MATH IA Final of International BACC PDF

Preview

Summary

Shows you how to do a IBDP math IA that gives you good marks...

Description

MATH EXPLORATION Investigating the Volume and Surface Area of The Gherkins Tower Rationale Ever since mankind learnt about the art of engineering shapes and making solid models for roofs to live under, the concept of infrastructure which includes Bridges, Buildings, water dams and many other things have caught the fancy of people and is of paramount importance. My Interest in Buildings and their peculiar shape has always been immense. Wanting to take up the study of buildings and bridges, this topic is of utmost interest and curiosity for me. The structures of Buildings and Bridges I have admired are of great extent like Burj Khalifa, The Shard, and The Golden Gate Bridge. However, if I had to single down to one building it would be the iconic Gherkins Tower in London. Therefore, I decided to undertake a study on the Surface Area and Volume of this iconic tower. The connection of Calculus to this topic magnetized me towards this as it being my favorite topic in Mathematics. This investigation is about the Volume and Curved Surface Area of the famous Gherkins Tower. Personally, I have had a profound interest in buildings and their shape since a very young age. I have always loved the outlandish designs, along with the aerodynamic shape that the building possesses.

Figure 1 Image of the Gherkins Tower

HOW I MODELLED THE DIAGRAM OF THE GHERKINS TOWER Firstly, I took a 3D image of the Gherkins Tower and then pasted it in AutoCAD. The image was in one scale, therefore applying the horizontal X Line (XL) on the base to get the points in one scale. Then to solve the curvature problem at the top of the Gherkins Tower marked by the points

(P15,P17 and P18), I drew tangent at the top and then started the next function in AutoCAD by applying PolyLine(PL) and then mirrored it to the left side as well. Then I researched and found that the height of the tower is 180m and scaled it accordingly which gave me the points shown below.

Figure 2- AutoCAD model of the tower and Figure 3-co-ordinates of the modeled tower

FINDING THE FUNCTIONS The three set of functions were found out using the Lagrange’s Interpolation formula. This formula helps in finding the functions from a set of given points in terms of

x∧ y .

Since I had

points which I found using AutoCAD where I modeled the building, Lagrange’s Interpolation Formula can be used to find out the function. I assumed that these functions were polynomials as Lagrange’s interpolation formula is only applicable for polynomials. Lagrange’s Interpolation formula is illustrated below:-

Figure 4- Lagrange's Interpolation Formula

¿ the aboveformula x i is therespective x co−ordinate∧ y i is the following y co−ordinate of the x i term . From the Image of the Gherkins tower which I modeled, I decided to take 3 points and interpolate them to find the resulting quadratic equation. Although the accuracy of a cubic polynomial would be greater, the polynomial won’t model the Gherkins tower as precisely as taking 3 points does. Therefore, I have decided to chosen to model a quadratic equation rather than cubic.

P1 =

( x − x 1 )( x −x2 ) ( x − x 0 )( x −x2 ) ( x − x 0 )( x − x 1 ) × y2 × y0 + × y 1+ ( x 0−x 1)( x 0 −x2 ) ( x1 −x 0 )( x 1−x 2 ) ( x 2− x 0 )( x 2− x 1)

x 2 + x ( − x2 − x1 ) +x 1 x 2

x 2+ x (−x 2−x 0 ) +x 0 x2

x 2+ x (−x 1−x 0 ) +x 0 x 1

× y 0+ 2 × y 1+ 2 × y2 x 0 2 + x 0( −x2 −x1 ) + x 1 x 2 x 1 + x1 (− x 2− x 0) +x 0 x 2 x 2 + x 2 ( − x 1− x 0) +x 0 x 1 y0 y1 y2 a= 2 + 2 + 2 x 0 + x 0 ( −x2 −x1 )+ x 1 x 2 x 1 + x1 (−x 2−x 0 ) + x 0 x 2 x 2 + x 2 (− x 1− x 0 ) + x 0 x 1 ¿

where a isthe co−efficient of x 2 b=

( − x 2− x 1 ) × y 0 (− x 2−x 0 ) × y 1 (− x 1−x 0 ) × y 2 + 2 + 2 x + x 0 ( −x 2−x 1 )+ x 1 x 2 x 1 + x 1 (−x 2−x 0 ) + x 0 x 2 x 2 + x 2 (− x 1− x 0 ) + x 0 x 1 2 0

where b is the co−efficient of x

x x (¿ ¿ 0 x 1 )× y 2 2 x 2 + x2(−x 1−x 0 )+ x 0 x 1 (¿ ¿ 0 x 2 )× y 1 +¿ 2 x 1 + x 1 (−x 2−x 0 ) + x 0 x 2 (x 1 x 2)× y 0 +¿ c= 2 x 0 +x 0 ( −x 2−x 1 )+x 1 x 2 where c is the co−efficient of constant term

¿(

( x −26.643)( x−28.076 ) ( x−23.077) ( x −28.076 ) ( x−23.077 ×0)+( × 27.665)+( ( 23.077 −26.643)( 23.077 −28.076 ) ( 26.643− 23.077) ( 26.643 −28.076 ) ( 28.076−23.077 2

y 1 ≈ 2.30985 0 x 1 −107.880000 x 1 +1241.160000 (6. d . p)

Coding Since expanding the entire Lagrange’s interpolation formula for finding out the equation is quite tedious, I have thought of writing a Scilab coding which can yield me an accurate result as well as make my process for finding the functions quicker as I just had to input the X-value and its corresponding Y-value and the coding gives the function to 6 d.p.

Middle section: - (Points P9, P11, P13)

i x Y 0 27.410 82.994 1 24.630 110.659 2 19.670 138.323 I then utilized the Coding and found out the equation of the middle section of the model. The 2 equation is:- y 2=−0.565119 x 2 +19.4573 x 2−25.7535

Top section: - (Points P15, P17, P18) i x Y 0 12.406 165.988 1 8.999 172.993 2 0 179.999 I then utilized the Coding and found out the equation of the top section of the model. The equation 2

is:- y 3=−0.102977 x 3 +0.148157 x3 179.999

Using the online desmos graphing calculator, I graphed the above found equation which I will use to evaluate the limits and then find the surface area of the Gherkins Tower. The equations that I found out were approximately accurate, as seen in the above Image the graph depicted by the green line shows it maximum point on (0,179.99) and the total vertical height of The Gherkins is 180 m and hence is nearly accurate 2 y 1=2.30985 x 1 −107.088 x 1 +1241.16 (Bottom section)

y 2=−0.565119 x 2 2 +19.4573 x 2−25.7535 (Middle section) 2

y 3=−0.102977 x 3 +0.148157 x3 179.999 (Top section)

The image on the right shows the image that I used to model the building using AutoCAD, and hence I decided to superimpose the equations on the image. The image below shows that the equations nearly imposed on the 3D image of The Gherkins used for modeling it. The graph can then be represented by a piecewise function:

{

2.30985 x 12 −107.088 x 1 +1241.16 23.069≤ x ≤ 29.572 f ( x ) −0.565119 x 22 +19.4573 x 2−25.753517.215 ≤ x ≤ 28.618 2 −0.102977 x 3 +0.148157 x 3 179.999 0 ≤ x ≤ 29.572

Since I have modeled the tower relative to the y-axis, the equation that I found out is in terms of x and hence I need to make it the subject of the equation. Therefore for that I have used the completing the square method.

Bottom section-

2 y 1=2.30985 x 1 −107.088 x 1 +1241.16

using completing square method ,the equation of x is as follows 107.088 2 ¿ −1241.16 2 √ 2.30985 ¿ 107.088 +( ) 2 √ 2.3095 y 1 +¿ √¿ ¿ ¿

Middle section-

√

(

2 y 2=−0.565119 x 2 +19.4573 x 2−25.7535

19.4573 19.4573 )−25.753− y 2 +( ) 2 √ 0.56119 2 √ 0.56119 =x 2 √ 0.565119

Top section-

2

y 3=−0.102977 x 3 +0.148157 x3 179.999

0.14857 2 ¿ − y3 2 √ 0.102977 ¿ 0.14857 ) +( 2√ 0.102977 179.999+¿ √¿ ¿

Derivation 1 The formula for Surface Area can be derived using the help of the surface area of a frustum. Frustum is just a part of the cone.

Figure 5-Diagram for a frustum of a cone

Surface Areaof the Frustum=2 π ×

Figure 6-Showing the Segment length

( r +2 R )× L=π (r + R ) L 1

2

1

2

Weuse Frustum for the Surface Area approximation as the Surface Area of Frustum very closely x (¿¿ k) ∆y ∆y 2 2 (∆ x ) 2 2 approximates the area swept by the band . k +( ¿¿ k ) (∆ x k ) +( ¿¿ k ) , as L=√ ¿ f ( x k−1 ) +f ¿ √ ¿ Frustum SA=π ¿

Therefore ,the area of the original surfaceis the ∑ of the areas swept bythe bands arcs .

1

Weir, Maurice D., et al. "Areas of Surfaces of Revolution." Thomas' Calculus, 13th ed., Pearson College Division, 2009.

x (¿¿ k) ∆y 2 2 f ( x k−1 ) + f ¿ √(∆ x k ) +( ¿¿ k ) … … .(1) π¿ n

∑¿ k=1

usingthe figure6 yk f ( ck ) = xk thenusing the definitionof differntiation, which isthe rate of change f ( ck )= '

∆ yk ∆ xk

'

f ( c k ) ∆ x k =∆ y k using the above function … ..(2) then substituing ∆ y k =f ' ( c k ) ∆ x k ∈the ( 1) equation x (¿¿ k ) 2 ' f ( c k ) ∆ xk ¿ ¿ (∆ x k )2 +¿ f ( x k−1 ) + f ¿ √¿ π¿ n

¿ ∑ k=1 by taking ∆ x k 2 common x (¿¿ k ) f ( x k−1 ) + f ¿

√( 1+ (f ( c )) ) ∆ x 2

'

k

π¿ n

¿∑ ¿ k=1

2 k

x (¿¿ k ) f ( x k−1 ) + f ¿

√( 1+ (f ( c )) ) ∆ x 2

'

k

k

π¿ n

¿∑ ¿ k=1

x ¿ (¿ k ¿)=f ( x )∧∆ x k =dx f ( x k−1 ) +f ¿ since ¿ b

S . A=∫ 2 πf (x ) √ 1+(f ' ( x ) )2 dx a

Calculating the Surface Area of The Gherkins Tower Using the above derived Surface Area formula, I will now be calculating the Surface Area for the Gherkins tower. b

S . A=∫ 2 πf ( y)√ 1+(f ' ( y ))2 dy a

Figure 6-Graph of the curves and the intersection points which will be used as limits

Since I have divided the Gherkins tower in 3 sections, the top section, middle section and the bottom section, I will have three equations which I have listed above as equations 1, 2, and 3. Thus, I will have total of three surface areas and the total Curved Surface Area (CSA) would be found out by the summation of three CSA separately. The blue curve (equation 2) above is not used for integration; however, it is used in order to find the intersection point with the red curve (equation 1) , which will give one of the limits for integration. The limits of integration are marked as the points above as the y co−ordinates . The limits are evaluated using the graph in order to find the surface area for the required portion of the Gherkins Tower; otherwise the surface area will be evaluated for irrelevant areas under the graph showed above. The limits used won’t be the x co−ordinates

as the tower is modeled relative to the y−axis .

The points marked above are the intersection points which would be used as limits, i.e. the ycoordinates of the points above.

√ y 1+ 0.0288+( 35.2 ) ¿ ¿ ¿

Bottom section: -

,

1 1 dx ) )× =(( 1.52 dy 2 √ y 1+ 0.0288

√ y 1 +0.0288 +(35.2) ((

1 1 )× ) 2 √ y1 +0.0288 1.52 ¿(¿¿ 1.52) √ 1+¿2 dx 2π ¿ 68.3

∫¿ 0

usingGraphic Display Calculator( GDC ) ¿ evaluate Area of Bottom Section :−¿

Bottom Section S . A=11554.9 m 2 ≈ 11600 m 2 (3 s . f .)

Middle Section: - This section is again the integration of the curve x 1 integration with the curve

√ y 1+ 0.0288+( 35.2 ) ¿ ¿ ¿

,

x 2 till the intersection of x 1

1 1 dx =(( ) )× dy 1.52 2 √ y 1+ 0.0288

but the limits will be the

with the curve x 3 .

√ y 1 +0.0288 +(35.2) ((

1 1 )× ) 2 √ y1 +0.0288 1.52 2 ¿(¿¿ 1.52) √ 1+¿ dx 2π ¿ 94.3

∫¿ 68.3

usingGraphic Display Calculator ( GDC ) ¿ evaluate Areaof Middle Section:−¿ Middle Section S . A=4754.53 m

2

≈ 4750 m 2 (3 s . f .)

Top Section: -

√ 180 − y 3+(0.231 ) =x 0.321

2 π (¿

3

√ 180− y 3 +( 0.231 ) 0.321

(

)

1 1 dx × × (−1) ) =( 0.321 dy 2 √ 180− y 3

,

√

) 1+(

(

)

2 1 1 × (−1) ) dx × 2 √ 180− y 3 0.321

180

∫¿ 94.3

u sing Graphic Display Calculator( GDC ) ¿ evaluate Area of Top Section :−¿

Top Section S . A=11166.3m

2

≈ 11200 m 2 (3 s . f .) Total Surface Area of the Gherkins Tower: - The total Surface Area can be found out by summing the Bottom, Middle, and Top Section S.A S . A total=Bottom S . A + Middle S . A + Top S . A ¿ 11600 + 4750 + 11200

≈ 27600 m2 (3. s . f ) Finding the Volume of the Gherkins Tower:Similarly like Surface Area, the volume would be found out for three sections and the total volume would be the sum of the three volumes. b

Volume formula is given by-

∫ π (f ( y ))2 dy a

Bottom section: -

√ y 1 +0.0288 +(35.2) =x 1.52

1

68.3

∫

π(

0

√ y 1 +0.0288 +(35.2) )2 1.52

dy

Bottom Section Volume=154291.0 m 3 ≈ 154000 m3 (3. s . f ) Middle Section:94.3

∫

π(

√ y 1 +0.0288 +(35.2) =x 1.52

√ y 1 +0.0288 +(35.2 ) )2 1.52

68.3

1

dy

Middle SectionVolume=69101.6 m 3 ≈ 69100 m3 (3. s . f ) Top Section:180

∫ 94.3

π(

√ 180 − y 3+(0.231 ) =x 0.321

√ 180 − y 3 +( 0.231) 0.321

3

2

) dy

Top SectionVolume=119551.7 m

3

≈ 120000 m3 (3. s . f ) Total Volume: - The total volume is found out using the summation of volume of all the sections. Volumetotal=Bottom V +MiddleV +Top V ¿ 154000 +69100 + 120000

≈ 343100 m3 (3. s . f )

Limitations and Improvements Firstly, the biggest limitation is in the curves as it is not a smooth curve like the exterior of the Gherkins; however, I have assumed the same for the curves that I have obtained. This could be resolved by using better graphing technology which can accurately model the curves as well as the Gherkins Tower. Moreover, I have used Scilab for coding which is giving me the answers to 6 decimal points even though the equation continues to have value beyond it. Admittedly, a graphing calculator or the mathematics tools available could’ve been used, but due to the limitations of the equipments at my disposal the next best alternative seemed the coding. Moreover, during the modeling of the Gherkins Tower in AutoCAD, I had chosen a 3D picture of Gherkins which took only the Gherkins in the frame. However, in reality the angle of the placement of the image plays a

major role in modeling of the shape using AutoCAD, therefore to avoid this I had to place it exactly at the origin.

There is some space that is left between the curves; however, I have taken it to be as smooth curves.

Conclusion Overall, this investigation was a very enriching experience for me personally as I was able to use mathematics in an area which I want to take up in the future. This exploration has opened up chances for me to model different types of buildings of different shapes and similarly carry out different needs. This exploration could be extended by me in the future because I will be able to use this investigation to perhaps derive answers to critical questions about the quantity of glass panes that are needed by the building’s exterior which may affect the Surface Area consequently. This would also give rise to new questions that I would be introduced to like why certain buildings are aerodynamic, what makes them special in consideration to other conventional forms of buildings. All in all I am very grateful to apply mathematics (Calculus) to the field that I am interested in. Also I am happy that I am able to make my individual work worthy of addressing issues for the construction industry as they would be able to minimize the cost and optimize the use of materials.

BIBLIOGRAPHY Weir, Maurice D., et al. "Areas of Surfaces of Revolution." Thomas' Calculus, 13th ed., Pearson College Division, 2009....