매개변수 곡면

This parametrization is almost trivial. We know it will have the form $\mathbf{r}(x,y)= \psi_1(x,y)\hat{\boldsymbol{\imath}}+ \psi_2(x,y)\hat{\boldsymbol{\jmath}}+ \psi_3(x,y)\hat{\mathbf{k}}$ where $\psi_1$ gives the $x$-component (i.e. $x$), $\psi_2$ gives the $y$-component (i.e. $y$), and $\psi_3$ gives the $z$-component (i.e. $e^{x+1}+xy$). So, $\mathbf{r}(x,y)= x\hat{\boldsymbol{\imath}}+ y\hat{\boldsymbol{\jmath}}+ (e^{x+1}+xy)\hat{\mathbf{k}}$

Our parametrization is

$$\begin{aligned} x(u,v) &= u+v \\ y(u,v) &= u^2+v^2 \\ z(u,v) &= u-v \end{aligned}$$

• Adding $x(u,v)$ and $z(u,v)$ gives $x(u,v)+z(u,v) = 2u$.
• Subtracting $z(u,v)$ from $x(u,v)$ gives $x(u,v)-z(u,v) = 2v$.

So $u=\frac{1}{2}\big(x(u,v)+z(u,v)\big)$ and $v=\frac{1}{2}\big(x(u,v)-z(u,v)\big)$. So on our surface

$$\begin{aligned} y(u,v) &= u^2+v^2 = \frac{1}{4}\big(x(u,v)+z(u,v)\big)^2 +\frac{1}{4}\big(x(u,v)-z(u,v)\big)^2 \\ &= \frac{1}{2} x(u,v)^2 + \frac{1}{2} z(u,v)^2 \end{aligned}$$

All points of our surface lie on $2y= x^2+z^2$. This is a parabolic bowl:

• no points have $y<0$ and
• the $y=Y$ (with $Y> 0$) cross-section is the circle $x^2+z^2=2Y$, $y=Y$
• the $x=0$ cross-section is the parabola $2y=z^2$, $x=0$
• the $z=0$ cross-section is the parabola $2y=x^2$, $z=0$

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 $(0,0,0)$.