매개변수 곡면

(a) In the sketch below, the point $(x,y,z)$ deviates from the centre $(2,2,4)$ by $\sin\theta$ units in the $\hat{\mathbf{k}}$ direction, and by $\cos\theta$ units in the $\sqrt \frac1{\sqrt 2}(\hat{\boldsymbol{\imath}}+\hat{\boldsymbol{\jmath}})$ direction. So, $(x,y,z)=(2+\frac1{\sqrt 2}\cos\theta~,~2+\frac1{\sqrt 2}\cos\theta~,~4+\sin\theta)$.

So, we can parametrize the circle as $(x,y,z)=(2+\frac1{\sqrt 2}\cos\theta~,~2+\frac1{\sqrt 2}\cos\theta~,~4+\sin\theta)$, with $0 \le \theta \le 2\pi$.

Remark: it's easy to check that this equation satisfies the two properties we desire. Since the $x$- and $y$ coordinates match, it's in the plane $x=y$. To check that it's a circle centred at $(2,2,4)$, we note the distance from $(x,y,z)$ to $(2,2,4)$ is:

$$\begin{aligned} d&=\sqrt{(x-2)^2+(y-2)^2+(z-4)^2}=\sqrt{\left(\frac1{\sqrt 2}\cos\theta\right)^2+\left(\frac1{\sqrt 2}\cos\theta\right)^2+\left(\sin\theta\right)^2}\\ &=\sqrt{\frac12\cos^2\theta+\frac12\cos^2\theta+\sin^2\theta}=\sqrt{\cos^2\theta+\sin^2\theta}=1 \end{aligned}$$

So, our points all have distance one from the same point — that is, they lie on a circle of radius 1.

(b) Consider a point $(x,y,z)=(2+\frac1{\sqrt 2}\cos\theta~,~2+\frac1{\sqrt 2}\cos\theta~,~4+\sin\theta)$, rotating $\phi$ radians about the line $x=y=4$.

The new position of the point has the same height, $z=4+\sin\theta$. Its distance from the line $x=y=4$ is also preserved: $R=\sqrt{(x-4)^2+(y-4)^2+(z-z)^2}=\sqrt{(\frac1{\sqrt 2}\cos\theta-2)^2+(\frac1{\sqrt 2}\cos\theta-2)^2+0)}=\cos\theta-2\sqrt2$.

The circle traced out by a point $(x,y,z)=(2+\frac1{\sqrt2}\cos\theta,2+\frac{1}{\sqrt 2} \cos \theta,4+\sin\theta)$ on the circle is centred at $(4,4,z)$ with radius $\sqrt2(4-x)$, so it has equation $x=4+\sqrt2(2-\sqrt2\cos\theta)\cos\phi$, $y=4+\sqrt2(2-\sqrt2\cos\theta)\sin\phi$, $z=4\sin\theta$.

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 $\mathbf{r}_0$.