When starting to think about what function I wanted to print to show level curves, I immediately thought of the magic carpet that they ride in the movie Aladdin. In my head this was a perfect example of a function in three-dimensional space because you could clearly see the plane and its curves. When the magic carpet is flying, you can imagine that when the sun is in the perfect spot, the shadow of the magic carpet would produce a shadow of one set of level curves.
When attempting to find a function that mimicked a magic carpet, I started doing some research. I figured some sort of sine or cosine function would give me a nice curvy function so I attempted to use f(x,y)= sin(x-y), but this function was a little too wavy and did not give me the shape I wanted. So, I kept researching and found my function of:
\[ f(x,y) =xy(\dfrac{x^{2}-y^{2}}{x^{2}+y^{2}})\]
Something fun I discovered while researching functions was that most people don't think of level curves in terms of magic carpets, but I think it's helpful to think of something you are familiar with when explaining something unfamiliar. The function that I chose resembles a magic carpet when it's a little too windy outside. The three-dimensional graph below shows the function as well:
My function continues to increase in all directions, but since I wanted it to look like a magic carpet, I restricted it to only show when x and y were both between -2 and 2. Because of my restrictions in the x and y direction, the function only moves between -1 and 1 in the z direction.
Since parameterized surfaces are functions in three-dimensions, they can be hard to visualize. In order to better visualize them we can draw them in two dimensions, we call these level curves. In order to find the level curves, you must set the function equal to a specific z-value and graph the new equation. You can do this for as many "heights", or z values, as you want or as you deem necessary. Since my functions z values range from -1 to 1, I chose 9 different z values to find 9 level curves. I calculated the functions for the following z values: -0.8, -0.6, -0.4, -0.2, 0, 0.2, 0.4, 0.6, and 0.8. The following functions resulted:
\[ -0.8 =xy(\dfrac{x^{2}-y^{2}}{x^{2}+y^{2}})\]
\[ -0.6 =xy(\dfrac{x^{2}-y^{2}}{x^{2}+y^{2}})\]
\[ -0.4 =xy(\dfrac{x^{2}-y^{2}}{x^{2}+y^{2}})\]
\[ -0.2 =xy(\dfrac{x^{2}-y^{2}}{x^{2}+y^{2}})\]
\[ 0 =xy(\dfrac{x^{2}-y^{2}}{x^{2}+y^{2}}), y=0, x=0, y=x, y=-x\]
\[ 0.2 =xy(\dfrac{x^{2}-y^{2}}{x^{2}+y^{2}})\]
\[ 0.4 =xy(\dfrac{x^{2}-y^{2}}{x^{2}+y^{2}})\]
\[ 0.6 =xy(\dfrac{x^{2}-y^{2}}{x^{2}+y^{2}})\]
\[ 0.8 =xy(\dfrac{x^{2}-y^{2}}{x^{2}+y^{2}})\]
When these functions are graphed on a graphing calculator in two dimensions, the following graph appears.
I enjoy looking at the level curves graph because you can easily tell multiple things about the function in three dimensions. You can easily tell that the function is centered around (0,0) and is perfectly symmetric. You can also see that the further the level curve is from the origin, the closer it is to the previous curve. The following image is what my print will look like with the level curves at their appropriate heights.
The dimensions of my print, including a stand for display, end up being about 54mm by 54mm by 48mm. I had to print with supports so the print time will be a bit longer.
Throughout the semester we have covered a variety of topics and how their mathematical orientation applies to real world scenarios. One topic we discussed, and I would like to revisit, is integration over a region in a plane which involves calculating a double integral. Integrating functions of two variables allows us to calculate the volume under the function in a 3D space. You can see a more in depth description and my previous example in my blog post, https://ukyma391.blogspot.com/2021/09/integration-for-over-regions-in-plane_27.html . I want to revisit this topic because in my previous attempt my volume calculations were incorrect, and my print lacked structural stability. I believed this print and calculation was the topic I could most improve on and wanted to give it another chance. What needed Improvement? The function used previously was f(x) = cos(xy) bounded on [-3,3] x [-1,3]. After solving for the estimated and actual volume, it was difficult to represent in a print...
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