参数曲线
There are a great many curves out there that we can’t even write down as a single equation in terms of only $x$ and $y$. So, to deal with some of these problems we introduce parametric equations. Instead of defining $y$ in terms of $x$ ($y = f\left( x \right)$) or $x$ in terms of $y$ ($x = h\left( y \right)$) we define both $x$ and $y$ in terms of a third variable called a parameter as follows,
$$
x = f\left( t \right) \\
y = g\left( t \right) \\
z = h\left( t \right)
$$
or
$$
\mathbf{r}(t)=\langle x(t), y(t), z(t)\rangle
$$
This third variable is usually denoted by $t$ (as we did here) but doesn’t have to be of course. Sometimes we will restrict the values of $t$ that we’ll use and at other times we won’t. This will often be dependent on the problem and just what we are attempting to do.
Each value of $t$ defines a point $\left( {x,y} \right) = \left( {f\left( t \right),g\left( t \right)} \right)$ that we can plot. The collection of points that we get by letting $t$ be all possible values is the graph of the parametric equations and is called the parametric curve.
Calculus II - Parametric Equations and Curves
Tangent equation
We have a 3Dparametric curve:
$$
\begin{array}{l}x=\varphi(t); \\ y=\psi(t); \\ z=\omega(t);\end{array}
$$
The tangent at $M\left(x_0, y_0, z_0\right)$ is:
$$
\frac{x-x_0}{\varphi^{\prime}\left(t_0\right)}=\frac{y-y_0}{\psi^{\prime}\left(t_0\right)}=\frac{z-z_0}{\omega^{\prime}\left(t_0\right)}
$$
Normal plane
The normal plane at $M$ is:
$$
\varphi^{\prime}\left(t_0\right)\left(x-x_0\right)+\psi^{\prime}\left(t_0\right)\left(y-y_0\right)+\omega^{\prime}\left(t_0\right)\left(z-z_0\right)=0
$$
参数曲面
A one-dimensional curve in space results from a vector function that relies upon one parameter, so a two-dimensional surface naturally involves the use of two parameters.
If
$$
x=x(s, t) \\
y=y(s, t) \\
z=z(s, t)
$$
are functions of independent parameters $ s $ and $ t $, then the terminal points of all vectors of the form:
$$
\mathbf{r}(s, t)=x(s, t) \mathbf{i}+y(s, t) \mathbf{j}+z(s, t) \mathbf{k}
$$
form a surface in space. The equations $ x=x(s, t)$, $y=y(s, t) $, and $ z=z(s, t) $ are the parametric equations for the surface, or a parametrization of the surface. In Preview Activity 11.6.1 we investigate how to parameterize a cylinder and a cone.
Normal vector
Assume $M(x_0, y_0, z_0)$ is a point on parametric surface $F(x_0, y_0, z_0)$:
The normal vector at $M$ is:
$$
\vec{n} =[F_x\left(x_0, y_0, z_0\right), F_y\left(x_0, y_0, z_0\right), F_z\left(x_0, y_0, z_0\right)]
$$
Tangent plane at $M$
$$
F_x\left(x_0, y_0, z_0\right)\left(x-x_0\right)+F_y\left(x_0, y_0, z_0\right)\left(y-y_0\right)+F_z\left(x_0, y_0, z_0\right)\left(z-z_0\right)=0
$$
Normal line equation
$$
\frac{x-x_0}{F_x\left(x_0, y_0, z_0\right)}=\frac{y-y_0}{F_y\left(x_0, y_0, z_0\right)}=\frac{z-z_0}{F_z\left(x_0, y_0, z_0\right)}
$$
椭圆的参数化表示
Preview Activity 11.6.1. Recall the standard parameterization of the unit circle that is given by$$
x(t)=\cos (t) \quad \text { and } \quad y(t)=\sin (t),
$$where $ 0 \leq t \leq 2 \pi $
a. Determine a parameterization of the circle of radius 1 in $\mathbb{R}^{3}$ that has its center at $ (0,0,1) $ and lies in the plane $ z=1 $.b. Determine a parameterization of the circle of radius $1$ in 3-space that has its center at $ (0,0,-1) $ and lies in the plane $ z=-1 $.
c. Determine a parameterization of the circle of radius 1 in 3-space that has its center at $ (0,0,5) $ and lies in the plane $ z=5 $.d. Taking into account your responses in (a), (b), and (c), describe the graph that results from the set of parametric equations
$$
x(s, t)=\cos (t), \quad y(s, t)=\sin (t), \quad \text { and } \quad z(s, t)=s,
$$where $ 0 \leq t \leq 2 \pi $ and $ -5 \leq s \leq 5 $. Explain your thinking.
e. Just as a cylinder can be viewed as a "stack" of circles of constant radius, a cone can be viewed as a stack of circles with varying radius. Modify the parametrizations of the circles above in order to construct the parameterization of a cone whose vertex lies at the origin, whose base radius is 4 , and whose height is 3 , where the base of the cone lies in the plane $ z=3 $. Use appropriate technology to plot the parametric equations you develop. (Hint: The cross sections parallel to the $ x y $-plane are circles, with the radii varying linearly as $ z $ increases.)
Surfaces Defined Parametrically and Surface Area
平面的点法线方程
一个平面 I上一点 $M_{0}\left(x_{0}, y_{0}, z_{0}\right) $ 和它的一个法线向量 $ n=(A, B, C) $ 为已知时,位置就完全确定了，下面我们来建立平面方程:
设 $ M(x, y, z) $ 是平面 $ \Gamma $ 上的任一点则向量 $ \overrightarrow{M_{0} M} $ 必与平面 $ \Gamma $ 的法线向量 $ n $ 垂直,即它们的数量积等于零
$$
n \cdot \overrightarrow{M_{0} M}=0
$$
因为
$$
\begin{array}{c}
n=(A, B, C) \\
\overrightarrow{M_{0} M}=\left(x-x_{0}, y-y_{0}, z-z_{0}\right)
\end{array}
$$
所以有
$$
A\left(x-x_{0}\right)+B\left(y-y_{0}\right)+C\left(z-z_{0}\right)=0
$$
上式即为平面方程。
- 例题
求过点 (2,-3,0) 且以 n=(1,-2,3) 为法线向量的平面的方程。
解: 根据平面的,点法式方程
$$
A\left(x-x_{0}\right)+B\left(y-y_{0}\right)+C\left(z-z_{0}\right)=0
$$
得所得平面的方程为
$$
(x-2)-2(y+3)+3 z=0
$$
即
$$
x-2 y+3 z-8=0
$$
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