Sierpinski Carpet involves understanding how much of the square is removed at each iteration. Starting with a square of area \(A\), the process removes the central section - which is \(1/9\) of the square. Therefore, \(8/9\) of the area remains after the first iteration.

For subsequent steps, this process repeats on each smaller segment, producing a consistent pattern. The area at each step can be expressed as a power of \((8/9)\) multiplied by the original area \(A\).
  • Initial area is \(A\)
  • After step 1, \((8/9) \times A\) is left
  • Step 2 results in \((8/9)^2 \times A\)
  • The pattern continues, halving the area repeatedly
This approach simplifies the calculation, allowing for easy determination of the area at any step \(N\).
Fractal Geometry
Fractal Geometry describes the study of figures that exhibit a repeating, non-regular pattern at every scale. The Sierpinski Carpet is an exemplary type of fractal geometry due to its self-similar structure, where each part resembles the whole structure. This field of study can encompass various natural and mathematical phenomena, from snowflakes to coastlines.

One fascinating aspect of fractal geometry is how it allows for figures with fractional dimensions. While traditional geometry confines objects to whole dimensions, fractals can exist in between dimensions, providing unprecedented insights into complex patterns.
  • Represents infinitely repeating structures
  • Explores non-integer dimensions
  • Bridges mathematics and nature
Understanding fractals such as the Sierpinski Carpet expands our comprehension of dimensionality and intricate designs.
Mathematical Formulas
Utilizing mathematical formulas is key to understanding and working with fractals like the Sierpinski Carpet. At the core of these formulas is the concept of self-similarity and repetitive scaling, which makes observing patterns and making predictions possible.

The area formula derived for the Sierpinski Carpet is based on the remaining portion after each step. Initially, the area is \(A\). After each removal, \((8/9)^N \times A\) represents the area at step \(N\).
  • Formulas illustrate recursive patterns
  • Area at step \(N\) given by \((8/9)^N \times A\)
  • Enables computation of complex patterns easily
These formulas are not only beautiful in their simplicity but also powerful tools in exploring infinite processes.

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Most popular questions from this chapter

Consider the construction of a Koch snowflake starting with a seed triangle having area \(A=81 .\) Let \(R\) denote the number of triangles added at a particular step, \(S\) the area of each added triangle, \(T\) the total new area added, and \(Q\) the area of the "snowflake" obtained at a particular step of the construction. Complete the missing entries in Table \(12-3 .\) $$ \begin{array}{l|c|c|c|c} & R & S & T & Q \\ \hline \text { Start } & 0 & 0 & 0 & 81 \\ \hline \text { Step 1 } & 3 & 9 & 27 & 108 \\ \hline \text { Step 2 } & 12 & 1 & 12 & 120 \\ \hline \text { Step 3 } & & & & \\ \hline \text { Step 4 } & & & & \\ \hline \text { Step 5 } & & & & \end{array} $$

Refer to a variation of the chaos game. In this game you start with a square \(A B C D\) with sides of length 27 as shown in Fig. \(12-41\) and a fair die that you will roll many times. When you roll a 1 , choose vertex \(A\); when you roll a 2, choose vertex \(B\); when you roll a 3 , choose vertex \(C\); and when you roll a 4 choose vertex \(D .\) (When you roll a 5 or \(a\) 6, disregard the roll and roll again.) A sequence of rolls will generate a sequence of points \(P_{1}, P_{2}, P_{3}, \ldots\) inside or on the boundary of the square according to the following rules. Start. Roll the die. Mark the chosen vertex and call it \(P_{1}\). Step 1. Roll the die again. From \(P_{1}\) move two-thirds of the way toward the new chosen vertex. Mark this point and call it \(P_{2}\) Steps \(2,3,\) etc. Each time you roll the die, mark the point two-thirds of the way between the previous point and the chosen vertex. Using a rectangular coordinate system with \(A\) at \((0,0), B\) at \((27,0), C\) at \((27,27),\) and \(D\) at \((0,27),\) find the sequence of rolls that would produce the given sequence of marked points. (a) \(P_{1}:(27,0), P_{2}:(27,18), P_{3}:(9,24), P_{4}:(3,8)\) (b) \(P_{1}:(0,27), P_{2}:(18,9), P_{3}:(24,3), P_{4}:(8,19)\) (c) \(P_{1}:(27,27), P_{2}:(9,9), P_{3}:(21,3), P_{4}:(7,19)\)

Refer to a variation of the Koch snowflake called the Koch antisnowflake. The Koch antisnowflake is much like the Koch snowflake, but it is based on a recursive rule that removes equilateral triangles. The recursive replacement rule for the Koch antisnowflake is as follows: Assume that the seed triangle of the Koch antisnowflake has area \(A=729 .\) Let \(R\) denote the number of triangles subtracted at a particular step, \(S\) the area of each subtracted triangle, \(T\) the total area subtracted, and \(Q\) the area of the shape obtained at a particular step of the construction. Complete the missing entries in Table \(12-12 .\) $$ \begin{array}{l|c|c|c|c} & R & S & T & Q \\ \hline \text { Start } & 0 & 0 & 0 & 729 \\ \hline \text { Step 1 } & 3 & 81 & 243 & 486 \\ \hline \text { Step 2 } & 12 & 9 & 108 & 378 \\ \hline \text { Step 3 } & & & & \\ \hline \text { Step 4 } & & & & \\ \hline \text { Step 5 } & & & & \\ \hline \end{array} $$

Refer to a variation of the chaos game. In this game you start with a square \(A B C D\) with sides of length 27 as shown in Fig. \(12-41\) and a fair die that you will roll many times. When you roll a 1 , choose vertex \(A\); when you roll a 2, choose vertex \(B\); when you roll a 3 , choose vertex \(C\); and when you roll a 4 choose vertex \(D .\) (When you roll a 5 or \(a\) 6, disregard the roll and roll again.) A sequence of rolls will generate a sequence of points \(P_{1}, P_{2}, P_{3}, \ldots\) inside or on the boundary of the square according to the following rules. Start. Roll the die. Mark the chosen vertex and call it \(P_{1}\). Step 1. Roll the die again. From \(P_{1}\) move two-thirds of the way toward the new chosen vertex. Mark this point and call it \(P_{2}\) Steps \(2,3,\) etc. Each time you roll the die, mark the point two-thirds of the way between the previous point and the chosen vertex. Using graph paper, find the points \(P_{1}, P_{2}, P_{3},\) and \(P_{4}\) corresponding to (a) the sequence of rolls 2,2,4,4 (b) the sequence of rolls 2,3,4,1 . (c) the sequence of rolls 1,3,4,1 .

Refer to the Menger sponge, \(a\) three dimensional cousin of the Sierpinski gasket. The Menger sponge is defined by the following recursive construction rule. Start. Start with a solid seed cube [Fig. \(12-42(a)\) ]. Step 1. Subdivide the seed cube into 27 equal subcubes and remove the central cube and the six cubes in the centers of each face. This leaves \(a\) "sponge" consisting of 20 solid subcubes, as shown in Fig. \(12-42(b) .\) Step 2. Subdivide each solid subcube into 27 subcubes and remove the central cube and the six cubes in the centers of each face. This gives the "sponge" shown in Fig. \(12-42(\mathrm{c}) .\) (Call the procedure of removing the central cube and the cubes in the center of each face procedure MS.) Steps \(3,4,\) etc. Apply procedure MS to each cube of the "sponge" obtained in the previous step. Assume that the seed cube of the Menger sponge has volume 1 . (a) Let \(C\) denote the total number of cubes removed at a particular step of the construction, \(U\) the volume of each removed cube, and \(V\) the volume of the sponge at that particular step of the construction. Complete the entries in the following table. $$ \begin{array}{l|l|l|l} & C & U & V \\ \hline \text { Start } & 0 & 0 & 1 \\ \hline \text { Step 1 } & 7 & \frac{1}{27} & \frac{20}{27} \\ \hline \text { Step 2 } & & & \\ \hline \text { Step 3 } & & & \\ \hline \text { Step 4 } & & & \\ \hline \text { Step } \boldsymbol{N} & & & \end{array} $$
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