• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month
Page
1. 1
1
2. 2
2
3. 3
3
4. 4
4
5. 5
5
6. 6
6
7. 7
7
8. 8
8
9. 9
9
10. 10
10
11. 11
11
12. 12
12
13. 13
13

# The Koch Snowflake Portfolio

Extracts from this document...

Introduction

The Koch Snowflake

Nn = the number of sides

ln = the length of a single side

Pn = the length of the perimeter

An = the area of the snowflake

1. Using an initial side length of 1, create a table that shows the values of Nn , ln , Pn and An for n = 0, 1, 2 and 3. Use exact values in your results. Explain the relationship between successive terms in the table for each quantity Nn , ln , Pn and An.

ll

l→     hh                       →    h

Equilateral triangle        →               Two halves                 →     Rectangle

If l is the length of an equilateral triangle and h is the height then the rectangle above has the area of:

(1 / 2) * l * h

hl

l / 2

Quoting the Pythagorean theorem,

c2 = a2 + b2

l2 = h2 + (l / 2)2

Thus,

h = √[(3 / 4)l2] = √(3) / 2 l

Hence the area of the equilateral triangle is

(1/2) * l * h

= √(3) / 4 l2

Stage 0:

Number of sides:

Is determined by counting the sides of the triangle.

N0 = 3

Length of a single side

Is stated through the question.

L0 = 1

Area of the snowflake

Is calculated through the formula established for the equilateral triangle (as stated on page 2).

A0 = √(3) / 4

Middle

1 = 0.5773502692

A2 = [(3 x 4) (1 / 9)2 √(3) / 4] + A1

A2 = [(12) (1 / 32)2 √(3) / 4] + A1

A2 = 0.0641500299 + A1

A2 = 0.0641500299 + 0.5773502692

A2 = 0.6415002991

Perimeter of the snowflake

Is calculated through the perimeter of an equilateral triangle formula:

l2 * N2 = P2

(1 / 9) * 48 = 5.333333333

P2 = 5.333333333

Stage 3:

Number of sides:

Is determined by multiplying N2 * 4 or proven through counting the sides.

N3 = 192

Length of a single side

Is determined through dividing l2 by 3.

l3 = 1 / 27

Area of the snowflake

Is calculated through the formula established for the equilateral triangle (as stated on page 2). As well as adding the A2 onto the answer.

A2 = 0.6415002991

A3 = [(3 * 4 * 4) (1 / 27)2 √(3) / 4] + A2

A3 = [(48) (1 / 33)2 √(3) / 4] + A2

A3 = 0.0285111244 + A2

A3 = 0.0285111244 + 0.6415002991

A3 = 0.6700114235

Perimeter of the snowflake

is calculated through the following formula:

l3 * N3 = P3

(1 / 27) * 192 =

P3 = 7.111111111

 Stage 0 Stage 1 Stage 2 Stage 3 Number of sides 3 12 48 192 Length of a single side 1 1/3or0.333333333 1/32or0.111111111 1/33or0.037037037 Length of the perimeter 3 4 5.333333333 7.111111111 Area of the snowflake 0.4330127019 0.5773502692 0.6415002991 0.6700114235

3. For each of the three graphs above develop a statement in terms if the four sets of values shown in this graph. Explain how you arrived at your generalizations. Verify that your generalizations apply consistently to the sets of values produced above.

Conclusion

7. Explain what happens to the perimeter and area, as n gets very large. What conclusions can you make about the area as n → infinity? Comment on your results.

As n gets very large the perimeter continues growing exponentially. There is not limit to the number that n can reach due to the calculation formula having no boundary. Eventually the snowflake would be made up of only sharp corners with no smooth lines connecting them, however this would not limit the infinite series because as long as the perimeter of the snowflake is connected the fractal will maintain its shape.

The area of the snowflake would also follow this trend. Although as n reached infinity the increase in area would become microscopic and possibly unmeasurable. However theoretically speaking the area would still increase.

This student written piece of work is one of many that can be found in our International Baccalaureate Maths 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

# Related International Baccalaureate Maths essays

1. ## The Koch Snowflake

of the previous triangle and so would be the other sides since it is a equilateral triangle. And so I was able to realize that there was a pattern that each time the side length would get smaller by,1-3. .

2. ## Math Portfolio - The Koch snowflake investigation.

Length of sides At stage n=0, there is an equilateral triangle which has 3 sides. Those 3 sides each are 1 unit. For the next stage where n=1, each side is divided into 3 parts. Thus the length of the new side is one-third the previous side.

1. ## Math Portfolio: trigonometry investigation (circle trig)

As the value of ? increases from to, cos? goes from -1 to 0. As the value of ? increases from to, cos? goes from 0 to 1. The maxima which are the maximum value for y or the highest point in y value, of this the maxima, the highest point in the y value

2. ## Investigating the Koch Snowflake

every single set of values plotted against the value of n, are in the following pages respectively. Number of Sides (Nn) One side of the figure from the previous stage becomes four sides in the following step, thus we begin with three sides, and the general expression for the number

1. ## MAths HL portfolio Type 1 - Koch snowflake

it is a geometric progression since every successive term after the 1st term is being multiplied by the common ratio 4. We can confirm this common ratio by taking two values from the table on the left. Ex. 1 N1 = 3 and N2 = 12 Ex.

2. ## The Koch Snowflake

/ ()) = 60� = 1/2 = tan 60 () x = Area of shaded small triangle = x 2 = Total area of Stage 2 = + = = Area of Stage 3 = Area of Stage 2 + Area of 48 small shaded triangles Height= cos -1 (()

1. ## The Koch Snowflake

So ln can be presented in terms of ln-1: ln=ln-1 To get next generation, each length should times one-third of the previous one. Thus we calculate length of each shape: l0==1 l1==1� l2==1��= l3==1���= The length of each side Then we can put our data into the diagram below: We

2. ## Koch Snowflake

Ln � Every successive quantity is 1/3rd of the previous length thus this is also a geometric sequence, i.e. the ratio between two successive terms remains the same. Pn � This also follows a geometric pattern as every successive quantity increases by a factor of 4/3 and the ratio here as well remains the same.

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