Presentation

Mathematical communication

Personal engagement

Reflection

Use of mathematics

Plastic pollution is a global problem that has been growing very rapidly as plastic production rates are intensifying and the amount of plastic discarded increases. Much of the plastic discarded ends up in the ocean. After watching the documentary series Seaspiracy on Netflix I have started to wonder how plastic pollution of the ocean can be minimized and how I personally contribute to this global problem. One of the ways to reduce the use of plastic is to minimize its use of it during manufacturing. The more efficient way would be to use other materials, for example, cardboard or glass, however, hard plastic is durable and cheap to manufacture, hence, there is much doubt that manufacture of it will be cut down any time soon. However, plastic can be used more effectively, for example, optimally shaped bottles for beverages can be produced to minimize the amount of plastic used for the given volume, which would not only be beneficial for the producers, as the costs of manufacture will be reduced because the less raw material is used, but it will also mean that overall amount of discarded plastic will be reduced this way decreasing plastic pollution.

image1.jpg

The most often used plastic attribute in our household is plastic water bottles, specifically, ‘Akvilė’ bottled water, hence, I have decided to calculate this water bottle’s surface area and compare it to an optimal cylinder of the same volume to investigate how effective is the shape of this plastic water bottle in using as little plastic as possible for the given volume.

First of all, in order to calculate the surface area of the plastic bottle seen in Figure 1 I had to measure out its dimensions. I did it using a ruler. The height of the bottle is 21.0 cm ± 0.05. The diameter of the bottom of the bottle is 5.5 cm ± 0.05. The volume of the plastic bottle is known as well – 500 cm3. Then, I had to determine which functions would describe my water bottle best.

image2.png

After that, I uploaded the picture to GeoGebra5. To maintain the proportions of the water bottle I graphed functions of y = 2.75 and y = -2.75, as well as, x = 21, as it can be seen in Figure 3. This way I could place the water bottle onto the Cartesian coordinate system while keeping the same proportions. The bottom half of the water bottle (x &lt; 0) was excluded from the calculations, as it only appears because of the photo angle and the distortion it causes. The inclusion of this part would generate a larger surface area and volume than in reality.

image3.png

Then, I placed points along the contour of the water bottle. I assume that y = f1(x) and y = f​​​​​​​5(x) are parabolas, y = f2(x), y = f4(x), y = f6(x) linear functions and y = f3(x) a cubic function. The points placed can be seen in Figure 4:

image4.2351ROTATE GRAPH 2

The coordinates for each point as well as the functions to which they belong can be seen in Table 1 below. Each coordinate is approximated to 4 d.p.

image5.jpg

Then, I can find the equations of the functions, as I have the coordinates of points that belong to the functions.\(y= f_1(x)function\) It is assumed that y = f1(x) is a parabola. The general formula for quadratic functions is as follows: \(y = ax^2 + bx + c, a ≠ 0\)​​​​​​​ It can be seen that in this equation there are 3 unknown values (a, b, and c) and it can be solved using a system of equations with the coordinates of the three points which belong to this function. This was my initial idea:\(\left\{2.7543=0^z\bullet a+0\bullet b+c\right.\\\ 3.1225=1.7235^2a+1.7235b+c\\3.0314=3.6507^2a+3.6507b+c\) However, after doing some research I have found that I can calculate my coefficients using mathematical matrices, which are sets of numbers arranged in rows and columns to form a rectangular array. Since matrices are no longer a part of the IB Mathematics Analysis &amp; Approaches SL curriculum I had to learn how to use them in order to apply this knowledge in my IA. The three points A (0; 2.7543), B (1.7235; 3.1225), and C (3.6507; 3.0314) can be used to arrange three matrices. In the first matrix we put the values of x (A matrix), in the second we put the coefficients a, b, and c (B matrix) and we get the values of y (C matrix). When multiplying matrices, every element of the first row in the first matrix is multiplied with every element of each column in the second matrix8 (in this case there is one column in the second matrix). Because the c is a constant, meaning that the value of c will not change when the value of x is changed, in the first matrix’s last column 1 is written. This more clearly can be seen below. \([x_1^2\ x_1\ 1] \\\ [x_2^2\ x_2\ 1]\\\ [x_3^2\ x_3\ 1]\)\(\atop\left[{a\\b\\c}\right]\)=\(\atop\left[{y_1\\y_2\\y_3}\right]\) This can more simply be written as: \([A][B]=[C]\) As I am trying to solve for matrix B and have the values for matrices A and C, I need to multiply the equation by the inverse of matrix B. Anything multiplied by its inverse will be equal to one. The work can be seen below: \([A]^{-1}[A][B]=[C][A]^{-1}\rightarrow1[B]=[C][A]^{-1}\rightarrow[B]=[C][A]^{-1}\) Using this, values of matrix B can easily be calculated. For further steps seen below, I used Graphing Digital Calculator (GDC). At first I edited the matrices using matrix → edit and then multiplied the matrices together to get the values of matrix B and the equation of the function y = f1(x). Elements of the matrix B are given approximately 4 d.p \(\left[\begin{matrix}0^2\\1.7235^2\\3.6507^2\end{matrix}\begin{matrix}0\\1.7235\\3.6507\end{matrix}\begin{matrix}1\\1\\1\end{matrix}\right]^{-1}\)\(\atop\left[{2.7543\\3.0314\\3.0314}\right]\)=\(\atop\left[{-0.0715\\0.0337\\2.7543}\right]\)→​​​​​​​ insert formula \(\rightarrow y=f_1(x)=-0.0715x^2+0.3368x+2.7543,x\in[0;3.6507]\) Then, I graphed the function to see if it fits the contour as can be seen in Figure 5:

image6.8457ROTATE BOTTLE GRAPH 3

As it can be seen from Figure 5, the function fits the contour of the water bottle very well, hence, I can calculate the equation of another function \(y=f_2(x)function\) I assumed that function y = f2(x) is linear. The general equation for linear functions is as follows: \(y=ax+b\)​​​​​​​ At first, I inserted the coordinates of point C (3.6507; 3.0314) into the equation to find the expression of b: \(3.0314=3.6407a+b\) 𝑏 = 3.0314 − 3.6507𝑎 Then, I inserted coordinates of another point (E (7.1047; 2.3442)) which belongs to the function y = f2(x) into the equation with the substitute for b. Value for a is given approximate to 5 d.p. and value for b is given approximately to 4 d.p.: \(2.3442=7.1047a+3.0314-3.6507a\) \(3.4540a=-0.6872\) \(a=\frac{-0.6872}{3.4540}=-0.19896\) \(b=3.0314-3.6507\bullet(-0.19896)=3.7578\) Now the equation for y = f2(x) can be written down: \(y=f_2(x)=-0.19896x+3.7578,x\in[3.6507;7.1047]\) Function y = f2(x) is then graphed to see how well it fits, as can be seen in Figure 6:

image7.png

\(y=f_3(x)function\) I assumed that y = f3(x) is a cubic function, because of its specific shape. The general formula for a cubic equation is given below: ​​​​​​​​​​​​​​\(y=ax^3+bx^2+cx+d,a\neq0\) In this case, there will be 4 unknown values a, b, c and d. Nevertheless, the equation for this function can also be calculated with the matrix method used earlier to calculate y = f1(x) on page 5. It will be slightly modified because 4 points (E (7.1047; 2.3442), F (8.0814; 2.6197), G (9.0000; 3.0000), H (9.8352; 2.9915)) are needed to find 4 unknown coefficients. The matrices will look as follows: //\(\left[\begin{matrix}x^3_1\\x^3_2\\x^3_3\\x^3_4\end{matrix}\begin{matrix}x^2_1\\x^2_2\\x^2_3\\x^2_4\end{matrix}\begin{matrix}x{_1}\\x_2\\x_3\\x_4\end{matrix}\begin{matrix}1\\1\\1\\1\end{matrix}\right]^{-1}\)\(\left[\begin{matrix}y_1\\y_2\\y_3\\y_4\end{matrix}\right]\)​​​​​​​​​​​​​​=\(\left[\begin{matrix}a\\b\\c\\d\end{matrix}\right]\) I insert x and y values of points E (7.1047; 2.3442), F (8.0814; 2.6197), G (9.0000; 3.0000), H (9.8352; 2.9915) to the formula given above and this way unknown values of coefficients can be calculated. For the steps seen below, I again am using GDC to edit the matrices (matrix → edit) and then multiply the inverse of the first matrix by the second matrix. a, b, c, and d values are given approximately to 4 d.p.: \(\left[\begin{matrix}7.1047^3\\8.0814^3\\9.0000^3\\9.8352^3\end{matrix}\begin{matrix}7.1047^2\\8.0814^2\\9.0000^2\\9.8352^2\end{matrix}\begin{matrix}7.1047\\8.0814\\9.0000\\9.8352\end{matrix}\begin{matrix}1\\1\\1\\1\end{matrix}\right]^{-1}\)\(\left[\begin{matrix}2.3442\\2.6197\\3.0000\\2.9915\end{matrix}\right]\)=\(\left[\begin{matrix}-0.1140\\2.8274\\-22.9048\\63.2513\end{matrix}\right] \)→​​​​​​​ \(\rightarrow y=h(x)=-0.1140x^3+2.8274x^2-22.9048x+63.2513,x\in[7.1047;9.8352\) Function y = f3(x) was then graphed to see how well it fits the graph. It can be seen in Figure 7 below:

image8.9478water bottle rotate 4

\(y=f_4(x)function\) The function y = f4(x) is assumed to be linear. The calculations for this functions are the same as for y = f2(x) function on page 6, but using points H (9.8352; 2.9915) and J (14.2968; 3.1037). Values for a and b are given approximately to 4 d.p.: \(2.9915=9.8352a+b\rightarrow b=2.9915-9.8352a\) \(3.1037=14.2968a+2.9915-9.8352a\rightarrow 4.4616a=0.1122\rightarrow a=\frac{0.1122}{4.4616}=0.0251\) \(b=2.9915-9.8352\bullet0.0251=2.7446\) \(y=f_4=0.0251x+2.7446,x\in[9.8352;2.9915]\) The graphed function is seen in Figure 8 below:

image9.4907FIG 8 ROTATE BOTTLE

y = f5(x) function As y = f5(x) function As y = f5(x) is assumed to be a parabola just as y = f1(x), the equation of this function is calculated in the same way as on page 5 using matrices with points J (14.2968; 3.1037), K (16.7392; 2.6180) and L (18.1964; 1.2700). a, b, and c values are given approximately to 4 d.p. \(\left[\begin{matrix}14.2968^2\\16.7392^2\\18.1964^2\end{matrix}\begin{matrix}14.2968\\16.7392\\18.1964\end{matrix}\begin{matrix}1\\1\\1\end{matrix}\right]^{-1}\)\(\atop\left[{3.1037\\2.6180\\1.2700}\right]\) = \(\atop\left[{-0.1862\\5.5808\\-38.6198}\right]\)→ \(\rightarrow y=f_5(x)=-0.1862x^2+5.5808x-38.6198,x\in [14.2968;18.1964]\) Graphed function is seen in Figure 9 below:

image10.2707water bottle fig 9

\(y=f_6(x)function\) The last function is linear, hence, the equation is calculated using the same method as for y = f​​​​​​​2(x) on page 6, but with the points L (18.1964; 1.2700) and M (18.5627; 1.2592). Values for a and b are given approximately to 4 d.p.: \(1.2700=18.1964a+b\rightarrow b=1.2700-18.1964a\) \(1.2592=18.5627a+1.2700-18.1964a\rightarrow0.3663a=-0.0108\rightarrow a=\frac{-0.0108}{0.3663}=-0.0295\) b = 1.2700 - 18.1964 - (-0.0295) = 1.8068 \(y=f_6(x)=-0.0295x+1.8068,x\in[18.1964;1.2592]\) Graphed function can be seen in Figure 10 below:

image11.3443figure 10 rotate bottle

Now that I have all of my functions to describe the shape of my bottle I can graph them to see if they fit the bottle well and all go through the points. This can be seen below in Figure 11:

image12.png

From Figure 11 it can be seen that all of the functions fit near perfectly to the contour of the water bottle and at this point, I can use the equations of the functions to find the surface area of my water bottle. The general formula9 to find the surface area is given below: \(A=2\pi\int_{a}^{b}{y\sqrt{1+\left(y'\right)^2}dx}\) This formula gives the surface area of a body rotated about the x-axis bound by 𝑎≤𝑥≤𝑏, hence, it is suitable to find the surface area of my water bottle. I placed my function equations into the formula to calculate the surface area using GDC (math → fnInt). All of the values were given approximate to 2 d.p.: ​​​​​​​\(A_1=2\pi\int^{3.6507}_0((-0.0715x^2+0.3368x+2.7543)\sqrt{1+((-0.0715x^2+0.3368x+2.7543)')^2})dx=\\70.94 cm^2\) \(A_2=2\pi\int^{7.1047}_{3.6507}((-0.19896x+3.7578)\sqrt{1+((-0.19896x+3.7578)")^2})dx=59.48cm^2\)​​​​​​​ \(A_3=2\pi\int^{9.8352}_{7.1047}((-0.1140x^3+2.8274x^2-22.9048x+63.2513)\sqrt{1+((-0.1140x^3+2.8274x^2-22.9048x+63.2513)")^2})dx=50.01cm^2\)\(A_4=2\pi\int^{14.2968}_{9.8352}((0.0251x+2.7446)\sqrt{1+((0.0251x+2.7446)")^2})dx=85.46cm^2\) \(A_5=2\pi\int^{19.1964}_{14.2968}((-0.1862x^2+5.5808x-38.6198)\sqrt{1+((-0.1862x^2+5.5808x-38.6198)")^2})dx=\\73.96\ cm^2\) \(A_6=2\pi\int^{18.5627}_{18.1964}((-0.0295x+1.8068)\sqrt{1+((0.0295x+1.8068)")^2})dx=2.91cm^2\) Then, I have to calculate the surface area of the bottom of the water bottle. I use the general formula for the area of a circle: \(A=\pi r^2\) The diameter of the bottom is 5.5 cm ± 0.05, hence, 𝑟 =\(\frac{5.5}2=2.75cm\ ±\)0.05. Now I can calculate the surface area of the bottom with my values. The result is given approximately to 2 d.p.: \(A_{(bottom)}=\pi\bullet2.75^2=23.76\ cm^2\) The total surface area of the water bottle can be calculated by summing up the values for surface areas found 
above: \(A_{(total)}=70.94+59.48+50.01+85.46+73.96+2.91+23.76=366.52cm^2\)

image13.5195figure 12 plastic bottle

I was not sure if my calculation have been accurate, hence, I have decided to estimate if my findings are accurate by comparing the total surface area of the water bottle to the surface area of a cylinder and a cone of the same dimensions. The volume of a cone will represent the minimum possible total surface area and the area of a cylinder the maximum possible total surface area. The parameters taken: h = 21.0 cm ± 0.05 and r = 2.75 cm ± 0.05. The surface area of the modeled cone represented in Figure 12 can be calculated using the following formula: \(A=\pi r(r+\sqrt{h^2+r^2}\) I use my parameters to calculate the surface area of the cone. The result is given at approximately 2 d.p. \(A_{cone}=2.75\pi(2.75+\sqrt{21.0^2+2.75^2}=206.73\ cm^2\)

image14.png

The surface area of the modeled cylinder seen in Figure 13 can be calculated using the following formula: \(A_{(cylinder)}=2\pi rh+2\pi r^2\) However, as I do not need to calculate the top of the bottle and only the bottom of the bottle, I will be using the following formula:​​​​​​​ \(A_{(cylinder)}=2\pi rh+2\pi r^2\) I use my parameters to calculate the surface area of the cylinder. The result is given approximately to 2 d.p.: \(A_{(cylinder)}=2\bullet2.75\pi\bullet21.0+\pi\bullet2.75^2=386.61cm^2\) Now I can compare my total surface area value with the values I calculated for the surface area of a cone and a cylinder: \(206.73cm^2&lt;366.52cm62&lt;386.61cm^2\) From this it can be seen that my calculated value is valid, as it is less than the value calculated for the cylinder and more than the value calculated for the cone. I have decided to also calculate the total volume of the bottle and compare it to the known volume of 500 cm3 to check my findings. I did it using the formula10 below: \(V=\pi\int_a^by^2\ dx\) The volume for every function needed to be calculated. I did it using GDC (math → fnInt). All of the values were given approximate to 2 d.p.: \(V_1=\pi\int_0^{3.6507}((-0.0715x^2+0.3368x+2.7543)^2)dx=106.92cm^3 \) \(V_2=\pi\int_{3.6507}^{7.1047}((-0.19896x+3.7578)^2)dx=78.82cm^3\) \(V_3=\pi\int_{7.1047}^{9.8352}((-0.1140x^3+2.8274x^2-22.9048x+63.2513)^2)dx=66.58cm^3\)​​​​​​​ \(V_4=\pi\int_{9.8352}^{14.2968}((0.0251x+2.7446)^2)dx=130.19cm^3\) \(V_5=\pi\int_{14.296}^{18.1964}((-0.1862x^2+5.5808x-38.6198)^2)dx=91.03cm^3\) \(V_6=\pi\int_{18.1964}^{18.5627}((-0.0295x+1.8068)^2)dx=1.84cm^3\)​​​​​​​ Now I can calculate the total volume by adding the values together:\(V_{(total)}=106.92+78.82+66.58+130.19+91.03+1.83=475.37cm^3\) While the actual volume of the bottle is 500 cm3, the volume I calculated is 475.37 cm3 which indicates rather small absolute uncertainty in the calculations. What may have contributed to this is the angle at which the photo was taken, which may have disrupted the actual dimensions of the water bottle. I also may have excluded a larger bottom part of the bottle than it was necessary. Moreover, the neck part of the bottle was not included in the calculations. Other reasons could be an approximation, distortions when placing the bottle onto the Cartesian coordinate system, and possible inaccuracies in the functions. Now that I know that my values are valid, I can further investigate if plastic is used efficiently for this volume. I will be comparing the surface area I got for the water bottle with the surface area for a cylinder of the same volume, as this shape can actually be used as a water bottle.

image15.png

An optimal 500 cm3 cylinder was modeled to minimize the total surface area for the given volume by optimization. The volume of a cylinder (the shape can be seen in Figure 14) can be calculated using the following formula: \(V=\pi r^2h\) ​​​​​​​​​​​​​​Now I can express h in terms of volume and r: \(h=\frac{V}{\pi r^2}\rightarrow h=\frac{500}{\pi r^2}\)​​​​​​​ This expression can be inserted into the adapted formula for the surface area of a cylinder used earlier:                                                      \(A_{(cyclinder)\ \ =\ 2\pi rh\ +\ \pi r^2\rightarrow\ A_{(cylinder)\ }(r)\ =\ 2\pi r\frac{500}{\pi r^2}+\pi r^2=\ \frac{1000}{r}\ +\ \pi r^2}\) Now I have to find the derivative of this function:                                                            \(A'_{(cylinder)}(r)=(\frac{1000}{r}+\pi r^2)'=-\frac{1000}{r^2}+2\pi r\) When this derivative is equal to 0, the surface area is at its minimum, hence, we can find the value of r for the minimum surface area. I will be using GDC to graph this function and then find the r-value at y = 0 using calc → zero. The result is given approximately to 2 d.p: \(-\frac{1000}{r^2}+2\pi r=0\rightarrow r ≈ 5.42\ cm\) To confirm that r = 5.42 is a minimum turning point where the surface area is at its minimum, a second derivative has to be calculated: \(\bar A_{(cylinder)}(r)=(-{\frac{1000}{r^2}+2\pi r})'=\frac{2000}{r^3}+2\pi\) \(A^-_{(cylinder)}(5.42)=\frac{2000}{5.42^3}+2\pi≈ 18.84\rightarrow 18.84&gt;0\rightarrow r=5.42is\ the \ minimum point\) Now I can calculate the surface area for this cylinder. The result is given approximately to 2 d.p.: \(A_{(cylinder)}=\frac{1000}{5.42}+5.42^2\pi≈ 276.79cm^2\)​​​​​​​

Table 2. Total surface area and volume of the water bottle and the optimal cylinder

image16.jpg

From Table 2, it can be seen that the optimized cylinder for the given volume of 500 cm3 has a significantly smaller total surface area than the ‘Akvilė’ water bottle (276.79 cm2 &lt; 366.52 cm2). This means that the shape of the water bottle is not effective in minimizing the use of plastic. The difference between the calculated volume of revolution (475.37 cm3) and the actual volume (500 cm3) is only 24.63 cm3 which shows that my calculations and the equations of the functions were relatively accurate and correct. As mentioned before, this inaccuracy could have occurred because of the angle of the photo, excluded the bottom of the bottle (x&lt;0), excluded the neck part of the bottle, approximations, and other distortions and inaccuracies. Nevertheless, the difference between the values is rather small and the comparison between the total surface area of a cone and a cylinder with the same parameters further confirmed the validity of my calculations\((206.73cm^2&lt;366.52cm62&lt;386.61cm^2)\)​​​​​​​ However, there are some further limitations to my calculations, for example, the thickness of the water bottle was not considered in the investigation. In my case, the walls of the water bottle were very thin, hence, the difference is not significant. Nevertheless, it is important to consider when calculating the volume, as when using functions to calculate the volume of revolution, wall thickness is not considered to make the volume larger than in reality. Moreover, the shape of the bottom of the water bottle was considered a circle, however, in reality, it has some bulges, which would be difficult to describe using functions, but would contribute to the total surface area, as well as, volume. Moreover, many approximations should be considered as a limitation, since most values were approximated to 4 or 5 d.p. The final values for the total surface area and volume were approximated to 2 d.p., the value of 𝜋 was also approximated. Overall, this investigation was useful in many ways. First of all, I proposed a shape of a water bottle that could be used for the same volume but with a significantly lower total surface area, which would mean that less plastic is used during manufacturing. Moreover, I have realized how mathematics can be used to solve real-life problems creatively, in this case, minimizing the use of plastic during manufacturing. This investigation also allowed me to further enhance my knowledge of functions, derivatives, and integrals. Furthermore, I have learned how to use mathematical matrices to make my calculations more simple and quick and felt very satisfied that I could apply my mathematical knowledge independently.

Mathematics AA SL's Sample Internal Assessment

Plastic Water Bottle: Finding Surface Area, Volume and Possible Optimal Shape

Table of content

Introduction

Exploration

Evaluation, conclusion, and reflections