접평면과 선형근사

Write $F(x,y,z) = x^2+y^2+(z-1)^2-1$ and $G(x,y,z) = x^2+y^2+(z+1)^2-1$. Let $S_1$ denote the surface $F(x,y,z)=0$ and $S_2$ denote the surface $G(x,y,z)=0$. First note that $F(0,0,0)=G(0,0,0)=0$ so that the point $(0,0,0)$ lies on both $S_1$ and $S_2$. The gradients of $F$ and $G$ are

$$\begin{aligned} { \boldsymbol{\nabla} } F(x,y,z) &=\left<\frac{\partial F}{\partial x}(x,y,z)\,,\, \frac{\partial F}{\partial y}(x,y,z)\,,\, \frac{\partial F}{\partial z}(x,y,z)\right> =\left< 2x\,,\,2y\,,\,2(z-1)\right> \\ { \boldsymbol{\nabla} } G(x,y,z) &=\left<\frac{\partial G}{\partial x}(x,y,z)\,,\, \frac{\partial G}{\partial y}(x,y,z)\,,\, \frac{\partial G}{\partial z}(x,y,z)\right> =\left< 2x\,,\,2y\,,\,2(z+1)\right> \end{aligned}$$

In particular,

$${ \boldsymbol{\nabla} } F(0,0,0)=\left< 0,0,-2\right>\qquad { \boldsymbol{\nabla} } G(0,0,0)=\left< 0,0,2\right>$$

so that the vector $\hat{\mathbf{k}}=-\frac{1}{2}{ \boldsymbol{\nabla} } F(0,0,0) =\frac{1}{2}{ \boldsymbol{\nabla} } G(0,0,0)$ is normal to both surfaces at $(0,0,0)$. So the tangent plane to both $S_1$ and $S_2$ at $(0,0,0)$ is

$$\hat{\mathbf{k}}\cdot\left< x-0,y-0,z-0\right>=0\qquad\text{or}\qquad z=0$$

Denote by $P$ the plane $z=0$. Thus $S_1$ is tangent to $P$ at $(0,0,0)$ and $P$ is tangent to $S_2$ at $(0,0,0)$. So it is reasonable to say that $S_1$ and $S_2$ are tangent at $(0,0,0)$.