UTM Ideals Varieties and Algorithm - Chapter 1 Section 3 Exercise 12

Problem:

Solution:

For part (a), thanks to K3DSurf, see a video plot!


There are two things worth notice in the video. It starts with a view that it is symmetric in $ x $, and it ends with a view showing as $ |x| $ increases, $ y $ and $ z $ increases, and $ y $ grows slower than $ z $ does, both are normal behavior.

Last but not least, notice the curve does not pass through the origin. This does NOT make sense. That seems to be a limitation to the tool, not a math problem.

Initially I do not know K3DSurf support plotting a curve, so I tried to create a surface by wrapping the curve as a tube. Conceptually, it is a collection of circle with the center on the curve and is on the plane orthogonal to the tangent vector.

To do that, I used this solution from MathOverflow.

First, a general point on the curve has coordinate $ (u, u^2, u^4) $.

Second, the tangent vector is $ (1, 2u, 4u^3) $

Third, I pick these two unit vectors orthogonal to the tangent vector

$ \frac{(2u, -1, 0)}{\sqrt{4u^2 + 1}} $
$ \frac{(4u^3, -1, 0)}{\sqrt{16u^6 + 1}} $

Now we can represent the parametric surface as

$ (u, u^2, u^4) + \cos v \frac{(2u, -1, 0)}{\sqrt{4u^2 + 1}} + \sin v \frac{(4u^3, -1, 0)}{\sqrt{16u^6 + 1}} $

K3DSurf also plot this formula well, of course, if K3D surf support parametric curve then it is much simpler to just plot the curve.

These are the parametric equations feeding the tool, and it does plot what we want except the tube look thin as the tool auto scales.

v   + 0.3 * cos(u) * 2 * v / sqrt(4 * v^2 + 1) + 0.3 * sin(u) * 3 * v^3 / sqrt(16 * v^6 + 1)
v^2 - 0.3 * cos(u) / sqrt(4 * v^2 + 1)
v^4                                            - 0.3 * sin(u) / sqrt(16 * v^6 + 1)

Part (b) is trivial, indeed I have already do that above, $ (u, u^2, u^4) $ is the parametrization.

For part (c), we have already derived the tangent vector $ (1, 2u, 4u^3) $, so the tangent surface is

$ (u, u^2, u^4) + v(1, 2u, 4u^3) $.