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 $(0,0,0)$.

Denote by $S$ the surface $G(x,y,z)=0$ and by $C$ the parametrized curve $\mathbf{r}(t)=\big(x(t),y(t),z(t)\big)$. To start, we'll find the tangent plane to $S$ at $\mathbf{r}_0$ and the tangent line to $C$ at $\mathbf{r}_0$.

• The tangent vector to $C$ at $\mathbf{r}_0$ is $\left( x'(t_0)\,,\,y'(t_0)\,,\,z'(t_0) \right)$, so the parametric equations for the tangent line to $C$ at $\mathbf{r}_0$ are

$$x-x_0 = t x'(t_0)\qquad y-y_0 = t x'(t_0)\qquad z-z_0 = t x'(t_0) \tag{$E_1$}$$

• The gradient $\left(\frac{\partial G}{\partial x}\big( x_0\,,\,y_0\,,\,z_0\big)\,,\, \frac{\partial G}{\partial y}\big( x_0\,,\,y_0\,,\,z_0\big)\,,\, \frac{\partial G}{\partial z}\big( x_0\,,\,y_0\,,\,z_0\big)\right)$ is a normal vector to the surface $S$ at $(x_0,y_0,z_0)$. So the tangent plane to the surface $S$ at $(x_0,y_0,z_0)$ is

$$\left(\frac{\partial G}{\partial x}\big( x_0\,,\,y_0\,,\,z_0\big)\,,\, \frac{\partial G}{\partial y}\big( x_0\,,\,y_0\,,\,z_0\big)\,,\, \frac{\partial G}{\partial z}\big( x_0\,,\,y_0\,,\,z_0\big)\right) \cdot \left( x-x_0\,,\, y-y_0\,,\,z-z_0\right) = 0$$

or

$$\frac{\partial G}{\partial x}\big( x_0\,,\,y_0\,,\,z_0\big)\ (x-x_0) +\frac{\partial G}{\partial y}\big( x_0\,,\,y_0\,,\,z_0\big)\ (y-y_0) +\frac{\partial G}{\partial z}\big( x_0\,,\,y_0\,,\,z_0\big)\ (z-z_0) = 0 \tag{$E_2$}$$

Next, we'll show that the tangent vector $\left( x'(t_0)\,,\,y'(t_0)\,,\,z'(t_0) \right)$ to $C$ at $\mathbf{r}_0$ and the normal vector $\left(\frac{\partial G}{\partial x}\big( x_0\,,\,y_0\,,\,z_0\big)\,,\, \frac{\partial G}{\partial y}\big( x_0\,,\,y_0\,,\,z_0\big)\,,\, \frac{\partial G}{\partial z}\big( x_0\,,\,y_0\,,\,z_0\big)\right)$ to $S$ at $\mathbf{r}_0$ are perpendicular to each other. To do so, we observe that, for every $t$, the point $\big(x(t),y(t),z(t)\big)$ lies on the surface $G(x,y,z)=0$ and so obeys

$$\begin{aligned} G\big(x(t),y(t),z(t)\big) =0 \end{aligned}$$

Differentiating this equation with respect to $t$ gives, by the chain rule,

$$\begin{aligned} 0&= \frac{\mathrm{d}}{\mathrm{d}t}G\big(x(t),y(t),z(t)\big) \\ &=\frac{\partial G}{\partial x}\big( x(t)\,,\,y(t)\,,\,z(t)\big)\ x'(t) +\frac{\partial G}{\partial y}\big( x(t)\,,\,y(t)\,,\,z(t)\big)\ y'(t) +\frac{\partial G}{\partial z}\big( x(t)\,,\,y(t)\,,\,z(t)\big)\ z'(t) \end{aligned}$$

Then setting $t=t_0$ gives

$$\frac{\partial G}{\partial x}\big( x_0\,,\,y_0\,,\,z_0\big)\ x'(t_0) +\frac{\partial G}{\partial y}\big( x_0\,,\,y_0\,,\,z_0\big)\ y'(t_0) +\frac{\partial G}{\partial z}\big( x_0\,,\,y_0\,,\,z_0\big)\ z'(t_0) = 0 \tag{$E_3$}$$

Finally, we are in a position to show that if $(x,y,z)$ is any point on the tangent line to $C$ at $\mathbf{r}_0$, then $(x,y,z)$ is also on the tangent plane to $S$ at $\mathbf{r}_0$. As $(x,y,z)$ is on the tangent line to $C$ at $\mathbf{r}_0$ then there is a $t$ such that, by $(E_1)$,

$$\begin{aligned} &\frac{\partial G}{\partial x}\big( x_0\,,\,y_0\,,\,z_0\big)\ \{x-x_0\} +\frac{\partial G}{\partial y}\big( x_0\,,\,y_0\,,\,z_0\big)\ \{y-y_0\} +\frac{\partial G}{\partial z}\big( x_0\,,\,y_0\,,\,z_0\big)\ \{z-z_0\} \\ &=\frac{\partial G}{\partial x}\big( x_0\,,\,y_0\,,\,z_0\big)\ \big\{ t\,x'(t_0)\big\} +\frac{\partial G}{\partial y}\big( x_0\,,\,y_0\,,\,z_0\big)\ \big\{ t\,y'(t_0)\big\} +\frac{\partial G}{\partial z}\big( x_0\,,\,y_0\,,\,z_0\big)\ \big\{ t\,z'(t_0)\big\}\\ &= t\left[\frac{\partial G}{\partial x}\big( x_0\,,\,y_0\,,\,z_0\big)\ x'(t_0) +\frac{\partial G}{\partial y}\big( x_0\,,\,y_0\,,\,z_0\big)\ y'(t_0) +\frac{\partial G}{\partial z}\big( x_0\,,\,y_0\,,\,z_0\big)\ z'(t_0) \right] =0 \end{aligned}$$

by $(E_3)$. That is, $(x,y,z)$ obeys the equation, $(E_2)$, of the tangent plane to $S$ at $\mathbf{r}_0$ and so is on that tangent plane. So the tangent line to $C$ at $\mathbf{r}_0$ is contained in the tangent plane to $S$ at $\mathbf{r}_0$.