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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.

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