Modelling the surface area of a ceramic pot

The polynomial function of the bottom section of the ceramic pot is first found. From the shape of the curve, it appears to be a quadratic function; the graph is concave downwards with one maximum point at (3.0,3.7). I could have chosen a function of a higher-degree polynomial, such as a cubic polynomial since the characteristics of the curve also apply to part of a cubic graph. But although the accuracy of the interpolated data points would have been greater with a cubic polynomial, this does not necessarily mean that the graph will be a smooth fit for the curve of the ceramic pot. Thus, I chose simplicity over accuracy and decided to use a quadratic function instead. The Lagrange interpolation formula finds the equation of a unique polynomial of order 𝑛 passing through (𝑛+1) data points. Hence, three data points were chosen since I am finding a polynomial of order 2. Their 𝑥 and 𝑦 coordinates are shown below.

Coordinates

The same method I illustrated to find the function of the bottom section of the ceramic pot was applied to find the function of the middle section. Similarly, the graph appears to be a quadratic function from the shape of the curve as it is concave downwards with one maximum point at (8.5,2.4). Hence, three points were chosen. Their 𝑥 and 𝑦 coordinates are shown below.

The three function equations were found analytically using the Lagrange interpolation formula. I assumed the functions are all polynomial functions as the Lagrange interpolation formula only applies to polynomial functions. I felt that this method was appropriate as I recorded down values of 𝑥 at unequal intervals which the formula takes into account. The formula is as follows that for a unique polynomial 𝑃(𝑥):

\(P(x)=\sum\limits_{i=0}\limits^{n}{y_i\ \frac{(x-x_0)...(x-x_{i-1})(x-x_{i+1})...(x-x_n)}{(x_i-x_0)...(x_i-x_{i-1})(x_i-x_{i+1})...(x_i-x_n)}}\)

The same method I illustrated to find the function of the bottom section of the ceramic pot was applied to find the function of the middle section. Similarly, the graph appears to be a quadratic function from the shape of the curve as it is concave upwards with one minimum point at (6.8,1.4). Hence, three points were chosen. Their 𝑥 and 𝑦 coordinates are shown below.

Coordinates

During Maths HL class, we were taught how to utilise integral calculus in order to find the volume of a solid of revolution in the interval [𝑎,𝑏]. The formula given to us was:

\(V=\int_{a}^{b}{\pi y^2dx}\)

Hence, I wondered if there was a similar way in which the surface area of a solid of revolution could be found through calculus. After doing some research, I found a formula that would allow me to find the surface area, 𝐴, of the ceramic pot in the interval [𝑎,𝑏]:

\(A=\int_{a}^{b}{2\pi y\ ds}\)

\(Where\ y=f(x)>\ 0,a\ \le x\le b\ and\ ds=\sqrt{1+\left(\frac{dy}{dx}\right)^2}\ dx\)

This formula only applies to solids of revolution, which is just as well since the ceramic pot I have has a circular cross-section as shown below in Figure 2. Thus, the aim of my exploration will be to learn how to mathematically derive the formula for the surface area of a solid of revolution and then, apply it to approximate the surface area of the ceramic pot.

A solid of revolution is made by rotating a continuous function  \(y=f(x)\) about the x-axis in the interval [𝑎,𝑏]. The solid of revolution can be divided into an infinite number of frustums, created by taking a line segment and rotating it around the x-axis, with equal width ∆dx. The number of frustums is taken as infinite as this will allow the solid to be modeled as closely as possible. An example of a solid revolution being divided into frustums at equal intervals of width ∆dx to reflect its shape is shown below with an image of my ceramic pot, although only to 4 frustums.

Personally, creating art has always been something I have been drawn to as it allows me to figuratively pour my feelings out onto paper. Taking visual arts as an IB subject was only natural for me; what was not natural for me was having to create art across a range of different art forms as an IB requirement, as I was only comfortable working with paint and pencil. It was only until I began to experiment ceramics at the start of IB that I became interested in other forms of art. The glazed ceramic pot shown below in Figure 1 was made and painted early into Grade. Though simplistic in design, it is of great significance to me and hence, I wanted to find a way in which I could tie it together with my Maths Exploration. Hence, I decided to centre my investigation on finding the surface area of my ceramic pot. This topic has wider implications; the surface area of a ceramic body determines the amount of glaze that needs to be applied to it prior to firing, to seal and protect the fired clay piece ("The Basics of Glaze - Kiln Arts", 2017). Ceramic glazes can also be pricey, especially if they have precious metal components. Thus, this exploration has practical applications as well if I want to roughly approximate the cost of coating future ceramics I make with a certain glaze and to determine if it would be an economically viable option to use that glaze.

• Introduction
  • Rationale
  • Aim 
• Exploring the formula for surface area
• Graphing the ceramic pot
• Finding the functions
• Calculating the surface area of the ceramic pot 
• Limitations and Improvements 
• Conclusion 

To calculate the surface area of the ceramic pot, the formula indicates that the equation of the curve must be found. Prior to finding the functions of the graph of the ceramic pot, I had to first record the coordinates of several points on the ceramic pot in order to plot out its graph. This was initially done by wrapping a tape measure made of ribbon around several points on the ceramic pot to find their circumferences. However, I had difficulty doing so as the tape measure kept slipping as the ceramic pot was too smooth. Hence, I decided instead to only measure the circumferences of the bottom and top tips of the ceramic pot as this was easier. After finding the circumferences, the formula for the area of a circle was used to find the radius of each circular cross-section:

𝐶𝑖𝑟𝑐𝑢𝑚𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑜𝑓 𝑎 𝑐𝑖𝑟𝑐𝑙𝑒 = 2𝜋𝑟

The radius of the circle, which then corresponds to its positive𝑥-coordinate on the graph, is then given by:

\(r=\frac{\ Circumference\ of\ a\ circle}{2\pi}\)

For example, the circumference of the bottom cross-section of the ceramic pot gives a radius of:

\(r=\frac{\ 13}{2\pi}\approx2.1\ cm\ (1\ d.p.)\)

The radii of the bottom and top cross sections of the ceramic pot are recorded below in Table 1.