Using Mathematica to Graph the Knotted Torus and Möbius Strip

Leonhard Euler and Karl Gauss were the pioneers in leading the way to the understanding of parametric surfaces. These men and many others paved the way for Mathematica, the "world's most powerful general computation system"[1]. Mathematica was first invented in 1988 by a group of mathematicians at Wolfram Research led by Stephen Wolfram. Even though I learned that Mathematica's syntax can be very tricky, once it is learned properly Mathematica's capabilities are very, very powerful.

To begin, I will give a formal definition on exactly what it means to parameterize a surface. A parametric surface is simply a mapping from the 2D plane to the 3D space. For curves in a two-dimension plane only one parameter is needed. Since surfaces are in three-dimension, two parameters, u and v, will be used instead of one. So the points (x, y, z) in the  plane will be parameterized by each point (u, v) in the  plane. So each point (u, v) will be sent to  and will be defined by: r(u,v) = x(u, v)i + y(u, v)j + z(u, v)k, which is called a parametric surface. The parametric equations of the surface are as follows, x = x(u, v), y = y(u, v), z = z(u, v).

With all the background information now being said, I will proceed with a couple of simple examples on how to come up with these parametric equations for surfaces in the three-dimensional system.
Parameterizing Simple Surfaces
The first surface I looked at was the right circular cylinder. Looking at the cylinder, the first aspect of the shape that is very useful for finding the parametric equations is that its cross section taken in the xy-space(can be any plane you choose parallel to the xy-plane)is a circles. So each slice of the cylinder taken in this particular direction will just be a circle. So I just need to find the basic equation of the circle and scale it up by a factor in the z-direction. Since I was working with a cylinder in this first example, it is easiest to find the parametric equations by using cylindrical coordinates. In cylindrical coordinates, x = rcos(u), so I let x = 2cos(u). Of course, any value may be chosen for the radius except for zero. Likewise, y = 2sin(u), and this would give us the equation of the form of the circle x2+ y2= r2, since cos2(u) + sin2(u)=1. And since we want a circular shape we want u to stretch from 0 to 2or 360°). Since we just want the z-coordinate to go vertical, I just scaled it by a factor of v, where v goes from -3 to 3. So with those parametric equations plugged in Mathematica like this:
ParametricPlot3D[{2Cosu , 2Sinu , v}, {u, 0, 2 Pi}, {v, -3, 3}] and you will obtain the right circular cylinder.

Sphere
The next shape I worked on was the sphere. It is quite obvious in this case, that I would be working with spherical coordinates this time. Spherical coordinates are defined by: x = sinφ cosθ, y= sinφ sinθ, z = cosOnce again, we can let be anything that we want it be. For simplicity's sake we will set  = 1, which gives us the distance from a point in three-dimensional space to the origin. The coordinate v works as the θ coordinate the same way it did in the previous example. So once again it will range from 0 to 2. The coordinate u will represent , the angle between the z-axis and any line from the origin to a point on the sphere. So this angle will range from 0 to So the parametric equations and the graph will look like this: ParametricPlot3D[{Sin[u]Cos[v], Sin[u]Sin[v], Cos[u]}, {u, 0, Pi}, {v, 0, 2 Pi}]

Surface of Revolution
Besides looking at just standard parametric surfaces, I also worked some on surfaces of revolution. A surface of revolution is just the graph of a continuous function revolved about an axis. One of the surfaces of revolution that I took a look at was the revolution of the function z = 1/x. For the function z = 1/x, I will once again work with cylindrical coordinates. Looking first at just the function z = 1/x, revolved about the z-axis, and it's graph can make finding the radius for this problem very easy. Notice that since this function is just a bunch of cross sections of circles whose radii keep getting smaller and smaller as x increases, we can find the radius of any particular cross section by finding the distance from the z-axis to the function z = 1/x. Since v is going to be the scaling factor in this problem, we can find the radius with respect to v. So the radius is equal to 1/v. Revolving the line z = 1/x about the z-axisNotice that since 1/v is our radius, v must be greater than zero. The parametric equations and graph look like this: ParametricPlot3D[{1/v Cos[u], 1/v Sin[u], v}, {u, 0, 2 Pi}, {v, 1,2}]

The next parametric surface I took a look at was the torus, which is just a basic doughnut-shaped graph. It is found by finding a way to change a rectangle in  into the torus in .A way of thinking of this is to find the parametric equations that would turn the rectangle first into a cylinder space (which we have already done) and then by attaching the two ends of the cylinder together as if it were a slinky. We will proceed with this example using cylindrical coordinates. The torus example in the case that I did will be in the xy-plane with the z-axis going right down the middle of the hole in the torus. The first step to finding the equations of the surface is to take a slice out of the torus. One slice of the torus is just a circle. We will parameterize the torus by two angular coordinates u and v, described as follows. The v coordinate determines the position on the circular slice of the torus. The u coordinate corresponds to the cylindrical coordinate θ, which tells you how far to spin around the z-axis. We first need to find a parametric equation for whatever v is equal to. This will give us an equation for the z-coordinate. To find v, we need to set up a right triangle inside a typical circular slice of the torus. Let's assume the circular slice is a circle of radius 3 whose center is 6 units away from the z-axis. V is shown in the diagram below. Notice that sin(v) = z/3. So writing this equation in respect to z gives us z = 3 sin(v). The u coordinate, which I will get to in a moment, gives us the rotation of each individual slice around the entire torus. To find v we need to set up a right triangle inside one of circle slices of the torus. In this right triangle, v is the angle theta we want to rotate, and we will pick a radius of 4 for the circle. Notice that sin(v)= z/4. So, writing this equation is respect to z gives us z = 4sin(v).

It remains to find equations for x and y in terms of the angular coordinates u and v. We know x = r cosu and y = r sinu, where r is the distance from the point (x, y, z) to the z-axis. Now from the above diagram, we see that r = 6 + 4 cosv. So the parametric equations for the torus look like this:
ParametricPlot3D[{(6 + 3 Cos[v]) Cos[u], (6
+ 3Cos[v]) Sin[u], 3 Sin[v]}, {u, 0, 2 Pi}, {v, 0, 2 Pi}]

The next parametric surface is a bit trickier and probably the most difficult one for me to see at first. The surface I looked at is found by twisting a rectangular piece of paper and then "gluing" the ends together. This surface is called the Möbius strip. The cross sections of the Möbius strip are called fibers. Each fiber is just a line segment. We need to determine the direction of the fiber as a function of u. First, we will set it up so that the midpoint of each fiber lies on the unit circle in the xy-plane. A typical point on the unit circle is of the form (cosu, sinu, 0) for 0 ≤ u ≤ 2π. We decided to use a gradual twist of the fibers over the small interval, 3π/4≤ u ≤ 5π/4.

To get the fiber direction, we will need to employ two different directional vectors, the radial vector, denoted G(u), and the vertical vector, denoted H. The fiber direction will be a linear combination of G(u) and H. The radial vector will always point inward from the points on the Möbius strip, so we need its coordinates to be: G(u) = (-Cos(u), -Sin(u), 0 ). The vertical vector will be parallel to the z-axis. If we were looking straight down at the xy-plane, H would be sticking straight out of the paper. So H will be equal to . If the fiber has length 2a, there is a typical point on the Möbius strip that has the form of (Cos(u), Sin(u), 0) +vF(u). We want the fiber direction to start rotating at u = 3π/4 gradually until it points down at u = 5π/4. Let ψ(u) denote the angle between F(u) and the vertical vector H = . So at the point 3π/4, the vertical vector will start being rotated by an angle ψ (going from 0 to π ). At u = 5π/4, the fiber will be pointing in the exact opposite direction as it was before when it was completely vertical. Such a function ψ(u) is given by the following:
0, 0 ≤ u ≤ 3π/4
2u - 3π/2, 3π/4 ≤ u ≤ 5π/4
π, 5π/4 ≤ u ≤ 2π
Once we have a formula for ψ , we obtain the following formula for the fiber direction: F(u) = cos(ψ(u)) G(u) + Sin(ψ(u))H . [u_]:= Piecewise [{{0,0u3 Pi/4}, {2u-3 Pi/2, 3 Pi/4u 5 Pi/4}, {Pi, 5 Pi/4u2 Pi}}]
And we define a below as a = .3
ParametricPlot3D[{Cos[u] (1 - v Sin[[u]]), Sin[u] (1 -v Sin[[u]]), v Cos[[u]]}, {u, 0, 2 Pi}, {v, -.3, .3}, PlotRange{{-2, 2}, {-2, 2}, {-1, 1}}, ViewPoint {0, 6, 4}, BoxRatios{2,2,1}, SphericalRegionTrue, PlotPoints {25, 5}

Knotted Torus
The final parametric surface that I worked on was the knotted torus. To work with this surface, extensive knowledge of vectors must be recalled. First we needed to come up with a curve in three-dimensional space so that we could turn this curve into the knotted torus surface. All we needed for the curve was a function for a line that went through the "regular" torus, but we needed the line to wrap around the torus several times to make a knot with itself. We let t be the parameter for this function. The line in rectangular coordinates needed to map to a function of the line in three dimensional space. We let t represent the x-coordinate and since y = mx (with m being the slope) is the equation of the line, we picked 2/3 to be the slope of the line and just plugged this information back into the equation for the torus we found earlier. So the equation for the curve becomes: ParametricPlot3D[{(3 + Cos[2t/3]) Cos[t], (3 + Cos[2t/3]) Sin[t], Sin[2t/3]}, {t, 0, 6 Pi}]

Now all we have to do turn this curve into a surface and we are through. This is where the use of tangent and normal vectors will come into play. We need to use the same method we used for the Möbius strip, where we write cosand sin as a linear combination with two vectors. The two vectors we need this time are the normal vector and the binormal vector since they will be perpendicular to the unit tangent vector. So the plane that those two vectors are in multiplied by Cosand Sinwhere goes from 0 to 2 So the first thing we need to find is the unit tangent vector of the position curve we found earlier. We will denote that position curve by p[t]. To find the unit tangent vector, we simply use the formula: T[t_]:= (p'[t]. p'[t])^ (-1/2)p'[t]. This is found by using the chain rule to this:
Next, we need to find two vectors perpendicular to our tangent vector, which we will denote T[t]. The normal vector n[t] is found by the formula:
And the binormal vector is found by finding the cross product of the unit tangent vector and the normal vector.

The cross product will give us a vector that is perpendicular to both. So now since we have two vectors that are in a plane where everything is perpendicular to the unit tangent vector, we can write a linear combination as we discussed earlier. We want a radius in our equation that is smaller than the unit circle so that the surface is easier to see in mathematica so we will choose 0.2. So the equation for the surface is the position vector, p[t], of the points of the curve added to the linear combination involving the normal and binormal vectors. The final equations submitted to mathematica, looked like this:
q1 = 3
q2 = 2
where q1 and q2 will generate various knots as long as they are both relatively prime.
ParametricPlot3D[p[t] + .2Cos[]n[t] + .2Sin[]B[t], {, 0, 2Pi}, {t, 0, 2 q2 Pi},BoxRatiosAutomatic, PlotPoints{25, 125}]
where the "Calculus`VectorAnalysis` is a program downloaded by mathematica so that it can compute cross products.
This equation will give us a picture of a knotted torus.

I had had some experience before with using mathematica, but until the project came along I had no idea how powerful of a tool it can be. Besides the ability to graph amazing surfaces, I saw that it had the power to calculate page long derivatives in a matter of seconds. I found, that even though I was very frustrated on numerous occasions trying to get the syntax of mathematica correct, that mathematica can be a very interesting tool to play around with to discover new surfaces that I always thought impossible to visualize when I was taking Calculus III. I enjoyed using mathematica and I hope to continue to learn even more about its capabilities in the future.

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