Consider a circle of radius 1 in the 2D plane. The common representation is below, where y is expressed as two radical expressions.
\[x^2 + y^2 = 1 \\ y = \pm \sqrt{1-x^2}\]
However, the above expression is quite confusing, and y is expressed as two separate functions in order to create the whole image of the circle. Recalling trigonometry, each point on the unit circle corresponding to an angle, \(\theta\) has coordinates defined by the ordered pair, \(\displaystyle(\cos( \theta), \sin( \theta))\)
The x-coordinate on the unit circle for an angle \(\theta\) is defined by \(\cos( \theta)\) and likewise, the y-coordinate is defined by \(\sin( \theta) \)
This means we can redefine our above function of y in terms of x to create the unit circle in terms of a new parameter we can call t.
We simply let \(x = \cos(t) \) and \(y = \sin(t)\) and we can generate a new function \(r(t)\) where
\[r(t)=(x(t), y(t))\]
This new function gives us, given our chosen parameterization:
\[r(t) = (\cos(t), \sin(t))\]
Here, \(r(t)\) takes a single variable, and maps it onto the 2D plane to give us our original circle of radius 1!
We can also parameterize an entire surface in 3-dimensions if we choose a parameterization of two variables.
I wanted to create a surface which had a similar undulating behavior to a drop of water, dropped into a larger pool. In this case, a wave oscillates out from a central point, but diminishes in amplitude as it propagates outward. To capture the the propagation in two directions, I needed two independent variables, which were mapped to a point in 3D space.
Such a function has the following form:
\[r(u, v) = (u, v, z(u, v))\]
The function of u and v I chose was:
\[z(u,v) = \frac{\sin( \sqrt{u^2 + v^2 + c^2})}{\sqrt{u^2 + v^2 +c^2}}\]
Here, c is a constant, and I chose c=1 for simplicity
I also gave a scaling factor of 11 to make the wave pattern more dramatic
\[z(u,v) = 11*(\frac{\sin( \sqrt{u^2 + v^2 + 1})}{\sqrt{u^2 + v^2 +1}})\]
I give the plot of my surface below:
We can also see the behavior of a surface in a lower dimension by taking a "level curve" of the function. In the level curve, we find all points on the curve that are associated with a given value. In general a level curve of a function of two variables is:
\[f(x,y) = k\]
Here, we will get a plot, of all x, y such that they produce the value k given the function \(\displaystyle f(x,y)\)
To create the level curves for our function \(\displaystyle r(u, v)\), we can choose the values of independent variables which exactly equal the given value k. Below I give the level curves of \(\displaystyle r(u,v)=1\), which is equaivalent to finding u and v such that \(\displaystyle z(u,v)=1\)
The level curves change as we change the value of k. Changing k from 1 to 0.1 gives a new plot!
Many values can simultaneously satisfy a given value of k. Also, we can see how choosing a value to define a level curve reduces our dimension of \(r(u,v)\) from 3 to 2 dimensions.
Due to the undulating nature of our "water droplet" curve, many values of u and v will simultaneously solve the level curve for a reasonable value of k.
Any level curve can be thought about as inserting a plane in 3D space which slices the surface horizontally. For our above two examples, we had one plane at z = 1 and another at z=0.1 and just recorded which points on the surface that collided with the plane!
Here is my object, created from this function and rendered in OpenScad, defined on the interval [-10,10] for both x and y. I chose level curves between -2 and 5 (inclusive). My object will have dimensions of 10 inches by 10 inches by 13 inches, but will be scaled down to a total width and depth of 3inches, and a total height of roughly 4 inches for ease of printing.
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