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# math portfolio type 1

Extracts from this document...

Introduction

The Koch snowflake

1. Using an initial side length of 1, create a table that shows the values of,, and 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,, and .

Let Nn= the length of sides.

When the n values are 0, 1, 2, and3, the number of sides will be;

 n number of sides 0 3 1 12 2 48 3 192

As coming of new triangle each side, the number of side increase. Each of one side will be separated into 4. Therefore the function to find the number of sides will be [= 4]. Don’t forget the first triangle’s number of side is 3.

Let ln= the length of single side

When the n values are 0, 1, 2, and3, the length of sides will be;

 n length of side 0 1 1 1/3 2 1/9 3 1/27

As coming of new triangle each side, the length of a single side will be decrease because it will be separated 3 part of it. In other words, each of side will be separated 1/3 each time. Therefore the function of to find the length of a single side will be [ln =]. First, the length of a single side is 1.

Let Pn= the length of the perimeter

Middle

= 4]

[ln =]

[]

[=]

[= 4]

Due to the n increase, the number of side also increases because each time the one side will be separated into 4.

Now we check that function is correct or not.

It has to be the same values in table below.

 n number of sides 0 3 1 12 2 48 3 192

[= 4]

= 4×3 = 12

= 4×12=48

=4×48=192

Therefore, this function is correct.

[ln =]

Due to the n increase, the number of side decreases because a single side will be separated each time, so the length of a single side has to decrease.

Now, check the functions.

 n length of side 0 1 1 1/3 2 1/9 3 1/27

[ln =]

=

=

=

So, this function is correct either.

[]

Due to increase the value of n, the length of the perimeter also increases because of the number of sides are increasing.

Now, check the function

 n length of the perimeter 0 3 1 4 2 16/3 3 64/9

[]

1×3＝3

12×＝4

48×

192×=

Thus, this function is correct as well.

[=]

Due to value of n increase, the area also increases because it keeps expanding. The changing of area is going to a little because of the area of coming triangle is smaller and smaller.

Now, check the function.

 n area of the snowflake 0 1 2 3

[=]

== =

====

====

The function is correct as well.

Conclusion

==

[]

=768×=

[=]

====

Diagram of one of side when n=4.

1. Calculate values for,, and . You need not verify these answers.

[= 4]

= 4=4×()=12288

[ln =]

====

[]

=768×=

[=]

==

[====]

+=

1.  Write down successive values of  in term of. What pattern emerges?

[=] have been proved. However, the value ofcan’t get from this function because of anddoes not exist in this report. Thus find another characteristic as transform the function, the values of 1, 2, 3, and 4.

+

+

These function can be more simply thus

It is obvious that the function has characteristic thus it can define.

Let

Therefore the value of will be

(sum of )

1.  Explain what happens to the perimeter and area as n gets very large. What conclusion can you make about the area as n →∞? Comment on your results.

The perimeter and area will be infinite. By the solution of area of figure could represent [(sum of)]. That means when value of n is infinite, area also infinite as well. Therefore, the value of perimeter and area of snowflake are infinite when the n value is infinite either.

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

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2. ## Koch Snowflake

n = 2 N2 = 3*42 = 3*16 = 48 For n = 3 N3 = 3*43 = 3*64 = 192 Thus, the generalizations for Nn made above apply consistently to the sets of values given in the table and also satisfy the graph made above.

1. ## The Koch Snowflake

+ An = An-1 + An = An-1 + Part 4: Stage 4 Nn 768 Ln Pn 9 An Nn = 3 x 4n = 3 x 44 = 768 Ln = 1 x (1/3)n = 1 x (1/3) 4 = Pn = 3 x (4/3)n = 3 x (4/3)

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

1 N1 = 3 and N2 = 12 Ex. 2 N2 = 12 and N3 = 48 General formula for Nn The general formula can be verified by putting in values from the table Using Graphamatica, I have plotted and drawn this Graph of n vs. Ln Scale X-axis - 1 step = 0.5 units Y-axis - 1 step

1. ## Investigating the Koch Snowflake

n-1((V3)/4) From the general expressions above we can find Nn, ln, Pn, and An, at n=4. Where that N4= 3(4)4=768, l4= 1/ (3)4= 1/81, Pn = 3(4/3)4=256/27, A4 = A4-1 + (1/3) (4/9) 4-1((V3)/4) = A3 + (1/3) (4/9)3((V3)/4)

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

It is directly proportional. Generalizations and predictions As we know the nth term for Nn, Ln, Pn and An we can now predict the results for terms till n=6. n Nn Ln Pn An 0 3 1 3 0.433012701 1 12 4 0.577350269 2 48 0.641500298 3 192 0.670011422 4

1. ## Mathematics Portfolio. In this portfolio project, the task at hand is to investigate the ...

The function needs two constants, one to use as a scaling factor increasing time, and secondly a power, to decrease the function exponentially as +Gx decreases. Therefore a power function will fit the data the best: +Gx = atb where a is a positive constant and b is a negative exponent.

2. ## The Koch Snowflake

had to use the formula A=,1-2.�??�h which is the formula to find the area of a triangle. In order to find the height of each triangle I had to use the Pythagorean theorem (a2+b2=c2) (Note that in the drawings of triangles and when I solved I used b instead of h for the height of the triangle.)

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