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Mathematics Exploration 2013

Mohd Nasiruddin B Md Zaini E12B Page 1

REVOLUTION OF SOLID

While attending a family day during the semester break, suddenly there was an upcoming guessing

the number of screws in a container contest and the gift quite worth it so I started to get the

information on how to estimate it via internet. But the problem is now the volume of the container

and the screws is quite difficult to be measure as it doesnt have definite shape like cylinder, square or

sphere and also it is not transparent where you cant count the number of screws in one row around

the jar and multiply it with tall of the screws in one line (how to win a jellybean contest, 2007/03/07).

furthermore, it has cylindrical shape part at the bottom which actually will make people confuse to

estimate as its cannot be filled with screws and only act as the base. Fortunately, the container is

symmetrical and it base and top is in circular shape. So, here one idea popped in my mind which is by

using the revolution of the solid(buku dlm dorm) where I can get the approximate volume of any

complex shape container and the screw only by knowing the cross sectional area of the container.

Thus I can get the number of screws that can be filled by dividing the volume of the container with

the volume of screw.The container was just like this.

And the screws


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The plan for the container and the screws ( mathematical modeling on solid of revolution/ SJSU)

1) container

Section 1

y = 0.013x3 - 0.196x2 + 0.317x + 6.346

by curve fitting

10cm

12.9cm

6.6cm

2cm

Section 2

a) Sections of a screws container (side view) b) Exterior profile of a screws container

0.2cm

c) Thickness of the container (top view)


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Section 2

0.36cm

0.38cm

0.72 cm

1.7cm

b) Interior profile of screw

section 1 section 3

a)Sections of screw

Curve fitting

Now I have scanned both the container and the screw in attempt to model them using Geogebra to fit

the equation of the upper part of the containers . In order to model those two items I have marked

every point on the upper side and compute it with x- coordinates and y-coordinates.


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a) Container (bowmandickson,2007)Volume of Section 1 (Outer diameter)

alphabet x-coordinates y-coordinates

A 0 6.37

B 1.01 6.41

C 2 6.32

D 3.02 5.98

E 4.01 5.25

F 4.99 4.65

G 6 4.00

H 7 3.5

I 7.99 2.94

J 9.01 2.94

K 10 2.94

Table 1: Table of x and y coordinates of the curve fitting for outer diameter


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y = 0.0131x3 - 0.1961x2 + 0.3178x + 6.1464

0

1

2

3

4

5

6

7

0 2 4 6 8 10

Table 2 : The plotted graph of x and y coordinates for the outer diameter curve fitting

Equation of curve: f(x) = 0.013x3 - 0.196x2 + 0.317x + 6.346

The equation for the curve is polynomial which is a cubic function and its been choosed as it fit to

the outer diameter of the container the best.

Since the thickness of the container is about 0.2 cm. so we need to consider the actual volume that

can be filled with screws which is by subtracting the value of y-coordinates by the thickness of the

container which is 0.2cm. Hence a new point of y-coordinates is plotted and the equation is obtained.

New coordinates (inner diameter)

Table 2: Table of x and y coordinates of new curve Table 3 : The plotted graph of x and y

fitting of inner diameter of the containers after coordinates for the inner diameter of

considering the thickness of the container curve fitting by using Microsoft excel

alphabet x-coordinates

(y-coordinates

0.2 cm)

A 0 6.17

B 1.01 6.21

C 2 6.12

D 3.02 5.78

E 4.01 5.05

F 4.99 4.45

G 6 3.80

H 7 3.30I 7.99 2.74

J 9.01 2.74

K 10 2.74

y = 0.0131x3 - 0.1961x2 + 0.3178x + 6.3464

0

1

2

3

4

5

6

7

0 2 4 6 8 10 12

y

x

Curve Fitting


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Equation of new curve f(x)= 0.013x3

- 0.196x2

+ 0.317x + 6.146 (cubic function)

From the equation of the new curve, we can get the area under the curve by using the integration.

Next, from the same equation we can get the volume of the curve by revolving the equation over the

x-axis using integration method or using mathematical software like autograph to revolve it moreclearly.

Formula of definite integral for finding volume (wiley plus)

V = )2

dx = )2

dx

Volume section 1 : V1 = )2 dx= (0.013x30.196x2 + 0.317x + 6.146)2 dx= 217.1

= 682.04 cm3

Using autograph to revolve it:


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Volume of section 2

V2 = 2h (Area of cylinder)= (

)2 (2)

= 19.22

= 60.3814 cm3

As we can see earlier that the base of the container at section 2 cannot be filled by the screws, so

here we need to subtract the volume of full container with the volume of section 2 which is in

cylindrical shape.

So, the net volume of the container is

Vnet = V1V2

= 682.04 - 60.3814

= 621.659 cm3

a)

Screw (bowmandickson,2007)

y = -0.048x2

- 0.645x + 0.364(section 1)

y = -1.010x2 + 4.185x - 4.202 (section 3)

y= 0.13(section 2)

- Curve fitting for screw


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ALPHABETS X Y

A 0 0.36

B 0.1 0.31

C 0.2 0.23

D 0.27 0.18

E 0.36 0.13F 0.45 0.13

G 0.6 0.13

H 0.74 0.13

I 0.87 0.13

J 0.99 0.13

K 1.12 0.13

L 1.27 0.13

M 1.39 0.13

N 1.5 0.13

O 1.66 0.13

P 1.79 0.13Q 1.93 0.13

R 2.06 0.13

S 2.19 0.12

T 2.27 0.1

U 2.35 0.44

V 2.44 0

Table 4 : Table of x and y coordinates of curve fitting for the screw

Table 5 : The plotted graph of x and y coordinates of curve fitting for section 1 region of screw

y = -0.0482x2 - 0.645x + 0.3644

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4


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From the points that had been fitted to the screw, there three different function for every section.

These function thus is combined to be a piecewise function.

f(x)= { ,

y = 1E-16x + 0.13

0

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

0.18

0.2

0 0.5 1 1.5 2 2.5

y = -1.0102x2 + 4.1852x - 4.2029

-0.05

0

0.05

0.1

0.15

0.2

2 2.1 2.2 2.3 2.4 2.5


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These function is rotate about the x-axis to find the total volume enclosed by the function to get the

volume of the screw. Im using the autograph software (at the right corner for every volume

calculation) to show briefly about the revolving of the function part by part until whole volume of

screw is obtained.

Volume section 1, V1

= )2 dx=

(-0.048x

20.645x + 0.364)

2dx

= ( 0.002304 x4 + 0.06192x3 + 0.0381081x20.46956x + 0.132496) dx= [ [ ( (0.36)5 + (0.36)4 + (0.36)3 (0.36)2 + (0.132496)(0.36) ]

(0) ] ]

= 0.02346 = 0.090823cm

3

Volume of section 2

V2 = r2 h (volume of cylinder)= (0.13)2 (1.7)= 0.02873 cm3= 0.090258cm3


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Volume of section 3

V3 =

)2 dx

= ( -1.010x2 + 4.185x - 4.202)dx= ( 1.0201 x48.4537x3 + 26.0022x235.1708x + 17.6568) dx= [ [ ( ) (2.44)5 (2.44)4 + (2.44)3 (2.44)2 + (17.6568)(2.44) ]

[ (

(2.06)5 -

(2.06)4 +

(2.06)3

(2.06)2 + 17.6568(2.06) ] ]

= 0.003646

= 0.011454cm3

Vtotal = V1 + V2 + V3

= 0.02346 + 0.02873 + 0.003646 = 0.00432 cm3= 0.013572 cm3


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The approximation of screws in the container

Eventhough the two volume had been obtained, but somehow I did found out that there is some

distance (air space) between the screws when it is filled in the container as the result of uneven

surfaces of the screws and the container itself. So, here I would make an approximation that 20% of

the volume of the container was filled by the air space as the result of the gaps between screws. (howto win a jellybean contest, 2007/03/07).

Real volume of container , Vreal

= Vnet - (Vnet x

)

= 621.659(621.659 x

)= 497.327 cm3

So, the total number of screws that can be filled is, n

1) Using simple divisionn =

=

= 36643.6

36644

2) Using arithmetic progression (Cirrito Fabio, Buckle & Dunbar, July 2004)Tn = a + (n-1) d , a= 0.013572cm

3, d= 0.013572cm

3,

497.327 = 0.013572 + (n-1)(0.013572)

Tn = 531.507 cm3

(real volume of container)

36642.6 = n - 1

n = 36643.6

36644


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So, the number of screws that can be filled in the container is about 36644!! Which quite a large

number and almost impossible to be guessed. But using this method, maybe I can be the first winner

for this game if im fast enough to do this calculation.

Limitation

However, note that there will be a slight limitation when calculating those volumes as the metre ruler

itself got its own systematic error which the calibration is not too accurate when measuring the side

length of those objects. Next, as I had mentioned earlier, there will be a slightly air space between the

screws due to uneven surface between them when placed in the container. So, there will be a slight

flaws there and 20% of the volume might be the volume of air space which is the total gaps between

the screws just a rough estimation.

Conclusion and real life application for the revolution of solid

The revolution of solid use the concept of calculus of which is integration which is used to finding the

area enclosed by a substance or volume of the substance from a known function. Nowadays, it had

been used in many area of engineering and heavy industry to model their product before casting it

into a solid object. For instance the manufacturing of bottle. Lately, there were various design of

bottle from the bottle producer as one of the factor to attract people to buy their product. But when

designing the complex shape of bottle, they also need to take a highlight on the minimum volume of

water that the new shape bottle can be filled. So there will revolved the function of the bottle first to

determined it. Lastly, when exploring the ways on how to revolve the solid object, I also had found

out that there is another complicated method to revolve it by using washer method which is very

significant in manufacturing engineering to produce part of gears and complicated mechanical

component. From that moment, I realize that almost every single part of the mechanical component is

tested first either it volumes or functionary by using computer software before it been cast into a solid

object.


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Bibliography

Source

http://howtutorial.com/howto/how-to-win-a-guess-the-number-of-jelly-beans-in-a-jar-contest/

http://www.wyzant.com/help/math/calculus/integration/finding_volume

http://www.engr.sjsu.edu/trhsu/Chapter%202%20Math%20Modeing.pdf

http://diggy.wordpress.com/2007/03/07/how-to-win-a-jellybean-counting-contest/

http://bowmandickson.com/2012/06/04/calculus-final-project-spotlight-3d-solid-modeling/

books
http://howtutorial.com/howto/how-to-win-a-guess-the-number-of-jelly-beans-in-a-jar-contest/http://howtutorial.com/howto/how-to-win-a-guess-the-number-of-jelly-beans-in-a-jar-contest/http://www.wyzant.com/help/math/calculus/integration/finding_volumehttp://www.wyzant.com/help/math/calculus/integration/finding_volumehttp://www.engr.sjsu.edu/trhsu/Chapter%202%20Math%20Modeing.pdfhttp://www.engr.sjsu.edu/trhsu/Chapter%202%20Math%20Modeing.pdfhttp://diggy.wordpress.com/2007/03/07/how-to-win-a-jellybean-counting-contest/http://diggy.wordpress.com/2007/03/07/how-to-win-a-jellybean-counting-contest/http://bowmandickson.com/2012/06/04/calculus-final-project-spotlight-3d-solid-modeling/http://bowmandickson.com/2012/06/04/calculus-final-project-spotlight-3d-solid-modeling/http://bowmandickson.com/2012/06/04/calculus-final-project-spotlight-3d-solid-modeling/http://diggy.wordpress.com/2007/03/07/how-to-win-a-jellybean-counting-contest/http://www.engr.sjsu.edu/trhsu/Chapter%202%20Math%20Modeing.pdfhttp://www.wyzant.com/help/math/calculus/integration/finding_volumehttp://howtutorial.com/howto/how-to-win-a-guess-the-number-of-jelly-beans-in-a-jar-contest/


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MATHEMATICS HL EXPLORATION

DECLARATION OF AUTHENTICITY

Name: MOHD NASIRUDDIN B MD ZAINI

Class : E12B

Title: Guessing the number of screws in a container by using revolution of solid

With this I declare that this work is my own.

.

For examiner use only

Criterion Achievement level

Criterion A: Communication assesses the organization and

coherence of the exploration0 1 2 3 4

Criterion B: Mathematical presentation assesses to what

extent the student is able to use appropriate math language

and multiple forms of math representation including the use

of appropriate ICT tools

0 1 2 3

Criterion C: Personal engagement assesses the extent to

which the student engages with the exploration and makes it

his or her own

0 1 2 3 4

Criterion D: Reflection assesses how the student reviews,

analyses and evaluates the exploration

0 1 2 3

Criterion E: Use of Mathematics assesses to what extent

and how well students use mathematics in the exploration0 1 2 3 4 5 6

TOTAL MARKS / 20


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