求两曲面交线在原点(0,0,0)处的曲率及参数化方法

Hey, let's work through this problem step by step—it's a nice mix of differential geometry and multivariable calculus!

First, confirm that $(0,0,0)$ lies on both surfaces: substituting into $S_1$ and $S_2$ gives $0=0$ for both, so we're good to go.

简便的曲率求解途径 #

You don't need explicit parameterization to compute the curvature here. Here's a straightforward method using gradients, Hessian matrices, and curve curvature properties:

Step 1: Compute surface gradients (normal vectors) at $(0,0,0)$ #

For $S_1: F_1(x,y,z) = x^{11}+y{10}+z^9+x3y^2+x2y^2+y3+z^2+x2+y^2+x+10y+2z$:

• $\nabla F_1(0,0,0) = (F_{1x}, F_{1y}, F_{1z}) = (1, 10, 2)$

For $S_2: F_2(x,y,z) = x^4-y2z^2+x3-y^2z+z$:

• $\nabla F_2(0,0,0) = (F_{2x}, F_{2y}, F_{2z}) = (0, 0, 1)$

Step 2: Find the tangent direction of the intersection curve #

The tangent vector $\mathbf{v}$ is the cross product of the two normals (since it lies in both tangent planes):
$$\mathbf{v} = \nabla F_1 \times \nabla F_2 = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \
1 & 10 & 2 \
0 & 0 & 1
\end{vmatrix} = (10, -1, 0)$$
Its magnitude is $|\mathbf{v}| = \sqrt{10^2 + (-1)^2} = \sqrt{101}$.

Step 3: Compute Hessian matrices and quadratic forms #

The Hessian of $F_1$ at $(0,0,0)$ (only non-zero second derivatives matter) is:
$$H_1 = \begin{pmatrix} 2 & 0 & 0 \ 0 & 2 & 0 \ 0 & 0 & 2 \end{pmatrix}$$
For the unit tangent vector $\mathbf{t} = \mathbf{v}/|\mathbf{v}| = (10, -1, 0)/\sqrt{101}$, the quadratic form is:
$$\mathbf{t}^T H_1 \mathbf{t} = \frac{10^2 \cdot 2 + (-1)^2 \cdot 2}{101} = \frac{202}{101} = 2$$

The Hessian of $F_2$ at $(0,0,0)$ is the zero matrix (all second derivatives are 0 at the origin), so its quadratic form is $0$.

Step 4: Solve for curvature #

For a curve on a surface $F=0$, differentiating $\nabla F \cdot \mathbf{r}'(s) = 0$ (where $s$ is arc length) gives:
$$\nabla F \cdot \mathbf{r}''(s) + \mathbf{t}^T H \mathbf{t} = 0$$
Since $\mathbf{r}''(s) = k\mathbf{n}$ (where $k$ is curvature and $\mathbf{n}$ is the unit principal normal), we have two equations:

1. $(1,10,2) \cdot (k\mathbf{n}) = -2$
2. $(0,0,1) \cdot (k\mathbf{n}) = 0$

The second equation tells us $\mathbf{n}$ has no $z$-component (lies in the $xy$-plane). The principal normal is perpendicular to $\mathbf{t}$, so $\mathbf{n}$ is parallel to $(1,10,0)$ (since $(1,10,0) \cdot (10,-1,0) = 0$). Taking $\mathbf{n} = (1,10,0)/\sqrt{101}$, substitute into the first equation:
$$k \cdot \frac{1 \cdot 1 + 10 \cdot 10 + 2 \cdot 0}{\sqrt{101}} = -2$$
$$k \cdot \frac{101}{\sqrt{101}} = -2$$
$$k = -\frac{2}{\sqrt{101}}$$
The magnitude of curvature is $\frac{2\sqrt{101}}{101} \approx 0.199$.

泰勒展开计算曲率的可行性 #

Absolutely! Here's how it works:

1. Local parameterization via Taylor expansion: Approximate the intersection curve near $(0,0,0)$ by setting $\mathbf{r}(t) = (10t + at^2 + o(t^2), -t + ct^2 + o(t^2), ft^3 + o(t^3))$ (using the tangent direction for first-order terms).
2. Substitute into surface equations: Plug this into $S_1$ and $S_2$, match coefficients of like powers of $t$ to solve for $a,c,f$.
3. Compute curvature: Use the formula for curvature in arbitrary parameters:
   $$k = \frac{|\mathbf{r}'(t) \times \mathbf{r}''(t)|}{|\mathbf{r}'(t)|^3}$$
   At $t=0$, calculate $\mathbf{r}'(0) = (10,-1,0)$, $\mathbf{r}''(0) = (2a, 2c, 0)$—their cross product has magnitude $202$. Dividing by $(\sqrt{101})^3$ gives the same curvature result as above.

显式参数化方法的可能性 #

A global explicit parameterization (using elementary functions) is unlikely—since both surfaces are high-degree polynomials, their intersection is a high-degree algebraic curve, which can't generally be expressed in terms of elementary functions.

However, local explicit parameterization via Taylor series exists (as we did above) thanks to the Implicit Function Theorem: since the gradients $\nabla F_1$ and $\nabla F_2$ are not parallel at $(0,0,0)$, the intersection curve is smooth near the origin, and we can write it as a Taylor series expansion to any desired order.

内容的提问来源于stack exchange，提问作者MasterJ