参数方程

发布于 2022-07-12  329 次阅读


Please refresh the page if equations are not rendered correctly.
---------------------------------------------------------------

如公式显示不正确,请刷新(F5)网页

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:
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

x-2 y+3 z-8=0

届ける言葉を今は育ててる
最后更新于 2022-11-28