曲线与曲面参数化及切平面计算的正确性验证与改进建议

Hey Nell, let's go through each of your points one by one to check correctness and offer some tweaks to make things clearer or fix small issues:

a) Curve Parameterization (Note: Critical Clarification First!) #

Wait a second—you labeled this as a curve, but the set $C = {(x, y, z) \in \mathbb{R}^3 ,|, x^2 + y^2 = e^3}$ is actually an infinite circular cylinder (a 2-dimensional surface), not a 1-dimensional curve. That's a key terminology mix-up!

That said, your parameterization logic is on the right track for representing this cylinder:

• You correctly identified that $r = e^{3/2}$ is constant (since $x^2 + y^2 = r^2 = e^3$)
• Your expression $C(t, z) = (e^{3/2} \cos(t), e^{3/2} \sin(t), z)$ is a valid parameterization for the cylinder

Improvements: #

• Since this is a 2D surface, you need two parameters: $t$ (for the circular cross-section) and $z$ (for the infinite length along the z-axis). Be sure to specify their ranges: $t \in [0, 2\pi)$ and $z \in \mathbb{R}$.
• Rename this to "Parameterization of Cylindrical Surface" to avoid terminology confusion.

b) Surface Parameterization #

Great work here—your parameterization is fully correct! Let's verify quickly:

• Plugging into the surface equation: $x^2 + y^2 = (e^{u/2} \cos(v))^2 + (e^{u/2} \sin(v))^2 = e^u (\cos^2 v + \sin^2 v) = e^u$, and since $z = u$, this gives $x^2 + y^2 = e^z$ which matches the definition of $S$.

Small Tweak: #

• Add parameter ranges for clarity: $u \in \mathbb{R}$ (since $z$ can be any real number) and $v \in [0, 2\pi)$ (to cover the full circular cross-section without redundancy).

c) Tangent Plane to $S$ at $(0, 1, 0)$ #

Perfect—your entire process is spot-on:

• Choosing $F(x, y, z) = x^2 + y^2 - e^z$ is the right approach for using the gradient method
• Calculating $\nabla F = (2x, 2y, -e^z)$ is correct, and evaluating at $(0,1,0)$ gives the normal vector $(0,2,-1)$
• The point-normal form $2(y - 1) - z = 0$ is valid

Minor Improvement: #

• You can simplify the plane equation to a more standard form: $2y - z = 2$ (just expand the left-hand side and rearrange terms).

d) Parameterization of Rotated Surface #

There are a couple of issues here to fix, but let's start with the core logic:

• The original curve is $C = {(x, y, z) \in \mathbb{R}^3 ,|, z = y^4}$ (this is a 1D curve lying in the y-z plane). When rotating around the z-axis, every point $(0, y, z)$ on $C$ sweeps out a circle in the plane $z = \text{constant}$, with radius $|y|$.

Problems with Your Current Parameterization: #

1. Too many parameters: A rotated surface is 2-dimensional, so you only need two parameters (not three: $r, \theta, z$).
2. Incomplete coverage: Your expression uses $y = z^{1/4}$, which only accounts for $y \geq 0$ (since the 4th root is non-negative). The original curve allows $y$ to be any real number, so we need to include negative $y$ values too.

Corrected Parameterization: #

A simpler, complete parameterization uses $r$ (radius of the circular cross-section) and $\theta$ (angle around the z-axis):
$$S(r, \theta) = (r \cos(\theta), r \sin(\theta), r^4)$$

• Here, $r \in \mathbb{R}$ (or $r \geq 0$ if you want to avoid redundant points, since $r$ and $-r$ with $\theta + \pi$ give the same point) and $\theta \in [0, 2\pi)$.
• Verification: For any point on $S$, $z = r^4 = (\sqrt{x^2 + y^2})4 = (x^2 + y²⁾2$, which is exactly the equation of the surface obtained by rotating $z = y^4$ around the z-axis.

Alternatively, if you prefer to use $z$ as a parameter, you can write:
$$S(z, \theta) = (\pm z^{1/4} \cos(\theta), \pm z^{1/4} \sin(\theta), z)$$
where $z \geq 0$ (since $z = y^4$ can't be negative) and $\theta \in [0, 2\pi)$. The $\pm$ accounts for both positive and negative $y$ values from the original curve.

备注：内容来源于stack exchange，提问作者Nell