• 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
  15. 15
    15
  16. 16
    16
  17. 17
    17
  18. 18
    18
  19. 19
    19
  20. 20
    20
  21. 21
    21
  22. 22
    22
  23. 23
    23
  24. 24
    24
  25. 25
    25

Investigating ratio of areas and volumes

Extracts from this document...

Introduction

Investigating Ratios of Areas and Volumes

Michael Zuber


Introduction:

This paper has been separated into three parts:

Part i investigates the ratio of the areas formed when y = xnis graphed between two arbitrary parameters x = a and x=b such that a<b.

Part ii investigates the ratio of the volumes formed around the x-axis when y = xn is graphed between two arbitrary parameters x = a and x=b such that a<b.

Part iii investigates the ratio of the volumes formed around the y-axis when y = xn is graphed between two arbitrary parameters x = a and x=b such that a<b.

To investigate these ratios the program Autograph 3.20 was used in order to obtain sets of results. All areas were found using Simpson’s rule at 50 divisions.


Part i:

Introduction:

In this investigation the ratio of the areas formed when y = xnis graphed between two arbitrary parameters x = a and x=b such that a<b will be investigated.

image00.png

This investigation will investigate the ratio area A: area B.

Area A is the area contained in between the graph and the y-axis between the arbitrary parameters x = a and x = b.

Area B is the area contained in between the graph and the x-axis between the arbitrary parameters x = a and x = b.

However, in investigating this there is a small problem. Using modern technology the area between the graph and the x-axis can easily be found. However, not all programs allow for the area in between the graph and the y-axis to be found.

This problem, however, can be resolved. Area A can be expressed in another form. Considering the inverse function of y = xn,  y = x1/n is found. Considering the graph of y = x1/nwe the following is obtained:


image01.png

...read more.

Middle

y = xn and the x-axis.

However, to find Volume A using modern technology is tricky.

Volume A can be expressed in another manner. Consider the horizontal line that meets the graph at the point x = b (y = bn):


image02.png

Volume A can be considered as the volume contained around the x-axis by the graph of this horizontal line (y = bn)between the points x = 0 and x = b subtracted by volume B and the volume contained around the x-axis by the graph of y = an between the points x = 0 ans x = a. This last volume will be considered as volume C.

This method will be used to find the ratio Volume A: Volume B.

Investigation:

y = x2between x = 0and x = 1:

Consider the graph of y = x2 between the points x = 0 and x = 1.

Volume B will equal the volume contained around the x-axis between the graph and the x-axis between the points x = 0 and x = 1.

Volume A will equal the volume contained around the x-axis between the graph of  y = 12 (= 1) and the x-axis between the points x = 0 and x = 1 subtracted by volume B. (Note: since in this case a = 0 volume C = 0).

Using the program Autograph 3.20 the following results are obtained:

Volume B = 0.2π

Volume under y = 1 = 1π

Volume A = 1 - 0.2π = 0.8π

Ratio volume B: volume A = 0.8π: 0.2π

= 4: 1

y = xnbetween x = 0and x = 1:

Now consider other functions of the type y = xn, n image03.png Z+ between the points x = 0 and x = 1.

Volume B will be the volume contained around the x-axis between the graph and the x-axis between the points x = 0 and x = 1.

Volume A will be the volume contained between the line y = 1n (= 1) and the x-axis between the points x = 0 and x = 1 subtracted by volume B (volume C = 0).

...read more.

Conclusion

y = 2 and the x-axis between the points x = 0n (= 0) and x = 2n subtracted by volume A (a = 0 therefore volume C = 0).

n

Area under horizontal line (in π)

Volume A (in π)

Volume B (in π)

Volume A / Volume B

2

16

8.00

8.00

1.0

3

32

19.20

12.80

1.5

4

64

42.65

21.35

2.0

5

128

91.35

36.65

2.5

6

256

191.70

64.30

3.0

7

512

397.50

114.50

3.5

8

1024

817.40

206.60

4.0

9

2048

1671.00

377.00

4.4

10

4096

3404.00

692.00

4.9

π

35.3

21.56

13.74

1.57

e

26.32

15.16

11.16

1.36

From these results it is evident that the conjecture holds true between the points x = 0 and x = 2.

y = xnbetween the points x = 1 and x = 2:

Volume A will equal the volume contained around the x-axis between the graph of y = x1/n and the x-axis between the points x = 1n (= 1)and x = 2n.

Volume B will equal the volume contained around the x-axis between the graph of y = 2 and the x-axis between the points x = 0 and x = 2n subtracted by volume A and volume C.

Volume C will equal the volume contained around the x-axis between the graph of y = 1 and the x-axis between the points x = 0 and x = 1n (=1).


n

Area under horizontal line (in π)

Volume A (in π)

Volume B (in π)

Volume C (in π)

Volume A / Volume B

2

16

7.5

7.5

1

1.0

3

32

18.6

12.4

1

1.5

4

64

42.66

20.34

1

2.1

5

128

91.41

35.59

1

2.6

6

256

191.9

63.1

1

3.0

7

512

398

113

1

3.5

8

1024

818.6

204.4

1

4.0

9

2048

1674

373

1

4.5

10

4096

3410

685

1

5.0

π

35.3

21.56

12.74

1

1.69

e

26.32

15.17

10.15

1

1.49

From these results it is evident that the conjecture holds true between the points x = 1 and x = 2.

y = xnbetween the points x = e and x = π:

Volume A will equal the volume contained around the x-axis between the graph of y = x1/n and the x-axis between the points x = enand x = πn.

Volume B will equal the volume contained around the x-axis between the graph of y = π and the x-axis between the points x = 0 and x = πn subtracted by volume A and volume C.

Volume C will equal the volume contained around the x-axis between the graph of y = e and the x-axis between the points x = 0 and x = en.


n

Area under horizontal line (in π)

Volume A (in π)

Volume B (in π)

Volume C (in π)

Volume A / Volume B

2

97.41

21.41

21.40

54.6

1.0

3

306

94.56

63.04

148.4

1.5

4

961.3

372.00

185.93

403.4

2.0

5

3020

1374.00

549.56

1097

2.5

6

9488

4881.00

1626.23

2981

3.0

7

29807

16880.00

4824.84

8102

3.5

8

93642

57300.00

14317.58

22024

4.0

9

294185

191700.00

42616.50

59869

4.5

10

924210

634600.00

126870.13

162740

5.0

π

359.9

115.40

73.50

171

1.57

e

221.7

63.19

46.51

112

1.36

From these results it is evident that the conjecture holds true between the points x = e and x = π.

Proving the conjecture:

Using calculus the conjecture can be proven.

Volume A:

image07.png


Volume B:

image08.png


Ratio volume A: volume B:

image09.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. The Fibonacci numbers and the golden ratio

    I plugged in the discriminant to the quadratic equation for finding the roots. The results for the positive root is, 1.618 and the negative root is, -0.618. This proves my conjecture. Look below to see my calculations and explanations. I divided both sides of this equation Un = Un-2 + Un-1, by Un-1.

  2. Math IA type 2. In this task I will be investigating Probabilities and investigating ...

    find the probability of Adam winning the deuce and add that value to the probability of Adam winning the non- deuce games to subsequently find the odds of him winning which is my eventual goal. Therefore the expression for the probability of Adam winning the non-deuce games is: Therefore after

  1. Mathematics (EE): Alhazen's Problem

    found by looking at where the corresponding chords are equal to one another, in other words where chord a - chord b = 0. By making a table showing chord a - chord b we could perhaps find possible solutions.

  2. Math IA Type 1 In this task I will investigate the patterns in the ...

    x3 and x5 are intersections of one linear function and the cubic function and x1, x4 and x6 are intersections of another linear function and the cubic function. Therefore the sum of the roots of the intersection of the first linear function and the cubic function would be x2+x3+x5= And

  1. Salida del sol en NY

    desea saber a qu� hora amanecer� el d�a de ma�ana, tendr� que resolver o encontrar la hora a partir de la aplicaci�n por separado de ambos modelos y despu�s sacar una hora promedio de cuando saldr�a el sol, sin embargo ser�a conveniente que se fijara en el d�a anterior y

  2. Investigating ratios of areas and volumes

    = 2 As we can see, the area under the curve rotates around the x-axis. Now, using the same procedure I will find the volume of revolution of within the same parameters. = As we can observe, comparing this graph with the graph of , the volume gets more narrow.

  1. Investigating Ratios of Areas and Volumes

    - 0 = 1.9048... - 1.9038... = 1.2699... 1.2699.../ 1.9038... = 2/3 Area of A and B is now defined as the following: Area A: y = xn, y = an, y = bn, and the y-axis Area B: y =x, x = a, y = b, and the x-axis

  2. The investigation given asks for the attempt in finding a rule which allows us ...

    only uses one side that is not shared; For example, only the lengths of 3 and 4 are by itself, while the values after and before are used again by the next trapezoids. Hence, from the expression above, all lengths beside the first and last are used twice, therefore, taking

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