• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

HL type 1 portfolio on the koch snowflake

Extracts from this document...

Introduction

HL TYPE 1

By Abhishek Chhabria

                                                                                    (Math HL)  

International Baccalaureate

B.D.Somani International School

INTRODUCTION

The Koch curve was named after Helge von Koch in 1904. The generation of this fractal is simple. We begin with a straight line of unit length and divide it into three equally sized parts. The middle section is replaced with and equilateral triangle and its base is removed. After one iterations, the length is increased by four-thirds. As this process is repeated, the length of the figure tends to infinity as the length of the side of each new triangle goes to zero. Assuming this could be iterated an infinite number of times, the result would be a figure which is infinitely wiggly, having no straight lines whatsoever. The construction is very simple but still looks really beautiful.

Throughout, let us consider-

image00.png

  1. Using an initial side length equal to 1, we deduce a table which shows the values of the above mentioned variables with respect to changes in ‘n’.

image01.pngimage76.pngimage86.pngimage94.png

     Stage 0                   Stage 1                  Stage 2                  Stage 3

image101.png

image80.png

image83.png

image87.png

image33.png

0

3

1

3

image02.png

1

12

image10.png

4

image20.png

2

48

image28.png

image38.png

image48.png

3

192

image58.png

image69.png

image74.pngimage75.png

Now we study the relationship between successive terms for each of the above geometric deductions of Koch’s snowflake.

Let us consider the term in the image77.pngstage to be image78.pngrespectively, according to the general term being studied, and ‘r’ to be the ratio between successive terms.

image79.png

  • For image80.png,

image81.png

...read more.

Middle

This statement generalizes the behavior of the graph.

To verify the generalization we derive the conjecture from the table’s values.

image106.png

Therefore, with reference to the graph, we are now convinced that the generalization applies consistently to the table’s values forimage80.png.

  • For image83.png, (second graph)

Again, we attempt at deriving a conjecture from our deductions in step1.

We enter values of the graph’s points in the statistics list of a Graphic Display Calculator (Texas Instrument).

Then, we undertake Exponential Regression image103.png

and the deductions from the calculator are as follows:

image107.png

This implies that the conjecture for image83.png is:

image108.png

“The conjectureimage108.png suggests that as we move right along the x-axis (0 onwards), i.e. as the value of ‘n’ increases by 1 unit, the corresponding value on the y-axis (1 onwards) gets multiplied by image109.png units and the graph slopes gently.”

This statement generalizes the behavior of the graph.

To verify the generalization we derive the conjecture from the table’s values.

image110.png

image111.png

Therefore, with reference to the graph, we are now convinced that the generalization applies consistently to the table’s values forimage83.png.

  • For image87.png, (third graph),

We attempt at deriving a conjecture from our deductions in step1.

We enter values of the graph’s points in the statistics list of a Graphic Display Calculator (Texas Instrument).

...read more.

Conclusion

Below are value tables ofimage54.png.

image55.pngimage56.png

 It is evident that the increase in image33.pngis a converging one which means as ‘n’ increases the difference between successive terms decreases.

Now as ‘n’ approaches infinity, that is image57.png, we compute:

image59.png

Therefore we can say that (recalling that the area of the original triangleimage60.png)

As image61.png

Verification:

Imagine drawing a circle around the original figure. No matter how large the perimeter gets, the area of the figure remains inside the circle.

image62.pngimage63.png

Percentage of original Area enclosed by curve of infinite perimeter is:

image64.png

Comments:

Clearly the perimeter will increase in the further stages and become infinite, but the area of the figure will be less than the area of the

circumcircle of the original equilateral triangle. This figure has

an infinite perimeter but a finite area!  The area enclosed by the closed curve of infinite length is actually only 60 percent more than that of the original area of the equilateral triangle we started in Stage 0. A remarkable property indeed. It is the significant property of a fractal shape that has self-similarity to an infinite depth. That is, you can enlarge a portion of the boundary to any extent and find shapes similar to the original figure.

7.image65.png In step 1 we found that image66.png.

The general expression found in step 3 isimage67.png

image68.png

Proof: (By Principle of Mathematical Induction)

image70.png

image71.png

image72.png

image73.png

...read more.

This student written piece of work is one of many that can be found in our GCSE Fencing Problem section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related GCSE Fencing Problem essays

  1. The Koch Snowflake

    with progression of n. Q.3) For each of the graphs above, develop a statement in term of n that generalizes the statement shown in its graph. Explain how you arrived at your generalizations. Verify that that your generalizations apply consistently to the sets of values produced in the table.

  2. Koch Snowflake Math Portfolio

    0 3 1 12 2 48 3 192 i. The trend seen in the above data and the graph is that with a successive increase in the stage, the number of sides increases by 4 times the previous number. The number of sides starts from 3, therefore, a = 3 (1st term)

  1. The Koch Snowflake

    3)/36 = ( 3)/3 A2 = 12 (1/2 x 1/9 x 1/9 x Sin60�) + ( 3)/3 = 12 (1/2 x 1/9� x ( 3)/2) + ( 3)/3 = 12 ( 3)/324 + ( 3)/3 = ( 3)/27 + ( 3)/3 = ( 3)/27 + 9 ( 3)/27 = 10

  2. Geography Investigation: Residential Areas

    In this hypothesis I am predicting that the outskirts of town will have more houses that are owned by the individual resident and that the property is not rented, may it be privately or rented from a housing association. For the initial enquiry into this hypothesis I will use all the residents I surveyed from every road.

  1. Area Under a Straight Line Graph - Calculate the area under a straight line ...

    Now I am going to find a generalisation for different gradients as well as intercepts. To do this I am going to do some more complex diagrams. More Complex Diagrams (changing the gradients and intercepts!): Equation Area of 'triangle' Area of 'rectangle' Total Area y = 2x (10 x 20)/2

  2. Koch’s Snowflake Investigation

    Instead of working out the area of each shape in cm�, I will count the number of triangles in each shape, as this is more efficient. In the first snowflake (see back of project) there were 81 triangles. The sides of the whole triangular shape are each 9 centimetres long,

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work