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

IB math project

Extracts from this document...

Introduction

1.

image00.png

image11.png

My conjecture, for the function image12.png bounded by between ‘a’ and ‘b’, is, the ratio of area A to area B is equivalent to the ratio of n to 1.

Test

  1. nimage26.pngΖ+
  2. Fraction
  3. Negatives
  4. Irrational Numbers

1) nimage26.pngΖ+

n=3

n=5

image40.png

image48.png

2) Fractions

image49.png

image50.png

3) Negatives

image01.png

For all values of n inimage02.png, no values exist in the region bounded by 0 and 1; therefore no conjecture can be inferred.

4) Irrational Numbers

image03.png

image04.png

2.

image05.png

Bounds 0 to 2

image06.png

Bounds 1 to 2

image07.png

TEST

Bounds 0 to 2

image08.png

Bounds 0 to 3

image09.png

Bounds 2 to 4

image10.png

My conjecture is valid for all functions of image12.png

...read more.

Middle

 and image16.png, then my conjecture, of n to 1, holds true for all ratios between areas A and B.

For Negative Bounds:

Bounds –1 to 0

image17.png

The conjecture of n:1 is the exact opposite for conjecture of the negatively bounded regions in quadrant III. (1:n)

If a set of positive bounds were to be changed to negative, then the ratio between A and B in the positive bounds would be equivalent to that of B and A in the negative bounds. (see diagram explanation below)

image18.png

image19.png

image20.png

3.

As mentioned in question 2, my conjecture, of n to 1, holds true for all ratios between areas A and B, if a<b; as long the function in question is image12.png.

Proof:

With the function in question being image21.png

...read more.

Conclusion

image37.png

4b.

image38.png

image39.png

image41.png

My conjecture, for the function image12.png revolved around the y-axis, bounded by between ‘0’ and ‘1’, is, the ratio of area A to area B is equivalent to the ratio of 1 to 2n. In correlation to the ratio of the function revolved around the x-axis, the ratio between area A and area B is the exact opposite.

Test

  1. n=3, n=5
  2. Fraction

Different Bounds

n=2

n=5

image42.png

image43.png

2) Fractions

image44.png

image45.png

3) Bounds

Bounds 0 to 2

Bounds 1 to 2

image44.png

image46.png

Proof:

With the function in question being image21.png rotated around the x-axis, I was able to calculate area A by finding the intregal using the shells method, and by using simple international methods I was able to determine area B. The ratio between area A and area B is 2n to 1 (refer to diagram on the right)

image47.png

...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

    �984�$��F�i5�m\B �-Q�WI��P|�h�#��hy�Y����zyO*��:o%Z��o�~�֭8�E[�J�����U�@�]"v��'��'�(r)@�ÐXZZ�F�7�x���-<x�_��""]{;Դ��Z9T'KENÕª0j�+�;���κ��[=v����ׯß1c&M�d�...�7�[O"�(�@� 6o�,)&�-� 4kÖ��'>�+(r)��X S�N�3gN}}� A���H�Ø1���S"���c*\��@��� RM�49p�[o1/2�o�3/4� b�H<x�e�l�� ��9R�@�Y�o)��U�q 5�Ñ3� z�'(r)\��y��^b�--�j��Nů���#}X�hNLY��ij'G� ���SOÍ5k��(tm)M�^b�z���eLJ��"�2�ÖZLz� a��� �:`��L�<y����wȺw�3/4c� �ݶm�z���]cÔ°qÞ dk'�_9zZ��...a(tm)w�A�CX��q�'�F��w�5j����1/2(tm)5p3�F�B�s=o�1/4^1/2z�(r)�B�u�J!"SØ"�Z"<m+��O>>�?F��c��(c)��:e��×`�8��4lcZ"<���ѣ�rTBQ��Q+J&�(r)����~�]1/45�BM0��8�0Ý"'#GN�0�o�3/4I �:��X�����u�CU�\�l���Æ-�h�^q(� (c)�b�#P9K�J�`�(r)5���(r)��B<�\�]Q-�g�B":�ȨY�Y�{ء���X�"%'�b�B...�" �f�|�zi�Y'ND�S����ȨY�z]"v�N�t �" �ft�T�����ȨY�V�s���V��ҧ6v���H�2jF�-"T�"��3�u�d(c)�pZT'��E�@D�(c)%�"�d(c)dP'\��Z�+TÇ�9l-}V��c�Ԣ hc2jvi...m�:A*���Csj� N��"�W�nÅ�[!"-GA�[�I�IQpr� 5"t�>�Ev�"^i�<1/2��h�d��R|Z�\���ai����w�?%d�,R$�H#���T���P'Y'�Z'�+ԥ�E��᧶�E �|y2j�u�"��"����"n^�H9K��9��e�^(r):���g9@Q?�Y�"1/2Q�e"<cdÔ�/���.�b-�k #PV��p�...˨%�1/4d���h�mI:�<�_A<\M-*�f'~~��(tm)��(c)�3I�Ð7&P'�H���Zb�+T� HÝ´ �bx s-���d'2j&���ͯ�q�)�N��RLez���S �QKr *�lÓ¦M ����?�m4Z+����'d���"�;o�i�$ �[�n�R�S�V5[Lg�C�l'&Kr���g���Yjq��*�f...�9A�iÓ¦ ...��&cB�V@"�5:�ÍQ�B" �aP ��i��G^��Z�H�� �0�"z�95V?5�N��Ȩ(tm)��(c)�> j"&P����ZF-Z�!Kc�)_���x�]:� �V"!�f...*-w���ѣ � �#$ �!ÌE�G`�SoC��5zj��H+ÊQ3��0(r)��K-,1.��"��Z$k*DQ1/4k���x�����`i3/4 5�:X"vm"v���'m ��-��d���Q"�up�QÚk�"'AF-�5�(�5��e� (r)^1/2:��T�I2j&�S7K��{�iX��WO4mUK�W �fX��4����c�U�0 ��-Q3�?~'��1,"�?�(:�t(tm)�,��oAFͰׯ_/�O�:8\=]4= ��e� +� 7l'2,"��gDȯ ß��a%�1/4� �"�Ý"K�KÐ¥"d�L�� 5�s�h�¤ï¿½ï¿½0���<|��]% �fRslaWp " 8�n@(tm)

  2. Statistics project. Comparing and analyzing the correlation of the number of novels read per ...

    the number of books read can be accounted for by the variation in the modal mark. In other words, approximately 70% of the variation is attributed to other factors. GIRLS No of books(x) Modal mark(y) (xy) (x�) (y�) 2 70 140 4 4900 3 70 210 9 4900 2 70

  1. Maths Project. Statistical Analysis of GCSE results at my secondary school summer 2010 ...

    8 34 m 34 78 McC 9 40 m 40 77 McG 10 34 m 34 76 McG 9 46 m 46 75 McM 9 34 m 34 74 McMi 9 46 f 46 73 Mi 10 46 f 46 72 Mi 11 40 m 40 71 Mi 10 34

  2. Artificial Intelligence &amp;amp; Math

    Either the possible threat is dismissed or acted upon defensively by police, leading to a decrease in successful cyber crimes. IT concepts well described and some developments; not enough detail, particularly of developments for explanation, and certainly no analysis. The Impact of the Issue The legislature will increase the number

  1. Investigating ratio of areas and volumes

    and values for Area A and Area B were rounded to 3 d.p.) From the above data the following conjecture can be made: For the graph of y = xn in between the points x = 0 and x = 1 the ratio area A: area B is: n: 1.

  2. SL Math IA: Fishing Rods

    Therefore it was no surprise that this function acts as the best fit for this data. The other cause for this septic function having the best correlation to the original data is due to the septic function being established by creating a system of equations using all of the data points.

  1. Gold Medal heights IB IA- score 15

    point at the height points of data point and the minimum point at the lowest data point. The regression model is limited because the stretch only represents majority of the average data and not the overall image of the correlation.

  2. Lacsap's fractions - IB portfolio

    using the fourth formula: k6= 0.5×62+0.5×6=18+3=21 k7= 0.5×72+0.5×7=24.5+3.5=28 All four formulae had the same results, thus it can be concluded that they are valid. GENERAL STATEMENT FOR THE DENOMINATOR Graph 2: It shows the relation between n (the row number)

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