As I’ve always been amazed at how math could be applied to the real world, I found it frustrating that I could not find a useful application form the complex branches of mathematics in my everyday life.

The idea for this Internal Assessment came from this desire, which led me to attempt to observe the world around me in mathematical terms. Ultimately, I realized that a fairly advanced area of mathematics such as calculus could be applied in the alimentary products that we often purchase and that are made in series by automated machines or employees. I thought that, while such products are expected to be consistent in their shape and constitution, they are obviously subject to a certain degree of imprecision that alters them.

I was curious about finding a way to measure how such products vary from each other and thus determine how reliable the factory that makes them actually is. Therefore, I decided to analyze donuts for this investigation, being a popular food that I often consume in my everyday life. Experimental analysis was conducted to measure the radius of the cross-section and actual radius of each of the 15 donuts, which will be then used to find the total volume of the samples. Assuming the donut’s shape to be a perfect torus, its volume V could be calculated using integration. A normal distribution was then plotted to find the probability of obtaining a “perfect donut” when selecting a donut at random from a pack of 15 donuts. In this investigation, a donut was considered to be “perfect” if it matched the producer’s claim of volume Vp, which was derived from the declared weight of each donut found at the back of the package. While there are multiple methods of using integration to find the volume of a torus, I decided to integrate the area of a washer between the top and bottom of the donut. 1.

INVESTIGATION a. The Volume of the Donuts In this section, a general formula for the volume of a torus was found using integration and then adopted to find the volume of each of the 15 donuts. Once the formula was found, the physical data of each donut had to be obtained and inserted into the formula.

While the donuts’ radius R was measured using a Vernier Caliper, the radius r of their cross-section was found by measuring the diameter S with the caliper and dividing it by 2. i. Finding an equation for the volume As mentioned in the Introduction, to find the volume of the donuts, their shape was assumed to be a perfect torus. However, I discovered that developing a formula for the volume of such shape is not as immediate as one might think. The following diagram shows how similar the donuts and the shape of a torus can be, therefore justifying my assumption: S Z R O r Y X ? Figure 1 shows an over positioning picture of a donut and a torus. where r is the radius of the cross-section perpendicular to the x-axis and R is the radius of the donut itself, also being the distance between the center O and the center of the cross-section perpendicular to the x-axis. Firstly, I calculated the area W of the cross-section of the torus perpendicular to the y-axis, also called a washer, which is represented in the following diagram: Y X W O ? Figure 1 shows the area W of the cross-section of the donut perpendicular to the y-axis, also called washer and the center O of the donut. The diagram was constructed using Autograph.

Note that, being the cross-section of the torus perpendicular to the y-axis, the area W (represented as a blue disc in the diagram) varies with respect to its position on the y-axis since the distance d between the outer edge (represented as a purple semicircle on the diagram) and inner edge (represented as a red semicircle on the diagram) of the cross-section of the torus perpendicular to the x-axis varies depending on its position on the y-axis.To calculate the area W, the distance d has to be found. Note that the cross-section of the torus perpendicular to the x-axis is a circle, which allows using the equation of the circle to calculate d. In the following diagram, the distance d is labeled as a yellow line: Y Ro Ri X ? Figure 2 shows the cross-section of the donut perpendicular to the x-axis, notice that it is a perfect circle. The diagram was constructed using Autograph. In the diagram, r is the radius of the circle, R is the distance between the center of the circle and the center O of the donut and Ri and Ro are respectively the inner and outer radius of the donut, being the distance between the center O of the donut and the inner edge for Ri and outer edge for Ro. Since the center of the donut O was chosen as the origin of the axes, the center of the circle lies on the x-axis but is moved from the origin O by distance R, the equation of the circle can be written as: (1) Using this equation, a formula for the inner radius Ri of the donut and the outer radius Ro of the donut could be found by rearranging the equation for x: Ri: Ro: Where y is the y-axis value which the washer area W is located at. Therefore, the area of the washer W can be found as the difference between the area Ao defined by the outer radius Ro and the area Ai defined by the inner radius Ri: Ro Ri O Ao Ai W ? Figure 3 shows the two-dimensional top view of the donut and the relationship between the area defined by the outer radius Ro and the area defined by the inner radius Ro.

In the diagram, the blue section W is the area of the washer. (2) r Once the equation for the cross-sectional area W was found, the volume V of the torus could be obtained by integrating Equation (2) for y between the y-axis values of r and –r, as shown in the diagram: Y -r X W ? Figure 4 shows the upper bound r and lower bound –r of the integral for the volume of the torus. The diagram was constructed using Autograph. By doing so, the area W of the blue washer in the diagram, displayed from a horizontal point of view in the diagram, is repeatedly integrated from the top of the donut to its bottom and varies as the inner and outer radii Ro and Ri change when passing along the red and purple edges: (3) To solve the integral of Equation (3), the following substitution will be used: Since y was substituted, the upper and lower bounds of the integral should also be changed accordingly: Therefore, the formula becomes: Using the double angle formulae, since: This means that: Therefore, (4) From the equation, it can be observed that, since the radius is squared, a slight variation in the radius r of the donuts will result in a considerable change in their volume V. This might be a limitation to the faithfulness of calculating the volume of a donut by assuming that it is a perfect torus because of the small imperfections in their shape. This has to be taken into account when finding the probability of obtaining a “perfect” donut out of the pack. ii. Finding the volume of each donut For each of the 15 donuts, the diameter S of the cross-section perpendicular to the x-axis and the donuts’ radius R were measured using a Vernier caliper.

Subsequently, the radius r of the cross-sectional area perpendicular to the x-axis was found by dividing the values of S by 2. Equation (4) was finally used to find the volume of each donut. The results are displayed in the following table: S (cm) r (cm) R (cm) V (cm3) 2.46 1.

23 2.94 87.80 2.58 1.29 2.95 96.

90 2.50 1.25 3.

07 94.69 2.46 1.23 3.08 91.

98 2.46 1.23 3.05 91.08 2.48 1.24 3.02 91.

66 2.62 1.31 2.95 99.93 2.52 1.26 2.92 91.

51 2.46 1.23 2.91 86.90 2.52 1.26 3.

07 96.21 2.52 1.26 3.01 94.33 2.60 1.

30 2.92 97.41 2.42 1.21 3.00 86.70 2.

48 1.24 2.92 88.62 2.46 1.23 3.09 92.28 ? Table 1 shows the radius R of each donut, the diameter S of the cross-section perpendicular to the x-axis and the derived radius r with respect to the volume V of each donut.

b. The Normal distribution of the Donuts In order to find the probability of selecting a donut that matches the description at the back of the packaging out of the sample of 15 that was selected, a normal distribution of the donuts’ volume V has to be plotted. i. Finding the Standard Deviation Prior to finding the standard deviation of the pack of donuts, the mean volume Vm was calculated from the values of the Volume V obtained in paragraph a: Subsequently, the square of each data point from the mean was calculated using the values from table 1: V (cm3) Square distance from Vm 87.80 22.42 96.

90 19.08 94.69 4.64 91.98 0.31 91.08 2.

10 91.66 0.76 99.93 54.71 91.

51 1.05 86.90 31.

70 96.21 13.50 94.33 3.

22 97.41 23.78 86.70 34.

02 88.62 15.27 92.28 0.

07 ? Table 2 shows the square distance from the mean volume with respect to the donuts’ volume V. The values of the square distance were then added together to obtain their sum Vs: Vs was then divided by the number n of data points, which correspond to the 15 donuts whose radius was measured. The square root was finally taken to find the standard deviation ?v: The obtained value for the standard deviation is considerably large and this could indicate that, in a random pack of 15 donuts, the buns vary significantly in their volume, which would lead to thinking that the producer’s machinery lacks precision. ii. Finding the Normal distribution Once the mean and the standard deviation of the volume of the donuts were obtained, a graph displaying the normal distribution could be plotted. The following graph was plotted using the Desmos graphic calculator: Vm ? Figure 5 shows the Normal Distribution of the donuts’ volume V with its mean Vm. The shape of the graph confirms the prediction made in paragraph 2, b, ii as the normal distribution appears to be flat and spread out.

This means that the volume of the donuts in the package varies noticeably, which is indicative of defects in their production. 2. The producer’s claim for the volume of the Donuts i. Finding the density of the donuts However, to further determine the reliability of the producer, I chose to In order to determine the volume that the producer declares its donuts to have, the indication for the weight of a single donut at the back of the packaging was used in combination with the density D of the donuts. In order to find the latter, which was assumed to be the same for every donut, the following equation was used: (5) Rearranging for D: (6) Where V is the volume of the donuts, m is the mass of the donuts and D is their density. However, while the mass of each donut is provided by the producer at the back of the package and the volume was calculated in paragraph a, the density had to be found manually. To do so, the donuts, whose volume V was calculated in paragraph a, were weighted using a weight scale to find their mass m.

The density value of each donut was therefore calculated: V (cm3) m (g) Density (g/cm3) 87.80 50.2 0.572 96.90 54.1 0.558 94.

69 53.5 0.565 91.98 52.4 0.570 91.

08 52.1 0.572 91.66 52.3 0.571 99.93 56.

6 0.566 91.51 52.6 0.575 86.90 50 0.575 96.21 53.

6 0.557 94.33 52.8 0.560 97.

41 55.9 0.574 86.70 49.5 0.

571 88.62 50.7 0.572 92.

28 52.6 0.570 ? Table 3 shows the density of each donut with respect to its volume and mass. From the table, the average density D was found to be: ii. Finding the producer’s claim for the volume of the Donuts Once the density D of the donuts’ material was found as the average of the density of the 15 donuts, the volume Vp that the producer claims each donut to have was calculated using Equation (5), where mp is the mass value of every donut declared at the back of the package: mp = 55g. When comparing the result with the mean volume of the 15 donuts in the package it can be observed that the producer claims the majority of the donuts to have a larger volume than what is offered in the sample package. 3.

The probability of getting a “perfect” donut Once the volume Vp declared by the producer was found, the probability of picking a perfect donut, being a donut that perfectly matches the theoretical volume value, was found through the normal distribution plotted in paragraph 2,b,ii: The probability was found using a Casio fx-9860gii graphic calculator: Considering the significantly small probability of obtaining a perfect donut, I realized that a margin of error should be taken into account due to the possibility of small defects in the calculation of the volume V resulting from considering each donut to be a perfect torus. Thus, an error of 5% within the volume Vp was allowed to the “perfect donut” as it seemed to be a realistic percentage of how much each donut’s volume could differ from the theoretical shape of a torus. This means that both an upper and lower bound of the normal distribution should be calculated to restrict the area of probability to 5% of the producer’s claim of the volume. First, the upper bound a of the “perfect volume” was calculated by adding 5% to the volume Vs and the lower bound b was calculated by subtracting 5% to the volume Vs:Upper bound a: Lower bound b: Then, the probability of randomly selecting a “perfect donut” from the pack of 15 donuts was found as: Using the graphic calculator: The result tells us that there are more than half of the chances of choosing a donut that lies within 5% of the volume declared by the producer, which is still a significantly small probability when considering that 43.

9% of the donuts of the sample package would not meet the producer’s claim. Therefore, I wanted to investigate if it was more likely to select a donut whose volume was larger than what declared by the producer or smaller. My prediction is that the probability Psmaller of selecting a donut whose volume is smaller than Vp is larger than the probability Plarger of selecting a donut whose volume is larger than Vp due to the rubbing of the donuts against one another when they are in close contact inside the package and the resulting reduction of their volume before they arrive at the customer: Using the graphic calculator: Using the graphic calculator: The results confirm the prediction that the probability of selecting a donut whose volume is smaller than is larger than the probability of selecting a donut whose volume is larger than 4. CONCLUSION From the investigation, it could be inferred that the reliability of the producer’s claim is relatively low. While the producer claims that each donut in a pack of 15 is 55g, meaning that its volume should measure 96.66 cm3, it was found that the probability of obtaining a perfect donut is only 5.84%.

However, this is understandable as it is practically impossible that all the 15 donuts of the pack have a shape that exactly matches a torus. When a margin of error of 5% was allowed, it could be observed that the probability of randomly selecting a perfect donut increases significantly and becomes 56.1%, which is still a poor result when compared to the remaining 43.9% probability of picking a donut that does not lie within the producer’s claim. In addition, the fact that the average volume Vm of the sample population of 15 donuts was found to be 92.53 cm3 and the volume Vp declared by the producer was calculated to be 96.66 cm3, further demonstrates how customers buying the product would be likely to pay more than they should as they get less “volume” of donuts. Moreover, the hypothesis that the probability Psmaller of selecting a donut whose volume is smaller than Vp was higher than the probability Plarger of selecting a donut whose volume is larger than Vp was confirmed by the results obtained in paragraph 4.

While this could be due to the rubbing of the donuts between each other in the package, which could have led to a reduction in their volume prior to measuring their diameter S and radius R, it could also demonstrate how the customers would obtain less “volume” of donuts than what was claimed when purchasing the product.