• 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
  14. 14
    14

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 ofimage00.png,image01.png,image91.png and image62.pngfor n=0, 1, 2 and 3. Use exact values in your results. Explain the relationship between successive terms in the table for each quantityimage00.png,image01.png,image91.png and image62.png.

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 [image00.png= 4image16.png]. 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 =image02.png]. First, the length of a single side is 1.

Let Pn= the length of the perimeter

...read more.

Middle

= 4image16.png]

[ln =image02.png]

[image13.png]

[image26.png=image27.png]

These functions are fit with graphs which had already done no2.

[image00.png= 4image16.png]

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

[image00.png= 4image16.png]

image119.png= 4×3 = 12

image120.png= 4×12=48

image121.png=4×48=192

Therefore, this function is correct.

[ln =image02.png]

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 =image02.png]

image122.png= image123.png

image124.png=image125.png

image126.png=image127.png

So, this function is correct either.

[image13.png]

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

[image13.png]

image128.png1×3=3

image129.png12×image123.png=4

image130.png48×image125.pngimage131.png

image132.png192×image127.png= image133.png

Thus, this function is correct as well.

[image26.png=image27.png]

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

image21.png

1

image31.png

2

image39.png

3

image48.png

[image26.png=image27.png]

image134.png=image135.png= image136.png=image137.png

image138.png=image139.png=image03.png=image04.png=image05.png

image06.png=image07.png=image08.png=image09.png=image10.png

The function is correct as well.

...read more.

Conclusion

image19.png=image20.png=image22.png

[image23.png]

image24.png=768×image22.png= image25.png

[image26.png=image27.png]

image28.png=image29.png=image30.png=image32.png=image33.png

Diagram of one of side when n=4.

image34.png

  1. Calculate values forimage35.png,image36.png,image37.png and image38.png. You need not verify these answers.

[image00.png= 4image16.png]

image35.png= 4image40.png=4×(image41.png)=12288

[ln =image02.png]

image42.png=image43.png=image44.png=image45.png=image46.png

[image23.png]

image47.png=768×image22.png= image25.png

[image26.png=image27.png]

image49.png=image50.png=image51.png

[image52.png=image53.png=image54.png=image56.png=image57.png]

image58.pngimage59.png+image60.png=image61.png

  1.  Write down successive values of image62.png in term ofimage63.png. What pattern emerges?

 [image26.png=image27.png] have been proved. However, the value ofimage65.pngcan’t get from this function because of image66.pngandimage67.pngdoes not exist in this report. Thus find another characteristic as transform the function, the values of 1, 2, 3, and 4.

image68.png

image69.pngimage70.png

image71.pngimage72.png+image73.png

image74.pngimage75.pngimage76.png+image77.png

These function can be more simply thus

image78.png

image79.png

image80.png

image81.png

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

Let image82.png

image83.png

image84.png

image86.png

image87.png

Therefore the value of image88.pngwill be

image89.png(sum of image90.png)

  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 [image89.png(sum ofimage90.png)]. 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.

...read more.

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

See related essaysSee related essays

Related International Baccalaureate Maths essays

  1. Extended Essay- Math

    ��F�_6Cx ÎgG"���#|mw��ucL��...�� w0�� ��8;����)�F��E���&� �Xï¿½Ø ï¿½_��wpF�0�{bM�x�>%��o�o���a8�jj� &S�,v�bFl.i��2� �@� i#� �F���V �"��ÅP���ÈHdL$����3�_�_ãq�)'�!1/4��{6~d�?2(c)���d��3/4�b1/2(c)���Ç1/4_��_��-��Z��VA�uкh ��fEs�2Zm��Ck!}�1/4G$ ��qG~���"��dM-`�w���.��(c)�<�/ ux�}��$!��Ȩ�jPp��r��-"�d���v���\;1ë·²_~�"��ѿh!(c)��!�x�_��q�^!�"�_4��;����"�[-�$�#-�x�GÞ³"PZ@�K� 8w� �`� c@"� �@.� � PN�P��� n�^p <#�)�S`|��;X� ' &��D )HR�t!�r��!(���1/2�>(*"* �Pt�"z�{�c�9���-��0 �...(tm)a-X-��a� v�1/2� 8N�s�� (r)���6�~?...��O� �(V"J��2B�By Q1�t�!T �u�...��CM�Pkh,� Í-A���CG��Ñ��zt�=�~�^DoaHn�Fc�q�a1��L�3�y�(tm)�|�b��X1��� ��bc"�-�-�c�4v��q�p:�]8 . -�+�5�n�Fq3�x"���7�{�#��|#3/4?��ů"M�.�?!(tm)p"PK�"<"��ii�hth�hBh�h�h.� �1/4��J$�D{"��I,#^" ��hÉ´'�F������i{h��~%'H�$}' )�T@j �&1/2&� c�"�� �� "�k���BO �7 �C�B_B����A"���Â�P����a...'�Q�qc8�a�F�{�sdY"lB�'�kȷ��L(&!&#&?�}L�LL3�Xf1f ��|�f�a�E2�2� KK%�M-)V"("k�Q�+��?�x� ����.�����s�� �boa�"���"#"�8G;�+N4�$�=g"�)��.f.-.?(r)C\W�^p��'� �(c)�5�C�+<1/4<f<Q<�<�yxYy�yCx�y"y��t��|�|��>����-��/ p � � �Xt�l|%D#�.(T,�'�(�'l#1/4W��� ���H�H(c)�]'UQ1QW���sb�bb)b�^�"�ģ�"ÅH`%�%B%�$F$aI�`�J�GR�"�U�J�4FZC:B�Z�(tm)

  2. 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)

  1. 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

  2. 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)

  1. 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.

  2. The Koch Snowflake

    2- Using Microsoft Excel, create graphs of the four sets of values plotted against the value of n. In your solution, include a printout for each graph. 3- Explain, in words, the relationship between successive terms of the sequence of Nn , ln , and Pn for n= 0, 1,

  1. Math Portfolio Type II

    Such an increasing and decreasing trend in the population is observed in the following years although the population seems to come closer to the sustainable limit, i.e 60000 all the time. In the 8th year, the population is exactly at its sustainable limit and henceforth, the population stabilizes by only varying slightly until the 18th year.

  2. Math Portfolio Type II Gold Medal heights

    and in order to give a proper understanding here are some graphs that represented modified graphs of .function as the One can see from the graphs that all of the functions have an asymptote at x=0. If b is positive the functions approach the asymptote while rapidly decreasing in y value.

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