Polar Equation Project: Burj Al Arab
By Junsung Lim
Buildings are defined as “structures with roof and wall”. Simply put, buildings protect us from threats, such as harsh weather, animals, etc. However, now that technology is improving day by day, architects have more freedom to express their artistic abilities in these structures. This year in Dr. Tong’s Honors Precalculus class at Concordia International School Shanghai, we had the opportunity to use our knowledge of polar equations to draw out buildings that have their unique features. My group member and I decided to draw out the Burj Al Arab tower using polar equations, and below at the left is the picture we used as a reference, and at the right is our final superimposed graph for this project.
My group member and I decided to choose this architecture because this was the best design that would reflect our learnings from this semester. We were able to identify lines, an ellipse, parabolic curves, and even a rotated ellipse in this building. Link to Desmos
Drawing the Graph: Lines
I would first like to introduce the process of drawing the several rows of lines that represent the windows on the building. I first found a rectangular equation, y=mx+b, so that the slope and the y-intercept match the window line. Then I converted the rectangular equation to a polar equation. Below is the process used to convert:
After writing all the polar equations for the individual lines, we realized that some sets of windows had the same slope and had the same distance between each other. We wanted to visually illustrate this in our equations, so we decided to set up terms. Using the slider tool in Desmos, we were able to identify specific terms where a=1, b=2, c=3, and d=4. Then, we set up the equations using the terms. This can be shown below:
The equation at the left shows sets of equations that were used to represent 4 lines at the building. The slope for these lines is all the same, m = -0.01, and the distance between each line are also the same. Therefore, by using the terms abcd, we were able to represent that the windows follow the same slope but just go down by an equal distance of 0.52 as you go down the equations.
Adjusting Domain of Theta:
While I was doing the project, I didn’t really understand the concept of adjusting the domain of theta and I just plugged in numbers until I got the line segment I wanted to acquire. It was not until the end of the project that I understood what was really happening.
The equation r = (-3.1–0.52a)/(sinθ + 0.01cosθ) with the domain of 85.8°≤θ≤102.2° is under the horizontal x-axis in polar equation. So, I asked myself, if the domain is 85.8°≤θ≤102.2°, shouldn’t the line be over the horzintonal x-axis, in between the polar grid degrees of 85.8° to 102.2°? After playing around with Desmos I figured out that the two equations below with different domain still gives the same segment:
To briefly explain what is happening, the first equation has a r-value of negative while the second equation has an r-value of positive. For example, plugging in 85.8° to the first equation gives negative/positive, making the r-value negative. Plugging in 256.8° to the second equation gives negative/negative, making the r-value positive. The difference between 256.8° and 85.8° is exactly 180°. We learned in class that coordinates (r, θ) and (-r, θ+180) gives equal points. Therefore, the two equations with the domain difference of 180° for θ still give the same segment.
Drawing the Graph: Ellipses
The Burj Al Arab building had one ellipse, and it can be seen at the right top corner of the building. Drawing an ellipse also follows a similar process: first write out a rectangular equation, then convert it into polar equations. Since the ellipse in this building had a horizontal major axis, the standard rectangular equation for this ellipse is:
Using the standard grids, I estimated the center of the ellipse, the major axis length, and the minor axis length. With these three pieces of information, I can easily develop a standard rectangular equation of this ellipse, based on what we learned in class. For example, the rectangular equation below had a center point of (1.6,5.6) and had major axis length of 0.9 and a minor axis length of 0.3 Below is the standard rectangular equation and then the polar equation of it:
From the rectangular equation, we substituted y = rsinθ and x = rcosθ into the rectangular equation. Then, by using the quadratic formula, you can come up with a polar equation for the ellipse. Below is the process I used to convert the rectangular ellipse equation to a polar equation:
Drawing the Graph: Parabola
Moving on, parabolas were used to represent the curves that shape the overall building shape on the right side of the building. Unfortunately, we couldn’t use one single parabola to match the overall curve, so we had to use several curves and lines so that the overall curve does not look awkward. Below are sets of equations representing the curve at the very right of the building:
I would first like to explain the process of coming up with the rectangular equations of the parabola. We first used standard grids of the Desmos graph to find three appropriate points along the curve we are trying to draw. After getting the three points, we used a website that uses systems of equations to convert these three points into a rectangular equation. We then got the form of y = ax²+ bx + c. After getting the rectangular equation, the process followed a similar pathway to drawing ellipses, substituting y = rsinθ and x = rcosθ. With the process, we were able to come up with the formulas above for the first, third, and fifth row in the example above.
Drawing the Graph: Rotated Ellipses
The most complicated shape to draw using Desmos was a rotated ellipse. This can be seen at the very left of the building. I would like to guide the process using visuals to help understand the process easily:
1) Locate the center and the length of the major axis, and minor axis of the rotated ellipse you want to draw. Then write a polar equation using the three pieces of information. The ellipse you draw with the equations won’t be flipped, but it is okay.
2) Add 45° to θ in the polar equation. This will flip the ellipse and put it somewhere else, but it is okay.
3) Locate the center and the length of the major axis, and minor axis of the rotated ellipse in step 2. Then write a polar equation using the three pieces of information.
4) Add -45° to θ in the polar equation. This will put your rotated ellipse at where you wanted.
Conclusion:
Although this project was called the “polar project”, polar equations were not the only things that were used in this project. For example, I had to be familiar with the rectangular equation of ellipses so that I can apply it in this project and quickly come up with an equation of it based on the points I have.
I also learned a lot of new things from this project too. For example, I was able to learn to use quadratic equations to convert from rectangular to polar equations. Furthermore, I was able to explain the relationship between the domain and how it affects the line. If it was not for this project, I wouldn’t have been familiar with this subject.
Overall, I learned a lot from this project, and I am glad that I got to be part of it.
This is a project from the Honors PreCalculus course at Concordia International School Shanghai.
