1 Parametric Equations and Polar Coordinates 参数方程与极坐标

本页译自 OpenStax《Calculus Volume 3》第 1 章 Parametric Equations and Polar Coordinates。公式经本地 MathJax 渲染，自定义宏已注入。

1.1 Parametric Equations 1.1 参数方程

1.1.1 Plot a curve described by parametric equations.
1.1.2 Convert the parametric equations of a curve into the form $y = f(x).$
1.1.3 Recognize the parametric equations of basic curves, such as a line and a circle.
1.1.4 Recognize the parametric equations of a cycloid.

1.1.1 描绘由参数方程描述的曲线。
1.1.2 将曲线的参数方程化为 $y = f(x)$ 的形式。
1.1.3 识别基本曲线（如直线与圆）的参数方程。
1.1.4 识别摆线的参数方程。

In this section we examine parametric equations and their graphs. In the two-dimensional coordinate system, parametric equations are useful for describing curves that are not necessarily functions. The parameter is an independent variable that both *x* and *y* depend on, and as the parameter increases, the values of *x* and *y* trace out a path along a plane curve. For example, if the parameter is *t* (a common choice), then *t* might represent time. Then *x* and *y* are defined as functions of time, and $\left( {x(t),y(t)} \right)$ can describe the position in the plane of a given object as it moves along a curved path.

本节我们研究参数方程及其图形。在二维坐标系中，参数方程适用于描述未必是函数的曲线。参数是 *x* 与 *y* 都依赖的独立变量；随着参数增大，*x* 与 *y* 的取值沿一条平面曲线描出一条路径。例如，若参数为 *t*（一种常见选择），则 *t* 可表示时间。于是 *x* 与 *y* 定义为时间的函数，且 $\left( {x(t),y(t)} \right)$ 可描述给定物体沿曲线路径运动时在平面中的位置。

Parametric Equations and Their Graphs 参数方程及其图形

Consider the orbit of Earth around the Sun. Our year lasts approximately 365.25 days, but for this discussion we will use 365 days. On January 1 of each year, the physical location of Earth with respect to the Sun is nearly the same, except for leap years, when the lag introduced by the extra $\frac{1}{4}$ day of orbiting time is built into the calendar. We call January 1 “day 1” of the year. Then, for example, day 31 is January 31, day 59 is February 28, and so on.

考虑地球绕太阳的轨道。我们的一年约为 365.25 天，但此处讨论取 365 天。每年 1 月 1 日，地球相对于太阳的实际位置几乎相同，闰年除外——闰年的日历吸收了多出的 $\frac{1}{4}$ 天公转时间造成的滞后。我们把 1 月 1 日称为一年中的“第 1 天”。于是，例如第 31 天是 1 月 31 日，第 59 天是 2 月 28 日，依此类推。

The number of the day in a year can be considered a variable that determines Earth’s position in its orbit. As Earth revolves around the Sun, its physical location changes relative to the Sun. After one full year, we are back where we started, and a new year begins. According to Kepler’s laws of planetary motion, the shape of the orbit is elliptical, with the Sun at one focus of the ellipse. We study this idea in more detail in Conic Sections.

一年中的天数可视为决定地球轨道位置的变量。地球绕太阳公转时，其实际位置相对于太阳不断变化。经过一整年后，我们又回到起点，新的一年由此开始。根据开普勒行星运动定律，轨道形状为椭圆，太阳位于椭圆的一个焦点上。我们将在“圆锥曲线”中更详细地研究这一思想。

Figure 1.2 depicts Earth’s orbit around the Sun during one year. The point labeled $F_{2}$ is one of the foci of the ellipse; the other focus is occupied by the Sun. If we superimpose coordinate axes over this graph, then we can assign ordered pairs to each point on the ellipse (Figure 1.3). Then each *x* value on the graph is a value of position as a function of time, and each *y* value is also a value of position as a function of time. Therefore, each point on the graph corresponds to a value of Earth’s position as a function of time.

图 1.2 描绘了地球一年内绕太阳的轨道。标有 $F_{2}$ 的点是椭圆的一个焦点；另一个焦点被太阳占据。如果我们将坐标轴叠加到该图形上，就能给椭圆上的每个点赋予有序对（图 1.3）。于是图形上的每个 *x* 值都是位置作为时间函数的值，每个 *y* 值也是位置作为时间函数的值。因此，图形上的每个点都对应地球位置作为时间函数的一个值。

We can determine the functions for $x(t)$ and $y(t),$ thereby parameterizing the orbit of Earth around the Sun. The variable $t$ is called an independent parameter and, in this context, represents time relative to the beginning of each year.

我们可以确定 $x(t)$ 与 $y(t)$ 的函数，从而对地球绕太阳的轨道进行参数化。变量 $t$ 称为独立参数，在此背景下表示相对于每年年初的时间。

A curve in the $\left( {x,y} \right)$ plane can be represented parametrically. The equations that are used to define the curve are called parametric equations.

$\left( {x,y} \right)$ 平面中的曲线可以用参数形式表示。用于定义该曲线的方程称为参数方程。

If *x* and *y* are continuous functions of *t* on an interval *I*, then the equations

若 *x* 与 *y* 是区间 *I* 上关于 *t* 的连续函数，则方程

$$x = x(t)\ \text{and}\ y = y(t)$$

are called parametric equations and *t* is called the parameter. The set of points $\left( {x,y} \right)$ obtained as *t* varies over the interval *I* is called the graph of the parametric equations. The graph of parametric equations is called a parametric curve or *plane curve*, and is denoted by *C*.

称为参数方程，*t* 称为参数。当 *t* 取遍区间 *I* 时得到的有序对集合 $\left( {x,y} \right)$ 称为参数方程的图形。参数方程的图形称为参数曲线或*平面曲线*，记作 *C*。

Notice in this definition that *x* and *y* are used in two ways. The first is as functions of the independent variable *t.* As *t* varies over the interval *I*, the functions $x(t)$ and $y(t)$ generate a set of ordered pairs $\left( {x,y} \right).$ This set of ordered pairs generates the graph of the parametric equations. In this second usage, to designate the ordered pairs, *x* and *y* are variables. It is important to distinguish the variables *x* and *y* from the functions $x(t)$ and $y(t).$

注意在这一定义中，*x* 与 *y* 有两种用法。第一种是作为独立变量 *t* 的函数。当 *t* 取遍区间 *I* 时，函数 $x(t)$ 与 $y(t)$ 生成有序对集合 $\left( {x,y} \right).$ 这个有序对集合生成参数方程的图形。在第二种用法中，为表示有序对，*x* 与 *y* 是变量。将变量 *x*、*y* 与函数 $x(t)$、$y(t)$ 区分开十分重要。

Graphing a Parametrically Defined Curve 绘制参数定义的曲线

Sketch the curves described by the following parametric equations:

描绘下列参数方程所描述的曲线：

1. $x(t) = t - 1,\quad y(t) = 2t + 4,\quad-3 \leq t \leq 2$

2. $x(t) = t^{2} - 3,\quad y(t) = 2t + 1,\quad-2 \leq t \leq 3$

3. $x(t) = 4\ \text{cos}\ t,\quad y(t) = 4\ \text{sin}\ t,\quad 0 \leq t \leq 2\pi$

Solution 解答

1. To create a graph of this curve, first set up a table of values. Since the independent variable in both $x(t)$ and $y(t)$ is *t*, let *t* appear in the first column. Then $x(t)$ and $y(t)$ will appear in the second and third columns of the table.

1. 要绘制该曲线图形，先建立数值表。由于在 $x(t)$ 与 $y(t)$ 中的独立变量都是 *t*，让 *t* 出现在第一列。于是 $x(t)$ 与 $y(t)$ 将出现在表的第二、三列。

| *t* | $x(t)$ | $y(t)$ |

|-----|--------|--------|

| −3 | −4 | −2 |

| −2 | −3 | 0 |

| −1 | −2 | 2 |

| 0 | −1 | 4 |

| 1 | 0 | 6 |

| 2 | 1 | 8 |

The second and third columns in this table provide a set of points to be plotted. The graph of these points appears in Figure 1.4. The arrows on the graph indicate the orientation of the graph, that is, the direction that a point moves on the graph as *t* varies from −3 to 2.

该表的第二、三列给出一组待描出的点。这些点的图形见图 1.4。图形上的箭头表示图形的定向，即当 *t* 从 −3 变到 2 时点在图形上移动的方向。

2. To create a graph of this curve, again set up a table of values.

2. 要绘制该曲线图形，再次建立数值表。

| *t* | $x(t)$ | $y(t)$ |

|-----|--------|--------|

| −2 | 1 | −3 |

| −1 | −2 | −1 |

| 0 | −3 | 1 |

| 1 | −2 | 3 |

| 2 | 1 | 5 |

| 3 | 6 | 7 |

The second and third columns in this table give a set of points to be plotted (Figure 1.5). The first point on the graph (corresponding to $t = -2)$ has coordinates $\left( {1,-3} \right),$ and the last point (corresponding to $t = 3)$ has coordinates $\left( {6,7} \right).$ As *t* progresses from −2 to 3, the point on the curve travels along a parabola. The direction the point moves is again called the orientation and is indicated on the graph.

该表的第二、三列给出一组待描出的点（图 1.5）。图形上的第一个点（对应 $t = -2$）坐标为 $\left( {1,-3} \right)$，最后一个点（对应 $t = 3$）坐标为 $\left( {6,7} \right).$ 当 *t* 从 −2 变到 3 时，曲线上的点沿一条抛物线移动。点移动的方向同样称为定向，并在图形上标出。

3. In this case, use multiples of $\pi\text{/}6$ for *t* and create another table of values:

3. 本例取 *t* 为 $\pi\text{/}6$ 的倍数，再建立数值表：

| *t* | $x(t)$ | $y(t)$ | | *t* | $x(t)$ | $y(t)$ |

|------------------|---------------------------|-------------------------|-----|-------------------|---------------------------|---------------------------|

| 0 | 4 | 0 | | $\frac{7\pi}{6}$ | $-2\sqrt{3} \approx -3.5$ | 2 |

| $\frac{\pi}{6}$ | $2\sqrt{3} \approx 3.5$ | $2$ | | $\frac{4\pi}{3}$ | −2 | $-2\sqrt{3} \approx -3.5$ |

| $\frac{\pi}{3}$ | $2$ | $2\sqrt{3} \approx 3.5$ | | $\frac{3\pi}{2}$ | 0 | −4 |

| $\frac{\pi}{2}$ | 0 | 4 | | $\frac{5\pi}{3}$ | 2 | $-2\sqrt{3} \approx -3.5$ |

| $\frac{2\pi}{3}$ | −2 | $2\sqrt{3} \approx 3.5$ | | $\frac{11\pi}{6}$ | $2\sqrt{3} \approx 3.5$ | 2 |

| $\frac{5\pi}{6}$ | $-2\sqrt{3} \approx -3.5$ | 2 | | $2\pi$ | 4 | 0 |

| $\pi$ | −4 | 0 | | | | |

The graph of this plane curve appears in the following graph.

该平面曲线的图形如下图所示。

This is the graph of a circle with radius 4 centered at the origin, with a counterclockwise orientation. The starting point and ending points of the curve both have coordinates $\left( {4,0} \right).$

这是半径为 4、以原点为圆心的圆，定向为逆时针。曲线的起点与终点坐标均为 $\left( {4,0} \right).$

Sketch the curve described by the parametric equations

描绘下列参数方程所描述的曲线

$$x(t) = 3t + 2,\quad y(t) = t^{2} - 1,\quad-3 \leq t \leq 2.$$

Eliminating the Parameter 消去参数

To better understand the graph of a curve represented parametrically, it is useful to rewrite the two equations as a single equation relating the variables *x* and *y.* Then we can apply any previous knowledge of equations of curves in the plane to identify the curve. For example, the equations describing the plane curve in Example 1.1b. are

为了更好地理解用参数表示的曲线图形，把两个方程改写为一个关联变量 *x* 与 *y* 的单一方程是有用的。这样我们就能运用先前关于平面曲线方程的知识来识别曲线。例如，例 1.1b 中描述平面曲线的方程为

$$x(t) = t^{2} - 3,\quad y(t) = 2t + 1,\quad-2 \leq t \leq 3.$$

Solving the second equation for *t* gives

对第二个方程解 *t* 得

$$t = \frac{y - 1}{2}.$$

This can be substituted into the first equation:

这可代入第一个方程：

$$x = \left( \frac{y - 1}{2} \right)^{2} - 3 = \frac{y^{2} - 2y + 1}{4} - 3 = \frac{y^{2} - 2y - 11}{4}.$$

This equation describes *x* as a function of *y.* These steps give an example of *eliminating the parameter*. The graph of this function is a parabola opening to the right. Recall that the plane curve started at $\left( {1,-3} \right)$ and ended at $\left( {6,7} \right).$ These terminations were due to the restriction on the parameter *t.*

该方程把 *x* 描述为 *y* 的函数。这些步骤给出了一个*消去参数*的例子。该函数的图形是开口向右的抛物线。回顾该平面曲线起点为 $\left( {1,-3} \right)$、终点为 $\left( {6,7} \right).$ 这些端点是由对参数 *t* 的限制造成的。

Eliminating the Parameter 消去参数

Eliminate the parameter for each of the plane curves described by the following parametric equations and describe the resulting graph.

对下列参数方程所描述的每个平面曲线消去参数，并说明所得图形的形状。

1. $x(t) = \sqrt{2t + 4},\quad y(t) = 2t + 1,\quad-2 \leq t \leq 6$

2. $x(t) = 4\ \text{cos}\ t,\quad y(t) = 3\ \text{sin}\ t,\quad 0 \leq t \leq 2\pi$

Solution 解答

1. To eliminate the parameter, we can solve either of the equations for *t.* For example, solving the first equation for *t* gives

1. 为消去参数，我们可以解两个方程之一求 *t*。例如，对第一个方程解 *t* 得

$$\begin{array}{rll}

x & = & \sqrt{2t + 4} \\

x^{2} & = & {2t + 4} \\

{x^{2} - 4} & = & {2t} \\

t & = & {\frac{x^{2} - 4}{2}.}

\end{array}$$

Note that when we square both sides it is important to observe that $x \geq 0.$ Substituting $t = \frac{x^{2} - 4}{2}$ this into $y(t)$ yields

注意，两边平方时必须留意 $x \geq 0.$ 将 $t = \frac{x^{2} - 4}{2}$ 代入 $y(t)$ 得

$$\begin{array}{rll}

{y(t)} & = & {2t + 1} \\

y & = & {2\left( \frac{x^{2} - 4}{2} \right) + 1} \\

y & = & {x^{2} - 4 + 1} \\

y & = & {x^{2} - 3.}

\end{array}$$

This is the equation of a parabola opening upward. There is, however, a domain restriction because of the limits on the parameter *t*. When $t = -2,$ $x = \sqrt{2(-2) + 4} = 0,$ and when $t = 6,$ $x = \sqrt{2(6) + 4} = 4.$ The graph of this plane curve follows.

这是一条开口向上的抛物线方程。但由于参数 *t* 的取值范围限制，存在定义域限制。当 $t = -2$ 时，$x = \sqrt{2(-2) + 4} = 0$；当 $t = 6$ 时，$x = \sqrt{2(6) + 4} = 4.$ 该平面曲线的图形如下。

2. Sometimes it is necessary to be a bit creative in eliminating the parameter. The parametric equations for this example are

2. 有时消去参数需要一些技巧。本例的参数方程为

$$x(t) = 4\ \text{cos}\ t\ \text{and}\ y(t) = 3\ \text{sin}\ t.$$

Solving either equation for *t* directly is not advisable because sine and cosine are not one-to-one functions. However, dividing the first equation by 4 and the second equation by 3 (and suppressing the *t*) gives us

直接解任一方程求 *t* 并不可取，因为正弦与余弦不是一一对应函数。但将第一个方程除以 4、第二个方程除以 3（并略去 *t*）可得

$$\text{cos}\ t = \frac{x}{4}\ \text{and}\ \text{sin}\ t = \frac{y}{3}.$$

Now use the Pythagorean identity $\text{cos}^{2}t + \text{sin}^{2}t = 1$ and replace the expressions for $\text{sin}\ t$ and $\text{cos}\ t$ with the equivalent expressions in terms of *x* and *y*. This gives

现在利用勾股恒等式 $\text{cos}^{2}t + \text{sin}^{2}t = 1$，并将 $\text{sin}\ t$ 与 $\text{cos}\ t$ 的表达式替换为关于 *x*、*y* 的等价表达式。于是得到

$$\begin{array}{rll}

{\left( \frac{x}{4} \right)^{2} + \left( \frac{y}{3} \right)^{2}} & = & 1 \\

{\frac{x^{2}}{16} + \frac{y^{2}}{9}} & = & 1.

\end{array}$$

This is the equation of a horizontal ellipse centered at the origin, with semimajor axis 4 and semiminor axis 3 as shown in the following graph.

这是一个以原点为中心的横向椭圆方程，半长轴为 4、半短轴为 3，如下图所示。

As *t* progresses from $0$ to $2\pi,$ a point on the curve traverses the ellipse once, in a counterclockwise direction. Recall from the section opener that the orbit of Earth around the Sun is also elliptical. This is a perfect example of using parameterized curves to model a real-world phenomenon.

当 *t* 从 $0$ 到 $2\pi$ 变化时，曲线上的一点沿逆时针方向绕椭圆一周。回顾本节开头，地球绕太阳的轨道也是椭圆形的。这是用参数曲线建模现实世界现象的绝佳例子。

Eliminate the parameter for the plane curve defined by the following parametric equations and describe the resulting graph.

对下列参数方程所定义的平面曲线消去参数，并说明所得图形的形状。

$$x(t) = 2 + \frac{3}{t},\quad y(t) = t - 1,\quad 2 \leq t \leq 6$$

So far we have seen the method of eliminating the parameter, assuming we know a set of parametric equations that describe a plane curve. What if we would like to start with the equation of a curve and determine a pair of parametric equations for that curve? This is certainly possible, and in fact it is possible to do so in many different ways for a given curve. The process is known as parameterization of a curve.

至此，我们已了解消去参数的方法，前提是已知描述某平面曲线的一组参数方程。如果我们想从一条曲线的方程出发，反求它的一组参数方程，该怎么办？这当然可行，而且事实上对一条给定曲线可以用许多不同的方式做到。这一过程称为曲线的参数化。

Parameterizing a Curve 对曲线进行参数化

Find two different pairs of parametric equations to represent the graph of $y = 2x^{2} - 3.$

求两组不同的参数方程来表示 $y = 2x^{2} - 3$ 的图形。

Solution 解答

First, it is always possible to parameterize a curve by defining $x(t) = t,$ then replacing *x* with *t* in the equation for $y(t).$ This gives the parameterization

首先，总可以通过令 $x(t) = t$，再在 $y(t)$ 的方程中用 *t* 替换 *x* 来对曲线进行参数化。这样就得到如下参数化

$$x(t) = t,\quad y(t) = 2t^{2} - 3.$$

Since there is no restriction on the domain in the original graph, there is no restriction on the values of *t.*

由于原图形对定义域没有限制，对 *t* 的取值也就没有限制。

We have complete freedom in the choice for the second parameterization. For example, we can choose $x(t) = 3t - 2.$ The only thing we need to check is that there are no restrictions imposed on *x*; that is, the range of $x(t)$ is all real numbers. This is the case for $x(t) = 3t - 2.$ Now since $y = 2x^{2} - 3,$ we can substitute $x(t) = 3t - 2$ for *x.* This gives

对第二种参数化我们有完全的自由。例如，可取 $x(t) = 3t - 2.$ 唯一需要检查的是对 *x* 没有附加限制，即 $x(t)$ 的值域为全体实数。对 $x(t) = 3t - 2$ 正是如此。既然 $y = 2x^{2} - 3,$ 我们可将 *x* 替换为 $x(t) = 3t - 2.$ 于是得到

$$\begin{array}{cl}

{y(t)} & {= 2\left( {3t - 2} \right)^{2} - 3} \\

& {= 2\left( {9t^{2} - 12t + 4} \right) - 3} \\

& {= 18t^{2} - 24t + 8 - 3} \\

& {= 18t^{2} - 24t + 5.}

\end{array}$$

Therefore, a second parameterization of the curve can be written as

因此，该曲线的第二种参数化可写为

$$x(t) = 3t - 2\ \text{and}\ y(t) = 18t^{2} - 24t + 5.$$

Find two different sets of parametric equations to represent the graph of $y = x^{2} + 2x.$

求两组不同的参数方程来表示 $y = x^{2} + 2x$ 的图形。

Cycloids and Other Parametric Curves 摆线与其他参数曲线

Imagine going on a bicycle ride through the country. The tires stay in contact with the road and rotate in a predictable pattern. Now suppose a very determined ant is tired after a long day and wants to get home. So he hangs onto the side of the tire and gets a free ride. The path that this ant travels down a straight road is called a cycloid (Figure 1.9). A cycloid generated by a circle (or bicycle wheel) of radius *a* is given by the parametric equations

想象一次穿越乡间的自行车骑行。轮胎始终与路面接触，并以可预测的模式转动。现在假设一只十分坚定的蚂蚁在漫长的一天后感到疲惫，想要回家。于是它挂在轮胎侧面，搭了一次免费的车。这只蚂蚁沿笔直道路行进的轨迹称为一条摆线（图 1.9）。由半径为 *a* 的圆（或自行车轮）生成的摆线由下列参数方程给出

$$x(t) = a\left( {t - \text{sin}\ t} \right),\quad y(t) = a\left( {1 - \text{cos}\ t} \right).$$

半径为 *a* 的圆（或自行车轮）所生成的摆线由参数方程 $$x(t) = a\left( {t - \text{sin}\ t} \right),\quad y(t) = a\left( {1 - \text{cos}\ t} \right).$$ 给出。

To see why this is true, consider the path that the center of the wheel takes. The center moves along the *x*-axis at a constant height equal to the radius of the wheel. If the radius is *a*, then the coordinates of the center can be given by the equations

要理解为何如此，考虑轮心的运动轨迹。轮心沿 *x* 轴运动，其恒定高度等于轮的半径。若半径为 *a*，则轮心的坐标可由下列方程给出

$$x(t) = at,\quad y(t) = a$$

轮心的坐标由 $$x(t) = at,\quad y(t) = a$$ 给出。

for any value of $t.$ Next, consider the ant, which rotates around the center along a circular path. If the bicycle is moving from left to right then the wheels are rotating in a clockwise direction. A possible parameterization of the circular motion of the ant (relative to the center of the wheel) is given by

对任意 $t$ 值成立。接下来考虑蚂蚁，它绕轮心沿圆形路径转动。若自行车从左向右运动，则车轮顺时针转动。蚂蚁（相对于轮心）的圆周运动的一个可能参数化由下式给出

$$x(t) = \text{−}a\ \text{sin}\ t,\quad y(t) = \text{−}a\ \text{cos}\ t.$$

蚂蚁（相对于轮心）的圆周运动由 $$x(t) = \text{−}a\ \text{sin}\ t,\quad y(t) = \text{−}a\ \text{cos}\ t.$$ 给出。

(The negative sign is needed to reverse the orientation of the curve. If the negative sign were not there, we would have to imagine the wheel rotating counterclockwise.) Adding these equations together gives the equations for the cycloid.

（需要负号来反转曲线的定向。若没有负号，我们就得想象车轮在逆时针转动。）将这些方程相加即可得到摆线的方程。

$$x(t) = a\left( {t - \text{sin}\ t} \right),\quad y(t) = a\left( {1 - \text{cos}\ t} \right).$$

由此得到摆线的参数方程 $$x(t) = a\left( {t - \text{sin}\ t} \right),\quad y(t) = a\left( {1 - \text{cos}\ t} \right).$$

Now suppose that the bicycle wheel doesn’t travel along a straight road but instead moves along the inside of a larger wheel, as in Figure 1.10. In this graph, the green circle is traveling around the blue circle in a counterclockwise direction. A point on the edge of the green circle traces out the red graph, which is called a hypocycloid.

现在假设自行车轮不是沿笔直道路行进，而是沿一个更大的轮子的内部运动，如图 1.10 所示。在该图形中，绿色的圆绕蓝色的圆沿逆时针方向运动。绿圆边缘上的一点描出红色的图形，称为 hypocycloid（内摆线）。

The general parametric equations for a hypocycloid are

内摆线的一般参数方程为

$$\begin{array}{l}

{x(t) = \left( {a - b} \right)\ \text{cos}\ t + b\ \text{cos}\left( \frac{a - b}{b} \right)\ t} \\

{y(t) = \left( {a - b} \right)\ \text{sin}\ t - b\ \text{sin}\left( \frac{a - b}{b} \right)\ t.}

\end{array}$$

These equations are a bit more complicated, but the derivation is somewhat similar to the equations for the cycloid. In this case we assume the radius of the larger circle is *a* and the radius of the smaller circle is *b.* Then the center of the wheel travels along a circle of radius $a - b.$ This fact explains the first term in each equation above. The period of the second trigonometric function in both $x(t)$ and $y(t)$ is equal to $\frac{2\pi b}{a - b}.$

这些方程稍微复杂一些，但其推导与摆线的方程有些相似。此情形下我们假设大圆的半径为 *a*、小圆的半径为 *b*。于是轮心沿半径为 $a - b$ 的圆运动。这一事实解释了上面每个方程中的第一项。在 $x(t)$ 与 $y(t)$ 中第二个三角函数的周期都等于 $\frac{2\pi b}{a - b}$。

The ratio $\frac{a}{b}$ is related to the number of cusps on the graph (cusps are the corners or pointed ends of the graph), as illustrated in Figure 1.11. This ratio can lead to some very interesting graphs, depending on whether or not the ratio is rational. Figure 1.10 corresponds to $a = 4$ and $b = 1.$ The result is a hypocycloid with four cusps. Figure 1.11 shows some other possibilities. The last two hypocycloids have irrational values for $\frac{a}{b}.$ In these cases the hypocycloids have an infinite number of cusps, so they never return to their starting point. These are examples of what are known as space-filling curves.

比值 $\frac{a}{b}$ 与图形上尖点（cusps，即图形的角点或尖端）的数目有关，如图 1.11 所示。该比值会产生一些非常有趣的图形，取决于比值是否为有理数。图 1.10 对应 $a = 4$ 且 $b = 1$，结果是一条有四个尖点的内摆线。图 1.11 展示了其他一些可能情况。最后两条内摆线的 $\frac{a}{b}$ 取无理数。在这些情形下，内摆线具有无穷多个尖点，因此永远不会回到起点。这些就是所谓的 space-filling curves（空间填充曲线）的例子。

The Witch of Agnesi 阿涅西箕舌线

Many plane curves in mathematics are named after the people who first investigated them, like the folium of Descartes or the spiral of Archimedes. However, perhaps the strangest name for a curve is the witch of Agnesi. Why a witch?

数学中许多平面曲线以最先研究它们的人命名，例如笛卡儿叶形线（folium of Descartes）或阿基米德螺线（spiral of Archimedes）。然而，或许最奇怪的曲线名称要数阿涅西箕舌线（witch of Agnesi）。为何叫“女巫”？

Maria Gaetana Agnesi (1718–1799) was one of the few recognized women mathematicians of eighteenth-century Italy. She wrote a popular book on analytic geometry, published in 1748, which included an interesting curve that had been studied by Fermat in 1630. The mathematician Guido Grandi showed in 1703 how to construct this curve, which he later called the “versoria,” a Latin term for a rope used in sailing. Agnesi used the Italian term for this rope, “versiera,” but in Latin, this same word means a “female goblin.” When Agnesi’s book was translated into English in 1801, the translator used the term “witch” for the curve, instead of rope. The name “witch of Agnesi” has stuck ever since.

玛丽亚·加埃塔纳·阿涅西（Maria Gaetana Agnesi，1718–1799）是十八世纪意大利为数不多的知名女数学家之一。她写了一本广为流传的解析几何著作，于 1748 年出版，其中收录了一条有趣的曲线，该曲线曾由费马于 1630 年研究过。数学家圭多·格兰迪（Guido Grandi）在 1703 年展示了如何构造这条曲线，他后来称之为“versoria”，这是一个拉丁词，指航行所用的绳索。阿涅西使用了该绳索的意大利语词“versiera”，但在拉丁语中，同一个词意为“女妖”。1801 年阿涅西的著作被译成英文时，译者用了“witch”（女巫）来称呼这条曲线，而非绳索。自此，“witch of Agnesi”（阿涅西箕舌线）这一名称便沿用下来。

The witch of Agnesi is a curve defined as follows: Start with a circle of radius *a* so that the points $(0,0)$ and $(0,2a)$ are points on the circle (Figure 1.12). Let *O* denote the origin. Choose any other point *A* on the circle, and draw the secant line *OA*. Let *B* denote the point at which the line *OA* intersects the horizontal line through $(0,2a).$ The vertical line through *B* intersects the horizontal line through *A* at the point *P*. As the point *A* varies, the path that the point *P* travels is the witch of Agnesi curve for the given circle.

阿涅西箕舌线定义如下：取一个半径为 *a* 的圆，使得点 $(0,0)$ 与 $(0,2a)$ 都在圆上（图 1.12）。记 *O* 为原点。在圆上另取一点 *A*，作割线（secant line）*OA*。记 *B* 为直线 *OA* 与过 $(0,2a)$ 的水平线的交点。过 *B* 的垂直线与过 *A* 的水平线相交于点 *P*。当点 *A* 变动时，点 *P* 所经过的轨迹就是给定圆的阿涅西箕舌线。

Witch of Agnesi curves have applications in physics, including modeling water waves and distributions of spectral lines. In probability theory, the curve describes the probability density function of the Cauchy distribution. In this project you will parameterize these curves.

阿涅西箕舌线在物理中有应用，包括模拟水波和谱线（spectral lines）的分布。在概率论中，该曲线描述了柯西分布（Cauchy distribution）的概率密度函数（probability density function）。在本项目中，你将对这些曲线进行参数化。

1. On the figure, label the following points, lengths, and angle:

1. 在图形上标出下列点、长度和角：

1. *C* is the point on the *x*-axis with the same *x*-coordinate as *A*.

1. *C* 是 *x* 轴上与 *A* 具有相同 *x* 坐标的点。

2. *x* is the *x*-coordinate of *P*, and *y* is the *y*-coordinate of *P*.

2. *x* 是 *P* 的 *x* 坐标，*y* 是 *P* 的 *y* 坐标。

3. *E* is the point $(0,a).$

3. *E* 是点 $(0,a).$

4. *F* is the point on the line segment *OA* such that the line segment *EF* is perpendicular to the line segment *OA*.

4. *F* 是线段 *OA* 上的一点，使得线段 *EF* 垂直于线段 *OA*。

5. *b* is the distance from *O* to *F*.

5. *b* 是从 *O* 到 *F* 的距离。

6. *c* is the distance from *F* to *A*.

6. *c* 是从 *F* 到 *A* 的距离。

7. *d* is the distance from *O* to *B*.

7. *d* 是从 *O* 到 *B* 的距离。

8. $\theta$ is the measure of angle $\text{∠}COA.$

8. $\theta$ 是角 $\text{∠}COA$ 的度量。

The goal of this project is to parameterize the witch using $\theta$ as a parameter. To do this, write equations for *x* and *y* in terms of only $\theta.$

本项目的目标是用 $\theta$ 作为参数对阿涅西箕舌线进行参数化。为此，将 *x* 和 *y* 写成仅含 $\theta$ 的方程。

2. Show that $d = \frac{2a}{\text{sin}\ \theta}.$

2. 证明 $d = \frac{2a}{\text{sin}\ \theta}.$

3. Note that $x = d\ \text{cos}\ \theta.$ Show that $x = 2a\ \text{cot}\ \theta.$ When you do this, you will have parameterized the *x*-coordinate of the curve with respect to $\theta.$ If you can get a similar equation for *y*, you will have parameterized the curve.

3. 注意 $x = d\ \text{cos}\ \theta.$ 证明 $x = 2a\ \text{cot}\ \theta.$ 完成这一步后，你就已用 $\theta$ 参数化了曲线的 *x* 坐标。若能得到 *y* 的类似方程，你就完成了该曲线的参数化。

4. In terms of $\theta,$ what is the angle $\text{∠}EOA?$

4. 用 $\theta$ 表示，角 $\text{∠}EOA$ 是多少？

5. Show that $b + c = 2a\ \text{cos}\left( {\frac{\pi}{2} - \theta} \right).$

5. 证明 $b + c = 2a\ \text{cos}\left( {\frac{\pi}{2} - \theta} \right).$

6. Show that $y = 2a\ \text{cos}\left( {\frac{\pi}{2} - \theta} \right)\ \text{sin}\ \theta.$

6. 证明 $y = 2a\ \text{cos}\left( {\frac{\pi}{2} - \theta} \right)\ \text{sin}\ \theta.$

7. Show that $y = 2a\ \text{sin}^{2}\theta.$ You have now parameterized the *y*-coordinate of the curve with respect to $\theta.$

7. 证明 $y = 2a\ \text{sin}^{2}\theta.$ 至此，你已用 $\theta$ 参数化了曲线的 *y* 坐标。

8. Conclude that a parameterization of the given witch curve is

8. 由此得出结论：给定箕舌线的一个参数化为

$$x = 2a\ \text{cot}\ \theta,y = 2a\ \text{sin}^{2}\theta, - \infty < \theta < \infty.$$

9. Use your parameterization to show that the given witch curve is the graph of the function $f(x) = \frac{8a^{3}}{x^{2} + 4a^{2}}.$

9. 利用你的参数化证明，给定箕舌线是函数 $f(x) = \frac{8a^{3}}{x^{2} + 4a^{2}}.$ 的图形。

Travels with My Ant: The Curtate and Prolate Cycloids 与我的蚂蚁同行：短摆线与长摆线

Earlier in this section, we looked at the parametric equations for a cycloid, which is the path a point on the edge of a wheel traces as the wheel rolls along a straight path. In this project we look at two different variations of the cycloid, called the curtate and prolate cycloids.

在本节前面，我们考察了摆线的参数方程，即轮缘上一点随轮子在笔直路径上滚动时所描出的轨迹。在本项目中，我们考察摆线的两种变体，称为短摆线（curtate cycloid）与长摆线（prolate cycloid）。

First, let’s revisit the derivation of the parametric equations for a cycloid. Recall that we considered a tenacious ant trying to get home by hanging onto the edge of a bicycle tire. We have assumed the ant climbed onto the tire at the very edge, where the tire touches the ground. As the wheel rolls, the ant moves with the edge of the tire (Figure 1.13).

首先，我们重述摆线参数方程的推导。回想一下，我们考虑了一只顽强的蚂蚁，它挂在自行车轮胎边缘试图回家。我们假设蚂蚁爬上了轮胎最外缘，即轮胎与地面接触之处。当轮子滚动时，蚂蚁随轮胎边缘一起运动（图 1.13）。

As we have discussed, we have a lot of flexibility when parameterizing a curve. In this case we let our parameter *t* represent the angle the tire has rotated through. Looking at Figure 1.13, we see that after the tire has rotated through an angle of *t*, the position of the center of the wheel, $C = \left( {x_{C},y_{C}} \right),$ is given by

如前所述，在参数化一条曲线时我们有很大的灵活性。此处令参数 *t* 表示轮胎转过的角度。观察图 1.13，我们看到轮胎转过角度 *t* 后，轮心的位置 $C = \left( {x_{C},y_{C}} \right)$ 由下式给出

$$x_{C} = at\ \text{and}\ y_{C} = a.$$

轮心位置为 $$x_{C} = at\ \text{and}\ y_{C} = a.$$

Furthermore, letting $A = \left( {x_{A},y_{A}} \right)$ denote the position of the ant, we note that

此外，记 $A = \left( {x_{A},y_{A}} \right)$ 为蚂蚁的位置，我们注意到

$$x_{C} - x_{A} = a\ \text{sin}\ t\ \text{and}\ y_{C} - y_{A} = a\ \text{cos}\ t.$$

Then

于是

$$\begin{array}{l}

{x_{A} = x_{C} - a\ \text{sin}\ t = at - a\ \text{sin}\ t = a(t - \text{sin}\ t)} \\

{y_{A} = y_{C} - a\ \text{cos}\ t = a - a\ \text{cos}\ t = a(1 - \text{cos}\ t).}

\end{array}$$

Note that these are the same parametric representations we had before, but we have now assigned a physical meaning to the parametric variable *t*.

注意，这些与我们之前得到的参数表示相同，但现在我们赋予了参数变量 *t* 一个物理意义。

After a while the ant is getting dizzy from going round and round on the edge of the tire. So he climbs up one of the spokes toward the center of the wheel. By climbing toward the center of the wheel, the ant has changed his path of motion. The new path has less up-and-down motion and is called a curtate cycloid (Figure 1.14). As shown in the figure, we let *b* denote the distance along the spoke from the center of the wheel to the ant. As before, we let *t* represent the angle the tire has rotated through. Additionally, we let $C = \left( {x_{C},y_{C}} \right)$ represent the position of the center of the wheel and $A = \left( {x_{A},y_{A}} \right)$ represent the position of the ant.

过了一会儿，蚂蚁在轮胎边缘转来转去感到头晕，于是它沿一根辐条向轮心爬去。通过向轮心爬，蚂蚁改变了它的运动轨迹。这条新轨迹上下起伏较小，称为短摆线（curtate cycloid，图 1.14）。如图所示，记 *b* 为沿辐条从轮心到蚂蚁的距离。如前，令 *t* 表示轮胎转过的角度。另外，令 $C = \left( {x_{C},y_{C}} \right)$ 表示轮心位置，$A = \left( {x_{A},y_{A}} \right)$ 表示蚂蚁位置。

1. What is the position of the center of the wheel after the tire has rotated through an angle of *t*?

1. 轮胎转过角度 *t* 后，轮心的位置是什么？

2. Use geometry to find expressions for $x_{C} - x_{A}$ and for $y_{C} - y_{A}.$

2. 利用几何方法，求出 $x_{C} - x_{A}$ 与 $y_{C} - y_{A}$ 的表达式。

3. On the basis of your answers to parts 1 and 2, what are the parametric equations representing the curtate cycloid?

3. 根据你在第 1、2 问中的答案，表示短摆线的参数方程是什么？

Once the ant’s head clears, he realizes that the bicyclist has made a turn, and is now traveling away from his home. So he drops off the bicycle tire and looks around. Fortunately, there is a set of train tracks nearby, headed back in the right direction. So the ant heads over to the train tracks to wait. After a while, a train goes by, heading in the right direction, and he manages to jump up and just catch the edge of the train wheel (without getting squished!).

等蚂蚁头脑清醒后，他意识到骑车人已经转弯，正朝着远离他家的方向行进。于是他跳下自行车轮胎四处张望。幸运的是，附近有一组铁轨，正指向回家的方向。于是蚂蚁走向铁轨旁等待。过了一会儿，一列火车顺着正确方向驶过，他设法跳起，正好抓住了火车轮子的边缘（没有被压扁！）。

The ant is still worried about getting dizzy, but the train wheel is slippery and has no spokes to climb, so he decides to just hang on to the edge of the wheel and hope for the best. Now, train wheels have a flange to keep the wheel running on the tracks. So, in this case, since the ant is hanging on to the very edge of the flange, the distance from the center of the wheel to the ant is actually greater than the radius of the wheel (Figure 1.15).

蚂蚁仍然担心头晕，但火车轮很滑且没有可攀爬的辐条，于是他决定就挂在轮子边缘听天由命。现在，火车轮有一个轮缘（flange）用以使轮子保持在轨道上运行。因此，在这种情况下，由于蚂蚁挂在轮缘的最外边缘，从轮心到蚂蚁的距离实际上大于轮的半径（图 1.15）。

The setup here is essentially the same as when the ant climbed up the spoke on the bicycle wheel. We let *b* denote the distance from the center of the wheel to the ant, and we let *t* represent the angle the tire has rotated through. Additionally, we let $C = \left( {x_{C},y_{C}} \right)$ represent the position of the center of the wheel and $A = \left( {x_{A},y_{A}} \right)$ represent the position of the ant (Figure 1.15).

这里的设定与蚂蚁在自行车轮上沿辐条爬升时基本相同。记 *b* 为从轮心到蚂蚁的距离，令 *t* 表示轮胎转过的角度。此外，令 $C = \left( {x_{C},y_{C}} \right)$ 表示轮心位置，$A = \left( {x_{A},y_{A}} \right)$ 表示蚂蚁位置（图 1.15）。

When the distance from the center of the wheel to the ant is greater than the radius of the wheel, his path of motion is called a prolate cycloid. A graph of a prolate cycloid is shown in the figure.

当从轮心到蚂蚁的距离大于轮的半径时，他的运动轨迹称为长摆线（prolate cycloid）。图中展示了长摆线的图形。

4. Using the same approach you used in parts 1– 3, find the parametric equations for the path of motion of the ant.

4. 运用你在第 1–3 问中采用的方法，求出蚂蚁运动轨迹的参数方程。

5. What do you notice about your answer to part 3 and your answer to part 4?

5. 你第 3 问与第 4 问的答案有什么相似之处？

Notice that the ant is actually traveling backward at times (the “loops” in the graph), even though the train continues to move forward. He is probably going to be *really* dizzy by the time he gets home!

注意，尽管火车继续向前运动，蚂蚁有时实际上是在向后行进（图形中的“圈”）。等他到家时，恐怕会*真的*晕头转向！

Section 1.1 Exercises 1.1 节习题

For the following exercises, sketch the curves below by eliminating the parameter *t*. Give the orientation of the curve.

对于下列习题，通过消去参数 *t* 画出下列曲线，并指出曲线的方向。

$x = t^{2} + 2t,$ $y = t + 1$

$x = t^{2} + 2t,$ $y = t + 1$（消去参数 $t$，画出曲线并指出方向）

2\.

$x = \text{cos}(t),y = \text{sin}(t),\left( {0,2\pi} \right\rbrack$

$x = \text{cos}(t),y = \text{sin}(t),\left( {0,2\pi} \right\rbrack$（消去参数 $t$，画出曲线并指出方向）

$x = 2t + 4,y = t - 1$

$x = 2t + 4,y = t - 1$（消去参数 $t$，画出曲线并指出方向）

4\.

$x = 3 - t,y = 2t - 3,1.5 \leq t \leq 3$

$x = 3 - t,y = 2t - 3,1.5 \leq t \leq 3$（消去参数 $t$，画出曲线并指出方向）

For the following exercises, eliminate the parameter and sketch the graphs.

对于下列习题，消去参数并画出图形。

$x = 2t^{2},\quad y = t^{4} + 1$

$x = 2t^{2},\quad y = t^{4} + 1$（消去参数并画出图形）

For the following exercises, use technology (CAS or calculator) to sketch the parametric equations.

对于下列习题，使用技术工具（CAS 或计算器）画出参数方程。

\[T\] $\begin{array}{ll}

\[T\]（使用技术工具） $\begin{array}{ll}

{x = t^{2} + t,} & {y = t^{2} - 1}

\end{array}$

\end{array}$（使用技术工具画出参数曲线）

\[T\] $\begin{array}{ll}

\[T\]（使用技术工具） $\begin{array}{ll}

{x = e^{\text{−}t},} & {y = e^{2t} - 1}

\end{array}$

\end{array}$（使用技术工具画出参数曲线）

8\.

\[T\] $\begin{array}{ll}

\[T\]（使用技术工具） $\begin{array}{ll}

{x = 3\ \text{cos}\ t,} & {y = 4\ \text{sin}\ t}

\end{array}$

\end{array}$（使用技术工具画出参数曲线）

\[T\] $\begin{array}{ll}

\[T\]（使用技术工具） $\begin{array}{ll}

{x = \text{sec}\ t,} & {y = \text{cos}\ t}

\end{array}$

\end{array}$（使用技术工具画出参数曲线）

For the following exercises, sketch the parametric equations by eliminating the parameter. Indicate any asymptotes of the graph.

对于下列习题，通过消去参数画出参数方程，并指出图形的任何渐近线。

10\.

10.

$x = e^{t},\quad y = e^{2t} + 1$

$x = e^{t},\quad y = e^{2t} + 1$（消去参数画出曲线，指出渐近线）

11.

$x = 6\ \text{sin}(2\theta),y = 4\ \text{cos}(2\theta)$

$x = 6\ \text{sin}(2\theta),y = 4\ \text{cos}(2\theta)$（消去参数画出曲线，指出渐近线）

12\.

12.

$\begin{array}{ll}

{x = \text{cos}\ \theta,} & {y = 2\ \text{sin}(2\theta)}

\end{array}$

\end{array}$（消去参数画出曲线，指出渐近线）

13.

$\begin{array}{ll}

{x = 3 - 2\ \text{cos}\ \theta,} & {y = -5 + 3\ \text{sin}\ \theta}

\end{array}$

\end{array}$（消去参数画出曲线，指出渐近线）

14\.

14.

$\begin{array}{ll}

{x = 4 + 2\ \text{cos}\ \theta,} & {y = -1 + \text{sin}\ \theta}

\end{array}$

\end{array}$（消去参数画出曲线，指出渐近线）

15.

$\begin{array}{ll}

{x = \text{sec}\ t,} & {y = \text{tan}\ t}

\end{array}$

\end{array}$（消去参数画出曲线，指出渐近线）

16\.

16.

$\begin{array}{ll}

{x = \text{ln}(2t),} & {y = t^{2}}

\end{array}$

\end{array}$（消去参数画出曲线，指出渐近线）

17.

$\begin{array}{ll}

{x = e^{t},} & {y = e^{2t}}

\end{array}$

\end{array}$（消去参数画出曲线，指出渐近线）

18\.

18.

$\begin{array}{ll}

{x = e^{-2t},} & {y = e^{3t}}

\end{array}$

\end{array}$（消去参数画出曲线，指出渐近线）

19.

$\begin{array}{ll}

{x = t^{3},} & {y = 3\ \text{ln}\ t}

\end{array}$

\end{array}$（消去参数画出曲线，指出渐近线）

20\.

20.

$\begin{array}{ll}

{x = 4\ \text{sec}\ \theta,} & {y = 3\ \text{tan}\ \theta}

\end{array}$

\end{array}$（消去参数画出曲线，指出渐近线）

For the following exercises, convert the parametric equations of a curve into rectangular form. No sketch is necessary. State the domain of the rectangular form.

对于下列习题，将曲线的参数方程化为直角坐标形式。无需画图。说明直角坐标形式的定义域。

21.

$\begin{array}{ll}

{x = t^{2} - 1,} & {y = \frac{t}{2}}

\end{array}$

\end{array}$（化为直角坐标形式并给出定义域，无需画图）

22\.

22.

$\begin{array}{ll}

{x = \frac{1}{\sqrt{t + 1}},} & {y = \frac{t}{1 + t},t > -1}

\end{array}$

\end{array}$（化为直角坐标形式并给出定义域，无需画图）

23.

$x = 4\ \text{cos}\ \theta,y = 3\ \text{sin}\ \theta,\theta \in \left( {0,2\pi} \right\rbrack$

$x = 4\ \text{cos}\ \theta,y = 3\ \text{sin}\ \theta,\theta \in \left( {0,2\pi} \right\rbrack$（化为直角坐标形式并给出定义域，无需画图）

24\.

24.

$\begin{array}{ll}

{x = \text{cosh}\ t,} & {y = \text{sinh}\ t}

\end{array}$

\end{array}$（化为直角坐标形式并给出定义域，无需画图）

25.

$\begin{array}{ll}

{x = 2t - 3,} & {y = 6t - 7}

\end{array}$

\end{array}$（化为直角坐标形式并给出定义域，无需画图）

26\.

26.

$\begin{array}{ll}

{x = t^{2},} & {y = t^{3}}

\end{array}$

\end{array}$（化为直角坐标形式并给出定义域，无需画图）

27.

$\begin{array}{ll}

{x = 1 + \text{cos}\ t,} & {y = 3 - \text{sin}\ t}

\end{array}$

\end{array}$（化为直角坐标形式并给出定义域，无需画图）

28\.

28.

$\begin{array}{ll}

{x = \sqrt{t},} & {y = 2t + 4}

\end{array}$

\end{array}$（化为直角坐标形式并给出定义域，无需画图）

29.

$\begin{array}{ll}

{x = \text{sec}\ t,} & {y = \text{tan}\ t,\pi \leq t < \frac{3\pi}{2}}

\end{array}$

\end{array}$（化为直角坐标形式并给出定义域，无需画图）

30\.

30.

$\begin{array}{ll}

{x = 2\ \text{cosh}\ t,} & {y = 4\ \text{sinh}\ t}

\end{array}$

\end{array}$（化为直角坐标形式并给出定义域，无需画图）

31.

$\begin{array}{ll}

{x = \text{cos}(2t),} & {y = \text{sin}\ t}

\end{array}$

\end{array}$（化为直角坐标形式并给出定义域，无需画图）

32\.

32.

$x = 4t + 3,y = 16t^{2} - 9$

$x = 4t + 3,y = 16t^{2} - 9$（化为直角坐标形式并给出定义域，无需画图）

33.

$\begin{array}{ll}

{x = t^{2},} & {y = 2\ \text{ln}\ t,t \geq 1}

\end{array}$

\end{array}$（化为直角坐标形式并给出定义域，无需画图）

34\.

34.

$\begin{array}{ll}

{x = t^{3},} & {y = 3\ \text{ln}\ t,t \geq 1}

\end{array}$

\end{array}$（化为直角坐标形式并给出定义域，无需画图）

35.

$\begin{array}{ll}

{x = t^{n},} & {y = n\ \text{ln}\ t,t \geq 1,}

\end{array}$ where *n* is a natural number

\end{array}$ 其中 *n* 为自然数（化为直角坐标形式并给出定义域，无需画图）

36\.

36.

$\begin{array}{l}

{x = \text{ln}(5t)} \\

{y = \text{ln}(t^{2})}

\end{array}$ where $1 \leq t \leq e$

\end{array}$ 其中 $1 \leq t \leq e$（化为直角坐标形式并给出定义域，无需画图）

37.

$\begin{array}{l}

{x = 2\ \text{sin}(8t)} \\

{y = 2\ \text{cos}(8t)}

\end{array}$

\end{array}$（化为直角坐标形式并给出定义域，无需画图）

38\.

38.

$\begin{array}{l}

{x = \text{tan}\ t} \\

{y = \text{sec}^{2}t - 1}

\end{array}$

\end{array}$（化为直角坐标形式并给出定义域，无需画图）

For the following exercises, the pairs of parametric equations represent lines, parabolas, circles, ellipses, or hyperbolas. Name the type of basic curve that each pair of equations represents.

对于下列习题，各组参数方程表示直线、抛物线、圆、椭圆或双曲线。说出每组方程所表示的基本曲线类型。

39.

$\begin{array}{l}

{x = 3t + 4} \\

{y = 5t - 2}

\end{array}$

\end{array}$（判断曲线类型）

40\.

40.

$\begin{array}{l}

{x - 4 = 5t} \\

{y + 2 = t}

\end{array}$

\end{array}$（判断曲线类型）

41.

$\begin{array}{l}

{x = 2t + 1} \\

{y = t^{2} - 3}

\end{array}$

\end{array}$（判断曲线类型）

42\.

42.

$\begin{array}{l}

{x = 3\ \text{cos}\ t} \\

{y = 3\ \text{sin}\ t}

\end{array}$

\end{array}$（判断曲线类型）

43.

$\begin{array}{l}

{x = 2\ \text{cos}(3t)} \\

{y = 2\ \text{sin}(3t)}

\end{array}$

\end{array}$（判断曲线类型）

44\.

44.

$\begin{array}{l}

{x = \text{cosh}\ t} \\

{y = \text{sinh}\ t}

\end{array}$

\end{array}$（判断曲线类型）

45.

$\begin{array}{l}

{x = 3\ \text{cos}\ t} \\

{y = 4\ \text{sin}\ t}

\end{array}$

\end{array}$（判断曲线类型）

46\.

46.

$\begin{array}{l}

{x = 2\ \text{cos}(3t)} \\

{y = 5\ \text{sin}(3t)}

\end{array}$

\end{array}$（判断曲线类型）

47.

$\begin{array}{l}

{x = 3\ \text{cosh}(4t)} \\

{y = 4\ \text{sinh}(4t)}

\end{array}$

\end{array}$（判断曲线类型）

48\.

48.

$\begin{array}{l}

{x = 2\ \text{cosh}\ t} \\

{y = 2\ \text{sinh}\ t}

\end{array}$

\end{array}$（判断曲线类型）

49\.

49.

Show that $\begin{array}{l}

证明 $\begin{array}{l}

{x = h + r\ \text{cos}\ \theta} \\

{y = k + r\ \text{sin}\ \theta}

\end{array}$ represents the equation of a circle.

\end{array}$ 表示圆的方程。

50\.

50.

Use the equations in the preceding problem to find a set of parametric equations for a circle whose radius is 5 and whose center is $\left( {-2,\ 3} \right).$

利用前一题中的方程，求半径为 5、圆心为 $\left( {-2,\ 3} \right).$ 的圆的参数方程。

For the following exercises, use a graphing utility to graph the curve represented by the parametric equations and identify the curve from its equation.

对于下列习题，使用绘图工具画出参数方程所表示的曲线，并根据方程辨识该曲线。

51.

\[T\] $\begin{array}{l}

\[T\]（使用技术工具） $\begin{array}{l}

{x = \theta + \text{sin}\ \theta} \\

{y = 1 - \text{cos}\ \theta}

\end{array}$

\end{array}$（使用绘图工具画出曲线并辨识）

52\.

52.

\[T\] $\begin{array}{l}

\[T\]（使用技术工具） $\begin{array}{l}

{x = 2t - 2\ \text{sin}\ t} \\

{y = 2 - 2\ \text{cos}\ t}

\end{array}$

\end{array}$（使用绘图工具画出曲线并辨识）

53.

\[T\] $\begin{array}{l}

\[T\]（使用技术工具） $\begin{array}{l}

{x = t - 0.5\ \text{sin}\ t} \\

{y = 1 - 1.5\ \text{cos}\ t}

\end{array}$

\end{array}$（使用绘图工具画出曲线并辨识）

54\.

54.

An airplane traveling horizontally at 100 m/s over flat ground at an elevation of 4000 meters must drop an emergency package on a target on the ground. The trajectory of the package is given by $x = 100t,y = -4.9t^{2} + 4000,t \geq 0$ where the origin is the point on the ground directly beneath the plane at the moment of release. How many horizontal meters before the target should the package be released in order to hit the target?

一架飞机在平坦地面上空 4000 米高度以 100 m/s 水平飞行，必须向地面目标投放应急物资。物资的轨迹由 $x = 100t,y = -4.9t^{2} + 4000,t \geq 0$ 给出，其中原点为投放瞬间飞机正下方的地面点。为使物资命中目标，应在距目标水平方向多远（米）处投放？

55.

The trajectory of a bullet is given by $x = v_{0}\left( {\text{cos}\ \alpha} \right)\ t,y = v_{0}\left( {\text{sin}\ \alpha} \right)\ t - \frac{1}{2}gt^{2}$ where $v_{0} = 500\ \text{m/s,}$ $g = 9.8 = 9.8{\ \text{m/s}}^{2},$ and $\alpha = 30\ \text{degrees}.$ When will the bullet hit the ground? How far from the gun will the bullet hit the ground?

子弹的轨迹由 $x = v_{0}\left( {\text{cos}\ \alpha} \right)\ t,y = v_{0}\left( {\text{sin}\ \alpha} \right)\ t - \frac{1}{2}gt^{2}$ 给出，其中 $v_{0} = 500\ \text{m/s,}$ $g = 9.8 = 9.8{\ \text{m/s}}^{2},$ $\alpha = 30\ \text{degrees}.$ 子弹何时落地？子弹落地时距枪口多远？

56\.

56.

\[T\] Use technology to sketch the curve represented by $x = \text{sin}(4t),y = \text{sin}(3t),0 \leq t \leq 2\pi.$

\[T\]（使用技术工具）使用技术工具画出由 $x = \text{sin}(4t),y = \text{sin}(3t),0 \leq t \leq 2\pi.$ 表示的曲线。

57.

\[T\] Use technology to sketch $x = 2\ \text{tan}(t),y = 3\ \text{sec}(t),\text{−}\pi < t < \pi.$

\[T\]（使用技术工具）使用技术工具画出 $x = 2\ \text{tan}(t),y = 3\ \text{sec}(t),\text{−}\pi < t < \pi.$

58\.

58.

Sketch the curve known as an *epitrochoid*, which gives the path of a point on a circle of radius *b* as it rolls on the outside of a circle of radius *a*. The equations are

画出称为*外摆线*（epitrochoid）的曲线，它表示半径为 *b* 的圆在半径为 *a* 的圆外侧滚动时，其上一点所经过的路径。方程为

$\begin{array}{l}

{x = (a + b)\text{cos}\ t - c \cdot \text{cos}\left\lbrack \frac{(a + b)t}{b} \right\rbrack} \\

{y = (a + b)\text{sin}\ t - c \cdot \text{sin}\left\lbrack \frac{(a + b)t}{b} \right\rbrack.}

\end{array}$

\end{array}$（使用技术工具画出外摆线）

Let $a = 1,b = 2,c = 1.$

令 $a = 1,b = 2,c = 1.$

59.

\[T\] Use technology to sketch the spiral curve given by $x = t\ \text{cos}(t),y = t\ \text{sin}(t)$ from $-2\pi \leq t \leq 2\pi.$

\[T\]（使用技术工具）使用技术工具画出由 $x = t\ \text{cos}(t),y = t\ \text{sin}(t)$ 给出、范围 $-2\pi \leq t \leq 2\pi.$ 的螺线。

60\.

60.

\[T\] Use technology to graph the curve given by the parametric equations $x = 2\ \text{cot}(t),y = 1 - \text{cos}(2t),\text{−}\pi\text{/}2 \leq t \leq \pi\text{/}2.$ This curve is known as the witch of Agnesi.

\[T\]（使用技术工具）使用技术工具画出由参数方程 $x = 2\ \text{cot}(t),y = 1 - \text{cos}(2t),\text{−}\pi\text{/}2 \leq t \leq \pi\text{/}2.$ 给出的曲线。该曲线称为阿涅西箕舌线（witch of Agnesi）。

61.

\[T\] Sketch the curve given by parametric equations $\begin{array}{l}

\[T\]（使用技术工具）画出由参数方程给出的曲线 $\begin{array}{l}

{x = \text{cosh}(t)} \\

{y = \text{sinh}(t),}

\end{array}$ where $-2 \leq t \leq 2.$

\end{array}$ 其中 $-2 \leq t \leq 2.$

---

1.2 Calculus of Parametric Curves 1.2 参数曲线的微积分

1.2.1 Determine derivatives and equations of tangents for parametric curves.
1.2.2 Find the area under a parametric curve.
1.2.3 Use the equation for arc length of a parametric curve.
1.2.4 Apply the formula for surface area to a volume generated by a parametric curve.

1.2.1 求参数曲线的导数与切线方程。
1.2.2 求参数曲线下的面积。
1.2.3 使用参数曲线的弧长公式。
1.2.4 将表面积公式应用于参数曲线生成的立体。

Now that we have introduced the concept of a parameterized curve, our next step is to learn how to work with this concept in the context of calculus. For example, if we know a parameterization of a given curve, is it possible to calculate the slope of a tangent line to the curve? How about the arc length of the curve? Or the area under the curve?

既然我们已经介绍了参数化曲线的概念，下一步就是学习如何在微积分中运用这一概念。例如，如果我们知道给定曲线的一种参数化，能否计算该曲线切线的斜率？曲线的弧长呢？或者曲线下的面积？

Another scenario: Suppose we would like to represent the location of a baseball after the ball leaves a pitcher’s hand. If the position of the baseball is represented by the plane curve $\left( {x(t),y(t)} \right),$ then we should be able to use calculus to find the speed of the ball at any given time. Furthermore, we should be able to calculate just how far that ball has traveled as a function of time.

另一种情形：假设我们想表示棒球离开投手手后的位置。如果棒球的位置由平面曲线 $\left( {x(t),y(t)} \right),$ 表示，那么我们应该能用微积分求出球在任意时刻的速度。此外，我们还应能计算球作为时间函数所经过的距离。

Derivatives of Parametric Equations 参数方程的导数

We start by asking how to calculate the slope of a line tangent to a parametric curve at a point. Consider the plane curve defined by the parametric equations

我们先来思考如何计算参数曲线在某点处切线的斜率。考虑由下列参数方程定义的平面曲线

$$x(t) = 2t + 3,\quad y(t) = 3t - 4,\quad-2 \leq t \leq 3.$$

The graph of this curve appears in Figure 1.16. It is a line segment starting at $\left( {-1,-10} \right)$ and ending at $(9,5).$

该曲线的图形见图 1.16。它是从点 $\left( {-1,-10} \right)$ 开始、到点 $(9,5)$ 结束的线段。

We can eliminate the parameter by first solving the equation $x(t) = 2t + 3$ for *t*:

我们可以先对方程 $x(t) = 2t + 3$ 解 *t* 来消去参数：

$$\begin{array}{rll}

{x(t)} & = & {2t + 3} \\

{x - 3} & = & {2t} \\

t & = & {\frac{x - 3}{2}.}

\end{array}$$

$$\begin{array}{rll} {x(t)} & = & {2t + 3} \\ {x - 3} & = & {2t} \\ t & = & {\frac{x - 3}{2}.} \end{array}$$

Substituting this into $y(t),$ we obtain

将此代入 $y(t)$，我们得到

$$\begin{array}{rll}

{y(t)} & = & {3t - 4} \\

y & = & {3\left( \frac{x - 3}{2} \right) - 4} \\

y & = & {\frac{3x}{2} - \frac{9}{2} - 4} \\

y & = & {\frac{3x}{2} - \frac{17}{2}.}

\end{array}$$

$$\begin{array}{rll} {y(t)} & = & {3t - 4} \\ y & = & {3\left( \frac{x - 3}{2} \right) - 4} \\ y & = & {\frac{3x}{2} - \frac{9}{2} - 4} \\ y & = & {\frac{3x}{2} - \frac{17}{2}.} \end{array}$$

The slope of this line is given by $\frac{dy}{dx} = \frac{3}{2}.$ Next we calculate $x^{\prime}(t)$ and $y^{\prime}(t).$ This gives $x^{\prime}(t) = 2$ and $y^{\prime}(t) = 3.$ Notice that $\frac{dy}{dx} = \frac{{dy}\text{/}{dt}}{{dx}\text{/}{dt}} = \frac{3}{2}.$ This is no coincidence, as outlined in the following theorem.

该直线的斜率由 $\frac{dy}{dx} = \frac{3}{2}$ 给出。接下来我们计算 $x^{\prime}(t)$ 与 $y^{\prime}(t)$。得到 $x^{\prime}(t) = 2$ 与 $y^{\prime}(t) = 3$。注意到 $\frac{dy}{dx} = \frac{{dy}\text{/}{dt}}{{dx}\text{/}{dt}} = \frac{3}{2}$。这并非巧合，如下述定理所述。

Derivative of Parametric Equations 参数方程的导数

Consider the plane curve defined by the parametric equations $x = x(t)$ and $y = y(t).$ Suppose that $x^{\prime}(t)$ and $y^{\prime}(t)$ exist, and assume that $x^{\prime}(t) \neq 0.$ Then the derivative $\frac{dy}{dx}$ is given by

考虑由参数方程 $x = x(t)$ 与 $y = y(t)$ 定义的平面曲线。设 $x^{\prime}(t)$ 与 $y^{\prime}(t)$ 存在，且 $x^{\prime}(t) \neq 0$。则导数 $\frac{dy}{dx}$ 由下式给出

$$\frac{dy}{dx} = \frac{{dy}\text{/}{dt}}{{dx}\text{/}{dt}} = \frac{y^{\prime}(t)}{x^{\prime}(t)}.$$ (1.1)

Proof 证明

This theorem can be proven using the Chain Rule. In particular, assume that the parameter *t* can be eliminated, yielding a differentiable function $y = F(x).$ Then $y(t) = F\left( {x(t)} \right).$ Differentiating both sides of this equation using the Chain Rule yields

本定理可用链式法则证明。具体而言，假设参数 *t* 可以消去，得到一个可微函数 $y = F(x)$。于是 $y(t) = F\left( {x(t)} \right)$。对该方程两边用链式法则求导，得到

$$y^{\prime}(t) = F^{\prime}\left( {x(t)} \right)x^{\prime}(t),$$

因此

$$F^{\prime}\left( {x(t)} \right) = \frac{y^{\prime}(t)}{x^{\prime}(t)}.$$

But $F^{\prime}\left( {x(t)} \right) = \frac{dy}{dx},$ which proves the theorem.

但 $F^{\prime}\left( {x(t)} \right) = \frac{dy}{dx}$，这就证明了定理。

□

Equation 1.1 can be used to calculate derivatives of plane curves, as well as critical points. Recall that a critical point of a differentiable function $y = f(x)$ is any point $x = x_{0}$ such that either $f^{\prime}\left( x_{0} \right) = 0$ or $f^{\prime}\left( x_{0} \right)$ does not exist. Equation 1.1 gives a formula for the slope of a tangent line to a curve defined parametrically regardless of whether the curve can be described by a function $y = f(x)$ or not.

公式 1.1 可用于计算平面曲线的导数，以及临界点。回顾：可微函数 $y = f(x)$ 的临界点是指满足 $f^{\prime}\left( x_{0} \right) = 0$ 或 $f^{\prime}\left( x_{0} \right)$ 不存在的任意点 $x = x_{0}$。公式 1.1 给出了由参数定义的曲线上切线斜率的计算公式，无论该曲线能否由函数 $y = f(x)$ 表示。

Finding the Derivative of a Parametric Curve 求参数曲线的导数

Calculate the derivative $\frac{dy}{dx}$ for each of the following parametrically defined plane curves, and locate any critical points on their respective graphs.

对下列各个由参数定义的平面曲线，计算导数 $\frac{dy}{dx}$，并指出各自图形上的临界点。

1. $x(t) = t^{2} - 3,\quad y(t) = 2t - 1,\quad-3 \leq t \leq 4$

2. $x(t) = 2t + 1,\quad y(t) = t^{3} - 3t + 4,\quad-2 \leq t \leq 2$

3. $x(t) = 5\ \text{cos}\ t,\quad y(t) = 5\ \text{sin}\ t,\quad 0 \leq t \leq 2\pi$

Solution 解

1. To apply Equation 1.1, first calculate $x^{\prime}(t)$ and $y^{\prime}(t)\text{:}$

1. 应用公式 1.1，先算出 $x^{\prime}(t)$ 和 $y^{\prime}(t)$：

$$\begin{array}{l}

{x^{\prime}(t) = 2t} \\

{y^{\prime}(t) = 2.}

\end{array}$$

$$\begin{array}{l} {x^{\prime}(t) = 2t} \\ {y^{\prime}(t) = 2.} \end{array}$$

Next substitute these into the equation:

接着将它们代入公式：

$$\begin{array}{l}

{\frac{dy}{dx} = \frac{{dy}\text{/}{dt}}{{dx}\text{/}{dt}}} \\

{\frac{dy}{dx} = \frac{2}{2t}} \\

{\frac{dy}{dx} = \frac{1}{t}.}

\end{array}$$

$$\begin{array}{l} {\frac{dy}{dx} = \frac{{dy}\text{/}{dt}}{{dx}\text{/}{dt}}} \\ {\frac{dy}{dx} = \frac{2}{2t}} \\ {\frac{dy}{dx} = \frac{1}{t}.} \end{array}$$

This derivative is undefined when $t = 0.$ Calculating $x(0)$ and $y(0)$ gives $x(0) = (0)^{2} - 3 = -3$ and $y(0) = 2(0) - 1 = -1,$ which corresponds to the point $\left( {-3,-1} \right)$ on the graph. The graph of this curve is a parabola opening to the right, and the point $\left( {-3,-1} \right)$ is its vertex as shown.

当 $t = 0$ 时该导数无定义。计算 $x(0)$ 与 $y(0)$ 得 $x(0) = (0)^{2} - 3 = -3$，$y(0) = 2(0) - 1 = -1$，对应图形上的点 $\left( {-3,-1} \right)$。该曲线的图形是一条向右开口的抛物线，点 $\left( {-3,-1} \right)$ 是它的顶点，如图所示。

2. To apply Equation 1.1, first calculate $x^{\prime}(t)$ and $y^{\prime}(t)\text{:}$

2. 应用公式 1.1，先算出 $x^{\prime}(t)$ 和 $y^{\prime}(t)$：

$$\begin{array}{l}

{x^{\prime}(t) = 2} \\

{y^{\prime}(t) = 3t^{2} - 3.}

\end{array}$$

$$\begin{array}{l} {x^{\prime}(t) = 2} \\ {y^{\prime}(t) = 3t^{2} - 3.} \end{array}$$

Next substitute these into the equation:

接着将它们代入公式：

$$\begin{array}{l}

{\frac{dy}{dx} = \frac{{dy}\text{/}{dt}}{{dx}\text{/}{dt}}} \\

{\frac{dy}{dx} = \frac{3t^{2} - 3}{2}.}

\end{array}$$

$$\begin{array}{l} {\frac{dy}{dx} = \frac{{dy}\text{/}{dt}}{{dx}\text{/}{dt}}} \\ {\frac{dy}{dx} = \frac{3t^{2} - 3}{2}.} \end{array}$$

This derivative is zero when $t = \pm 1.$ When $t = -1$ we have

当 $t = \pm 1$ 时该导数为零。当 $t = -1$ 时，有

$$x(-1) = 2(-1) + 1 = -1\ \text{and}\ y(-1) = (-1)^{3} - 3(-1) + 4 = -1 + 3 + 4 = 6,$$

which corresponds to the point $\left( {-1,6} \right)$ on the graph. When $t = 1$ we have

对应图形上的点 $\left( {-1,6} \right)$。当 $t = 1$ 时，有

$$x(1) = 2(1) + 1 = 3\ \text{and}\ y(1) = (1)^{3} - 3(1) + 4 = 1 - 3 + 4 = 2,$$

which corresponds to the point $\left( {3,2} \right)$ on the graph. The point $\left( {3,2} \right)$ is a relative minimum and the point $\left( {-1,6} \right)$ is a relative maximum, as seen in the following graph.

对应图形上的点 $\left( {3,2} \right)$。点 $\left( {3,2} \right)$ 是相对极小点，点 $\left( {-1,6} \right)$ 是相对极大点，如下图所示。

3. To apply Equation 1.1, first calculate $x^{\prime}(t)$ and $y^{\prime}(t)\text{:}$

3. 应用公式 1.1，先算出 $x^{\prime}(t)$ 和 $y^{\prime}(t)$：

$$\begin{array}{l}

{x^{\prime}(t) = -5\ \text{sin}\ t} \\

{y^{\prime}(t) = 5\ \text{cos}\ t.}

\end{array}$$

$$\begin{array}{l} {x^{\prime}(t) = -5\ \text{sin}\ t} \\ {y^{\prime}(t) = 5\ \text{cos}\ t.} \end{array}$$

Next substitute these into the equation:

接着将它们代入公式：

$$\begin{array}{l}

{\frac{dy}{dx} = \frac{{dy}\text{/}{dt}}{{dx}\text{/}{dt}}} \\

{\frac{dy}{dx} = \frac{5\ \text{cos}\ t}{-5\ \text{sin}\ t}} \\

{\frac{dy}{dx} = \text{−}\text{cot}\ t.}

\end{array}$$

$$\begin{array}{l} {\frac{dy}{dx} = \frac{{dy}\text{/}{dt}}{{dx}\text{/}{dt}}} \\ {\frac{dy}{dx} = \frac{5\ \text{cos}\ t}{-5\ \text{sin}\ t}} \\ {\frac{dy}{dx} = \text{−}\text{cot}\ t.} \end{array}$$

This derivative is zero when $\text{cos}\ t = 0$ and is undefined when $\text{sin}\ t = 0.$ This gives $t = 0,\frac{\pi}{2},\pi,\frac{3\pi}{2},\text{and}\ 2\pi$ as critical points for *t.* Substituting each of these into $x(t)$ and $y(t),$ we obtain

当 $\text{cos}\ t = 0$ 时该导数为零，当 $\text{sin}\ t = 0$ 时无定义。由此得到 $t = 0,\frac{\pi}{2},\pi,\frac{3\pi}{2},\text{and}\ 2\pi$ 为 *t* 的临界点。将其中每一个代入 $x(t)$ 与 $y(t)$，我们得到

$t$	$x(t)$	$y(t)$
0	5	0
$\frac{\pi}{2}$	0	5
$\pi$	−5	0
$\frac{3\pi}{2}$	0	−5
$2\pi$	5	0

参数 $t$ 对应的 $x(t)$ 与 $y(t)$ 值
$t$	$x(t)$	$y(t)$
0	5	0
$\frac{\pi}{2}$	0	5
$\pi$	−5	0
$\frac{3\pi}{2}$	0	−5
$2\pi$	5	0

These points correspond to the sides, top, and bottom of the circle that is represented by the parametric equations (Figure 1.19). On the left and right edges of the circle, the derivative is undefined, and on the top and bottom, the derivative equals zero.

这些点对应于该参数方程所表示（图 1.19）的圆的左、右、上、下各处。在圆的左右边缘，导数无定义；在上下边缘，导数等于零。

Calculate the derivative ${dy}\text{/}{dx}$ for the plane curve defined by the equations

计算由下列方程定义的平面曲线的导数 ${dy}\text{/}{dx}$：

$$x(t) = t^{2} - 4t,\quad y(t) = 2t^{3} - 6t,\quad-2 \leq t \leq 3$$

Finding a Tangent Line 求切线

Find an equation of the tangent line to the curve defined by the equations

求由下列方程定义的曲线的切线方程：

$$x(t) = t^{2} - 3,\quad y(t) = 2t - 1,\quad-3 \leq t \leq 4\ \text{when}\ t = 2.$$

Solution 解

First find the slope of the tangent line using Equation 1.1, which means calculating $x^{\prime}(t)$ and $y^{\prime}(t)\text{:}$

首先用公式 1.1 求该切线的斜率，即计算 $x^{\prime}(t)$ 与 $y^{\prime}(t)$：

$$\begin{array}{l}

{x^{\prime}(t) = 2t} \\

{y^{\prime}(t) = 2.}

\end{array}$$

$$\begin{array}{l} {x^{\prime}(t) = 2t} \\ {y^{\prime}(t) = 2.} \end{array}$$

Next substitute these into the equation:

接着将它们代入公式：

$$\begin{array}{l}

{\frac{dy}{dx} = \frac{{dy}\text{/}{dt}}{{dx}\text{/}{dt}}} \\

{\frac{dy}{dx} = \frac{2}{2t}} \\

{\frac{dy}{dx} = \frac{1}{t}.}

\end{array}$$

$$\begin{array}{l} {\frac{dy}{dx} = \frac{{dy}\text{/}{dt}}{{dx}\text{/}{dt}}} \\ {\frac{dy}{dx} = \frac{2}{2t}} \\ {\frac{dy}{dx} = \frac{1}{t}.} \end{array}$$

When $t = 2,$ $\frac{dy}{dx} = \frac{1}{2},$ so this is the slope of the tangent line. Calculating $x(2)$ and $y(2)$ gives

当 $t = 2$ 时，$\frac{dy}{dx} = \frac{1}{2}$，因此这就是切线的斜率。计算 $x(2)$ 与 $y(2)$ 得

$$x(2) = (2)^{2} - 3 = 1\ \text{and}\ y(2) = 2(2) - 1 = 3,$$

which corresponds to the point $\left( {1,3} \right)$ on the graph (Figure 1.20). Now use the point-slope form of the equation of a line to find an equation of the tangent line:

对应图形上的点 $\left( {1,3} \right)$（图 1.20）。现用直线方程的点斜式求该切线方程：

$$\begin{array}{rll}

{y - y_{0}} & = & {m\left( {x - x_{0}} \right)} \\

{y - 3} & = & {\frac{1}{2}\left( {x - 1} \right)} \\

{y - 3} & = & {\frac{1}{2}x - \frac{1}{2}} \\

y & = & {\frac{1}{2}x + \frac{5}{2}.}

\end{array}$$

$$\begin{array}{rll} {y - y_{0}} & = & {m\left( {x - x_{0}} \right)} \\ {y - 3} & = & {\frac{1}{2}\left( {x - 1} \right)} \\ {y - 3} & = & {\frac{1}{2}x - \frac{1}{2}} \\ y & = & {\frac{1}{2}x + \frac{5}{2}.} \end{array}$$

Find an equation of the tangent line to the curve defined by the equations

求由下列方程定义的曲线的切线方程：

$$x(t) = t^{2} - 4t,\quad y(t) = 2t^{3} - 6t,\quad-2 \leq t \leq 10\ \text{when}\ t = 5.$$

Second-Order Derivatives 二阶导数

Our next goal is to see how to take the second derivative of a function defined parametrically. The second derivative of a function $y = f(x)$ is defined to be the derivative of the first derivative; that is,

我们下一步的目标是了解如何求由参数定义的函数的二阶导数。函数 $y = f(x)$ 的二阶导数定义为它的一阶导数的导数；即，

$$\frac{d^{2}y}{dx^{2}} = \frac{d}{dx}\left\lbrack \frac{dy}{dx} \right\rbrack.$$

Since $\frac{dy}{dx} = \frac{{dy}\text{/}{dt}}{{dx}\text{/}{dt}},$ we can replace the $y$ on both sides of this equation with $\frac{dy}{dx}.$ This gives us

由于 $\frac{dy}{dx} = \frac{{dy}\text{/}{dt}}{{dx}\text{/}{dt}}$，我们可以将方程两边中的 $y$ 替换为 $\frac{dy}{dx}$。于是得到

$$\frac{d^{2}y}{dx^{2}} = \frac{d}{dx}\left( \frac{dy}{dx} \right) = \frac{\left( {d\text{/}{dt}} \right)\left( {{dy}\text{/}{dx}} \right)}{{dx}\text{/}{dt}}.$$ (1.2)

If we know ${dy}\text{/}{dx}$ as a function of *t,* then this formula is straightforward to apply.

若已知 ${dy}\text{/}{dx}$ 为 *t* 的函数，则该公式可直接应用。

Finding a Second Derivative 求二阶导数

Calculate the second derivative ${d^{2}y}\text{/}{dx^{2}}$ for the plane curve defined by the parametric equations $x(t) = t^{2} - 3,y(t) = 2t - 1,-3 \leq t \leq 4.$

计算由参数方程 $x(t) = t^{2} - 3,y(t) = 2t - 1,-3 \leq t \leq 4$ 定义的平面曲线的二阶导数 ${d^{2}y}\text{/}{dx^{2}}$。

Solution 解

From Example 1.4 we know that $\frac{dy}{dx} = \frac{2}{2t} = \frac{1}{t}.$ Using Equation 1.2, we obtain

由例 1.4 已知 $\frac{dy}{dx} = \frac{2}{2t} = \frac{1}{t}$。利用公式 1.2，得到

$$\frac{d^{2}y}{dx^{2}} = \frac{\left( {d\text{/}{dt}} \right)\left( {{dy}\text{/}{dx}} \right)}{{dx}\text{/}{dt}} = \frac{\left( {d\text{/}{dt}} \right)\left( {1\text{/}t} \right)}{2t} = \frac{\text{−}t^{-2}}{2t} = - \frac{1}{2t^{3}}.$$

Calculate the second derivative ${d^{2}y}\text{/}{dx^{2}}$ for the plane curve defined by the equations

计算由下列方程定义的平面曲线的二阶导数 ${d^{2}y}\text{/}{dx^{2}}$：

$$x(t) = t^{2} - 4t,\quad y(t) = 2t^{3} - 6t,\quad-2 \leq t \leq 3$$

and locate any critical points on its graph.

并指出其图形上的临界点。

Integrals Involving Parametric Equations 含参数方程的积分

Now that we have seen how to calculate the derivative of a plane curve, the next question is this: How do we find the area under a curve defined parametrically? Recall the cycloid defined by the equations $x(t) = t - \text{sin}\ t,\quad y(t) = 1 - \text{cos}\ t.$ Suppose we want to find the area of the shaded region in the following graph.

既然已经会求平面曲线的导数，下一个问题是：如何求由参数方程定义的曲线下方的面积？回忆由方程 $x(t) = t - \text{sin}\ t,\quad y(t) = 1 - \text{cos}\ t.$ 定义的摆线。假设我们想求下列图形中阴影区域的面积。

To derive a formula for the area under the curve defined by the functions

为推导由下列函数定义的曲线下面积的公式，

$$x = x(t),\quad y = y(t),\quad a \leq t \leq b,$$

we assume that $x(t)$ is increasing on the interval $t~ \in ~\lbrack a,~b\rbrack$ and $x(t)$ is differentiable and start with an equal partition of the interval $a \leq t \leq b.$ Suppose $t_{0} = a < t_{1} < t_{2} < \text{⋯} < t_{n} = b$ and consider the following graph.

我们假设 $x(t)$ 在区间 $t~ \in ~\lbrack a,~b\rbrack$ 上递增，且 $x(t)$ 可导，并从区间 $a \leq t \leq b$ 的等分划分开始。设 $t_{0} = a < t_{1} < t_{2} < \text{⋯} < t_{n} = b$，并考虑下列图形。

We use rectangles to approximate the area under the curve. The height of the $i$th rectangle is $y\left( t_{i–1} \right)$, so an approximation to the area is

我们用矩形逼近曲线下的面积。第 $i$ 个矩形的高度为 $y\left( t_{i–1} \right)$，因此面积的一个近似值为

$$\begin{array}{l}

\begin{matrix}

& {\sum\limits_{i = 1}^{n}y\left( t_{i - 1} \right)\left( x\left( t_{i} \right) - x\left( t_{i - 1} \right) \right)} \\

= & {\sum\limits_{i = 1}^{n}y\left( t_{i - 1} \right)\frac{\left( x\left( t_{i} \right) - x\left( t_{i - 1} \right) \right)}{\left( t_{i} - t_{i - 1} \right)}\left( t_{i} - t_{i - 1} \right)} \\

& \left. \rightarrow\int_{a}^{b}y(t)x'(t)dt~\text{as~max}\left\{ \left( t_{i} - t_{i - 1} \right) \right\}\rightarrow 0 \right.

\end{matrix}

\end{array}$$

当划分的最大步长趋于 0 时，上述黎曼和收敛到积分 $\int_{a}^{b}y(t)x'(t)\,dt$。

This follows from results obtained in Calculus 1 for the function $y\left( t_{i - 1} \right)\frac{\left( x\left( t_{i} \right) - x\left( t_{i - 1} \right) \right)}{\left( t_{i} - t_{i - 1} \right)}.$

这源于微积分 1 中关于函数 $y\left( t_{i - 1} \right)\frac{\left( x\left( t_{i} \right) - x\left( t_{i - 1} \right) \right)}{\left( t_{i} - t_{i - 1} \right)}$ 的结果。

Then a Riemann sum for the area is

于是面积的黎曼和为

$$A_{n} = \sum\limits_{i = 1}^{n}y\left( \overset{—}{t_{i}} \right)\ \left( x\left( t_{i} \right) - x\left( t_{i - 1} \right) \right).$$

Multiplying and dividing each area by $t_{i} - t_{i - 1}$ gives

将每个面积乘以再除以 $t_{i} - t_{i - 1}$ 得

$$A_{n} = {\sum\limits_{i = 1}^{n}{y\left( {x\left( {\overset{–}{t}}_{i} \right)} \right)\ \left( \frac{x\left( t_{i} \right) - x\left( t_{i - 1} \right)}{t_{i} - t_{i - 1}} \right)}}\left( {t_{i} - t_{i - 1}} \right) = {\sum\limits_{i = 1}^{n}{y\left( {x\left( {\overset{–}{t}}_{i} \right)} \right)\ \left( \frac{x\left( t_{i} \right) - x\left( t_{i - 1} \right)}{\text{Δ}t} \right)}}\text{Δ}t.$$

Taking the limit as $n$ approaches infinity gives

令 $n$ 趋于无穷取极限得

$$A = \underset{n\rightarrow\infty}{\text{lim}}A_{n} = {\int_{a}^{b}{y(t)x^{\prime}(t)\ dt}}.$$

If $x$ is a decreasing function for $a \leq t \leq b$, a similar derivation will show that the area is given by $\begin{matrix}

{- \int_{a}^{b}y(t)x'(t)dt} & =

\end{matrix}\int_{a}^{b}y(t)x'(t)dt$

若 $x$ 在 $a \leq t \leq b$ 上为递减函数，类似的推导表明该面积仍由 $\int_{a}^{b}y(t)x'(t)\,dt$ 给出（取绝对值即可）。

This leads to the following theorem.

由此得到以下定理。

Area under a Parametric Curve 参数曲线下的面积

Consider the non-self-intersecting plane curve defined by the parametric equations

考虑由参数方程定义的非自交平面曲线

$$x = x(t),\quad y = y(t),\quad a \leq t \leq b$$

and assume that $x(t)$ is differentiable. The area under this curve is given by

并假设 $x(t)$ 可导。该曲线下的面积为

$$A = {\int_{a}^{b}{y(t)x^{\prime}(t)\ dt}}.$$ (1.3)

Finding the Area under a Parametric Curve 求参数曲线下的面积

Find the area under the curve of the cycloid defined by the equations

求由下列方程定义的摆线曲线下的面积

$$x(t) = t - \text{sin}\ t,\quad y(t) = 1 - \text{cos}\ t,\quad 0 \leq t \leq 2\pi.$$

Solution 解

Using Equation 1.3, we have

利用公式 1.3，有

$$\begin{array}{cl}

A & {= {\int_{a}^{b}{y(t)x^{\prime}(t)\ dt}}} \\

& {= {\int_{0}^{2\pi}{\left( {1 - \text{cos}\ t} \right)\left( {1 - \text{cos}\ t} \right)\ dt}}} \\

& {= {\int_{0}^{2\pi}{(1 - 2\ \text{cos}\ t + \text{cos}^{2}t)dt}}} \\

& {= {\int_{0}^{2\pi}{\left( {1 - 2\ \text{cos}\ t + \frac{1 + \text{cos}\ 2t}{2}} \right)\ dt}}} \\

& {= {\int_{0}^{2\pi}{\left( {\frac{3}{2} - 2\ \text{cos}\ t + \frac{\text{cos}\ 2t}{2}} \right)\ dt}}} \\

& {= \left. {\frac{3t}{2} - 2\ \text{sin}\ t + \frac{\text{sin}\ 2t}{4}} \right|_{0}^{2\pi}} \\

& {= 3\pi.}

\end{array}$$

由公式 1.3 积分并化简，最终得到该摆线一个拱形下的面积为 $3\pi$。

Find the area under the curve of the hypocycloid defined by the equations

求由下列方程定义的 hypocycloid（内摆线）曲线下的面积

$$x(t) = 3\ \text{cos}\ t + \text{cos}\ 3t,\quad y(t) = 3\ \text{sin}\ t - \text{sin}\ 3t,\quad 0 \leq t \leq \pi.$$

Arc Length of a Parametric Curve 参数曲线的弧长

In addition to finding the area under a parametric curve, we sometimes need to find the arc length of a parametric curve. In the case of a line segment, arc length is the same as the distance between the endpoints. If a particle travels from point *A* to point *B* along a curve, then the distance that particle travels is the arc length. To develop a formula for arc length, we start with an approximation by line segments as shown in the following graph.

除了求参数曲线下的面积，有时还需要求参数曲线的弧长。对于线段，弧长等于两端点间的距离。若质点沿一条曲线从点 *A* 运动到点 *B*，则该质点经过的距离就是弧长。为推导弧长公式，我们从用线段逼近开始，如下列图形所示。

Given a plane curve defined by the functions $x = x(t),y = y(t),a \leq t \leq b,$ we start by partitioning the interval $\lbrack a,b\rbrack$ into *n* equal subintervals: $t_{0} = a < t_{1} < t_{2} < \text{⋯} < t_{n} = b.$ The width of each subinterval is given by $\text{Δ}t = {{(b - a)}\text{/}n}.$ We can calculate the length of each line segment:

给定由函数 $x = x(t),y = y(t),a \leq t \leq b$ 定义的平面曲线，我们首先将区间 $\lbrack a,b\rbrack$ 划分为 *n* 个相等的子区间：$t_{0} = a < t_{1} < t_{2} < \text{⋯} < t_{n} = b.$ 每个子区间的宽度为 $\text{Δ}t = {{(b - a)}\text{/}n}.$ 于是可计算每条线段的长度：

$$\begin{array}{l}

{d_{1} = \sqrt{\left( {x\left( t_{1} \right) - x\left( t_{0} \right)} \right)^{2} + \left( {y\left( t_{1} \right) - y\left( t_{0} \right)} \right)^{2}}} \\

{d_{2} = \sqrt{\left( {x\left( t_{2} \right) - x\left( t_{1} \right)} \right)^{2} + \left( {y\left( t_{2} \right) - y\left( t_{1} \right)} \right)^{2}}\ \text{etc}.}

\end{array}$$

各线段长度依次为 $d_{1}, d_{2}, \dots$，其中 $d_{k}$ 由相邻两点间的距离公式计算。

Then add these up. We let *s* denote the exact arc length and $s_{n}$ denote the approximation by *n* line segments:

将它们相加。记 *s* 为精确弧长，$s_{n}$ 为 *n* 条线段的近似值：

$$s \approx {\sum\limits_{k = 1}^{n}s_{k}} = {\sum\limits_{k = 1}^{n}\sqrt{\left( {x\left( t_{k} \right) - x\left( t_{k - 1} \right)} \right)^{2} + \left( {y\left( t_{k} \right) - y\left( t_{k - 1} \right)} \right)^{2}}}.$$ (1.4)

If we assume that $x(t)$ and $y(t)$ are differentiable functions of *t,* then the Mean Value Theorem (Introduction to the Applications of Derivatives) applies, so in each subinterval $\lbrack t_{k - 1},t_{k}\rbrack$ there exist ${\hat{t}}_{k}$ and ${\widetilde{t}}_{k}$ such that

若假设 $x(t)$ 和 $y(t)$ 是 *t* 的可导函数，则中值定理（导数应用导论）适用，故在每个子区间 $\lbrack t_{k - 1},t_{k}\rbrack$ 内存在 ${\hat{t}}_{k}$ 和 ${\widetilde{t}}_{k}$ 使得

$$\begin{array}{l}

{x\left( t_{k} \right) - x\left( t_{k - 1} \right) = x^{\prime}\left( {\hat{t}}_{k} \right)\left( {t_{k} - t_{k - 1}} \right) = x^{\prime}\left( {\hat{t}}_{k} \right)\text{Δ}t} \\

{y\left( t_{k} \right) - y\left( t_{k - 1} \right) = y^{\prime}\left( {\widetilde{t}}_{k} \right)\left( {t_{k} - t_{k - 1}} \right) = y^{\prime}\left( {\widetilde{t}}_{k} \right)\text{Δ}t.}

\end{array}$$

由中值定理，相邻函数值之差可写为导数与步长之积：$x\left( t_{k} \right) - x\left( t_{k - 1} \right) = x^{\prime}\left( {\hat{t}}_{k} \right)\text{Δ}t$，$y\left( t_{k} \right) - y\left( t_{k - 1} \right) = y^{\prime}\left( {\widetilde{t}}_{k} \right)\text{Δ}t$。

Therefore Equation 1.4 becomes

因此公式 1.4 变为

$$\begin{array}{cl}

s & {\approx {\sum\limits_{k = 1}^{n}s_{k}}} \\

& {= {\sum\limits_{k = 1}^{n}\sqrt{\left( {x^{\prime}\left( {\hat{t}}_{k} \right)\text{Δ}t} \right)^{2} + \left( {y^{\prime}\left( {\widetilde{t}}_{k} \right)\text{Δ}t} \right)^{2}}}} \\

& {= {\sum\limits_{k = 1}^{n}\sqrt{\left( {x^{\prime}\left( {\hat{t}}_{k} \right)} \right)^{2}\left( {\text{Δ}t} \right)^{2} + \left( {y^{\prime}\left( {\widetilde{t}}_{k} \right)} \right)^{2}\left( {\text{Δ}t} \right)^{2}}}} \\

& {= \left( {\sum\limits_{k = 1}^{n}\sqrt{\left( {x^{\prime}\left( {\hat{t}}_{k} \right)} \right)^{2} + \left( {y^{\prime}\left( {\widetilde{t}}_{k} \right)} \right)^{2}}} \right)\text{Δ}t.}

\end{array}$$

将中值定理的结果代入并提取公因子 $\text{Δ}t$，得到以导数表示的弧长黎曼和。

This is a Riemann sum that approximates the arc length over a partition of the interval $\lbrack a,b\rbrack.$ If we further assume that the derivatives are continuous and let the number of points in the partition increase without bound, the approximation approaches the exact arc length. This gives

这是在区间 $\lbrack a,b\rbrack$ 划分上逼近弧长的黎曼和。若进一步假设导数连续，并令划分中的点数无界增加，则近似值趋近于精确弧长。于是得到

$$\begin{array}{cl}

s & {= \underset{n\rightarrow\infty}{\text{lim}}{\sum\limits_{k = 1}^{n}s_{k}}} \\

& {= \underset{n\rightarrow\infty}{\text{lim}}\left( {\sum\limits_{k = 1}^{n}\sqrt{\left( {x^{\prime}\left( {\hat{t}}_{k} \right)} \right)^{2} + \left( {y^{\prime}\left( {\widetilde{t}}_{k} \right)} \right)^{2}}} \right)\text{Δ}t} \\

& {= {\int_{a}^{b}{\sqrt{\left( {x^{\prime}(t)} \right)^{2} + \left( {y^{\prime}(t)} \right)^{2}}dt}}.}

\end{array}$$

取极限后，黎曼和化为定积分 $s = {\int_{a}^{b}{\sqrt{\left( {x^{\prime}(t)} \right)^{2} + \left( {y^{\prime}(t)} \right)^{2}}dt}}$。

When taking the limit, the values of ${\hat{t}}_{k}$ and ${\widetilde{t}}_{k}$ are both contained within the same ever-shrinking interval of width $\text{Δ}t,$ so they must converge to the same value.

取极限时，${\hat{t}}_{k}$ 和 ${\widetilde{t}}_{k}$ 都落在宽度为 $\text{Δ}t$ 的同一不断收缩的区间内，故它们必收敛到同一值。

We can summarize this method in the following theorem.

可将此法总结为以下定理。

Arc Length of a Parametric Curve 参数曲线的弧长

Consider the plane curve defined by the parametric equations

考虑由参数方程定义的平面曲线

$$x = x(t),\quad y = y(t),\quad t_{1} \leq t \leq t_{2}$$

and assume that $x(t)$ and $y(t)$ are differentiable functions of *t.* Then the arc length of this curve is given by

并假设 $x(t)$ 和 $y(t)$ 是 *t* 的可导函数。则该曲线的弧长为

$$s = {\int_{t_{1}}^{t_{2}}{\sqrt{\left( \frac{dx}{dt} \right)^{2} + \left( \frac{dy}{dt} \right)^{2}}dt}}.$$ (1.5)

At this point a side derivation leads to a previous formula for arc length. In particular, suppose the parameter can be eliminated, leading to a function $y = F(x).$ Then $y(t) = F\left( {x(t)} \right)$ and the Chain Rule gives $y^{\prime}(t) = F^{\prime}\left( {x(t)} \right)x^{\prime}(t).$ Substituting this into Equation 1.5 gives

此处的一个旁支推导可回到先前的弧长公式。具体地，假设能消去参数，得到函数 $y = F(x).$ 于是 $y(t) = F\left( {x(t)} \right)$，由链式法则得 $y^{\prime}(t) = F^{\prime}\left( {x(t)} \right)x^{\prime}(t).$ 将其代入公式 1.5 得

$$\begin{array}{cl}

s & {= {\int_{t_{1}}^{t_{2}}{\sqrt{\left( \frac{dx}{dt} \right)^{2} + \left( \frac{dy}{dt} \right)^{2}}dt}}} \\

& {= {\int_{t_{1}}^{t_{2}}{\sqrt{\left( \frac{dx}{dt} \right)^{2} + \left( {F^{\prime}(x)\frac{dx}{dt}} \right)^{2}}dt}}} \\

& {= {\int_{t_{1}}^{t_{2}}{\sqrt{\left( \frac{dx}{dt} \right)^{2}\left( {1 + \left( {F^{\prime}(x)} \right)^{2}} \right)}dt}}} \\

& {= {\int_{t_{1}}^{t_{2}}{x^{\prime}(t)\sqrt{1 + \left( \frac{dy}{dx} \right)^{2}}dt}}.}

\end{array}$$

消参后，$\frac{dy}{dt} = F^{\prime}(x)\frac{dx}{dt}$，代入并提取 $\frac{dx}{dt}$ 得到以 $x$ 为变量的弧长积分形式。

Here we have assumed that $x^{\prime}(t) > 0,$ which is a reasonable assumption. The Chain Rule gives $dx = x^{\prime}(t)\ dt,$ and letting $a = x\left( t_{1} \right)$ and $b = x\left( t_{2} \right)$ we obtain the formula

此处假设了 $x^{\prime}(t) > 0$，这是合理的。由链式法则得 $dx = x^{\prime}(t)\ dt$，并令 $a = x\left( t_{1} \right)$、$b = x\left( t_{2} \right)$，便得到公式

$$s = {\int_{a}^{b}{\sqrt{1 + \left( \frac{dy}{dx} \right)^{2}}dx}},$$

which is the formula for arc length obtained in the Introduction to the Applications of Integration.

这正是积分应用导论中求弧长的公式。

Finding the Arc Length of a Parametric Curve 求参数曲线的弧长

Find the arc length of the semicircle defined by the equations

求由下列方程定义的半圆的弧长

$$x(t) = 3\ \text{cos}\ t,\quad y(t) = 3\ \text{sin}\ t,\quad 0 \leq t \leq \pi.$$

Solution 解

The values $t = 0$ to $t = \pi$ trace out the red curve in Figure 1.23. To determine its length, use Equation 1.5:

当 $t = 0$ 到 $t = \pi$ 时描绘出图 1.23 中的红色曲线。为确定其长度，使用公式 1.5：

$$\begin{array}{cl}

s & {= {\int_{t_{1}}^{t_{2}}{\sqrt{\left( \frac{dx}{dt} \right)^{2} + \left( \frac{dy}{dt} \right)^{2}}dt}}} \\

& {= {\int_{0}^{\pi}{\sqrt{\left( {-3\ \text{sin}\ t} \right)^{2} + \left( {3\ \text{cos}\ t} \right)^{2}}dt}}} \\

& {= {\int_{0}^{\pi}{\sqrt{9\ \text{sin}^{2}t + 9\ \text{cos}^{2}t}\ dt}}} \\

& {= {\int_{0}^{\pi}{\sqrt{9\left( {\text{sin}^{2}t + \text{cos}^{2}t} \right)}dt}}} \\

& {= {\int_{0}^{\pi}{3dt}} = \left. {3t} \right|_{0}^{\pi} = 3\pi.}

\end{array}$$

代入导数并利用 $\text{sin}^{2}t + \text{cos}^{2}t = 1$，积分得半圆弧长为 $3\pi$。

Note that the formula for the arc length of a semicircle is $\pi r$ and the radius of this circle is 3. This is a great example of using calculus to derive a known formula of a geometric quantity.

注意半圆的弧长公式为 $\pi r$，而该圆的半径为 3。这是用微积分推导已知几何量公式的一个很好的例子。

Find the arc length of the curve defined by the equations

求由下列方程定义的曲线的弧长

$$x(t) = 3t^{2},\quad y(t) = 2t^{3},\quad 1 \leq t \leq 3.$$

We now return to the problem posed at the beginning of the section about a baseball leaving a pitcher’s hand. Ignoring the effect of air resistance (unless it is a curve ball!), the ball travels a parabolic path. Assuming the pitcher’s hand is at the origin and the ball travels left to right in the direction of the positive *x*-axis, the parametric equations for this curve can be written as

现在回到本节开头提出的问题：棒球离开投手之手。忽略空气阻力（除非是曲线球！），球沿抛物线路径运动。假设投手的手在原点，球沿正 *x* 轴方向自左向右运动，则该曲线的参数方程可写为

$$x(t) = 140t,\quad y(t) = -16t^{2} + 2t$$

where *t* represents time. We first calculate the distance the ball travels as a function of time. This distance is represented by the arc length. We can modify the arc length formula slightly. First rewrite the functions $x(t)$ and $y(t)$ using *v* as an independent variable, so as to eliminate any confusion with the parameter *t:*

其中 *t* 表示时间。我们先计算球随时间运动的距离，该距离由弧长表示。可对弧长公式稍作修改：首先用 *v* 作为自变量重写函数 $x(t)$ 和 $y(t)$，以免与参数 *t* 混淆：

$$x(v) = 140v,\quad y(v) = -16v^{2} + 2v.$$

Then we write the arc length formula as follows:

于是弧长公式写为

$$\begin{array}{cl}

{s(t)} & {= {\int_{0}^{t}{\sqrt{\left( \frac{dx}{dv} \right)^{2} + \left( \frac{dy}{dv} \right)^{2}}dv}}} \\

& {= {\int_{0}^{t}{\sqrt{140^{2} + \left( {-32v + 2} \right)^{2}}dv}}.}

\end{array}$$

将 $x(v), y(v)$ 对 $v$ 求导后代入弧长公式，得到以时间 $t$ 为上限的弧长积分 $s(t)$。

The variable *v* acts as a dummy variable that disappears after integration, leaving the arc length as a function of time *t.* To integrate this expression we can use a formula from Appendix A,

变量 *v* 是一个哑变量，积分后消失，使弧长成为时间 *t* 的函数。为积分此式，可利用附录 A 中的公式，

$${\int{\sqrt{a^{2} + u^{2}}du}} = \frac{u}{2}\sqrt{a^{2} + u^{2}} + \frac{a^{2}}{2}\text{ln}\left| {u + \sqrt{a^{2} + u^{2}}} \right| + C.$$

We set $a = 140$ and $u = -32v + 2.$ This gives $du = -32dv,$ so $dv = - \frac{1}{32}du.$ Therefore

令 $a = 140$，$u = -32v + 2.$ 于是 $du = -32dv,$ 故 $dv = - \frac{1}{32}du.$ 因此

$$\begin{array}{cl}

{\int{\sqrt{140^{2} + \left( {-32v + 2} \right)^{2}}dv}} & {= - \frac{1}{32}{\int{\sqrt{a^{2} + u^{2}}du}}} \\

& {= - \frac{1}{32}\left\lbrack \begin{array}{l}

{\frac{\left( {-32v + 2} \right)}{2}\sqrt{140^{2} + \left( {-32v + 2} \right)^{2}}} \\

{+ \frac{140^{2}}{2}\text{ln}\left| {\left( {-32v + 2} \right) + \sqrt{140^{2} + \left( {-32v + 2} \right)^{2}}} \right|}

\end{array} \right\rbrack + C}

\end{array}$$

将 $a=140$、$u=-32v+2$ 及 $dv=-\frac{1}{32}du$ 代入积分公式，得到关于 $v$ 的原函数。

and

且

$$\begin{array}{cl}

{s(t)} & {= - \frac{1}{32}\left\lbrack {\frac{\left( {-32t + 2} \right)}{2}\sqrt{140^{2} + \left( {-32t + 2} \right)^{2}} + \frac{140^{2}}{2}\text{ln}\left| {\left( {-32t + 2} \right) + \sqrt{140^{2} + \left( {-32t + 2} \right)^{2}}} \right|} \right\rbrack} \\

& {\mspace{11mu} + \frac{1}{32}\left\lbrack {\sqrt{140^{2} + 2^{2}} + \frac{140^{2}}{2}\text{ln}\left| {2 + \sqrt{140^{2} + 2^{2}}} \right|} \right\rbrack} \\

& {= \left( {\frac{t}{2} - \frac{1}{32}} \right)\sqrt{1024t^{2} - 128t + 19604} - \frac{1225}{4}\text{ln}\left| {\left( {-32t + 2} \right) + \sqrt{1024t^{2} - 128t + 19604}} \right|} \\

& {\mspace{11mu} + \frac{\sqrt{19604}}{32} + \frac{1225}{4}\text{ln}\left( {2 + \sqrt{19604}} \right).}

\end{array}$$

代入上下限 $0$ 到 $t$ 后化简，得到球运动的弧长关于时间 $t$ 的显式表达式 $s(t)$。

This function represents the distance traveled by the ball as a function of time. To calculate the speed, take the derivative of this function with respect to *t.* While this may seem like a daunting task, it is possible to obtain the answer directly from the Fundamental Theorem of Calculus:

该函数表示球随时间运动的距离。为求速度，对此函数关于 *t* 求导。这看似艰巨，但可直接由微积分基本定理得到答案：

$$\frac{d}{dx}{\int_{a}^{x}{f(u)\ du}} = f(x).$$

Therefore

因此

$$\begin{array}{cl}

{s^{\prime}(t)} & {= \frac{d}{dt}\left\lbrack {s(t)} \right\rbrack} \\

& {= \frac{d}{dt}\left\lbrack {\int_{0}^{t}{\sqrt{140^{2} + \left( {-32v + 2} \right)^{2}}dv}} \right\rbrack} \\

& {= \sqrt{140^{2} + \left( {-32t + 2} \right)^{2}}} \\

& {= \sqrt{1024t^{2} - 128t + 19604}} \\

& {= 2\sqrt{256t^{2} - 32t + 4901}.}

\end{array}$$

由微积分基本定理，对弧长积分上限求导即得速度 $s^{\prime}(t) = \sqrt{1024t^{2} - 128t + 19604} = 2\sqrt{256t^{2} - 32t + 4901}$。

One third of a second after the ball leaves the pitcher’s hand, the distance it travels is equal to

球离开投手之手三分之一秒后，其运动距离等于

$$\begin{array}{cl}

{s\left( \frac{1}{3} \right)} & {= \left( {\frac{1\text{/}3}{2} - \frac{1}{32}} \right)\sqrt{1024\left( \frac{1}{3} \right)^{2} - 128\left( \frac{1}{3} \right) + 19604}} \\

& {\mspace{11mu} - \frac{1225}{4}\text{ln}\left| {\left( {-32\left( \frac{1}{3} \right) + 2} \right) + \sqrt{1024\left( \frac{1}{3} \right)^{2} - 128\left( \frac{1}{3} \right) + 19604}} \right|} \\

& {\mspace{11mu} + \frac{\sqrt{19604}}{32} + \frac{1225}{4}\text{ln}\left( {2 + \sqrt{19604}} \right)} \\

& {\approx 46.69\ \text{feet}.}

\end{array}$$

将 $t = \frac{1}{3}$ 代入 $s(t)$，得到此时球运动的距离约为 $46.69\ \text{feet}$（英尺）。

This value is just over three quarters of the way to home plate. The speed of the ball is

该值略超过到达本垒距离的四分之三。球的速度为

$$s^{\prime}\left( \frac{1}{3} \right) = 2\sqrt{256\left( \frac{1}{3} \right)^{2} - 16\left( \frac{1}{3} \right) + 4901} \approx 140.34\ \text{ft/s}.$$

This speed translates to approximately 95 mph—a major-league fastball.

此速度约为 95 英里/小时——是大联盟级别的快球速度。

Surface Area Generated by a Parametric Curve 由参数曲线生成的表面积

Recall the problem of finding the surface area of a volume of revolution. In Curve Length and Surface Area, we derived a formula for finding the surface area of a volume generated by a function $y = f(x)$ from $x = a$ to $x = b,$ revolved around the *x*-axis:

回顾求旋转体表面积的问题。在“曲线长度与表面积”一节中，我们推导了用于求由函数 $y = f(x)$ 从 $x = a$ 到 $x = b$ 绕 $x$ 轴旋转所成旋转体表面积的公式：

$$S = 2\pi{\int_{a}^{b}{f(x)\sqrt{1 + \left( {f^{\prime}(x)} \right)^{2}}dx}}.$$

We now consider a volume of revolution generated by revolving a parametrically defined curve $x = x(t),y = y(t),a \leq t \leq b$ around the *x*-axis as shown in the following figure.

现在考虑由参数定义的曲线 $x = x(t),y = y(t),a \leq t \leq b$ 绕 $x$ 轴旋转所成的旋转体，如下图所示。

The analogous formula for a parametrically defined curve is

对参数定义的曲线，相应的公式如下：

$$S = 2\pi{\int_{a}^{b}{y(t)\sqrt{\left( {x^{\prime}(t)} \right)^{2} + \left( {y^{\prime}(t)} \right)^{2}}dt}}$$ (1.6)

provided that $y(t)$ is not negative on $\lbrack a,b\rbrack.$

其中要求 $y(t)$ 在 $\lbrack a,b\rbrack$ 上非负。

Finding Surface Area 求表面积

Find the surface area of a sphere of radius *r* centered at the origin.

求以原点为中心、半径为 $r$ 的球面的表面积。

Solution 解

We start with the curve defined by the equations

我们从由下列方程定义的曲线入手：

$$x(t) = r\ \text{cos}\ t,\quad y(t) = r\ \text{sin}\ t,\quad 0 \leq t \leq \pi.$$

This generates an upper semicircle of radius *r* centered at the origin as shown in the following graph.

这生成以原点为中心、半径为 $r$ 的上半圆，如下图所示。

When this curve is revolved around the *x*-axis, it generates a sphere of radius *r*. To calculate the surface area of the sphere, we use Equation 1.6:

当该曲线绕 $x$ 轴旋转时，生成半径为 $r$ 的球面。为计算该球面的表面积，我们使用方程 1.6：

$$\begin{array}{cl}S & {= 2\pi{\int_{a}^{b}{y(t)\sqrt{\left( {x^{\prime}(t)} \right)^{2} + \left( {y^{\prime}(t)} \right)^{2}}dt}}} \\ & {= 2\pi{\int_{0}^{\pi}{r\ \text{sin}\ t\sqrt{\left( {\text{−}r\ \text{sin}\ t} \right)^{2} + \left( {r\ \text{cos}\ t} \right)^{2}}dt}}} \\ & {= 2\pi{\int_{0}^{\pi}{r\ \text{sin}\ t\sqrt{r^{2}\text{sin}^{2}t + r^{2}\text{cos}^{2}t}\ dt}}} \\ & {= 2\pi{\int_{0}^{\pi}{r\ \text{sin}\ t\sqrt{r^{2}\left( {\text{sin}^{2}t + \text{cos}^{2}t} \right)}dt}}} \\ & {= 2\pi{\int_{0}^{\pi}{r^{2}\text{sin}\ t\ dt}}} \\ & {= 2\pi r^{2}(\left. {\text{−}\text{cos}\ t} \right|_{0}^{\pi})} \\ & {= 2\pi r^{2}\left( {\text{−}\text{cos}\ \pi + \text{cos}\ 0} \right)} \\ & {= 4\pi r^{2}.}\end{array}$$

This is, in fact, the formula for the surface area of a sphere.

事实上，这正是球面的表面积公式。

Find the surface area generated when the plane curve defined by the equations

求由下列方程定义的平面曲线

$$x(t) = t^{3},\quad y(t) = t^{2},\quad 0 \leq t \leq 1$$

is revolved around the *x*-axis.

绕 $x$ 轴旋转时生成的表面积。

Section 1.2 Exercises 1.2 节习题

For the following exercises, each set of parametric equations represents a line. Without eliminating the parameter, find the slope of each line.

在以下习题中，每组参数方程都表示一条直线。不消去参数，求各直线的斜率。

62\.

62.

$\begin{array}{ll}{x = 3 + t,} & {y = 1 - t}\end{array}$

63.

$\begin{array}{ll}{x = 8 + 2t,} & {y = 1}\end{array}$

64\.

64.

$\begin{array}{ll}{x = 4 - 3t,} & {y = -2 + 6t}\end{array}$

65.

$\begin{array}{ll}{x = -5t + 7,} & {y = 3t - 1}\end{array}$

For the following exercises, determine the slope of the tangent line, then find an equation of the tangent line at the given value of the parameter.

在以下习题中，确定切线的斜率，再求该参数给定值处切线的方程。

66\.

66.

$\begin{array}{ll}{x = 3\ \text{sin}\ t,} & {y = 3\ \text{cos}\ t,\quad t = \frac{\pi}{4}}\end{array}$

67.

$\begin{array}{ll}{x = \text{cos}\ t,} & {y = 8\ \text{sin}\ t,}\end{array}t = \frac{\pi}{2}$

68\.

68.

$\begin{array}{ll}{x = 2t,} & {y = t^{3},\quad t = -1}\end{array}$

69.

$\begin{array}{ll}{x = t + \frac{1}{t},} & {y = t - \frac{1}{t},\quad t = 1}\end{array}$

70\.

70.

$\begin{array}{ll}{x = \sqrt{t},} & {y = 2t,\quad t = 4}\end{array}$

For the following exercises, find all points on the curve that have the given slope.

在以下习题中，求曲线上具有给定斜率的所有点。

71.

$\begin{array}{ll}{x = 4\ \text{cos}\ t,} & {y = 4\ \text{sin}\ t,}\end{array}$ slope = 0.5

$\begin{array}{ll}{x = 4\ \text{cos}\ t,} & {y = 4\ \text{sin}\ t,}\end{array}$ 斜率 = 0.5

72\.

72.

$\begin{array}{ll}{x = 2\ \text{cos}\ t,} & {y = 8\ \text{sin}\ t,\ \text{slope} = -1}\end{array}$

73.

$\begin{array}{ll}{x = t + \frac{1}{t},} & {y = t - \frac{1}{t},\ \text{slope} = 1}\end{array}$

74\.

74.

$\begin{array}{ll}{x = 2 + \sqrt{t},} & {y = 2 - 4t,\ \text{slope} = 0}\end{array}$

For the following exercises, write the equation of the tangent line in Cartesian coordinates for the given parameter *t*.

在以下习题中，对给定的参数 $t$，用直角坐标写出切线的方程。

75.

$\begin{array}{ll}{x = e^{\sqrt{t}},} & {y = 1 - \text{ln}\ t^{2},\quad t = 1}\end{array}$

76\.

76.

$\begin{array}{ll}{x = t\ \text{ln}\ t,} & {y = \text{sin}^{2}t,}\end{array}t = \frac{\pi}{4}$

77.

$\begin{array}{ll}{x = e^{t},} & {y = {(t - 1)}^{2},\quad\text{at}(1,1)}\end{array}$

78\.

78.

For $x = \text{sin}(2t),y = 2\ \text{sin}\ t$ where $0 \leq t < 2\pi.$ Find all values of *t* at which a horizontal tangent line exists.

对 $x = \text{sin}(2t),y = 2\ \text{sin}\ t$，其中 $0 \leq t < 2\pi.$ 求存在水平切线的所有 $t$ 值。

79.

For $x = \text{sin}(2t),y = 2\ \text{sin}\ t$ where $0 \leq t < 2\pi.$ Find all values of *t* at which a vertical tangent line exists.

对 $x = \text{sin}(2t),y = 2\ \text{sin}\ t$，其中 $0 \leq t < 2\pi.$ 求存在垂直切线的所有 $t$ 值。

80\.

80.

Find all points on the curve $x = 4\ \text{sin}(t),y = 4\ \text{cos}(t)$ that have the slope of $0.5$

求曲线 $x = 4\ \text{sin}(t),y = 4\ \text{cos}(t)$ 上斜率为 $0.5$ 的所有点。

81.

Find $\frac{dy}{dx}$ for $x = \text{sin}(t),y = \text{cos}(t).$

对 $x = \text{sin}(t),y = \text{cos}(t)$，求 $\frac{dy}{dx}$。

82\.

82.

Find an equation of the tangent line to $x = \text{sin}(t),y = \text{cos}(t)$ at $t = \frac{\pi}{4}.$

求曲线 $x = \text{sin}(t),y = \text{cos}(t)$ 在 $t = \frac{\pi}{4}$ 处的切线方程。

83.

For the curve $x = 4t,y = 3t - 2,$ find the slope and concavity of the curve at $t = 3.$

对曲线 $x = 4t,y = 3t - 2,$ 求其在 $t = 3$ 处的斜率与凹性。

84\.

84.

For the parametric curve whose equation is $x = 4\ \text{cos}\ \theta,y = 4\ \text{sin}\ \theta,$ find the slope and concavity of the curve at $\theta = \frac{\pi}{4}.$

对参数曲线 $x = 4\ \text{cos}\ \theta,y = 4\ \text{sin}\ \theta,$ 求其在 $\theta = \frac{\pi}{4}$ 处的斜率与凹性。

85.

Find the slope and concavity for the curve whose equation is $x = 2 + \text{sec}\ \theta,y = 1 + 2\ \text{tan}\ \theta$ at $\theta = \frac{\pi}{6}.$

对方程为 $x = 2 + \text{sec}\ \theta,y = 1 + 2\ \text{tan}\ \theta$ 的曲线，求其在 $\theta = \frac{\pi}{6}$ 处的斜率与凹性。

86\.

86.

Find all points on the curve $x = t + 4,y = t^{3} - 3t$ at which there are vertical and horizontal tangents.

求曲线 $x = t + 4,y = t^{3} - 3t$ 上具有垂直切线和水平切线的所有点。

87.

Find all points on the curve $x = \text{sec}\ \theta,y = \text{tan}\ \theta$ at which horizontal and vertical tangents exist.

求曲线 $x = \text{sec}\ \theta,y = \text{tan}\ \theta$ 上存在水平切线和垂直切线的所有点。

For the following exercises, find ${d^{2}y}\text{/}{dx^{2}.}$

在以下习题中，求 ${d^{2}y}\text{/}{dx^{2}.}$

88\.

88.

$\begin{array}{ll}{x = t^{4} - 1,} & {y = t - t^{2}}\end{array}$

89.

$\begin{array}{ll}{x = \text{sin}\left( {\pi t} \right),} & {y = \text{cos}\left( {\pi t} \right)}\end{array}$

90\.

90.

$\begin{array}{ll}{x = e^{\text{−}t},} & {y = t}\end{array}e^{2t}$

For the following exercises, find points on the curve at which tangent line is horizontal or vertical.

在以下习题中，求曲线上切线为水平或垂直的点。

91.

$\begin{array}{ll}{x = t(t^{2} - 3),} & {y = 3(t^{2} - 3)}\end{array}$

92\.

92.

$\begin{array}{ll}{x = \frac{3t}{1 + t^{3}},} & {y = \frac{3t^{2}}{1 + t^{3}}}\end{array}$

For the following exercises, find ${dy}\text{/}{dx}$ at the value of the parameter.

在以下习题中，求在给定参数值处的 ${dy}\text{/}{dx}$。

93.

$\begin{array}{ll}{x = \text{cos}\ t,} & {y = \text{sin}\ t,\quad t = \frac{3\pi}{4}}\end{array}$

94\.

94.

$\begin{array}{ll}{x = \sqrt{t},} & {y = 2t + 4,\quad t = 9}\end{array}$

95.

$\begin{array}{ll}{x = 4\ \text{cos}\left( {2\pi s} \right),} & {y = 3\ \text{sin}}\end{array}\left( {2\pi s} \right),\quad s = - \frac{1}{4}$

For the following exercises, find ${d^{2}y}\text{/}{dx^{2}}$ at the given point without eliminating the parameter.

在以下习题中，不消去参数，求给定点处的 ${d^{2}y}\text{/}{dx^{2}}$。

96\.

96.

$\begin{array}{lll}{x = \frac{1}{2}t^{2},} & {y = \frac{1}{3}t^{3},} & {t = 2}\end{array}$

97.

$x = \sqrt{t},\quad y = 2t + 4,\quad t = 1$

98\.

98.

Find *t* intervals on which the curve $x = 3t^{2},y = t^{3} - t$ is concave up as well as concave down.

求曲线 $x = 3t^{2},y = t^{3} - t$ 上凹（凹向上）与下凹（凹向下）的区间。

99.

Determine the concavity of the curve $x = 2t + \text{ln}\ t,y = 2t - \text{ln}\ t.$

判断曲线 $x = 2t + \text{ln}\ t,y = 2t - \text{ln}\ t$ 的凹性。

100\.

100.

Sketch and find the area under one arch of the cycloid $x = r\left( {\theta - \text{sin}\ \theta} \right),y = r\left( {1 - \text{cos}\ \theta} \right).$

画出摆线 $x = r\left( {\theta - \text{sin}\ \theta} \right),y = r\left( {1 - \text{cos}\ \theta} \right)$ 的一拱，并求其下面积。

101.

Find the area bounded by the curve $x = \text{cos}\ t,y = e^{t},0 \leq t \leq \frac{\pi}{2}$ and the lines $y = 1$ and $x = 0.$

求由曲线 $x = \text{cos}\ t,y = e^{t},0 \leq t \leq \frac{\pi}{2}$ 与直线 $y = 1$、$x = 0$ 所围成的面积。

102\.

102.

Find the area enclosed by the ellipse $x = a\ \text{cos}\ \theta,y = b\ \text{sin}\ \theta,0 \leq \theta < 2\pi.$

求由椭圆 $x = a\ \text{cos}\ \theta,y = b\ \text{sin}\ \theta,0 \leq \theta < 2\pi$ 所围成的面积。

103.

Find the area of the region bounded by $x = 2\ \text{sin}^{2}\theta,y = 2\ \text{sin}^{2}\theta\ \text{tan}\ \theta,$ for $0 \leq \theta \leq \frac{\pi}{2}.$

求由 $x = 2\ \text{sin}^{2}\theta,y = 2\ \text{sin}^{2}\theta\ \text{tan}\ \theta,$ 在 $0 \leq \theta \leq \frac{\pi}{2}$ 内所围成区域的面积。

For the following exercises, find the area of the regions bounded by the parametric curves and the indicated values of the parameter.

在以下习题中，求由参数曲线及所给参数值所围成区域的面积。

104\.

104.

$x = 2\ \text{cot}\ \theta,y = 2\ \text{sin}^{2}\theta,0 \leq \theta \leq \pi$

105.

\[T\] $x = 2a\ \text{cos}\ t - a\ \text{cos}(2t),y = 2a\ \text{sin}\ t - a\ \text{sin}(2t),0 \leq t < 2\pi$

106\.

106.

\[T\] $x = a\ \text{sin}(2t),y = b\ \text{sin}(t),0 \leq t < 2\pi$ (the “hourglass”)

\[T\] $x = a\ \text{sin}(2t),y = b\ \text{sin}(t),0 \leq t < 2\pi$（“沙漏形”）

107.

\[T\] $x = 2a\ \text{cos}\ t - a\ \text{sin}(2t),y = b\ \text{sin}\ t,0 \leq t < 2\pi$ (the “teardrop”)

\[T\] $x = 2a\ \text{cos}\ t - a\ \text{sin}(2t),y = b\ \text{sin}\ t,0 \leq t < 2\pi$（“泪滴形”）

For the following exercises, find the arc length of the curve on the indicated interval of the parameter.

在以下习题中，求参数给定区间上曲线的弧长。

108\.

108.

$x = 4t + 3,\quad y = 3t - 2,\quad 0 \leq t \leq 2$

109.

$\begin{array}{lll}{x = \frac{1}{3}t^{3},} & {y = \frac{1}{2}t^{2},} & {0 \leq t \leq 1}\end{array}$

110\.

110.

$\begin{array}{lll}{x = \text{cos}(2t),} & {y = \text{sin}(2t),} & {0 \leq t \leq \frac{\pi}{2}}\end{array}$

111.

$\begin{array}{lll}{x = 1 + t^{2},} & {y = \left( {1 + t} \right)^{3},} & {0 \leq t \leq 1}\end{array}$

112\.

112.

$\begin{array}{lll}{x = e^{t}\text{cos}\ t,} & {y = e^{t}\text{sin}\ t,} & {0 \leq t \leq \frac{\pi}{2}}\end{array}$ (Use a CAS for this and express the answer as a decimal rounded to three places.)

$\begin{array}{lll}{x = e^{t}\text{cos}\ t,} & {y = e^{t}\text{sin}\ t,} & {0 \leq t \leq \frac{\pi}{2}}\end{array}$（使用 CAS 计算，并将答案表示为保留三位小数的小数。）

113.

$x = a\ \text{cos}^{3}\theta,y = a\ \text{sin}^{3}\theta$ on the interval $\lbrack 0,2\pi)$ (the hypocycloid)

$x = a\ \text{cos}^{3}\theta,y = a\ \text{sin}^{3}\theta$ 在区间 $\lbrack 0,2\pi)$ 上（内摆线）

114\.

114.

Find the length of one arch of the cycloid $x = 4\left( {t - \text{sin}\ t} \right),y = 4\left( {1 - \text{cos}\ t} \right).$

求摆线 $x = 4\left( {t - \text{sin}\ t} \right),y = 4\left( {1 - \text{cos}\ t} \right)$ 一拱的长度。

115.

Find the distance traveled by a particle with position $\left( {x,y} \right)$ as *t* varies in the given time interval: $\begin{array}{lll}{x = \text{sin}^{2}t,} & {y = \text{cos}^{2}t,} & {0 \leq t \leq 3\pi}\end{array}.$

求一个粒子在给定时间区间内随 $t$ 变化时，其位置 $\left( {x,y} \right)$ 所经过的路程：$\begin{array}{lll}{x = \text{sin}^{2}t,} & {y = \text{cos}^{2}t,} & {0 \leq t \leq 3\pi}\end{array}.$

116\.

116.

Find the length of one arch of the cycloid $x = \theta - \text{sin}\ \theta,y = 1 - \text{cos}\ \theta.$

求摆线 $x = \theta - \text{sin}\ \theta,y = 1 - \text{cos}\ \theta$ 一拱的长度。

117.

Show that the total length of the ellipse $x = 4\ \text{sin}\ \theta,y = 3\ \text{cos}\ \theta$ is $L = 16{\int_{0}^{\pi\text{/}2}{\sqrt{1 - e^{2}\text{sin}^{2}\theta}\ d\theta,}}$ where $e = \frac{c}{a}$ and $c = \sqrt{a^{2} - b^{2}}.$

证明椭圆 $x = 4\ \text{sin}\ \theta,y = 3\ \text{cos}\ \theta$ 的总长度为 $L = 16{\int_{0}^{\pi\text{/}2}{\sqrt{1 - e^{2}\text{sin}^{2}\theta}\ d\theta,}}$ 其中 $e = \frac{c}{a}$，$c = \sqrt{a^{2} - b^{2}}$。

118\.

118.

Find the length of the curve $x = e^{t} - t,y = 4e^{t\text{/}2},-8 \leq t \leq 3.$

求曲线 $x = e^{t} - t,y = 4e^{t\text{/}2},-8 \leq t \leq 3$ 的长度。

For the following exercises, find the area of the surface obtained by rotating the given curve about the *x*-axis.

在以下习题中，求给定曲线绕 $x$ 轴旋转所得曲面的面积。

119.

$\begin{array}{lll}{x = t^{3},} & {y = t^{2},} & {0 \leq t \leq 1}\end{array}$

120\.

120.

$\begin{array}{lll}{x = a\ \text{cos}^{3}\theta,} & {y = a\ \text{sin}^{3}\theta,} & {0 \leq \theta \leq}\end{array}\frac{\pi}{2}$

121.

\[T\] Use a CAS to find the area of the surface generated by rotating $x = t + t^{3},y = t - \frac{1}{t^{2}},1 \leq t \leq 2$ about the *x*-axis. (Answer to three decimal places.)

\[T\] 使用 CAS 求由曲线 $x = t + t^{3},y = t - \frac{1}{t^{2}},1 \leq t \leq 2$ 绕 $x$ 轴旋转所成曲面的面积。（答案保留三位小数。）

122\.

122.

Find the surface area obtained by rotating $x = 3t^{2},y = 2t^{3},0 \leq t \leq 5$ about the *y*-axis.

求曲线 $x = 3t^{2},y = 2t^{3},0 \leq t \leq 5$ 绕 $y$ 轴旋转所得曲面的面积。

123.

Find the area of the surface generated by revolving $x = t^{2},y = 2t,0 \leq t \leq 4$ about the *x*-axis.

求曲线 $x = t^{2},y = 2t,0 \leq t \leq 4$ 绕 $x$ 轴旋转所成曲面的面积。

124\.

124.

Find the surface area generated by revolving $x = t^{2},y = 2t^{2},0 \leq t \leq 1$ about the *y*-axis.

求曲线 $x = t^{2},y = 2t^{2},0 \leq t \leq 1$ 绕 $y$ 轴旋转所成曲面的面积。

1.3 Polar Coordinates 1.3 极坐标

1.3.1 Locate points in a plane by using polar coordinates.
1.3.2 Convert points between rectangular and polar coordinates.
1.3.3 Sketch polar curves from given equations.
1.3.4 Convert equations between rectangular and polar coordinates.
1.3.5 Identify symmetry in polar curves and equations.

1.3.1 用极坐标在平面上定位点。
1.3.2 在直角坐标与极坐标之间转换点。
1.3.3 由给定方程绘制极坐标曲线。
1.3.4 在直角坐标与极坐标之间转换方程。
1.3.5 识别极坐标曲线与方程中的对称性。

The rectangular coordinate system (or Cartesian plane) provides a means of mapping points to ordered pairs and ordered pairs to points. This is called a *one-to-one mapping* from points in the plane to ordered pairs. The polar coordinate system provides an alternative method of mapping points to ordered pairs. In this section we see that in some circumstances, polar coordinates can be more useful than rectangular coordinates.

直角坐标（或笛卡尔平面）提供了一种将点映射到有序数对、并将有序数对映射回点的方法。这称为从平面上的点到有序数对的一种*一一映射*。极坐标系提供了将点映射到有序数对的另一种方法。在本节中我们将看到，在某些情况下，极坐标比直角坐标更有用。

Defining Polar Coordinates 定义极坐标

To find the coordinates of a point in the polar coordinate system, consider Figure 1.27. The point $P$ has Cartesian coordinates $\left( {x,y} \right).$ The line segment connecting the origin to the point $P$ measures the distance from the origin to $P$ and has length $r.$ The angle between the positive $x$-axis and the line segment has measure $\theta.$ This observation suggests a natural correspondence between the coordinate pair $\left( {x,y} \right)$ and the values $r$ and $\theta.$ This correspondence is the basis of the polar coordinate system. Note that every point in the Cartesian plane has two values (hence the term *ordered pair*) associated with it. In the polar coordinate system, each point also has two values associated with it: $r$ and $\theta.$

为了在极坐标系中确定一点坐标，考虑图 1.27。点 $P$ 的直角坐标为 $\left( {x,y} \right)$。连接原点与点 $P$ 的线段给出了从原点到 $P$ 的距离，其长度为 $r$。正 $x$ 轴与该线段之间的夹角为 $\theta$。这一观察提示了坐标对 $\left( {x,y} \right)$ 与数值 $r$、$\theta$ 之间的一种自然对应。这种对应正是极坐标系的基础。注意，直角平面上的每一点都带有与之关联的两个数值（因此有"有序对"这个术语）。在极坐标系中，每一点也带有两个与之关联的数值：$r$ 和 $\theta$。

Using right-triangle trigonometry, the following equations are true for the point $P\text{:}$

利用直角三角形三角学，对点 $P$ 而言下列方程成立：

$$\text{cos}\ \theta = \frac{x}{r}\ \text{so}\ x = r\ \text{cos}\ \theta$$ $$\text{sin}\ \theta = \frac{y}{r}\ \text{so}\ y = r\ \text{sin}\ \theta.$$

Furthermore,

此外，

$$r^{2} = x^{2} + y^{2}\ \text{and}\ \text{tan}\ \theta = \frac{y}{x}.$$

Each point $\left( {x,y} \right)$ in the Cartesian coordinate system can therefore be represented as an ordered pair $\left( {r,\theta} \right)$ in the polar coordinate system. The first coordinate is called the radial coordinate and the second coordinate is called the angular coordinate. Every point in the plane can be represented in this form.

因此，直角坐标系中的每个点 $\left( {x,y} \right)$ 都可以在极坐标系中表示为有序对 $\left( {r,\theta} \right)$。第一个坐标称为径向坐标，第二个坐标称为角坐标。平面上每一点都可以这种形式表示。

Note that the equation $\text{tan}\ \theta = {y\text{/}x}$ has an infinite number of solutions for any ordered pair $\left( {x,y} \right).$ However, if we restrict the solutions to values between $0$ and $2\pi$ then we can assign a unique solution to the quadrant in which the original point $\left( {x,y} \right)$ is located. Then the corresponding value of *r* is positive, so $r^{2} = x^{2} + y^{2}.$

注意，方程 $\text{tan}\ \theta = {y\text{/}x}$ 对任意有序对 $\left( {x,y} \right)$ 都有无穷多个解。但是，如果我们将解限制在 $0$ 到 $2\pi$ 之间，就能为原有点 $\left( {x,y} \right)$ 所在的象限指定唯一的解。此时对应的 *r* 取正值，从而 $r^{2} = x^{2} + y^{2}.$

Converting Points between Coordinate Systems 坐标系之间的点转换

Given a point $P$ in the plane with Cartesian coordinates $\left( {x,y} \right)$ and polar coordinates $\left( {r,\theta} \right),$ the following conversion formulas hold true:

设平面上一点 $P$ 的直角坐标为 $\left( {x,y} \right)$、极坐标为 $\left( {r,\theta} \right)$，则下列转换公式成立：

$$x = r\ \text{cos}\ \theta\ \text{and}\ y = r\ \text{sin}\ \theta,$$ (1.7) $$r^{2} = x^{2} + y^{2}\ \text{and}\ \text{tan}\ \theta = \frac{y}{x}.$$ (1.8)

These formulas can be used to convert from rectangular to polar or from polar to rectangular coordinates.

这些公式可用于在直角坐标与极坐标之间相互转换。

Converting between Rectangular and Polar Coordinates 直角坐标与极坐标之间的转换

Convert each of the following points into polar coordinates.

将下列各点转换为极坐标。

1. $(1,1)$

2. $(-3,4)$

3. $\left( {0,3} \right)$

4. $(5\sqrt{3},-5)$

Convert each of the following points into rectangular coordinates.

将下列各点转换为直角坐标。

5. $(3,{\pi\text{/}3})$

6. $(2,{{3\pi}\text{/}2})$

7. $(6,{{-5\pi}\text{/}6})$

Solution 解

1. Use $x = 1$ and $y = 1$ in Equation 1.8:

1. 在公式 1.8 中取 $x = 1$、$y = 1$：

$$\begin{array}{lcclccl}

\begin{array}{cll}

r^{2} & = & {x^{2} + y^{2}} \\

& = & {1^{2} + 1^{2}} \\

r & = & \sqrt{2}

\end{array} & & & \text{and} & & & \begin{array}{cll}

{\text{tan}\ \theta} & = & \frac{y}{x} \\

& = & {\frac{1}{1} = 1} \\

\theta & = & {\frac{\pi}{4}.}

\end{array}

\end{array}$$

Therefore this point can be represented as $\left( {\sqrt{2},\frac{\pi}{4}} \right)$ in polar coordinates.

因此，该点在极坐标中可表示为 $\left( {\sqrt{2},\frac{\pi}{4}} \right)$。

2. Use $x = -3$ and $y = 4$ in Equation 1.8:

2. 在公式 1.8 中取 $x = -3$、$y = 4$：

$$\begin{matrix}

\begin{matrix}

r^{2} & = & {x^{2} + y^{2}} \\

& = & {(-3)^{2} + (4)^{2}} \\

r & = & 5

\end{matrix} & & & \text{and} & & & \begin{matrix}

{\text{tan}\ \theta} & = & \frac{y}{x} \\

& = & {- \frac{4}{3}} \\

\theta & = & {\pi - \arctan\left( \frac{4}{3} \right)} \\

& \approx & {2.21.}

\end{matrix}

\end{matrix}$$

The point $( - 3,~4)$ lies in Quadrant $\text{II}$. Subtract the value of the reference angle, $\text{arctan}\left( \frac{4}{3} \right)$, from $\pi$ to find the radian measure of $\theta$.

点 $( - 3,~4)$ 位于第二象限。从 $\pi$ 中减去参考角 $\text{arctan}\left( \frac{4}{3} \right)$ 的值，得到 $\theta$ 的弧度。

Therefore this point can be represented as $\left( {5,2.21} \right)$ in polar coordinates.

因此，该点在极坐标中可表示为 $\left( {5,2.21} \right)$。

3. Use $x = 0$ and $y = 3$ in Equation 1.8:

3. 在公式 1.8 中取 $x = 0$、$y = 3$：

$$\begin{array}{lcclccl}

\begin{array}{cll}

r^{2} & = & {x^{2} + y^{2}} \\

& = & {(3)^{2} + (0)^{2}} \\

& = & {9 + 0} \\

r & = & 3

\end{array} & & & \text{and} & & & \begin{array}{cll}

{\text{tan}\ \theta} & = & \frac{y}{x} \\

& = & {\frac{3}{0}.}

\end{array}

\end{array}$$

Direct application of the second equation leads to division by zero. Graphing the point $\left( {0,3} \right)$ on the rectangular coordinate system reveals that the point is located on the positive *y*-axis. The angle between the positive *x*-axis and the positive *y*-axis is $\frac{\pi}{2}.$ Therefore this point can be represented as $\left( {3,\frac{\pi}{2}} \right)$ in polar coordinates.

直接套用第二个方程会导致除以零。在直角坐标系上描出点 $\left( {0,3} \right)$ 可知该点位于正 *y* 轴上。正 *x* 轴与正 *y* 轴之间的夹角为 $\frac{\pi}{2}$。因此该点在极坐标中可表示为 $\left( {3,\frac{\pi}{2}} \right)$。

4. Use $x = 5\sqrt{3}$ and $y = -5$ in Equation 1.8:

4. 在公式 1.8 中取 $x = 5\sqrt{3}$、$y = -5$：

$$\begin{array}{lcclccl}

\begin{array}{cll}

r^{2} & = & {x^{2} + y^{2}} \\

& = & {\left( {5\sqrt{3}} \right)^{2} + (-5)^{2}} \\

& = & {75 + 25} \\

r & = & 10

\end{array} & & & \text{and} & & & \begin{array}{cll}

{\text{tan}\ \theta} & = & \frac{y}{x} \\

& = & {\frac{-5}{5\sqrt{3}} = - \frac{\sqrt{3}}{3}} \\

\theta & = & {- \frac{\pi}{6}.}

\end{array}

\end{array}$$

Therefore this point can be represented as $\left( {10, - \frac{\pi}{6}} \right)$ in polar coordinates.

因此，该点在极坐标中可表示为 $\left( {10, - \frac{\pi}{6}} \right)$。

5. Use $r = 3$ and $\theta = \frac{\pi}{3}$ in Equation 1.7:

5. 在公式 1.7 中取 $r = 3$、$\theta = \frac{\pi}{3}$：

$$\begin{array}{lcclccl}

\begin{array}{cll}

x & = & {r\ \text{cos}\ \theta} \\

& = & {3\ \text{cos}\left( \frac{\pi}{3} \right)} \\

& = & {3\left( \frac{1}{2} \right) = \frac{3}{2}}

\end{array} & & & \text{and} & & & \begin{array}{cll}

y & = & {r\ \text{sin}\ \theta} \\

& = & {3\ \text{sin}\left( \frac{\pi}{3} \right)} \\

& = & {3\left( \frac{\sqrt{3}}{2} \right) = \frac{3\sqrt{3}}{2}.}

\end{array}

\end{array}$$

Therefore this point can be represented as $\left( {\frac{3}{2},\ \frac{3\sqrt{3}}{2}} \right)$ in rectangular coordinates.

因此，该点在直角坐标中可表示为 $\left( {\frac{3}{2},\ \frac{3\sqrt{3}}{2}} \right)$。

6. Use $r = 2$ and $\theta = \frac{3\pi}{2}$ in Equation 1.7:

6. 在公式 1.7 中取 $r = 2$、$\theta = \frac{3\pi}{2}$：

$$\begin{array}{lcclccl}

\begin{array}{cll}

x & = & {r\ \text{cos}\ \theta} \\

& = & {2\ \text{cos}\left( \frac{3\pi}{2} \right)} \\

& = & {2(0) = 0}

\end{array} & & & \text{and} & & & \begin{array}{cll}

y & = & {r\ \text{sin}\ \theta} \\

& = & {2\ \text{sin}\left( \frac{3\pi}{2} \right)} \\

& = & {2(-1) = -2.}

\end{array}

\end{array}$$

Therefore this point can be represented as $\left( {0,-2} \right)$ in rectangular coordinates.

因此，该点在直角坐标中可表示为 $\left( {0,-2} \right)$。

7. Use $r = 6$ and $\theta = - \frac{5\pi}{6}$ in Equation 1.7:

7. 在公式 1.7 中取 $r = 6$、$\theta = - \frac{5\pi}{6}$：

$$\begin{array}{lcclccl}

\begin{array}{cll}

x & = & {r\ \text{cos}\ \theta} \\

& = & {6\ \text{cos}\left( {- \frac{5\pi}{6}} \right)} \\

& = & {6\left( {- \frac{\sqrt{3}}{2}} \right)} \\

& = & {-3\sqrt{3}}

\end{array} & & & \text{and} & & & \begin{array}{cll}

y & = & {r\ \text{sin}\ \theta} \\

& = & {6\ \text{sin}\left( {- \frac{5\pi}{6}} \right)} \\

& = & {6\left( {- \frac{1}{2}} \right)} \\

& = & -3.

\end{array}

\end{array}$$

Therefore this point can be represented as $\left( {-3\sqrt{3},-3} \right)$ in rectangular coordinates.

因此，该点在直角坐标中可表示为 $\left( {-3\sqrt{3},-3} \right)$。

Convert $\left( {-8,-8} \right)$ into polar coordinates and $\left( {4,\frac{2\pi}{3}} \right)$ into rectangular coordinates.

将 $\left( {-8,-8} \right)$ 转换为极坐标，将 $\left( {4,\frac{2\pi}{3}} \right)$ 转换为直角坐标。

The polar representation of a point is not unique. For example, the polar coordinates $\left( {2,\frac{\pi}{3}} \right)$ and $\left( {2,\frac{7\pi}{3}} \right)$ both represent the point $\left( {1,\sqrt{3}} \right)$ in the rectangular system. Also, the value of $r$ can be negative. Therefore, the point with polar coordinates $\left( {-2,\frac{4\pi}{3}} \right)$ also represents the point $\left( {1,\sqrt{3}} \right)$ in the rectangular system, as we can see by using Equation 1.8:

一个点的极坐标表示并不唯一。例如，极坐标 $\left( {2,\frac{\pi}{3}} \right)$ 与 $\left( {2,\frac{7\pi}{3}} \right)$ 在直角坐标中都表示点 $\left( {1,\sqrt{3}} \right)$。而且 $r$ 的取值可以为负。因此，极坐标 $\left( {-2,\frac{4\pi}{3}} \right)$ 在直角坐标中也表示点 $\left( {1,\sqrt{3}} \right)$，这可由公式 1.8 看出：

$$\begin{array}{lcclccl}

\begin{array}{cll}

x & = & {r\ \text{cos}\ \theta} \\

& = & {-2\ \text{cos}\left( \frac{4\pi}{3} \right)} \\

& = & {-2\left( {- \frac{1}{2}} \right) = 1}

\end{array} & & & \text{and} & & & \begin{array}{cll}

y & = & {r\ \text{sin}\ \theta} \\

& = & {-2\ \text{sin}\left( \frac{4\pi}{3} \right)} \\

& = & {-2\left( {- \frac{\sqrt{3}}{2}} \right) = \sqrt{3}.}

\end{array}

\end{array}$$

Every point in the plane has an infinite number of representations in polar coordinates. However, each point in the plane has only one representation in the rectangular coordinate system.

平面上每一点在极坐标下都有无穷多种表示。然而，平面上每一点在直角坐标系下只有唯一的表示。

Note that the polar representation of a point in the plane also has a visual interpretation. In particular, $r$ is the directed distance that the point lies from the origin, and $\theta$ measures the angle that the line segment from the origin to the point makes with the positive $x$-axis. Positive angles are measured in a counterclockwise direction and negative angles are measured in a clockwise direction. The polar coordinate system appears in the following figure.

注意，平面上一点在极坐标下的表示还有直观的几何意义。具体地，$r$ 是该点离原点的有向距离，$\theta$ 度量的是从原点到该点的线段与正 $x$ 轴所成的角。正角按逆时针方向度量，负角按顺时针方向度量。极坐标系如图 1.28 所示。

The line segment starting from the center of the graph going to the right (called the positive *x*-axis in the Cartesian system) is the polar axis. The center point is the pole, or origin, of the coordinate system, and corresponds to $r = 0.$ The innermost circle shown in Figure 1.28 contains all points a distance of 1 unit from the pole, and is represented by the equation $r = 1.$ Then $r = 2$ is the set of points 2 units from the pole, and so on. The line segments emanating from the pole correspond to fixed angles. To plot a point in the polar coordinate system, start with the angle. If the angle is positive, then measure the angle from the polar axis in a counterclockwise direction. If it is negative, then measure it clockwise. If the value of $r$ is positive, move that distance along the terminal ray of the angle. If it is negative, move along the ray that is opposite the terminal ray of the given angle.

从图形中心向右出发的线段（在直角坐标中称为正 *x* 轴）就是极轴。中心点就是极点（即原点），对应于 $r = 0$。图 1.28 中最内层的圆包含所有离极点距离为 1 的点，由方程 $r = 1$ 表示。接着，$r = 2$ 是离极点 2 个单位的点的集合，依此类推。从极点发出的射线对应于固定的角。要在极坐标系中描点，先从角开始：若角为正，则从极轴起按逆时针方向量取该角；若角为负，则按顺时针方向量取。若 $r$ 为正，沿该角的终边射线移动相应距离；若 $r$ 为负，则沿与该角终边射线相反方向的射线移动。

Plotting Points in the Polar Plane 在极坐标平面上描点

Plot each of the following points on the polar plane.

在极坐标平面上描出下列各点。

1. $\left( {2,\frac{\pi}{4}} \right)$

2. $\left( {-3,\frac{2\pi}{3}} \right)$

3. $\left( {4,\frac{5\pi}{4}} \right)$

Solution 解

The three points are plotted in the following figure.

这三个点已描于下图中。

Plot $\left( {4,\frac{5\pi}{3}} \right)$ and $\left( {-3, - \frac{7\pi}{2}} \right)$ on the polar plane.

在极坐标平面上描出 $\left( {4,\frac{5\pi}{3}} \right)$ 与 $\left( {-3, - \frac{7\pi}{2}} \right)$。

Polar Curves 极坐标曲线

Now that we know how to plot points in the polar coordinate system, we can discuss how to plot curves. In the rectangular coordinate system, we can graph a function $y = f(x)$ and create a curve in the Cartesian plane. In a similar fashion, we can graph a curve that is generated by a function $r = f(\theta).$

既然我们已经知道如何在极坐标系中描点，就可以讨论如何绘制曲线了。在直角坐标系中，我们可以画出函数 $y = f(x)$ 的图形，并在笛卡尔平面上得到一条曲线。类似地，我们也可以画出由函数 $r = f(\theta)$ 所生成的曲线。

The general idea behind graphing a function in polar coordinates is the same as graphing a function in rectangular coordinates. Start with a list of values for the independent variable $(\theta$ in this case) and calculate the corresponding values of the dependent variable $r.$ This process generates a list of ordered pairs, which can be plotted in the polar coordinate system. Finally, connect the points, and take advantage of any patterns that may appear. The function may be periodic, for example, which indicates that only a limited number of values for the independent variable are needed.

在极坐标中绘制函数图形的基本思想，与在直角坐标系中绘制函数图形相同。先列出自变量（此处为 $\theta$）的取值，再计算对应的因变量 $r$ 的值。这个过程生成一组有序对，可在极坐标系中描出。最后把这些点连起来，并利用可能出现的任何规律。例如，函数可能是周期的，这意味着只需要有限个自变量取值即可。

Plotting a Curve in Polar Coordinates 在极坐标系中绘制曲线

1. Create a table with two columns. The first column is for $\theta,$ and the second column is for $r.$

1. 创建一个两列的表格。第一列为 $\theta$，第二列为 $r$。

2. Create a list of values for $\theta.$

2. 列出 $\theta$ 的取值。

3. Calculate the corresponding $r$ values for each $\theta.$

3. 对每个 $\theta$ 计算对应的 $r$ 值。

4. Plot each ordered pair $\left( {r,\theta} \right)$ on the coordinate axes.

4. 在坐标轴上描出每个有序对 $\left( {r,\theta} \right)$。

5. Connect the points and look for a pattern.

5. 将这些点连起来并寻找规律。

Watch this video for more information on sketching polar curves.

关于绘制极坐标曲线的更多信息，可观看相关视频。（原文含外链视频，已按要求去除外部链接）

Graphing a Function in Polar Coordinates 在极坐标中绘制函数图形

Graph the curve defined by the function $r = 4\ \text{sin}\ \theta.$ Identify the curve and rewrite the equation in rectangular coordinates.

画出由函数 $r = 4\ \text{sin}\ \theta$ 定义的曲线。识别该曲线，并将方程改写为直角坐标形式。

Solution 解

Because the function is a multiple of a sine function, it is periodic with period $2\pi,$ so use values for $\theta$ between 0 and $2\pi.$ The result of steps 1–3 appear in the following table. Figure 1.30 shows the graph based on this table.

由于该函数是正弦函数的倍数，它是周期为 $2\pi$ 的周期函数，因此取 $\theta$ 在 0 到 $2\pi$ 之间的值。第 1–3 步的结果如下表所示。图 1.30 给出了根据该表绘出的图形。

| $\theta$ | $r = 4\ \text{sin}\ \theta$ | | $\theta$ | $r = 4\ \text{sin}\ \theta$ | |------------------|-----------------------------|-----|-------------------|-----------------------------| | 0 | 0 | | $\pi$ | 0 | | $\frac{\pi}{6}$ | $2$ | | $\frac{7\pi}{6}$ | $-2$ | | $\frac{\pi}{4}$ | $2\sqrt{2} \approx 2.8$ | | $\frac{5\pi}{4}$ | $-2\sqrt{2} \approx -2.8$ | | $\frac{\pi}{3}$ | $2\sqrt{3} \approx 3.4$ | | $\frac{4\pi}{3}$ | $-2\sqrt{3} \approx -3.4$ | | $\frac{\pi}{2}$ | $4$ | | $\frac{3\pi}{2}$ | $-4$ | | $\frac{2\pi}{3}$ | $2\sqrt{3} \approx 3.4$ | | $\frac{5\pi}{3}$ | $-2\sqrt{3} \approx -3.4$ | | $\frac{3\pi}{4}$ | $2\sqrt{2} \approx 2.8$ | | $\frac{7\pi}{4}$ | $-2\sqrt{2} \approx -2.8$ | | $\frac{5\pi}{6}$ | $2$ | | $\frac{11\pi}{6}$ | $-2$ | | | | | $2\pi$ | 0 |

$\theta$ 与 $r = 4\ \text{sin}\ \theta$ 对应取值表
$\theta$	$r = 4\ \text{sin}\ \theta$	$\theta$	$r = 4\ \text{sin}\ \theta$
0	0	$\pi$	0
$\frac{\pi}{6}$	$2$	$\frac{7\pi}{6}$	$-2$
$\frac{\pi}{4}$	$2\sqrt{2} \approx 2.8$	$\frac{5\pi}{4}$	$-2\sqrt{2} \approx -2.8$
$\frac{\pi}{3}$	$2\sqrt{3} \approx 3.4$	$\frac{4\pi}{3}$	$-2\sqrt{3} \approx -3.4$
$\frac{\pi}{2}$	$4$	$\frac{3\pi}{2}$	$-4$
$\frac{2\pi}{3}$	$2\sqrt{3} \approx 3.4$	$\frac{5\pi}{3}$	$-2\sqrt{3} \approx -3.4$
$\frac{3\pi}{4}$	$2\sqrt{2} \approx 2.8$	$\frac{7\pi}{4}$	$-2\sqrt{2} \approx -2.8$
$\frac{5\pi}{6}$	$2$	$\frac{11\pi}{6}$	$-2$
		$2\pi$	0

This is the graph of a circle. The equation $r = 4\ \text{sin}\ \theta$ can be converted into rectangular coordinates by first multiplying both sides by $r.$ This gives the equation $r^{2} = 4r\ \text{sin}\ \theta.$ Next use the facts that $r^{2} = x^{2} + y^{2}$ and $y = r\ \text{sin}\ \theta.$ This gives $x^{2} + y^{2} = 4y.$ To put this equation into standard form, subtract $4y$ from both sides of the equation and complete the square:

这是一个圆。将方程 $r = 4\ \text{sin}\ \theta$ 化为直角坐标的方法是先两边同乘 $r$，得到 $r^{2} = 4r\ \text{sin}\ \theta.$ 再利用 $r^{2} = x^{2} + y^{2}$ 与 $y = r\ \text{sin}\ \theta$，得到 $x^{2} + y^{2} = 4y.$ 为了化为标准形式，从方程两边减去 $4y$ 并配方：

$$\begin{array}{rll} {x^{2} + y^{2} - 4y} & = & 0 \\ {x^{2} + \left( {y^{2} - 4y} \right)} & = & 0 \\ {x^{2} + \left( {y^{2} - 4y + 4} \right)} & = & {0 + 4} \\ {x^{2} + \left( {y - 2} \right)^{2}} & = & 4. \end{array}$$

This is the equation of a circle with radius 2 and center $\left( {0,2} \right)$ in the rectangular coordinate system.

这是在直角坐标系下，半径为 2、圆心为 $\left( {0,2} \right)$ 的圆的方程。

Create a graph of the curve defined by the function $r = 4 + 4\ \text{cos}\ \theta.$

画出由函数 $r = 4 + 4\ \text{cos}\ \theta$ 定义的曲线。

The graph in Example 1.12 was that of a circle. The equation of the circle can be transformed into rectangular coordinates using the coordinate transformation formulas in Equation 1.8. Example 1.14 gives some more examples of functions for transforming from polar to rectangular coordinates.

例 1.12 中的图形是一个圆。利用式 1.8 中的坐标转换公式，可将该圆的方程化为直角坐标形式。例 1.14 给出了更多从极坐标转换为直角坐标的函数示例。

Transforming Polar Equations to Rectangular Coordinates 将极坐标方程化为直角坐标方程

Rewrite each of the following equations in rectangular coordinates and identify the graph.

将下列各个方程化为直角坐标形式，并指出其图形。

1. $\theta = \frac{\pi}{3}$

2. $r = 3$

3. $r = 6\ \text{cos}\ \theta - 8\ \text{sin}\ \theta$

Solution 解

1. Take the tangent of both sides. This gives $\text{tan}\ \theta = \text{tan}({\pi\text{/}3}) = \sqrt{3}.$ Since $\text{tan}\ \theta = {y\text{/}x}$ we can replace the left-hand side of this equation by ${y\text{/}x}.$ This gives ${y\text{/}x} = \sqrt{3},$ which can be rewritten as $y = x\sqrt{3}.$ This is the equation of a straight line passing through the origin with slope $\sqrt{3}.$ In general, any polar equation of the form $\theta = K$ represents a straight line through the pole with slope equal to $\text{tan}\ K.$

1. 对两边取正切，得 $\text{tan}\ \theta = \text{tan}({\pi\text{/}3}) = \sqrt{3}.$ 由于 $\text{tan}\ \theta = {y\text{/}x}$，可用 ${y\text{/}x}$ 替换方程左端，得到 ${y\text{/}x} = \sqrt{3}$，即 $y = x\sqrt{3}.$ 这是一条过原点、斜率为 $\sqrt{3}$ 的直线方程。一般地，任何形如 $\theta = K$ 的极坐标方程都表示一条过极点、斜率为 $\text{tan}\ K$ 的直线。

2. First, square both sides of the equation. This gives $r^{2} = 9.$ Next replace $r^{2}$ with $x^{2} + y^{2}.$ This gives the equation $x^{2} + y^{2} = 9,$ which is the equation of a circle centered at the origin with radius 3. In general, any polar equation of the form $r = k$ where *k* is a positive constant represents a circle of radius *k* centered at the origin. (*Note*: when squaring both sides of an equation it is possible to introduce new points unintentionally. This should always be taken into consideration. However, in this case we do not introduce new points. For example, $\left( {-3,\frac{\pi}{3}} \right)$ is the same point as $\left( {3,\frac{4\pi}{3}} \right).)$

2. 先对方程两边平方，得 $r^{2} = 9.$ 再用 $x^{2} + y^{2}$ 替换 $r^{2}$，得到 $x^{2} + y^{2} = 9$，即以原点为圆心、半径为 3 的圆的方程。一般地，任何形如 $r = k$（$k$ 为正常数）的极坐标方程都表示以原点为圆心、半径为 $k$ 的圆。（注：对方程两边平方可能会无意中引入新的点，这一点应始终注意。不过此处并未引入新点。例如 $\left( {-3,\frac{\pi}{3}} \right)$ 与 $\left( {3,\frac{4\pi}{3}} \right)$ 是同一个点。）

3. Multiply both sides of the equation by $r.$ This leads to $r^{2} = 6r\ \text{cos}\ \theta - 8r\ \text{sin}\ \theta.$ Next use the formulas

3. 将方程两边同乘 $r$，得到 $r^{2} = 6r\ \text{cos}\ \theta - 8r\ \text{sin}\ \theta.$ 再利用公式

$$r^{2} = x^{2} + y^{2},\quad x = r\ \text{cos}\ \theta,\quad y = r\ \text{sin}\ \theta.$$

$$r^{2} = x^{2} + y^{2},\quad x = r\ \text{cos}\ \theta,\quad y = r\ \text{sin}\ \theta.$$ 即 $r^{2} = x^{2} + y^{2}$、$x = r\cos\theta$、$y = r\sin\theta$。

This gives

得到

$$\begin{array}{rll} r^{2} & = & {6\left( {r\ \text{cos}\ \theta} \right) - 8\left( {r\ \text{sin}\ \theta} \right)} \\ {x^{2} + y^{2}} & = & {6x - 8y.} \end{array}$$

To put this equation into standard form, first move the variables from the right-hand side of the equation to the left-hand side, then complete the square.

为了化为标准形式，先将右边的变量移到左边，然后配方。

$$\begin{array}{rll} {x^{2} + y^{2}} & = & {6x - 8y} \\ {x^{2} - 6x + y^{2} + 8y} & = & 0 \\ {\left( {x^{2} - 6x} \right) + \left( {y^{2} + 8y} \right)} & = & 0 \\ {\left( {x^{2} - 6x + 9} \right) + \left( {y^{2} + 8y + 16} \right)} & = & {9 + 16} \\ {\left( {x - 3} \right)^{2} + \left( {y + 4} \right)^{2}} & = & 25. \end{array}$$

This is the equation of a circle with center at $\left( {3,-4} \right)$ and radius 5. Notice that the circle passes through the origin since the center is 5 units away.

这是圆心在 $\left( {3,-4} \right)$、半径为 5 的圆的方程。注意该圆经过原点，因为圆心到原点的距离恰好为 5。

Rewrite the equation $r = \text{sec}\ \theta\ \text{tan}\ \theta$ in rectangular coordinates and identify its graph.

将方程 $r = \text{sec}\ \theta\ \text{tan}\ \theta$ 化为直角坐标形式，并指出其图形。

We have now seen several examples of drawing graphs of curves defined by polar equations. A summary of some common curves is given in the tables below. In each equation, *a* and *b* are arbitrary constants.

现在我们已经见过几个由极坐标方程定义的曲线作图示例。下面几张表汇总了一些常见曲线。在每个方程中，$a$ 和 $b$ 均为任意常数。

A cardioid is a special case of a limaçon (pronounced “lee-mah-son”), in which $a = b$ or $a = \text{−}b.$ The rose is a very interesting curve. Notice that the graph of $r = 3\ \text{sin}\ 2\theta$ has four petals. However, the graph of $r = 3\ \text{sin}\ 3\theta$ has three petals as shown.

心形线是蚶线（limaçon，读作“lee-mah-son”）的一种特殊情况，此时 $a = b$ 或 $a = \text{−}b$。玫瑰线是一种非常有趣的曲线。注意 $r = 3\ \text{sin}\ 2\theta$ 的图形有四个花瓣，而 $r = 3\ \text{sin}\ 3\theta$ 的图形有三个花瓣，如下图所示。

If the coefficient of $\theta$ is even, the graph has twice as many petals as the coefficient. If the coefficient of $\theta$ is odd, then the number of petals equals the coefficient. You are encouraged to explore why this happens. Even more interesting graphs emerge when the coefficient of $\theta$ is not an integer. For example, if it is rational, then the curve is closed; that is, it eventually ends where it started (Figure 1.34(a)). However, if the coefficient is irrational, then the curve never closes (Figure 1.34(b)). Although it may appear that the curve is closed, a closer examination reveals that the petals just above the positive *r* axis are slightly thicker. This is because the petal does not quite match up with the starting point.

若 $\theta$ 的系数为偶数，则图形花瓣数为该系数的两倍；若为奇数，则花瓣数等于该系数。鼓励读者探究其中的原因。当 $\theta$ 的系数不是整数时，会出现更有趣的图形。例如，若为有理数，则曲线是闭合的，即最终回到起点（图 1.34(a)）；若为无理数，则曲线永不闭合（图 1.34(b)）。尽管曲线看上去似乎是闭合的，但仔细观察会发现，正 $r$ 轴上方的花瓣略厚一些，这是因为花瓣并未与起点完全重合。

Chapter Opener: Describing a Spiral 章首语：描述一条螺线

Recall the chambered nautilus introduced in the chapter opener. This creature displays a spiral when half the outer shell is cut away. It is possible to describe a spiral using rectangular coordinates. Figure 1.35 shows a spiral in rectangular coordinates. How can we describe this curve mathematically?

回想章首引入的鹦鹉螺（chambered nautilus）。当外壳被切去一半时，这种生物会显现出一条螺线。螺线可以用直角坐标来描述。图 1.35 给出了直角坐标系中的一条螺线。我们该如何从数学上描述这条曲线？

Solution 解

As the point *P* travels around the spiral in a counterclockwise direction, its distance *d* from the origin increases. Assume that the distance *d* is a constant multiple *k* of the angle $\theta$ that the line segment *OP* makes with the positive *r*-axis. Therefore $d\left( {P,O} \right) = k\theta,$ where $O$ is the origin. Now use the distance formula and some trigonometry:

当点 *P* 沿螺线逆时针运动时，它到原点的距离 *d* 不断增大。设距离 *d* 为角度 $\theta$ 的常数倍 *k*，其中 $\theta$ 是线段 *OP* 与正 $r$ 轴所成的角。于是 $d\left( {P,O} \right) = k\theta$，其中 $O$ 为原点。现在利用距离公式与一些三角学知识：

$$\begin{array}{rll} {d\left( {P,O} \right)} & = & {k\theta} \\ \sqrt{\left( {x - 0} \right)^{2} + \left( {y - 0} \right)^{2}} & = & {k\ \text{arctan}\left( \frac{y}{x} \right)} \\ \sqrt{x^{2} + y^{2}} & = & {k\ \text{arctan}\left( \frac{y}{x} \right)} \\ {\text{arctan}\left( \frac{y}{x} \right)} & = & \frac{\sqrt{x^{2} + y^{2}}}{k} \\ y & = & {x\ \text{tan}\left( \frac{\sqrt{x^{2} + y^{2}}}{k} \right).} \end{array}$$

Although this equation describes the spiral, it is not possible to solve it directly for either *x* or *y*. However, if we use polar coordinates, the equation becomes much simpler. In particular, $d\left( {P,O} \right) = r,$ and $\theta$ is the second coordinate. Therefore the equation for the spiral becomes $r = k\theta.$ Note that when $\theta = 0$ we also have $r = 0,$ so the spiral emanates from the origin. We can remove this restriction by adding a constant to the equation. Then the equation for the spiral becomes $r = a + k\theta$ for arbitrary constants $a$ and $k.$ This is referred to as an Archimedean spiral, after the Greek mathematician Archimedes.

尽管这个方程描述了螺线，却无法直接解出 *x* 或 *y*。然而，若使用极坐标，方程会简单得多。具体而言，$d\left( {P,O} \right) = r$，而 $\theta$ 是第二个坐标。于是螺线方程变为 $r = k\theta$。注意当 $\theta = 0$ 时也有 $r = 0$，故螺线从原点发出。我们可以在方程中加上一个常数来去掉这一限制，于是螺线方程变为 $r = a + k\theta$，其中 $a$ 和 $k$ 为任意常数。这称为阿基米德螺线（Archimedean spiral），以希腊数学家阿基米德命名。

Another type of spiral is the logarithmic spiral, described by the function $r = a \cdot b^{\theta}.$ A graph of the function $r = 1.2\left( 1.25^{\theta} \right)$ is given in Figure 1.36. This spiral describes the shell shape of the chambered nautilus.

另一种螺线是对数螺线（logarithmic spiral），由函数 $r = a \cdot b^{\theta}$ 描述。函数 $r = 1.2\left( 1.25^{\theta} \right)$ 的图形见图 1.36。这条螺线描述了鹦鹉螺的壳形。

Suppose a curve is described in the polar coordinate system via the function $r = f(\theta).$ Since we have conversion formulas from polar to rectangular coordinates given by

设一条曲线在极坐标系中由函数 $r = f(\theta)$ 描述。由于我们有从极坐标到直角坐标的转换公式

$$\begin{array}{l} {x = r\ \text{cos}\ \theta} \\ {y = r\ \text{sin}\ \theta,} \end{array}$$

it is possible to rewrite these formulas using the function

利用该函数可将这些公式改写为

$$\begin{array}{l} {x = f(\theta)\ \text{cos}\ \theta} \\ {y = f(\theta)\ \text{sin}\ \theta.} \end{array}$$

This step gives a parameterization of the curve in rectangular coordinates using $\theta$ as the parameter. For example, the spiral formula $r = a + b\theta$ from Figure 1.31 becomes

这一步给出了该曲线在直角坐标系下的一种参数化，其中以 $\theta$ 为参数。例如，图 1.31 中的螺线公式 $r = a + b\theta$ 变为

$$\begin{array}{l} {x = \left( {a + b\theta} \right)\ \text{cos}\ \theta} \\ {y = \left( {a + b\theta} \right)\ \text{sin}\ \theta.} \end{array}$$

Letting $\theta$ range from $\text{−}\infty$ to $\infty$ generates the entire spiral.

让 $\theta$ 从 $\text{−}\infty$ 取到 $\infty$，即可生成整条螺线。

Symmetry in Polar Coordinates 极坐标下的对称性

When studying symmetry of functions in rectangular coordinates (i.e., in the form $y = f(x)),$ we talk about symmetry with respect to the *y*-axis and symmetry with respect to the origin. In particular, if $f\left( {\text{−}x} \right) = f(x)$ for all $x$ in the domain of $f,$ then $f$ is an even function and its graph is symmetric with respect to the *y*-axis. If $f\left( {\text{−}x} \right) = \text{−}f(x)$ for all $x$ in the domain of $f,$ then $f$ is an odd function and its graph is symmetric with respect to the origin. By determining which types of symmetry a graph exhibits, we can learn more about the shape and appearance of the graph. Symmetry can also reveal other properties of the function that generates the graph. Symmetry in polar curves works in a similar fashion.

在研究直角坐标下函数的对称性时（即形式为 $y = f(x)$ 的函数），我们讨论关于 *y* 轴的对称性与关于原点的对称性。特别地，若对 $f$ 定义域内所有 $x$ 都有 $f\left( {\text{−}x} \right) = f(x)$，则 $f$ 为偶函数，其图形关于 *y* 轴对称。若对 $f$ 定义域内所有 $x$ 都有 $f\left( {\text{−}x} \right) = \text{−}f(x)$，则 $f$ 为奇函数，其图形关于原点对称。通过确定图形具有哪些对称性，我们可更多地了解图形的形状与外观。对称性还能揭示生成该图形的函数的其他性质。极坐标曲线中的对称性以类似的方式起作用。

Symmetry in Polar Curves and Equations 极坐标曲线与方程的对称性

Consider a curve generated by the function $r = f(\theta)$ in polar coordinates.

考虑极坐标下由函数 $r = f(\theta)$ 生成的曲线。

1. The curve is symmetric about the polar axis if for every point $\left( {r,\theta} \right)$ on the graph, the point $\left( {r,\text{−}\theta} \right)$ is also on the graph. Similarly, the equation $r = f(\theta)$ is unchanged by replacing $\theta$ with $\text{−}\theta.$

1. 若图形上每一点 $\left( {r,\theta} \right)$ 都有点 $\left( {r,\text{−}\theta} \right)$ 也在图形上，则曲线关于极轴对称。类似地，将方程中的 $\theta$ 替换为 $\text{−}\theta$ 后，方程 $r = f(\theta)$ 保持不变。

2. The curve is symmetric about the pole if for every point $\left( {r,\theta} \right)$ on the graph, the point $\left( {r,\pi + \theta} \right)$ is also on the graph. Similarly, the equation $r = f(\theta)$ is unchanged when replacing $r$ with $\text{−}r,$ or $\theta$ with $\pi + \theta.$

2. 若图形上每一点 $\left( {r,\theta} \right)$ 都有点 $\left( {r,\pi + \theta} \right)$ 也在图形上，则曲线关于极点对称。类似地，将 $r$ 替换为 $\text{−}r$，或将 $\theta$ 替换为 $\pi + \theta$ 后，方程 $r = f(\theta)$ 保持不变。

3. The curve is symmetric about the vertical line $\theta = \frac{\pi}{2}$ if for every point $\left( {r,\theta} \right)$ on the graph, the point $\left( {r,\pi - \theta} \right)$ is also on the graph. Similarly, the equation $r = f(\theta)$ is unchanged when $\theta$ is replaced by $\pi - \theta.$

3. 若图形上每一点 $\left( {r,\theta} \right)$ 都有点 $\left( {r,\pi - \theta} \right)$ 也在图形上，则曲线关于直线 $\theta = \frac{\pi}{2}$ 对称。类似地，将 $\theta$ 替换为 $\pi - \theta$ 后，方程 $r = f(\theta)$ 保持不变。

The following table shows examples of each type of symmetry.

下表给出了各类对称性的示例。

Using Symmetry to Graph a Polar Equation 利用对称性绘制极坐标方程图形

Find the symmetry of the rose defined by the equation $r = 3\ \text{sin}\left( {2\theta} \right)$ and create a graph.

求由方程 $r = 3\ \text{sin}\left( {2\theta} \right)$ 定义的玫瑰线的对称性，并绘制图形。

Solution 解

Suppose the point $\left( {r,\theta} \right)$ is on the graph of $r = 3\ \text{sin}\left( {2\theta} \right).$

设点 $\left( {r,\theta} \right)$ 在 $r = 3\ \text{sin}\left( {2\theta} \right)$ 的图形上。

1. To test for symmetry about the polar axis, first try replacing $\theta$ with $\text{−}\theta.$ This gives $r = 3\ \text{sin}\left( {2\left( {\text{−}\theta} \right)} \right) = -3\ \text{sin}\left( {2\theta} \right).$ Since this changes the original equation, this test is not satisfied. However, returning to the original equation and replacing $r$ with $\text{−}r$ and $\theta$ with $\pi - \theta$ yields

1. 为检验关于极轴的对称性，首先尝试将 $\theta$ 替换为 $\text{−}\theta$。得到 $r = 3\ \text{sin}\left( {2\left( {\text{−}\theta} \right)} \right) = -3\ \text{sin}\left( {2\theta} \right)$。由于这改变了原方程，该检验不满足。然而，回到原方程，将 $r$ 替换为 $\text{−}r$、将 $\theta$ 替换为 $\pi - \theta$，得到

$$\begin{array}{l}

{\text{−}r = 3\ \text{sin}\left( {2\left( {\pi - \theta} \right)} \right)} \\

{\text{−}r = 3\ \text{sin}\left( {2\pi - 2\theta} \right)} \\

{\text{−}r = 3\ \text{sin}\left( {-2\theta} \right)} \\

{\text{−}r = -3\ \text{sin}\ 2\theta.}

\end{array}$$

Multiplying both sides of this equation by $-1$ gives $r = 3\ \text{sin}\ 2\theta,$ which is the original equation. This demonstrates that the graph is symmetric with respect to the polar axis.

将等式两边同乘 $-1$ 得到 $r = 3\ \text{sin}\ 2\theta$，即原方程。这表明图形关于极轴对称。

2. To test for symmetry with respect to the pole, first replace $r$ with $\text{−}r,$ which yields $\text{−}r = 3\ \text{sin}\left( {2\theta} \right).$ Multiplying both sides by −1 gives $r = -3\ \text{sin}\left( {2\theta} \right),$ which does not agree with the original equation. Therefore the equation does not pass the test for this symmetry. However, returning to the original equation and replacing $\theta$ with $\theta + \pi$ gives

2. 为检验关于极点的对称性，首先将 $r$ 替换为 $\text{−}r$，得到 $\text{−}r = 3\ \text{sin}\left( {2\theta} \right)$。两边同乘 −1 得 $r = -3\ \text{sin}\left( {2\theta} \right)$，与原方程不符。因此该方程未通过此项对称检验。然而，回到原方程，将 $\theta$ 替换为 $\theta + \pi$，得到

$$\begin{array}{cl}

r & {= 3\ \text{sin}\left( {2\left( {\theta + \pi} \right)} \right)} \\

& {= 3\ \text{sin}\left( {2\theta + 2\pi} \right)} \\

& {= 3\left( {\text{sin}\ 2\theta\ \text{cos}\ 2\pi + \text{cos}\ 2\theta\ \text{sin}\ 2\pi} \right)} \\

& {= 3\ \text{sin}\ 2\theta.}

\end{array}$$

Since this agrees with the original equation, the graph is symmetric about the pole.

由于这与原方程一致，图形关于极点对称。

3. To test for symmetry with respect to the vertical line $\theta = \frac{\pi}{2},$ first replace both $r$ with $\text{−}r$ and $\theta$ with $\text{−}\theta.$

3. 为检验关于直线 $\theta = \frac{\pi}{2}$ 的对称性，同时将 $r$ 替换为 $\text{−}r$、将 $\theta$ 替换为 $\text{−}\theta$。

$$\begin{array}{l}

{\text{−}r = 3\ \text{sin}\left( {2\left( {\text{−}\theta} \right)} \right)} \\

{\text{−}r = 3\ \text{sin}\left( {-2\theta} \right)} \\

{\text{−}r = -3\ \text{sin}\ 2\theta.}

\end{array}$$

Multiplying both sides of this equation by $-1$ gives $r = 3\ \text{sin}\ 2\theta,$ which is the original equation. Therefore the graph is symmetric about the vertical line $\theta = \frac{\pi}{2}.$

将等式两边同乘 $-1$ 得到 $r = 3\ \text{sin}\ 2\theta$，即原方程。因此图形关于直线 $\theta = \frac{\pi}{2}$ 对称。

This graph has symmetry with respect to the polar axis, the origin, and the vertical line going through the pole. To graph the function, tabulate values of $\theta$ between 0 and $\pi\text{/}2$ and then reflect the resulting graph.

该图形关于极轴、原点以及过极点的竖直线均对称。为绘制该函数，先列出 $\theta$ 在 0 到 $\pi\text{/}2$ 之间的值表，再对所得图形作反射。

| $\theta$ | $r$ |

|-----------------|-----------------------------------|

| $0$ | $0$ |

| $\frac{\pi}{6}$ | $\frac{3\sqrt{3}}{2} \approx 2.6$ |

| $\frac{\pi}{4}$ | $3$ |

| $\frac{\pi}{3}$ | $\frac{3\sqrt{3}}{2} \approx 2.6$ |

| $\frac{\pi}{2}$ | $0$ |

This gives one petal of the rose, as shown in the following graph.

这给出玫瑰线的一瓣，如下图所示。

Reflecting this image into the other three quadrants gives the entire graph as shown.

将该图形反射到其余三个象限，即得完整图形，如图所示。

Determine the symmetry of the graph determined by the equation $r = 2\ \text{cos}\left( {3\theta} \right)$ and create a graph.

确定由方程 $r = 2\ \text{cos}\left( {3\theta} \right)$ 定义的图形的对称性，并绘制图形。

Section 1.3 Exercises 1.3 节习题

In the following exercises, plot the point whose polar coordinates are given by first constructing the angle $\theta$ and then marking off the distance *r* along the ray.

在以下习题中，先作角 $\theta$，再沿射线标出距离 *r*，从而描出给定极坐标的点。

125.

$\left( {3,\frac{\pi}{6}} \right)$

126\.

126.

$\left( {-2,\frac{5\pi}{3}} \right)$

127.

$\left( {0,\frac{7\pi}{6}} \right)$

128\.

128.

$\left( {-4,\frac{3\pi}{4}} \right)$

129.

$\left( {1,\frac{\pi}{4}} \right)$

130\.

130.

$\left( {2,\frac{5\pi}{6}} \right)$

131.

$\left( {1,\frac{\pi}{2}} \right)$

For the following exercises, consider the polar graph below. Give two sets of polar coordinates for each point.

在以下习题中，考虑下面的极坐标图形。给出每点的两组极坐标。

132\.

132.

Coordinates of point *A*.

点 *A* 的坐标。

133.

Coordinates of point *B*.

点 *B* 的坐标。

134\.

134.

Coordinates of point *C*.

点 *C* 的坐标。

135.

Coordinates of point *D*.

点 *D* 的坐标。

For the following exercises, the rectangular coordinates of a point are given. Find two sets of polar coordinates for the point in $\left( {0,2\pi} \right\rbrack.$ Round to three decimal places.

在以下习题中，给定一点的直角坐标。求该点在 $\left( {0,2\pi} \right\rbrack$ 内的两组极坐标。结果保留三位小数。

136\.

136.

$\left( {2,\ 2} \right)$

137.

$\left( {3,-4} \right)$

138\.

138.

$\left( {8,\ 15} \right)$

139.

$\left( {-6,\ 8} \right)$

140\.

140.

$\left( {4,\ 3} \right)$

141.

$\left( {3,\text{−}\sqrt{3}} \right)$

For the following exercises, find rectangular coordinates for the given point in polar coordinates.

在以下习题中，求给定极坐标点的直角坐标。

142\.

142.

$\left( {2,\frac{5\pi}{4}} \right)$

143.

$\left( {-2,\frac{\pi}{6}} \right)$

144\.

144.

$\left( {5,\frac{\pi}{3}} \right)$

145.

$\left( {1,\frac{7\pi}{6}} \right)$

146\.

146.

$\left( {-3,\frac{3\pi}{4}} \right)$

147.

$\left( {0,\frac{\pi}{2}} \right)$

148\.

148.

$\left( {-4.5,6.5} \right)$

For the following exercises, determine whether the graphs of the polar equation are symmetric with respect to the $x$-axis, the $y$-axis, or the origin.

在以下习题中，判断极坐标方程的图形是否关于 $x$ 轴、$y$ 轴或原点对称。

149.

$r = 3\ \text{sin}(2\theta)$

150\.

150.

$r^{2} = 9\ \text{cos}\ \theta$

151.

$r = \text{cos}\left( \frac{\theta}{5} \right)$

152\.

152.

$r = 2\ \text{sec}\ \theta$

153.

$r = 1 + \text{cos}\ \theta$

For the following exercises, describe the graph of each polar equation. Confirm each description by converting into a rectangular equation.

在以下习题中，描述各极坐标方程的图形。通过转换为直角坐标方程来确认每个描述。

154\.

154.

$r = 3$

155.

$\theta = \frac{\pi}{4}$

156\.

156.

$r = \text{sec}\ \theta$

157.

$r = \text{csc}\ \theta$

For the following exercises, convert the rectangular equation to polar form and sketch its graph.

在以下习题中，将直角坐标方程转换为极坐标形式，并描绘其图形。

158\.

158.

$x^{2} + y^{2} = 16$

159.

$x^{2} - y^{2} = 16$

160\.

160.

$x = 8$

For the following exercises, convert the rectangular equation to polar form and sketch its graph.

在以下习题中，将直角坐标方程转换为极坐标形式，并描绘其图形。

161.

$3x - y = 2$

162\.

162.

$y^{2} = 4x$

For the following exercises, convert the polar equation to rectangular form and sketch its graph.

在以下习题中，将极坐标方程转换为直角坐标形式，并描绘其图形。

163.

$r = 4\ \text{sin}\ \theta$

164\.

164.

$r = 6\ \text{cos}\ \theta$

165.

$r = \theta$

166\.

166.

$r = \text{cot}\ \theta\ \text{csc}\ \theta$

For the following exercises, sketch a graph of the polar equation and identify any symmetry.

在以下习题中，描绘极坐标方程的图形，并指出任何对称性。

167.

$r = 1 + \text{sin}\ \theta$

168\.

168.

$r = 3 - 2\ \text{cos}\ \theta$

169.

$r = 2 - 2\ \text{sin}\ \theta$

170\.

170.

$r = 5 - 4\ \text{sin}\ \theta$

171.

$r = 3\ \text{cos}\left( {2\theta} \right)$

172\.

172.

$r = 3\ \text{sin}\left( {2\theta} \right)$

173.

$r = 2\ \text{cos}\left( {3\theta} \right)$

174\.

174.

$r = 3\ \text{cos}\left( \frac{\theta}{2} \right)$

175.

$r^{2} = 4\ \text{cos}\left( {2\theta} \right)$

176\.

176.

$r^{2} = 4\ \text{sin}\ \theta$

177.

$r = 2\theta$

178\.

178.

\[T\] The graph of $r = 2\ \text{cos}(2\theta)\text{sec}(\theta).$ is called a *strophoid.* Use a graphing utility to sketch the graph, and, from the graph, determine the asymptote.

\[T\] 该图形称为环索线（strophoid）。使用绘图工具描绘图形，并从图形中确定渐近线。

179.

\[T\] Use a graphing utility and sketch the graph of $r = \frac{6}{2\ \text{sin}\ \theta - 3\ \text{cos}\ \theta}.$

\[T\] 使用绘图工具描绘 $r = \frac{6}{2\ \text{sin}\ \theta - 3\ \text{cos}\ \theta}$ 的图形。

180\.

180.

\[T\] Use a graphing utility to graph $r = \frac{1}{1 - \text{cos}\ \theta}.$

\[T\] 使用绘图工具描绘 $r = \frac{1}{1 - \text{cos}\ \theta}$ 的图形。

181.

\[T\] Use technology to graph $r = e^{\text{sin}(\theta)} - 2\ \text{cos}\left( {4\theta} \right).$

\[T\] 使用技术手段描绘 $r = e^{\text{sin}(\theta)} - 2\ \text{cos}\left( {4\theta} \right)$ 的图形。

182\.

182.

\[T\] Use technology to plot $r = \text{sin}\left( \frac{3\theta}{7} \right)$ (use the interval $0 \leq \theta \leq 14\pi).$

\[T\] 使用技术手段描绘 $r = \text{sin}\left( \frac{3\theta}{7} \right)$ 的图形（使用区间 $0 \leq \theta \leq 14\pi$）。

183.

Without using technology, sketch the polar curve $\theta = \frac{2\pi}{3}.$

不使用技术手段，描绘极坐标曲线 $\theta = \frac{2\pi}{3}$ 的图形。

184\.

184.

\[T\] Use a graphing utility to plot $r = \theta\ \text{sin}\ \theta$ for $\text{−}\pi \leq \theta \leq \pi.$

\[T\] 使用绘图工具在 $\text{−}\pi \leq \theta \leq \pi$ 上描绘 $r = \theta\ \text{sin}\ \theta$。

185.

\[T\] Use technology to plot $r = e^{-0.1\theta}$ for $-10 \leq \theta \leq 10.$

\[T\] 使用技术手段在 $-10 \leq \theta \leq 10$ 上描绘 $r = e^{-0.1\theta}$ 的图形。

186\.

186.

\[T\] There is a curve known as the “*Black Hole*.” Use technology to plot $r = e^{-0.01\theta}$ for $-100 \leq \theta \leq 100.$

\[T\] 存在一条称为“*黑洞*”（Black Hole）的曲线。使用技术手段在 $-100 \leq \theta \leq 100$ 上描绘 $r = e^{-0.01\theta}$ 的图形。

187.

\[T\] Use the results of the preceding two problems to explore the graphs of $r = e^{-0.001\theta}$ and $r = e^{-0.0001\theta}$ for $|\theta| > 100.$

\[T\] 利用前两题的结果，探究 $|\theta| > 100$ 时 $r = e^{-0.001\theta}$ 与 $r = e^{-0.0001\theta}$ 的图形。

---

1.4 Area and Arc Length in Polar Coordinates 1.4 极坐标下的面积与弧长

1.4.1 Apply the formula for area of a region in polar coordinates.
1.4.2 Determine the arc length of a polar curve.

1.4.1 应用极坐标下区域面积的公式。
1.4.2 确定极坐标曲线的弧长。

In the rectangular coordinate system, the definite integral provides a way to calculate the area under a curve. In particular, if we have a function $y = f(x)$ defined from $x = a$ to $x = b$ where $f(x) > 0$ on this interval, the area between the curve and the *x*-axis is given by $A = {\int_{a}^{b}{f(x)\ dx}}.$ This fact, along with the formula for evaluating this integral, is summarized in the Fundamental Theorem of Calculus. Similarly, the arc length of this curve is given by $L = {\int_{a}^{b}{\sqrt{1 + \left( {f^{\prime}(x)} \right)^{2}}dx}}.$ In this section, we study analogous formulas for area and arc length in the polar coordinate system.

在直角坐标系下，定积分提供了计算曲线下面积的方法。特别地，若有定义于 $x = a$ 到 $x = b$ 的函数 $y = f(x)$，且在此区间上 $f(x) > 0$，则曲线与 *x* 轴之间的面积由 $A = {\int_{a}^{b}{f(x)\ dx}}$ 给出。这一事实连同该积分的计算公式，由微积分基本定理概括。类似地，该曲线的弧长由 $L = {\int_{a}^{b}{\sqrt{1 + \left( {f^{\prime}(x)} \right)^{2}}dx}}$ 给出。本节中，我们研究极坐标下面积与弧长的类似公式。

Areas of Regions Bounded by Polar Curves 极坐标曲线所围区域的面积

We have studied the formulas for area under a curve defined in rectangular coordinates and parametrically defined curves. Now we turn our attention to deriving a formula for the area of a region bounded by a polar curve. Recall that the proof of the Fundamental Theorem of Calculus used the concept of a Riemann sum to approximate the area under a curve by using rectangles. For polar curves we use the Riemann sum again, but the rectangles are replaced by sectors of a circle.

我们已经学习了在直角坐标系中以及参数方程定义的曲线下求面积的公式。现在将注意力转向推导极坐标曲线所围区域的面积公式。回想一下，微积分基本定理的证明中利用了黎曼和的概念，借助矩形来逼近曲线下的面积。对于极坐标曲线，我们同样使用黎曼和，只是用圆的扇形代替了矩形。

Consider a curve defined by the function $r = f(\theta),$ where $\alpha \leq \theta \leq \beta.$ Our first step is to partition the interval $\lbrack\alpha,\beta\rbrack$ into *n* equal-width subintervals. The width of each subinterval is given by the formula $\text{Δ}\theta = {{(\beta - \alpha)}\text{/}n},$ and the *i*th partition point $\theta_{i}$ is given by the formula $\theta_{i} = \alpha + i\text{Δ}\theta.$ Each partition point $\theta = \theta_{i}$ defines a line with slope $\text{tan}\theta_{i}$ passing through the pole as shown in the following graph.

考虑由函数 $r = f(\theta)$ 定义的曲线，其中 $\alpha \leq \theta \leq \beta$。我们的第一步是将区间 $\lbrack\alpha,\beta\rbrack$ 划分为 $n$ 个等宽的子区间。每个子区间的宽度由公式 $\text{Δ}\theta = {{(\beta - \alpha)}\text{/}n}$ 给出，第 $i$ 个分点 $\theta_{i}$ 由公式 $\theta_{i} = \alpha + i\text{Δ}\theta$ 给出。每个分点 $\theta = \theta_{i}$ 确定一条过极点、斜率为 $\text{tan}\theta_{i}$ 的直线，如下图所示。

The line segments are connected by arcs of constant radius. This defines sectors whose areas can be calculated by using a geometric formula. The area of each sector is then used to approximate the area between successive line segments. We then sum the areas of the sectors to approximate the total area. This approach gives a Riemann sum approximation for the total area. The formula for the area of a sector of a circle is illustrated in the following figure.

这些线段由定半径的圆弧连接。由此得到一些扇形，其面积可用几何公式计算。每个扇形的面积用来逼近相邻线段之间的面积。然后将各扇形的面积相加，逼近总面积。这种方法给出了总面积的黎曼和逼近。圆扇形的面积公式如下图所示。

Recall that the area of a circle is $A = \pi r^{2}.$ When measuring angles in radians, 360 degrees is equal to $2\pi$ radians. Therefore a fraction of a circle can be measured by the central angle $\theta.$ The fraction of the circle is given by $\frac{\theta}{2\pi},$ so the area of the sector is this fraction multiplied by the total area:

回想一下，圆的面积为 $A = \pi r^{2}$。当用弧度计量角度时，360 度等于 $2\pi$ 弧度。因此圆的一部分可用圆心角 $\theta$ 来度量。该部分占整个圆的比例为 $\frac{\theta}{2\pi}$，于是扇形的面积就是这个比例乘以总面积：

$$A = \left( \frac{\theta}{2\pi} \right)\ \pi r^{2} = \frac{1}{2}\theta r^{2}.$$

Since the radius of a typical sector in Figure 1.39 is given by $r_{i} = f\left( \theta_{i} \right),$ the area of the *i*th sector is given by

由于在图 1.39 中，典型扇形的半径由 $r_{i} = f\left( \theta_{i} \right)$ 给出，因此第 $i$ 个扇形的面积为

$$A_{i} = \frac{1}{2}\left( {\text{Δ}\theta} \right)\left( {f\left( \theta_{i} \right)} \right)^{2}.$$

Therefore a Riemann sum that approximates the area is given by

因此，逼近该面积的黎曼和由下式给出

$$A_{n} = {\sum\limits_{i = 1}^{n}A_{i}} \approx {\sum\limits_{i = 1}^{n}{\frac{1}{2}\left( {\text{Δ}\theta} \right)\left( {f\left( \theta_{i} \right)} \right)^{2}}}.$$

We take the limit as $n\rightarrow\infty$ to get the exact area:

令 $n\rightarrow\infty$ 取极限，得到精确面积：

$$A = \underset{n\rightarrow\infty}{\text{lim}}A_{n} = \frac{1}{2}{\int_{\alpha}^{\beta}{\left( {f(\theta)} \right)^{2}d\theta}}.$$

This gives the following theorem.

由此得到以下定理。

Area of a Region Bounded by a Polar Curve 极坐标曲线所围区域的面积

Suppose $f$ is continuous and nonnegative on the interval $\alpha \leq \theta \leq \beta$ with $0 < \beta - \alpha \leq 2\pi.$ The area of the region bounded by the graph of $r = f(\theta)$ between the radial lines $\theta = \alpha$ and $\theta = \beta$ is

设 $f$ 在区间 $\alpha \leq \theta \leq \beta$ 上连续且非负，且 $0 < \beta - \alpha \leq 2\pi$。由曲线 $r = f(\theta)$ 的图形与射线 $\theta = \alpha$ 和 $\theta = \beta$ 所围区域的面积为

$$A = \frac{1}{2}{\int_{\alpha}^{\beta}{\left\lbrack {f(\theta)} \right\rbrack^{2}d\theta}} = \frac{1}{2}{\int_{\alpha}^{\beta}{r^{2}d\theta}}.$$ (1.9)

Finding an Area of a Polar Region 求极坐标区域的面积

Find the area of one petal of the rose defined by the equation $r = 3\ \text{sin}\left( {2\theta} \right).$

求由方程 $r = 3\ \text{sin}\left( {2\theta} \right)$ 定义的玫瑰线的一个花瓣的面积。

Solution 解

The graph of $r = 3\ \text{sin}\left( {2\theta} \right)$ follows.

$r = 3\ \text{sin}\left( {2\theta} \right)$ 的图形如下。

When $\theta = 0$ we have $r = 3\ \text{sin}\left( {2(0)} \right) = 0.$ The next value for which $r = 0$ is $\theta = \pi\text{/}2.$ This can be seen by solving the equation $3\ \text{sin}(2\theta) = 0$ for $\theta.$ Therefore the values $\theta = 0$ to $\theta = \pi\text{/}2$ trace out the first petal of the rose. To find the area inside this petal, use Equation 1.9 with $f(\theta) = 3\ \text{sin}\left( {2\theta} \right),$ $\alpha = 0,$ and $\beta = \pi\text{/}2\text{:}$

当 $\theta = 0$ 时，有 $r = 3\ \text{sin}\left( {2(0)} \right) = 0$。使 $r = 0$ 的下一个值是 $\theta = \pi\text{/}2$。这可通过解方程 $3\ \text{sin}(2\theta) = 0$ 求 $\theta$ 看出。因此，从 $\theta = 0$ 到 $\theta = \pi\text{/}2$ 描出玫瑰线的第一片花瓣。要求该花瓣内部的面积，在公式 1.9 中取 $f(\theta) = 3\ \text{sin}\left( {2\theta} \right)$、$\alpha = 0$、$\beta = \pi\text{/}2$：

$$\begin{array}{cl} A & {= \frac{1}{2}{\int_{\alpha}^{\beta}{\left\lbrack {f(\theta)} \right\rbrack^{2}d\theta}}} \\ & {= \frac{1}{2}{\int_{0}^{\pi\text{/}2}{\left\lbrack {3\ \text{sin}\left( {2\theta} \right)} \right\rbrack^{2}d\theta}}} \\ & {= \frac{1}{2}{\int_{0}^{\pi\text{/}2}{9\ \text{sin}^{2}\left( {2\theta} \right)\ d\theta}}.} \end{array}$$

To evaluate this integral, use the formula $\text{sin}^{2}\alpha = \left( {1 - \text{cos}(2\alpha)} \right)\text{/}2$ with $\alpha = 2\theta\text{:}$

为了计算这个积分，使用公式 $\text{sin}^{2}\alpha = \left( {1 - \text{cos}(2\alpha)} \right)\text{/}2$，其中取 $\alpha = 2\theta$：

$$\begin{array}{cl} A & {= \frac{1}{2}{\int_{0}^{\pi\text{/}2}{9\ \text{sin}^{2}\left( {2\theta} \right)\ d\theta}}} \\ & {= \frac{9}{2}{\int_{0}^{\pi\text{/}2}{\frac{\left( {1 - \text{cos}\left( {4\theta} \right)} \right)}{2}d\theta}}} \\ & {= \frac{9}{4}\left( {\int_{0}^{\pi\text{/}2}{1 - \text{cos}\left( {4\theta} \right)\ d\theta}} \right)} \\ & {= \frac{9}{4}\left( \left. {\theta - \frac{\text{sin}\left( {4\theta} \right)}{4}} \right) \right._{0}^{\pi\text{/}2}} \\ & {= \frac{9}{4}\left( {\frac{\pi}{2} - \frac{\text{sin}\ 2\pi}{4}} \right) - \frac{9}{4}\left( {0 - \frac{\text{sin}\ 4(0)}{4}} \right)} \\ & {= \frac{9\pi}{8}.} \end{array}$$

Find the area inside the cardioid defined by the equation $r = 1 - \text{cos}\ \theta.$

求由方程 $r = 1 - \text{cos}\ \theta$ 定义的心形线内部的面积。

Example 1.16 involved finding the area inside one curve. We can also use Area of a Region Bounded by a Polar Curve to find the area between two polar curves. However, we often need to find the points of intersection of the curves and determine which function defines the outer curve or the inner curve between these two points.

例 1.16 处理的是求一条曲线内部的面积。我们也可以用"极坐标曲线所围区域的面积"来求两条极坐标曲线之间的面积。不过，常常需要先求出曲线的交点，并判断在这两个交点之间哪个函数定义的是外侧曲线、哪个是内侧曲线。

Finding the Area between Two Polar Curves 求两条极坐标曲线之间的面积

Find the area outside the cardioid $r = 2 + 2\ \text{sin}\ \theta$ and inside the circle $r = 6\ \text{sin}\ \theta.$

求心形线 $r = 2 + 2\ \text{sin}\ \theta$ 外部、圆 $r = 6\ \text{sin}\ \theta$ 内部的面积。

Solution 解

First draw a graph containing both curves as shown.

首先画出包含两条曲线的图形，如下所示。

To determine the limits of integration, first find the points of intersection by setting the two functions equal to each other and solving for $\theta\text{:}$

为确定积分限，先将两个函数令其相等，对 $\theta$ 求解：

$$\begin{array}{rll} {6\ \text{sin}\ \theta} & = & {2 + 2\ \text{sin}\ \theta} \\ {4\ \text{sin}\ \theta} & = & 2 \\ {\text{sin}\ \theta} & = & {\frac{1}{2}.} \end{array}$$

This gives the solutions $\theta = \frac{\pi}{6}$ and $\theta = \frac{5\pi}{6},$ which are the limits of integration. The circle $r = 3\ \text{sin}\ \theta$ is the red graph, which is the outer function, and the cardioid $r = 2 + 2\ \text{sin}\ \theta$ is the blue graph, which is the inner function. To calculate the area between the curves, start with the area inside the circle between $\theta = \frac{\pi}{6}$ and $\theta = \frac{5\pi}{6},$ then subtract the area inside the cardioid between $\theta = \frac{\pi}{6}$ and $\theta = \frac{5\pi}{6}\text{:}$

由此得到解 $\theta = \frac{\pi}{6}$ 和 $\theta = \frac{5\pi}{6}$，这就是积分限。圆 $r = 3\ \text{sin}\ \theta$（红色图形）是外侧函数，心形线 $r = 2 + 2\ \text{sin}\ \theta$（蓝色图形）是内侧函数。要计算两曲线之间的面积，先求圆在 $\theta = \frac{\pi}{6}$ 到 $\theta = \frac{5\pi}{6}$ 之间的面积，再减去心形线在同一区间内的面积：

$$\begin{array}{cl} A & {= \text{circle} - \text{cardioid}} \\ & {= \frac{1}{2}{\int_{\pi\text{/}6}^{{5\pi}\text{/}6}{\left\lbrack {6\ \text{sin}\ \theta} \right\rbrack^{2}d\theta}} - \frac{1}{2}{\int_{\pi\text{/}6}^{{5\pi}\text{/}6}{\left\lbrack {2 + 2\ \text{sin}\ \theta} \right\rbrack^{2}d\theta}}} \\ & {= \frac{1}{2}{\int_{\pi\text{/}6}^{{5\pi}\text{/}6}{36\ \text{sin}^{2}\theta\ d\theta}} - \frac{1}{2}{\int_{\pi\text{/}6}^{{5\pi}\text{/}6}{\left\lbrack 4 + 8\ \text{sin}\ \theta + 4\ \text{sin}^{2}\theta \right\rbrack\ d\theta}}} \\ & {= 18{\int_{\pi\text{/}6}^{{5\pi}\text{/}6}{\frac{1 - \text{cos}\left( {2\theta} \right)}{2}d\theta}} - 2{\int_{\pi\text{/}6}^{{5\pi}\text{/}6}{\left\lbrack 1 + 2\ \text{sin}\ \theta + \frac{1 - \text{cos}\left( {2\theta} \right)}{2} \right\rbrack d\theta}}} \\ & {= 9\left\lbrack {\theta - \frac{\text{sin}\left( {2\theta} \right)}{2}} \right\rbrack_{\pi\text{/}6}^{5\pi\text{/}6} - 2\left\lbrack {\frac{3\theta}{2} - 2\ \text{cos}\ \theta - \frac{\text{sin}\left( {2\theta} \right)}{4}} \right\rbrack_{\pi\text{/}6}^{5\pi\text{/}6}} \\ & {= 9\left( {\frac{5\pi}{6} - \frac{\text{sin}\ 2(5\pi\text{/}6)}{2}} \right) - 9\left( {\frac{\pi}{6} - \frac{\text{sin}\ 2(\pi\text{/}6)}{2}} \right)} \\ & {\mspace{7mu}\text{−}\left( {3\left( \frac{5\pi}{6} \right) - 4\ \text{cos}\ \frac{5\pi}{6} - \frac{\text{sin}\ 2(5\pi\text{/}6)}{2}} \right) + \left( {3\left( \frac{\pi}{6} \right) - 4\ \text{cos}\ \frac{\pi}{6} - \frac{\text{sin}\ 2(\pi\text{/}6)}{2}} \right)} \\ & {= 4\pi.} \end{array}$$

Find the area inside the circle $r = 4\ \text{cos}\ \theta$ and outside the circle $r = 2.$

求圆 $r = 4\ \text{cos}\ \theta$ 内部、圆 $r = 2$ 外部的面积。

In Example 1.17 we found the area inside the circle and outside the cardioid by first finding their intersection points. Notice that solving the equation directly for $\theta$ yielded two solutions: $\theta = \frac{\pi}{6}$ and $\theta = \frac{5\pi}{6}.$ However, in the graph there are three intersection points. The third intersection point is the origin. The reason why this point did not show up as a solution is because the origin is on both graphs but for different values of $\theta.$ For example, for the cardioid we get

在例 1.17 中，我们先求出交点，再计算圆内部、心形线外部的面积。注意，直接对 $\theta$ 解方程只得到两个解：$\theta = \frac{\pi}{6}$ 和 $\theta = \frac{5\pi}{6}$。然而，图形中实际有三个交点。第三个交点是原点。这个点没有作为解出现的原因是：原点同时位于两条图形上，但对应于不同的 $\theta$ 值。例如，对心形线有

$$\begin{array}{rll} {2 + 2\ \text{sin}\ \theta} & = & 0 \\ {\text{sin}\ \theta} & = & {-1,} \end{array}$$

so the values for $\theta$ that solve this equation are $\theta = \frac{3\pi}{2} + 2n\pi,$ where *n* is any integer. For the circle we get

因此，满足该方程的 $\theta$ 值为 $\theta = \frac{3\pi}{2} + 2n\pi$，其中 $n$ 为任意整数。对圆则有

$$6\ \text{sin}\ \theta = 0.$$

The solutions to this equation are of the form $\theta = n\pi$ for any integer value of *n.* These two solution sets have no points in common. Regardless of this fact, the curves intersect at the origin. This case must always be taken into consideration.

该方程的解为 $\theta = n\pi$，其中 $n$ 为任意整数。这两个解集没有公共点。尽管如此，两条曲线仍在原点相交。这种情况必须始终加以考虑。

Arc Length in Polar Curves 极坐标曲线的弧长

Here we derive a formula for the arc length of a curve defined in polar coordinates.

这里我们来推导一条由极坐标定义的曲线的弧长公式。

In rectangular coordinates, the arc length of a parameterized curve $\left( {x(t),y(t)} \right)$ for $a \leq t \leq b$ is given by

在直角坐标中，参数曲线 $\left( {x(t),y(t)} \right)$ 在区间 $a \leq t \leq b$ 上的弧长由下式给出：

$$L = {\int_{a}^{b}{\sqrt{\left( \frac{dx}{dt} \right)^{2} + \left( \frac{dy}{dt} \right)^{2}}dt}}.$$

弧长为 $$L = {\int_{a}^{b}{\sqrt{\left( \frac{dx}{dt} \right)^{2} + \left( \frac{dy}{dt} \right)^{2}}dt}}.$$

In polar coordinates we define the curve by the equation $r = f(\theta),$ where $\alpha \leq \theta \leq \beta.$ In order to adapt the arc length formula for a polar curve, we use the equations

在极坐标中，我们用方程 $r = f(\theta)$ 定义曲线，其中 $\alpha \leq \theta \leq \beta.$ 为了把弧长公式用于极坐标曲线，我们采用下列关系式

$$x = r\ \text{cos}\ \theta = f(\theta)\ \text{cos}\ \theta\ \text{and}\ y = r\ \text{sin}\ \theta = f(\theta)\ \text{sin}\ \theta,$$

即 $$x = r\ \text{cos}\ \theta = f(\theta)\ \text{cos}\ \theta\ \text{and}\ y = r\ \text{sin}\ \theta = f(\theta)\ \text{sin}\ \theta,$$

and we replace the parameter *t* by $\theta.$ Then

且我们把参数 *t* 替换为 $\theta$。于是

$$\begin{array}{l}{\frac{dx}{d\theta} = f^{\prime}(\theta)\ \text{cos}\ \theta - f(\theta)\ \text{sin}\ \theta} \\{\frac{dy}{d\theta} = f^{\prime}(\theta)\ \text{sin}\ \theta + f(\theta)\ \text{cos}\ \theta.}\end{array}$$

对 $\theta$ 求导，得到 $$\begin{array}{l}{\frac{dx}{d\theta} = f^{\prime}(\theta)\ \text{cos}\ \theta - f(\theta)\ \text{sin}\ \theta} \\{\frac{dy}{d\theta} = f^{\prime}(\theta)\ \text{sin}\ \theta + f(\theta)\ \text{cos}\ \theta.}\end{array}$$

We replace $dt$ by $d\theta,$ and the lower and upper limits of integration are $\alpha$ and $\beta,$ respectively. Then the arc length formula becomes

将 $dt$ 换为 $d\theta$，积分的下限与上限分别为 $\alpha$ 与 $\beta$。于是弧长公式变为

$$\begin{array}{cl}L & {= {\int_{a}^{b}{\sqrt{\left( \frac{dx}{dt} \right)^{2} + \left( \frac{dy}{dt} \right)^{2}}dt}}} \\ & {= {\int_{\alpha}^{\beta}{\sqrt{\left( \frac{dx}{d\theta} \right)^{2} + \left( \frac{dy}{d\theta} \right)^{2}}d\theta}}} \\ & {= {\int_{\alpha}^{\beta}{\sqrt{\left( {f^{\prime}(\theta)\ \text{cos}\ \theta - f(\theta)\ \text{sin}\ \theta} \right)^{2} + \left( {f^{\prime}(\theta)\ \text{sin}\ \theta + f(\theta)\ \text{cos}\ \theta} \right)^{2}}d\theta}}} \\ & {= {\int_{\alpha}^{\beta}{\sqrt{\left( {f^{\prime}(\theta)} \right)^{2}\left( {\text{cos}^{2}\theta + \text{sin}^{2}\theta} \right) + \left( {f(\theta)} \right)^{2}\left( {\text{cos}^{2}\theta + \text{sin}^{2}\theta} \right)}d\theta}}} \\ & {= {\int_{\alpha}^{\beta}{\sqrt{\left( {f^{\prime}(\theta)} \right)^{2} + \left( {f(\theta)} \right)^{2}}d\theta}}} \\ & {= {\int_{\alpha}^{\beta}{\sqrt{r^{2} + \left( \frac{dr}{d\theta} \right)^{2}}d\theta}}.}\end{array}$$

因此弧长公式化为 $$\begin{array}{cl}L & {= {\int_{a}^{b}{\sqrt{\left( \frac{dx}{dt} \right)^{2} + \left( \frac{dy}{dt} \right)^{2}}dt}}} \\ & {= {\int_{\alpha}^{\beta}{\sqrt{\left( \frac{dx}{d\theta} \right)^{2} + \left( \frac{dy}{d\theta} \right)^{2}}d\theta}}} \\ & {= {\int_{\alpha}^{\beta}{\sqrt{\left( {f^{\prime}(\theta)\ \text{cos}\ \theta - f(\theta)\ \text{sin}\ \theta} \right)^{2} + \left( {f^{\prime}(\theta)\ \text{sin}\ \theta + f(\theta)\ \text{cos}\ \theta} \right)^{2}}d\theta}}} \\ & {= {\int_{\alpha}^{\beta}{\sqrt{\left( {f^{\prime}(\theta)} \right)^{2}\left( {\text{cos}^{2}\theta + \text{sin}^{2}\theta} \right) + \left( {f(\theta)} \right)^{2}\left( {\text{cos}^{2}\theta + \text{sin}^{2}\theta} \right)}d\theta}}} \\ & {= {\int_{\alpha}^{\beta}{\sqrt{\left( {f^{\prime}(\theta)} \right)^{2} + \left( {f(\theta)} \right)^{2}}d\theta}}} \\ & {= {\int_{\alpha}^{\beta}{\sqrt{r^{2} + \left( \frac{dr}{d\theta} \right)^{2}}d\theta}}.}\end{array}$$

This gives us the following theorem.

由此得到如下定理。

Arc Length of a Curve Defined by a Polar Function 由极坐标函数定义的曲线的弧长

Let $f$ be a function whose derivative is continuous on an interval $\alpha \leq \theta \leq \beta.$ The length of the graph of $r = f(\theta)$ from $\theta = \alpha$ to $\theta = \beta$ is

设 $f$ 是在区间 $\alpha \leq \theta \leq \beta$ 上导数连续的函数。图像 $r = f(\theta)$ 从 $\theta = \alpha$ 到 $\theta = \beta$ 的长度为

$$L = {\int_{\alpha}^{\beta}{\sqrt{\left\lbrack {f(\theta)} \right\rbrack^{2} + \left\lbrack {f^{\prime}(\theta)} \right\rbrack^{2}}d\theta = {\int_{\alpha}^{\beta}{\sqrt{r^{2} + \left( \frac{dr}{d\theta} \right)^{2}}d\theta.}}}}$$ (1.10)

即 $$L = {\int_{\alpha}^{\beta}{\sqrt{\left\lbrack {f(\theta)} \right\rbrack^{2} + \left\lbrack {f^{\prime}(\theta)} \right\rbrack^{2}}d\theta = {\int_{\alpha}^{\beta}{\sqrt{r^{2} + \left( \frac{dr}{d\theta} \right)^{2}}d\theta.}}}}$$ (1.10)

Finding the Arc Length of a Polar Curve 求极坐标曲线的弧长

Find the arc length of the cardioid $r = 2 + 2\mspace{2mu}\text{cos}\mspace{2mu}\theta.$

求心形线 $r = 2 + 2\mspace{2mu}\text{cos}\mspace{2mu}\theta$ 的弧长。

Solution 解

When $\theta = 0,r = 2 + 2\mspace{2mu}\text{cos}\mspace{2mu} 0 = 4.$ Furthermore, as $\theta$ goes from $0$ to $2\textit{π}\text{,}$ the cardioid is traced out exactly once. Therefore these are the limits of integration. Using $f{(\theta)} = 2 + 2\mspace{2mu}\text{cos}\mspace{2mu}\theta,$ $\alpha = 0,$ and $\beta = 2\textit{π}\text{,}$ Equation 1.10 becomes

当 $\theta = 0$ 时，$r = 2 + 2\mspace{2mu}\text{cos}\mspace{2mu} 0 = 4$。此外，当 $\theta$ 从 $0$ 变到 $2\textit{π}$ 时，心形线恰好被描绘一次。因此这些就是积分的上下限。取 $f{(\theta)} = 2 + 2\mspace{2mu}\text{cos}\mspace{2mu}\theta$、$\alpha = 0$、$\beta = 2\textit{π}$，由方程 (1.10) 得

$$\begin{array}{cl}L & {= {\int_{\alpha}^{\beta}\sqrt{\left\lbrack {f(\theta)} \right\rbrack^{2} + \left\lbrack {f^{\prime}(\theta)} \right\rbrack^{2}}}\ d\theta} \\ & {= {\int_{0}^{2\pi}\sqrt{\left\lbrack {2 + 2\mspace{2mu}\text{cos}\mspace{2mu}\theta} \right\rbrack^{2} + \left\lbrack {- 2\mspace{2mu}\text{sin}\mspace{2mu}\theta} \right\rbrack^{2}}}\ d\theta} \\ & {= {\int_{0}^{2\pi}\sqrt{4 + 8\mspace{2mu}\text{cos}\mspace{2mu}\theta + 4\mspace{2mu}\text{cos}^{2}\mspace{2mu}\theta + 4\mspace{2mu}\text{sin}^{2}\mspace{2mu}\theta}}\ d\theta} \\ & {= {\int_{0}^{2\pi}\sqrt{4 + 8\mspace{2mu}\text{cos}\mspace{2mu}\theta + 4\left( {\text{cos}^{2}\mspace{2mu}\theta + \text{sin}^{2}\mspace{2mu}\theta} \right)}}\ d\theta} \\ & {= {\int_{0}^{2\pi}\sqrt{8 + 8\mspace{2mu}\text{cos}\mspace{2mu}\theta}}\ d\theta} \\ & {= 2{\int_{0}^{2\pi}\sqrt{2 + 2\mspace{2mu}\text{cos}\mspace{2mu}\theta}}\ d\theta.}\end{array}$$

代入后，弧长积分化为 $$\begin{array}{cl}L & {= {\int_{\alpha}^{\beta}\sqrt{\left\lbrack {f(\theta)} \right\rbrack^{2} + \left\lbrack {f^{\prime}(\theta)} \right\rbrack^{2}}}\ d\theta} \\ & {= {\int_{0}^{2\pi}\sqrt{\left\lbrack {2 + 2\mspace{2mu}\text{cos}\mspace{2mu}\theta} \right\rbrack^{2} + \left\lbrack {- 2\mspace{2mu}\text{sin}\mspace{2mu}\theta} \right\rbrack^{2}}}\ d\theta} \\ & {= {\int_{0}^{2\pi}\sqrt{4 + 8\mspace{2mu}\text{cos}\mspace{2mu}\theta + 4\mspace{2mu}\text{cos}^{2}\mspace{2mu}\theta + 4\mspace{2mu}\text{sin}^{2}\mspace{2mu}\theta}}\ d\theta} \\ & {= {\int_{0}^{2\pi}\sqrt{4 + 8\mspace{2mu}\text{cos}\mspace{2mu}\theta + 4\left( {\text{cos}^{2}\mspace{2mu}\theta + \text{sin}^{2}\mspace{2mu}\theta} \right)}}\ d\theta} \\ & {= {\int_{0}^{2\pi}\sqrt{8 + 8\mspace{2mu}\text{cos}\mspace{2mu}\theta}}\ d\theta} \\ & {= 2{\int_{0}^{2\pi}\sqrt{2 + 2\mspace{2mu}\text{cos}\mspace{2mu}\theta}}\ d\theta.}\end{array}$$

Next, using the identity $\text{cos}\left( {2\alpha} \right) = 2\mspace{2mu}\text{cos}^{2}\alpha - 1,$ add 1 to both sides and multiply by 2. This gives $2 + 2\mspace{2mu}\text{cos}\left( {2\alpha} \right) = 4\mspace{2mu}\text{cos}^{2}\alpha.$ Substituting $\alpha = {\theta\text{/}2}$ gives $2 + 2\mspace{2mu}\text{cos}\mspace{2mu}\theta = 4\mspace{2mu}\text{cos}^{2}({\theta\text{/}{2),}}$ so the integral becomes

接下来，利用恒等式 $\text{cos}\left( {2\alpha} \right) = 2\mspace{2mu}\text{cos}^{2}\alpha - 1$，两边加 1 再乘以 2，得到 $2 + 2\mspace{2mu}\text{cos}\left( {2\alpha} \right) = 4\mspace{2mu}\text{cos}^{2}\alpha$。代入 $\alpha = {\theta\text{/}2}$ 得 $2 + 2\mspace{2mu}\text{cos}\mspace{2mu}\theta = 4\mspace{2mu}\text{cos}^{2}({\theta\text{/}{2),}}$，于是积分变为

$$\begin{array}{cl}L & {= 2{\int_{0}^{2\pi}\sqrt{2 + 2\ \text{cos}\ \theta}}d\theta} \\ & {= 2{\int_{0}^{2\pi}{\sqrt{4\ \text{cos}^{2}\left( \frac{\theta}{2} \right)}d\theta}}} \\ & {= 2{\int_{0}^{2\pi}{\left. 2 \middle| {\text{cos}\left( \frac{\theta}{2} \right)} \right|d\theta}}.}\end{array}$$

利用三角恒等式化简，得 $$\begin{array}{cl}L & {= 2{\int_{0}^{2\pi}\sqrt{2 + 2\ \text{cos}\ \theta}}d\theta} \\ & {= 2{\int_{0}^{2\pi}{\sqrt{4\ \text{cos}^{2}\left( \frac{\theta}{2} \right)}d\theta}}} \\ & {= 2{\int_{0}^{2\pi}{\left. 2 \middle| {\text{cos}\left( \frac{\theta}{2} \right)} \right|d\theta}}.}\end{array}$$

The absolute value is necessary because the cosine is negative for some values in its domain. To resolve this issue, change the limits from $0$ to $\pi$ and double the answer. This strategy works because cosine is positive between $0$ and $\frac{\pi}{2}.$ Thus,

绝对值是必要的，因为余弦在其定义域的某些取值上为负。为此，把积分限由 $0$ 到 $\pi$ 改写并将结果乘 2。这一做法成立是因为余弦在 $0$ 到 $\frac{\pi}{2}$ 之间为正。于是

$$\begin{array}{cl}L & {= 4{\int_{0}^{2\pi}{\left| {\text{cos}\left( \frac{\theta}{2} \right)} \right|d\theta}}} \\ & {= 8{\int_{0}^{\pi}{\text{cos}\left( \frac{\theta}{2} \right)\ d\theta}}} \\ & {= 8\left. \left( {2\ \text{sin}\left( \frac{\theta}{2} \right)} \right. \right)_{0}^{\pi}} \\ & {= 16.}\end{array}$$

将积分限改为 $0$ 到 $\pi$ 并把结果乘 2，得 $$\begin{array}{cl}L & {= 4{\int_{0}^{2\pi}{\left| {\text{cos}\left( \frac{\theta}{2} \right)} \right|d\theta}}} \\ & {= 8{\int_{0}^{\pi}{\text{cos}\left( \frac{\theta}{2} \right)\ d\theta}}} \\ & {= 8\left. \left( {2\ \text{sin}\left( \frac{\theta}{2} \right)} \right. \right)_{0}^{\pi}} \\ & {= 16.}\end{array}$$

Find the total arc length of $r = 3\ \text{sin}\ \theta.$

求 $r = 3\ \text{sin}\ \theta$ 的总弧长。

Section 1.4 Exercises 1.4 节习题

For the following exercises, determine a definite integral that represents the area.

在以下习题中，确定表示面积的定积分。

188\.

188.

Region enclosed by $r = 4$

由 $r = 4$ 围成的区域

189.

Region enclosed by $r = 3\ \text{sin}\ \theta$

由 $r = 3\ \text{sin}\ \theta$ 围成的区域

190\.

190.

Region in the first quadrant within the cardioid $r = 1 + \text{sin}\ \theta$

在第一象限内、心形线 $r = 1 + \text{sin}\ \theta$ 之内的区域

191.

Region enclosed by one petal of $r = 8\ \text{sin}(2\theta)$

由 $r = 8\ \text{sin}(2\theta)$ 的一瓣围成的区域

192\.

192.

Region enclosed by one petal of $r = \text{cos}(3\theta)$

由 $r = \text{cos}(3\theta)$ 的一瓣围成的区域

193.

Region below the polar axis and enclosed by $r = 1 - \text{sin}\ \theta$

极轴下方、由 $r = 1 - \text{sin}\ \theta$ 围成的区域

194\.

194.

Region in the first quadrant enclosed by $r = 2 - \text{cos}\ \theta$

在第一象限内由 $r = 2 - \text{cos}\ \theta$ 围成的区域

195.

Region enclosed by the inner loop of $r = 2 - 3\ \text{sin}\ \theta$

由 $r = 2 - 3\ \text{sin}\ \theta$ 的内环围成的区域

196\.

196.

Region enclosed by the inner loop of $r = 3 - 4\ \text{cos}\ \theta$

由 $r = 3 - 4\ \text{cos}\ \theta$ 的内环围成的区域

197.

Region enclosed by $r = 1 - 2\ \text{cos}\ \theta$ and outside the inner loop

由 $r = 1 - 2\ \text{cos}\ \theta$ 围成且在它的内环之外的区域

198\.

198.

Region common to $r = 3\ \text{sin}\ \theta\ \text{and}\ r = 2 - \text{sin}\ \theta$

$r = 3\ \text{sin}\ \theta$ 与 $r = 2 - \text{sin}\ \theta$ 的公共区域

199.

Region common to $r = 2\ \text{and}\ r = 4\ \text{cos}\ \theta$

$r = 2$ 与 $r = 4\ \text{cos}\ \theta$ 的公共区域

200\.

200.

Region common to $r = 3\ \text{cos}\ \theta\ \text{and}\ r = 3\ \text{sin}\ \theta$

$r = 3\ \text{cos}\ \theta$ 与 $r = 3\ \text{sin}\ \theta$ 的公共区域

For the following exercises, find the area of the described region.

在以下习题中，求所述区域的面积。

201.

Enclosed by $r = 6\ \text{sin}\ \theta$

由 $r = 6\ \text{sin}\ \theta$ 围成

202\.

202.

Above the polar axis enclosed by $r = 2 + \text{sin}\ \theta$

极轴上方、由 $r = 2 + \text{sin}\ \theta$ 围成的区域

203.

Below the polar axis and enclosed by $r = 2 - \text{cos}\ \theta$

极轴下方、由 $r = 2 - \text{cos}\ \theta$ 围成的区域

204\.

204.

Enclosed by one petal of $r = 4\ \text{cos}\left( {3\theta} \right)$

由 $r = 4\ \text{cos}\left( {3\theta} \right)$ 的一瓣围成

205.

Enclosed by one petal of $r = 3\ \text{cos}\left( {2\theta} \right)$

由 $r = 3\ \text{cos}\left( {2\theta} \right)$ 的一瓣围成

206\.

206.

Enclosed by $r = 1 + \text{sin}\ \theta$

由 $r = 1 + \text{sin}\ \theta$ 围成

207.

Enclosed by the inner loop of $r = 3 + 6\ \text{cos}\ \theta$

由 $r = 3 + 6\ \text{cos}\ \theta$ 的内环围成

208\.

208.

Enclosed by $r = 2 + 4\ \text{cos}\ \theta$ and outside the inner loop

由 $r = 2 + 4\ \text{cos}\ \theta$ 围成且在内环之外

209.

Common interior of $r = 4\ \text{sin}\left( {2\theta} \right)\ \text{and}\ r = 2$

$r = 4\ \text{sin}\left( {2\theta} \right)$ 与 $r = 2$ 的内部公共区域

210\.

210.

Common interior of $r = 3 - 2\ \text{sin}\ \theta\ \text{and}\ r = -3 + 2\ \text{sin}\ \theta$

$r = 3 - 2\ \text{sin}\ \theta$ 与 $r = -3 + 2\ \text{sin}\ \theta$ 的内部公共区域

211.

Common interior of $r = 6\ \text{sin}\ \theta\ \text{and}\ r = 3$

$r = 6\ \text{sin}\ \theta$ 与 $r = 3$ 的内部公共区域

212\.

212.

Inside $r = 1 + \text{cos}\ \theta$ and outside $r = \text{cos}\ \theta$

在 $r = 1 + \text{cos}\ \theta$ 内部且在 $r = \text{cos}\ \theta$ 外部

213.

Common interior of $r = 2 + 2\ \text{cos}\ \theta\ \text{and}\ r = 2\ \text{sin}\ \theta$

$r = 2 + 2\ \text{cos}\ \theta$ 与 $r = 2\ \text{sin}\ \theta$ 的内部公共区域

For the following exercises, find a definite integral that represents the arc length.

在以下习题中，求表示弧长的定积分。

214\.

214.

$r = 4\ \text{cos}\ \theta\ \text{on the interval}\ 0 \leq \theta \leq \frac{\pi}{2}$

$r = 4\ \text{cos}\ \theta$，区间 $0 \leq \theta \leq \frac{\pi}{2}$

215.

$r = 1 + \text{sin}\ \theta$ on the interval $0 \leq \theta \leq 2\pi$

$r = 1 + \text{sin}\ \theta$，区间 $0 \leq \theta \leq 2\pi$

216\.

216.

$r = 2\ \text{sec}\ \theta\ \text{on the interval}\ 0 \leq \theta \leq \frac{\pi}{3}$

$r = 2\ \text{sec}\ \theta$，区间 $0 \leq \theta \leq \frac{\pi}{3}$

217.

$r = e^{\theta}\text{on the interval}\ 0 \leq \theta \leq 1$

$r = e^{\theta}$，区间 $0 \leq \theta \leq 1$

For the following exercises, find the length of the curve over the given interval.

在以下习题中，求给定区间上曲线的长度。

218\.

218.

$r = 6\ \text{on the interval}\ 0 \leq \theta \leq \frac{\pi}{2}$

$r = 6$，区间 $0 \leq \theta \leq \frac{\pi}{2}$

219.

$r = e^{3\theta}\text{on the interval}\ 0 \leq \theta \leq 2$

$r = e^{3\theta}$，区间 $0 \leq \theta \leq 2$

220\.

220.

$r = 6\ \text{cos}\ \theta\ \text{on the interval}\ 0 \leq \theta \leq \frac{\pi}{2}$

$r = 6\ \text{cos}\ \theta$，区间 $0 \leq \theta \leq \frac{\pi}{2}$

221.

$r = 8 + 8\ \text{cos}\ \theta\ \text{on the interval}\ 0 \leq \theta \leq \pi$

$r = 8 + 8\ \text{cos}\ \theta$，区间 $0 \leq \theta \leq \pi$

222\.

222.

$r = 1 - \text{sin}\ \theta\ \text{on the interval}\ 0 \leq \theta \leq 2\pi$

$r = 1 - \text{sin}\ \theta$，区间 $0 \leq \theta \leq 2\pi$

For the following exercises, use the integration capabilities of a calculator to approximate the length of the curve.

在以下习题中，利用计算器的积分功能近似求曲线的长度。

223.

\[T\] $r = 3\theta\ \text{on the interval}\ 0 \leq \theta \leq \frac{\pi}{2}$

\[T\] $r = 3\theta$，区间 $0 \leq \theta \leq \frac{\pi}{2}$

224\.

224.

\[T\] $r = \frac{2}{\theta}\ \text{on the interval}\ \pi \leq \theta \leq 2\pi$

\[T\] $r = \frac{2}{\theta}$，区间 $\pi \leq \theta \leq 2\pi$

225.

\[T\] $r = \text{sin}^{2}\left( \frac{\theta}{2} \right)\ \text{on the interval}\ 0 \leq \theta \leq \pi$

\[T\] $r = \text{sin}^{2}\left( \frac{\theta}{2} \right)$，区间 $0 \leq \theta \leq \pi$

226\.

226.

\[T\] $r = 2\theta^{2}\ \text{on the interval}\ 0 \leq \theta \leq \pi$

\[T\] $r = 2\theta^{2}$，区间 $0 \leq \theta \leq \pi$

227.

\[T\] $r = \text{sin}\left( {3\ \text{cos}\ \theta} \right)\ \text{on the interval}\ 0 \leq \theta \leq \pi$

\[T\] $r = \text{sin}\left( {3\ \text{cos}\ \theta} \right)$，区间 $0 \leq \theta \leq \pi$

For the following exercises, use the familiar formula from geometry to find the area of the region described and then confirm by using the definite integral.

在以下习题中，先用几何中熟悉的公式求所述区域的面积，再用定积分验证。

228\.

228.

$r = 3\ \text{sin}\ \theta\ \text{on the interval}\ 0 \leq \theta \leq \pi$

$r = 3\ \text{sin}\ \theta$，区间 $0 \leq \theta \leq \pi$

229.

$r = \text{sin}\ \theta + \text{cos}\ \theta\ \text{on the interval}\ 0 \leq \theta \leq \pi$

$r = \text{sin}\ \theta + \text{cos}\ \theta$，区间 $0 \leq \theta \leq \pi$

230\.

230.

$r = 6\ \text{sin}\ \theta + 8\ \text{cos}\ \theta\ \text{on the interval}\ 0 \leq \theta \leq \pi$

$r = 6\ \text{sin}\ \theta + 8\ \text{cos}\ \theta$，区间 $0 \leq \theta \leq \pi$

For the following exercises, use the familiar formula from geometry to find the length of the curve and then confirm using the definite integral.

在以下习题中，先用几何中熟悉的公式求曲线的长度，再用定积分验证。

231.

$r = 3\ \text{sin}\ \theta\ \text{on the interval}\ 0 \leq \theta \leq \pi$

$r = 3\ \text{sin}\ \theta$，区间 $0 \leq \theta \leq \pi$

232\.

232.

$r = \text{sin}\ \theta + \text{cos}\ \theta\ \text{on the interval}\ 0 \leq \theta \leq \pi$

$r = \text{sin}\ \theta + \text{cos}\ \theta$，区间 $0 \leq \theta \leq \pi$

233.

$r = 6\ \text{sin}\ \theta + 8\ \text{cos}\ \theta\ \text{on the interval}\ 0 \leq \theta \leq \pi$

$r = 6\ \text{sin}\ \theta + 8\ \text{cos}\ \theta$，区间 $0 \leq \theta \leq \pi$

234\.

234.

Verify that if $y = r\ \text{sin}\ \theta = f(\theta)\text{sin}\ \theta$ then $\frac{dy}{d\theta} = f\prime(\theta)\text{sin}\ \theta + f(\theta)\text{cos}\ \theta.$

验证：若 $y = r\ \text{sin}\ \theta = f(\theta)\text{sin}\ \theta$，则 $\frac{dy}{d\theta} = f\prime(\theta)\text{sin}\ \theta + f(\theta)\text{cos}\ \theta.$

For the following exercises, find the slope of a tangent line to a polar curve $r = f(\theta).$ Let $x = r\ \text{cos}\ \theta = f(\theta)\text{cos}\ \theta$ and $y = r\ \text{sin}\ \theta = f(\theta)\text{sin}\ \theta,$ so the polar equation $r = f(\theta)$ is now written in parametric form.

在以下习题中，求极坐标曲线 $r = f(\theta)$ 的切线斜率。令 $x = r\ \text{cos}\ \theta = f(\theta)\text{cos}\ \theta$，$y = r\ \text{sin}\ \theta = f(\theta)\text{sin}\ \theta$，于是极坐标方程 $r = f(\theta)$ 被写成参数形式。

235.

Use the definition of the derivative $\frac{dy}{dx} = \frac{{dy}\text{/}{d\theta}}{{dx}\text{/}{d\theta}}$ and the product rule to derive the derivative of a polar equation.

用导数定义 $\frac{dy}{dx} = \frac{{dy}\text{/}{d\theta}}{{dx}\text{/}{d\theta}}$ 与乘积法则推导极坐标方程的导数。

236\.

236.

$r = 1 - \text{sin}\ \theta;$ $\left( {\frac{1}{2},\frac{\pi}{6}} \right)$

$r = 1 - \text{sin}\ \theta$；$\left( {\frac{1}{2},\frac{\pi}{6}} \right)$

237.

$r = 4\ \text{cos}\ \theta;$ $\left( {2,\frac{\pi}{3}} \right)$

$r = 4\ \text{cos}\ \theta$；$\left( {2,\frac{\pi}{3}} \right)$

238\.

238.

$r = 8\ \text{sin}\ \theta;$ $\left( {4,\frac{5\pi}{6}} \right)$

$r = 8\ \text{sin}\ \theta$；$\left( {4,\frac{5\pi}{6}} \right)$

239.

$r = 4 + \text{sin}\ \theta;$ $\left( {3,\frac{3\pi}{2}} \right)$

$r = 4 + \text{sin}\ \theta$；$\left( {3,\frac{3\pi}{2}} \right)$

240\.

240.

$r = 6 + 3\ \text{cos}\ \theta;$ $\left( {3,\pi} \right)$

$r = 6 + 3\ \text{cos}\ \theta$；$\left( {3,\pi} \right)$

241.

$r = 4\ \text{cos}\left( {2\theta} \right);$ tips of the leaves

$r = 4\ \text{cos}\left( {2\theta} \right)$；叶尖

242\.

242.

$r = 2\ \text{sin}\left( {3\theta} \right);$ tips of the leaves

$r = 2\ \text{sin}\left( {3\theta} \right)$；叶尖

243.

$r = 2\theta;$ $\left( {\frac{\pi}{2},\frac{\pi}{4}} \right)$

$r = 2\theta$；$\left( {\frac{\pi}{2},\frac{\pi}{4}} \right)$

244\.

244.

Find the points on the interval $\text{−}\pi \leq \theta \leq \pi$ at which the cardioid $r = 1 - \text{cos}\ \theta$ has a vertical or horizontal tangent line.

求区间 $\text{−}\pi \leq \theta \leq \pi$ 上，心形线 $r = 1 - \text{cos}\ \theta$ 具有垂直或水平切线的点。

245.

For the cardioid $r = 1 + \text{sin}\ \theta,$ find the slope of the tangent line when $\theta = \frac{\pi}{3}.$

对心形线 $r = 1 + \text{sin}\ \theta$，求当 $\theta = \frac{\pi}{3}$ 时切线的斜率。

For the following exercises, find the slope of the tangent line to the given polar curve at the point given by the value of $\theta.$

在以下习题中，求给定极坐标曲线在给定 $\theta$ 值处的点的切线斜率。

246\.

246.

$r = 3\ \text{cos}\ \theta,\theta = \frac{\pi}{3}$

$r = 3\ \text{cos}\ \theta$，$\theta = \frac{\pi}{3}$

247.

$r = \theta,$ $\theta = \frac{\pi}{2}$

$r = \theta$，$\theta = \frac{\pi}{2}$

248\.

248.

$r = \text{ln}\ \theta,$ $\theta = e$

$r = \text{ln}\ \theta$，$\theta = e$

249.

\[T\] Use technology: $r = 2 + 4\ \text{cos}\ \theta$ at $\theta = \frac{\pi}{6}$

\[T\] 使用技术工具：$r = 2 + 4\ \text{cos}\ \theta$ 在 $\theta = \frac{\pi}{6}$ 处

For the following exercises, find the points at which the following polar curves have a horizontal or vertical tangent line.

在以下习题中，求下列极坐标曲线具有水平或垂直切线的点。

250\.

250.

$r = 4\ \text{cos}\ \theta$

251.

$r^{2} = 4\ \text{cos}\left( {2\theta} \right)$

252\.

252.

$r = 2\ \text{sin}(2\theta)$

253.

The cardioid $r = 1 + \text{sin}\ \theta$

心形线 $r = 1 + \text{sin}\ \theta$

254\.

254.

Show that the curve $r = \text{sin}\ \theta\ \text{tan}\ \theta$ (called a *cissoid of Diocles*) has the line $x = 1$ as a vertical asymptote.

证明曲线 $r = \text{sin}\ \theta\ \text{tan}\ \theta$（称为 *cissoid of Diocles*）以直线 $x = 1$ 为垂直渐近线。

1.5 Conic Sections 1.5 圆锥曲线

1.5.1 Identify the equation of a parabola in standard form with given focus and directrix.
1.5.2 Identify the equation of an ellipse in standard form with given foci.
1.5.3 Identify the equation of a hyperbola in standard form with given foci.
1.5.4 Recognize a parabola, ellipse, or hyperbola from its eccentricity value.
1.5.5 Write the polar equation of a conic section with eccentricity $e$.
1.5.6 Identify when a general equation of degree two is a parabola, ellipse, or hyperbola.

1.5.1 给定焦点和准线，识别抛物线的标准方程。
1.5.2 给定焦点，识别椭圆的标准方程。
1.5.3 给定焦点，识别双曲线的标准方程。
1.5.4 由离心率的值识别抛物线、椭圆或双曲线。
1.5.5 写出离心率为 $e$ 的圆锥曲线的极坐标方程。
1.5.6 判断一个二元二次方程何时表示抛物线、椭圆或双曲线。

Conic sections have been studied since the time of the ancient Greeks, and were considered to be an important mathematical concept. As early as 320 BCE, such Greek mathematicians as Menaechmus, Appollonius, and Archimedes were fascinated by these curves. Appollonius wrote an entire eight-volume treatise on conic sections in which he was, for example, able to derive a specific method for identifying a conic section through the use of geometry. Since then, important applications of conic sections have arisen (for example, in astronomy), and the properties of conic sections are used in radio telescopes, satellite dish receivers, and even architecture. In this section we discuss the three basic conic sections, some of their properties, and their equations.

圆锥曲线自古希腊时代起就已被研究，并被视为重要的数学概念。早在公元前 320 年，Menaechmus、Appollonius、Archimedes 等希腊数学家就对这些曲线着迷。Appollonius 写了整整八卷关于圆锥曲线的专著，在其中例如能够通过几何方法导出识别圆锥曲线的具体方法。此后，圆锥曲线的重要应用相继出现（例如在天文学中），其性质被用于射电望远镜、卫星天线接收器，甚至建筑学中。本节我们讨论三种基本圆锥曲线、它们的一些性质及其方程。

Conic sections get their name because they can be generated by intersecting a plane with a cone. A cone has two identically shaped parts called nappes. One nappe is what most people mean by “cone,” having the shape of a party hat. A right circular cone can be generated by revolving a line passing through the origin around the *y*-axis as shown.

圆锥曲线得名于它们可由平面与圆锥相交生成。圆锥有两片形状相同的部分，称为锥叶（nappes）。大多数人所说的“圆锥”指其中一片，形如派对帽。一个直立正圆锥可由过原点的直线绕 *y*-轴旋转生成，如图所示。

Conic sections are generated by the intersection of a plane with a cone (Figure 1.44). If the plane intersects both nappes, then the conic section is a hyperbola. If the plane is parallel to the generating line, the conic section is a parabola. If the plane is perpendicular to the axis of revolution, the conic section is a circle. If the plane intersects one nappe at an angle to the axis (other than $90\text{°}),$ then the conic section is an ellipse.

圆锥曲线由平面与圆锥的相交生成（图 1.44）。若平面与两片锥叶都相交，则圆锥曲线为双曲线。若平面平行于母线，则圆锥曲线为抛物线。若平面垂直于旋转轴，则圆锥曲线为圆。若平面与一片锥叶以不等于 $90\text{°}$ 的角度相交，则圆锥曲线为椭圆。

Parabolas 抛物线

A parabola is generated when a plane intersects a cone parallel to the generating line. In this case, the plane intersects only one of the nappes. A parabola can also be defined in terms of distances.

抛物线在平明与圆锥相交且平行于母线时生成。此时平面只与一片锥叶相交。抛物线也可以用距离来定义。

A parabola is the set of all points whose distance from a fixed point, called the focus, is equal to the distance from a fixed line, called the directrix. The point halfway between the focus and the directrix is called the vertex of the parabola.

抛物线是所有满足以下条件的点的集合：到定点（称为焦点）的距离等于到定直线（称为准线）的距离。焦点与准线正中间的点称为抛物线的顶点。

A graph of a typical parabola appears in Figure 1.45. Using this diagram in conjunction with the distance formula, we can derive an equation for a parabola. Recall the distance formula: Given point *P* with coordinates $\left( {x_{1},y_{1}} \right)$ and point *Q* with coordinates $\left( {x_{2},{\ \text{y}}_{2}} \right),$ the distance between them is given by the formula

典型抛物线的图形如图 1.45 所示。结合此图与距离公式，我们可以导出抛物线的方程。回顾距离公式：给定点 *P*（坐标为 $\left( {x_{1},y_{1}} \right)$）与点 *Q*（坐标为 $\left( {x_{2},{\ \text{y}}_{2}} \right)$），它们之间的距离由下式给出

$$d\left( {P,Q} \right) = \sqrt{\left( {x_{2} - x_{1}} \right)^{2} + \left( {y_{2} - y_{1}} \right)^{2}}.$$

Then from the definition of a parabola and Figure 1.45, we get

由抛物线的定义及图 1.45，我们得到

$$\begin{array}{rll} {d\left( {F,P} \right)} & = & {d\left( {P,Q} \right)} \\ \sqrt{\left( {0 - x} \right)^{2} + \left( {p - y} \right)^{2}} & = & {\sqrt{\left( {x - x} \right)^{2} + \left( {\text{−}p - y} \right)^{2}}.} \end{array}$$

Squaring both sides and simplifying yields

两边平方并化简得

$$\begin{array}{rll} {x^{2} + \left( {p - y} \right)^{2}} & = & {0^{2} + \left( {\text{−}p - y} \right)^{2}} \\ {x^{2} + p^{2} - 2py + y^{2}} & = & {p^{2} + 2py + y^{2}} \\ {x^{2} - 2py} & = & {2py} \\ x^{2} & = & {4py.} \end{array}$$

Now suppose we want to relocate the vertex. We use the variables $\left( {h,k} \right)$ to denote the coordinates of the vertex. Then if the focus is directly above the vertex, it has coordinates $\left( {h,k + p} \right)$ and the directrix has the equation $y = k - p.$ Going through the same derivation yields the formula $\left( {x - h} \right)^{2} = 4p\left( {y - k} \right).$ Solving this equation for *y* leads to the following theorem.

现假设我们要移动顶点。我们用变量 $\left( {h,k} \right)$ 表示顶点坐标。那么若焦点在顶点正上方，其坐标为 $\left( {h,k + p} \right)$，准线方程为 $y = k - p.$ 经过同样的推导可得公式 $\left( {x - h} \right)^{2} = 4p\left( {y - k} \right).$ 解出 *y* 即得如下定理。

Equations for Parabolas 抛物线的方程

Given a parabola opening upward with vertex located at $\left( {h,k} \right)$ and focus located at $\left( {h,k + p} \right),$ where *p* is a constant, the equation for the parabola is given by

给定一个开口向上的抛物线，其顶点位于 $\left( {h,k} \right)$，焦点位于 $\left( {h,k + p} \right)$，其中 *p* 为常数，则该抛物线方程为

$$y = \frac{1}{4p}\left( {x - h} \right)^{2} + k.$$ (1.11)

This is the standard form of a parabola.

这是抛物线的标准形式。

We can also study the cases when the parabola opens down or to the left or the right. The equation for each of these cases can also be written in standard form as shown in the following graphs.

我们也可以研究抛物线向下、向左或向右开口的情形。这些情形的方程也可写成如下图形所示的标准形式。

In addition, the equation of a parabola can be written in the general form, though in this form the values of *h*, *k*, and *p* are not immediately recognizable. The general form of a parabola is written as

此外，抛物线方程也可写成一般形式，但在该形式中 *h*、*k*、*p* 的值不能立即识别。抛物线的一般形式写作

$$ax^{2} + bx + cy + d = 0\quad\text{or}\quad ay^{2} + bx + cy + d = 0.$$

The first equation represents a parabola that opens either up or down. The second equation represents a parabola that opens either to the left or to the right. To put the equation into standard form, use the method of completing the square.

第一个方程表示向上或向下开口的抛物线。第二个方程表示向左或向右开口的抛物线。要化为标准形式，使用配方法。

Converting the Equation of a Parabola from General into Standard Form 将抛物线的一般方程化为标准形式

Put the equation $x^{2} - 4x - 8y + 12 = 0$ into standard form and graph the resulting parabola.

将方程 $x^{2} - 4x - 8y + 12 = 0$ 化为标准形式，并画出所得抛物线。

Solution 解

Since *y* is not squared in this equation, we know that the parabola opens either upward or downward. Therefore we need to solve this equation for *y,* which will put the equation into standard form. To do that, first add $8y$ to both sides of the equation:

由于本方程中 *y* 未被平方，可知抛物线向上或向下开口。因此我们需要将此方程解出 *y*，从而化为标准形式。为此，先在方程两边加上 $8y$：

$$8y = x^{2} - 4x + 12.$$

The next step is to complete the square on the right-hand side. Start by grouping the first two terms on the right-hand side using parentheses:

下一步是对右端配方。先利用括号将右端前两项分组：

$$8y = \left( {x^{2} - 4x} \right) + 12.$$

Next determine the constant that, when added inside the parentheses, makes the quantity inside the parentheses a perfect square trinomial. To do this, take half the coefficient of *x* and square it. This gives $\left( \frac{-4}{2} \right)^{2} = 4.$ Add 4 inside the parentheses and subtract 4 outside the parentheses, so the value of the equation is not changed:

接下来确定在括号内加上一个常数后，使括号内成为完全平方三项式。为此，取 *x* 系数的一半再平方。得 $\left( \frac{-4}{2} \right)^{2} = 4.$ 在括号内加 4，在括号外减 4，从而方程的值不变：

$$8y = \left( {x^{2} - 4x + 4} \right) + 12 - 4.$$

Now combine like terms and factor the quantity inside the parentheses:

现在合并同类项并对括号内因式分解：

$$8y = \left( {x - 2} \right)^{2} + 8.$$

Finally, divide by 8:

最后除以 8：

$$y = \frac{1}{8}\left( {x - 2} \right)^{2} + 1.$$

This equation is now in standard form. Comparing this to Equation 1.11 gives $h = 2,$ $k = 1,$ and $p = 2.$ The parabola opens up, with vertex at $\left( {2,1} \right),$ focus at $\left( {2,3} \right),$ and directrix $y = -1.$ The graph of this parabola appears as follows.

此方程现已为标准形式。与式 (1.11) 比较得 $h = 2,$ $k = 1,$ 且 $p = 2.$ 抛物线开口向上，顶点在 $\left( {2,1} \right)$，焦点在 $\left( {2,3} \right)$，准线为 $y = -1.$ 该抛物线的图形如下。

Put the equation $2y^{2} - x + 12y + 16 = 0$ into standard form and graph the resulting parabola.

将方程 $2y^{2} - x + 12y + 16 = 0$ 化为标准形式，并画出所得抛物线。

The axis of symmetry of a vertical (opening up or down) parabola is a vertical line passing through the vertex. The parabola has an interesting reflective property. Suppose we have a satellite dish with a parabolic cross section. If a beam of electromagnetic waves, such as light or radio waves, comes into the dish in a straight line from a satellite (parallel to the axis of symmetry), then the waves reflect off the dish and collect at the focus of the parabola as shown.

竖直（向上或向下）抛物线的对称轴是过顶点的竖直线。抛物线有一个有趣的反射性质。假设我们有一个具有抛物线截面的卫星天线。如果一束电磁波（如光波或无线电波）从卫星沿直线进入天线（平行于对称轴），则波会从天线反射并汇聚于抛物线的焦点，如图所示。

Consider a parabolic dish designed to collect signals from a satellite in space. The dish is aimed directly at the satellite, and a receiver is located at the focus of the parabola. Radio waves coming in from the satellite are reflected off the surface of the parabola to the receiver, which collects and decodes the digital signals. This allows a small receiver to gather signals from a wide angle of sky. Flashlights and headlights in a car work on the same principle, but in reverse: the source of the light (that is, the light bulb) is located at the focus and the reflecting surface on the parabolic mirror focuses the beam straight ahead. This allows a small light bulb to illuminate a wide angle of space in front of the flashlight or car.

考虑一个设计用来收集太空中卫星信号的抛物面天线。天线直接对准卫星，接收器位于抛物线的焦点处。来自卫星的无线电波从抛物面反射到接收器，接收器收集并解码数字信号。这使得一个小型接收器能够从广阔的天空中收集信号。手电筒和汽车前灯的工作原理相同，只是方向相反：光源（即灯泡）位于焦点处，抛物面镜的反射面将光束聚焦于正前方。这使得一个小型灯泡能够照亮手电筒或汽车前方广阔的空间。

Ellipses 椭圆

An ellipse can also be defined in terms of distances. In the case of an ellipse, there are two foci (plural of focus), and two directrices (plural of directrix). We look at the directrices in more detail later in this section.

椭圆也可以用距离来定义。就椭圆而言，有两个焦点（focus 的复数）和两条准线（directrix 的复数）。我们将在本节后面更详细地讨论准线。

An *ellipse* is the set of all points for which the sum of their distances from two fixed points (the foci) is constant.

椭圆是满足"到两个定点（焦点）的距离之和为常数"的所有点的集合。

A graph of a typical ellipse is shown in Figure 1.48. In this figure the foci are labeled as $F$ and $F^{\prime}.$ Both are the same fixed distance from the origin, and this distance is represented by the variable *c*. Therefore the coordinates of $F$ are $\left( {c,0} \right)$ and the coordinates of $F^{\prime}$ are $\left( {\text{-}c,0} \right).$ The points $P$ and $P^{\prime}$ are located at the ends of the major axis of the ellipse, and have coordinates $\left( {a,0} \right)$ and $\left( {\text{-}a,0} \right),$ respectively. The major axis is always the longest distance across the ellipse, and can be horizontal or vertical. Thus, the length of the major axis in this ellipse is 2*a.* Furthermore, $P$ and $P^{\prime}$ are called the vertices of the ellipse. The points $Q$ and $Q^{\prime}$ are located at the ends of the minor axis of the ellipse, and have coordinates $\left( {0,b} \right)$ and $\left( {0,\text{-}b} \right),$ respectively. The minor axis is the shortest distance across the ellipse. The minor axis is perpendicular to the major axis.

典型椭圆的图形如图 1.48 所示。图中两个焦点标为 $F$ 和 $F^{\prime}$，它们到原点的固定距离相同，该距离用变量 $c$ 表示。因此 $F$ 的坐标为 $\left( {c,0} \right)$，$F^{\prime}$ 的坐标为 $\left( {\text{-}c,0} \right)$。点 $P$ 和 $P^{\prime}$ 位于椭圆长轴的两端，坐标分别为 $\left( {a,0} \right)$ 和 $\left( {\text{-}a,0} \right)$。长轴总是椭圆中最长的距离，可以是水平的或垂直的。因此本椭圆长轴的长度为 2*a*。此外，$P$ 和 $P^{\prime}$ 称为椭圆的顶点。点 $Q$ 和 $Q^{\prime}$ 位于椭圆短轴的两端，坐标分别为 $\left( {0,b} \right)$ 和 $\left( {0,\text{-}b} \right)$。短轴是椭圆中最短的距离，且与长轴垂直。

According to the definition of the ellipse, we can choose any point on the ellipse and the sum of the distances from this point to the two foci is constant. Suppose we choose the point *P.* Since the coordinates of point *P* are $\left( {a,0} \right),$ the sum of the distances is

根据椭圆的定义，我们可以取椭圆上任意一点，该点到两个焦点的距离之和为常数。假设我们取点 $P$。由于点 $P$ 的坐标为 $\left( {a,0} \right)$，距离之和为

$$d\left( {P,F} \right) + d\left( {P,F^{\prime}} \right) = \left( {a - c} \right) + \left( {a + c} \right) = 2a.$$

Therefore the sum of the distances from an arbitrary point *A* with coordinates $\left( {x,y} \right)$ is also equal to 2*a.* Using the distance formula, we get

因此，任意一点 $A$（坐标为 $\left( {x,y} \right)$）的距离之和也等于 2*a*。利用距离公式，我们得到

$$\begin{array}{rll} {d\left( {A,F} \right) + d\left( {A,F^{\prime}} \right)} & = & {2a} \\ {\sqrt{\left( {x - c} \right)^{2} + y^{2}} + \sqrt{\left( {x + c} \right)^{2} + y^{2}}} & = & {2a.} \end{array}$$

Subtract the second radical from both sides and square both sides:

将第二项根式移到等式另一边，并对两边平方：

$$\begin{array}{rll} \sqrt{\left( {x - c} \right)^{2} + y^{2}} & = & {2a - \sqrt{\left( {x + c} \right)^{2} + y^{2}}} \\ {\left( {x - c} \right)^{2} + y^{2}} & = & {4a^{2} - 4a\sqrt{\left( {x + c} \right)^{2} + y^{2}} + \left( {x + c} \right)^{2} + y^{2}} \\ {x^{2} - 2cx + c^{2} + y^{2}} & = & {4a^{2} - 4a\sqrt{\left( {x + c} \right)^{2} + y^{2}} + x^{2} + 2cx + c^{2} + y^{2}} \\ {\text{-}2cx} & = & {4a^{2} - 4a\sqrt{\left( {x + c} \right)^{2} + y^{2}} + 2cx.} \end{array}$$

Now isolate the radical on the right-hand side and square again:

现在将右边的根式单独留在一边，再次平方：

$$\begin{array}{rll} {- 2cx} & = & {4a^{2} - 4a\sqrt{\left( {x + c} \right)^{2} + y^{2}} + 2cx} \\ {4a\sqrt{\left( {x + c} \right)^{2} + y^{2}}} & = & {4a^{2} + 4cx} \\ \sqrt{\left( {x + c} \right)^{2} + y^{2}} & = & {a + \frac{cx}{a}} \\ {\left( {x + c} \right)^{2} + y^{2}} & = & {a^{2} + 2cx + \frac{c^{2}x^{2}}{a^{2}}} \\ {x^{2} + 2cx + c^{2} + y^{2}} & = & {a^{2} + 2cx + \frac{c^{2}x^{2}}{a^{2}}} \\ {x^{2} + c^{2} + y^{2}} & = & {a^{2} + \frac{c^{2}x^{2}}{a^{2}}.} \end{array}$$

Isolate the variables on the left-hand side of the equation and the constants on the right-hand side:

将变量移到等式左边，常数移到右边：

$$\begin{array}{rll} & & \\ {x^{2} - \frac{c^{2}x^{2}}{a^{2}} + y^{2}} & = & {a^{2} - c^{2}} \\ {\frac{\left( {a^{2} - c^{2}} \right)x^{2}}{a^{2}} + y^{2}} & = & {a^{2} - c^{2}.} \end{array}$$

Divide both sides by $a^{2} - c^{2}.$ This gives the equation

两边同除以 $a^{2} - c^{2}$，得到方程

$$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{a^{2} - c^{2}} = 1.$$

If we refer back to Figure 1.48, then the length of each of the two green line segments is equal to *a*. This is true because the sum of the distances from the point *Q* to the foci $F\ \text{and}\ F^{\prime}$ is equal to 2*a*, and the lengths of these two line segments are equal. This line segment forms a right triangle with hypotenuse length *a* and leg lengths *b* and *c*. From the Pythagorean theorem, $c^{2} = b^{2} = a^{2}$ and $b^{2} + a^{2} = c^{2}.$ Therefore the equation of the ellipse becomes

回到图 1.48，两条绿色线段的长度都等于 $a$。这是因为点 $Q$ 到焦点 $F\ \text{and}\ F^{\prime}$ 的距离之和为 2*a*，且这两条线段长度相等。这条线段与斜边长度为 $a$、两直角边长度为 $b$ 和 $c$ 的直角三角形有关。由勾股定理，$c^{2} = b^{2} = a^{2}$ 且 $b^{2} + a^{2} = c^{2}$。因此椭圆的方程变为

$$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1.$$

Finally, if the center of the ellipse is moved from the origin to a point $\left( {h,k} \right),$ we have the following standard form of an ellipse.

最后，若将椭圆中心从原点移到点 $\left( {h,k} \right)$，便得到椭圆的如下标准形式。

Equation of an Ellipse in Standard Form 椭圆的标准方程

Consider the ellipse with center $\left( {h,k} \right),$ a horizontal major axis with length 2*a*, and a vertical minor axis with length 2*b*. Then the equation of this ellipse in standard form is

考虑中心为 $\left( {h,k} \right)$、长轴水平且长度为 2*a*、短轴垂直且长度为 2*b* 的椭圆。则该椭圆的标准方程为

$$\frac{\left( {x - h} \right)^{2}}{a^{2}} + \frac{\left( {y - k} \right)^{2}}{b^{2}} = 1$$ (1.12)

and the foci are located at $\left( {h \pm c,k} \right),$ where $c^{2} = a^{2} - b^{2}.$ The equations of the directrices are $x = h \pm \frac{a^{2}}{c}.$

焦点位于 $\left( {h \pm c,k} \right)$，其中 $c^{2} = a^{2} - b^{2}$。准线方程为 $x = h \pm \frac{a^{2}}{c}.$

If the major axis is vertical, then the equation of the ellipse becomes

若长轴为垂直方向，则椭圆方程变为

$$\frac{\left( {x - h} \right)^{2}}{b^{2}} + \frac{\left( {y - k} \right)^{2}}{a^{2}} = 1$$ (1.13)

and the foci are located at $\left( {h,k \pm c} \right),$ where $c^{2} = a^{2} - b^{2}.$ The equations of the directrices in this case are $y = k \pm \frac{a^{2}}{c}.$

焦点位于 $\left( {h,k \pm c} \right)$，其中 $c^{2} = a^{2} - b^{2}$。此时准线方程为 $y = k \pm \frac{a^{2}}{c}.$

If the major axis is horizontal, then the ellipse is called horizontal, and if the major axis is vertical, then the ellipse is called vertical. The equation of an ellipse is in general form if it is in the form $Ax^{2} + By^{2} + Cx + Dy + E = 0,$ where *A* and *B* are either both positive or both negative. To convert the equation from general to standard form, use the method of completing the square.

若长轴为水平方向，则称该椭圆为水平椭圆；若长轴为垂直方向，则称其为垂直椭圆。当椭圆方程形如 $Ax^{2} + By^{2} + Cx + Dy + E = 0$（其中 $A$ 与 $B$ 同为正或同为负）时，称为一般式。要将方程由一般式化为标准式，可使用配方法。

Finding the Standard Form of an Ellipse 求椭圆的标准形式

Put the equation $9x^{2} + 4y^{2} - 36x + 24y + 36 = 0$ into standard form and graph the resulting ellipse.

将方程 $9x^{2} + 4y^{2} - 36x + 24y + 36 = 0$ 化为标准形式，并画出所得椭圆。

Solution 解

First subtract 36 from both sides of the equation:

首先在等式两边同时减去 36：

$$9x^{2} + 4y^{2} - 36x + 24y = -36.$$

Next group the *x* terms together and the *y* terms together, and factor out the common factor:

接下来将含 $x$ 的项与含 $y$ 的项分别归组，并提取公因子：

$$\begin{array}{rll} {\left( {9x^{2} - 36x} \right) + \left( {4y^{2} + 24y} \right)} & = & -36 \\ {9\left( {x^{2} - 4x} \right) + 4\left( {y^{2} + 6y} \right)} & = & -36. \end{array}$$

We need to determine the constant that, when added inside each set of parentheses, results in a perfect square. In the first set of parentheses, take half the coefficient of *x* and square it. This gives $\left( \frac{-4}{2} \right)^{2} = 4.$ In the second set of parentheses, take half the coefficient of *y* and square it. This gives $\left( \frac{6}{2} \right)^{2} = 9.$ Add these inside each pair of parentheses. Since the first set of parentheses has a 9 in front, we are actually adding 36 to the left-hand side. Similarly, we are adding 36 to the second set as well. Therefore the equation becomes

我们需要确定一个数，加进每组括号内后恰好构成完全平方。第一组括号中，取 $x$ 系数的一半再平方，得 $\left( \frac{-4}{2} \right)^{2} = 4$。第二组括号中，取 $y$ 系数的一半再平方，得 $\left( \frac{6}{2} \right)^{2} = 9$。将这两个数分别加入各组括号内。由于第一组括号前有系数 9，实际上我们向左边加了 36；同理，第二组也加了 36。于是方程变为

$$\begin{array}{l} \\ {9\left( {x^{2} - 4x + 4} \right) + 4\left( {y^{2} + 6y + 9} \right) = -36 + 36 + 36} \\ {9\left( {x^{2} - 4x + 4} \right) + 4\left( {y^{2} + 6y + 9} \right) = 36.} \end{array}$$

Now factor both sets of parentheses and divide by 36:

现在将两组括号分别因式分解，并两边同除以 36：

$$\begin{array}{rll} & & \\ {9\left( {x - 2} \right)^{2} + 4\left( {y + 3} \right)^{2}} & = & 36 \\ {\frac{9\left( {x - 2} \right)^{2}}{36} + \frac{4\left( {y + 3} \right)^{2}}{36}} & = & 1 \\ {\frac{\left( {x - 2} \right)^{2}}{4} + \frac{\left( {y + 3} \right)^{2}}{9}} & = & 1. \end{array}$$

The equation is now in standard form. Comparing this to Equation 1.14 gives $h = 2,$ $k = -3,$ $a = 3,$ and $b = 2.$ This is a vertical ellipse with center at $\left( {2,-3} \right),$ major axis 6, and minor axis 4. The graph of this ellipse appears as follows.

方程现已化为标准形式。与方程 1.14 比较可得 $h = 2$、$k = -3$、$a = 3$、$b = 2$。这是一个中心在 $\left( {2,-3} \right)$、长轴为 6、短轴为 4 的垂直椭圆。该椭圆的图形如下。

Put the equation $9x^{2} + 16y^{2} + 18x - 64y - 71 = 0$ into standard form and graph the resulting ellipse.

将方程 $9x^{2} + 16y^{2} + 18x - 64y - 71 = 0$ 化为标准形式，并画出所得椭圆。

According to Kepler’s first law of planetary motion, the orbit of a planet around the Sun is an ellipse with the Sun at one of the foci as shown in Figure 1.50(a). Because Earth’s orbit is an ellipse, the distance from the Sun varies throughout the year. A commonly held misconception is that Earth is closer to the Sun in the summer. In fact, in summer for the northern hemisphere, Earth is farther from the Sun than during winter. The difference in season is caused by the tilt of Earth’s axis in the orbital plane. Comets that orbit the Sun, such as Halley’s Comet, also have elliptical orbits, as do moons orbiting the planets and satellites orbiting Earth.

根据开普勒行星运动第一定律，行星绕太阳运行的轨道是一个椭圆，太阳位于其中一个焦点上，如图 1.50(a) 所示。由于地球轨道是椭圆，地球到太阳的距离在一年中会变化。一个常见的误解是地球在夏季离太阳更近；实际上，对北半球而言，夏季地球离太阳比冬季更远。季节的差异是由地球自转轴在轨道平面内的倾斜造成的。绕太阳运行的彗星（如哈雷彗星）以及绕行星运行的卫星和绕地球运行的卫星，其轨道也都是椭圆。

Ellipses also have interesting reflective properties: A light ray emanating from one focus passes through the other focus after mirror reflection in the ellipse. The same thing occurs with a sound wave as well. The National Statuary Hall in the U.S. Capitol in Washington, DC, is a famous room in an elliptical shape as shown in Figure 1.50(b). This hall served as the meeting place for the U.S. House of Representatives for almost fifty years. The location of the two foci of this semi-elliptical room are clearly identified by marks on the floor, and even if the room is full of visitors, when two people stand on these spots and speak to each other, they can hear each other much more clearly than they can hear someone standing close by. Legend has it that John Quincy Adams had his desk located on one of the foci and was able to eavesdrop on everyone else in the House without ever needing to stand. Although this makes a good story, it is unlikely to be true, because the original ceiling produced so many echoes that the entire room had to be hung with carpets to dampen the noise. The ceiling was rebuilt in 1902 and only then did the now-famous whispering effect emerge. Another famous whispering gallery—the site of many marriage proposals—is in Grand Central Station in New York City.

椭圆还具有有趣的反射性质：从一个焦点发出的光线，经椭圆镜面反射后会通过另一个焦点；声波也同样如此。华盛顿特区的美国国会大厦中的国家雕像厅是一间著名的椭圆形房间，如图 1.50(b) 所示。该大厅曾作为美国众议院的会议场所近五十年。这间半椭圆形房间的两个焦点位置由地面上的标记清晰标出；即便满厅都是参观者，当两个分别站在这两个位置上的人彼此交谈时，他们能比听到近旁的人更清楚地听到对方。传说约翰·昆西·亚当斯将办公桌设在其中一个焦点处，无需起身便能窃听众议院中其他所有人。尽管这是个好故事，但它不太可能属实，因为原来的天花板产生太多回声，整个房间不得不挂满地毯来消减噪音。天花板于 1902 年重建，直到那时才出现如今著名的窃窃私语效应。另一个著名的窃窃私语廊——许多求婚事件的发生地——位于纽约市的中央车站。

Hyperbolas 双曲线

A hyperbola can also be defined in terms of distances. In the case of a hyperbola, there are two foci and two directrices. Hyperbolas also have two asymptotes.

双曲线也可以按距离来定义。对于双曲线，有两个焦点和两条准线。双曲线还有两条渐近线。

A hyperbola is the set of all points where the difference between their distances from two fixed points (the foci) is constant.

双曲线是所有满足"到两个定点（焦点）的距离之差为常数"的点的集合。

A graph of a typical hyperbola appears as follows.

一条典型双曲线的图形如下。

The derivation of the equation of a hyperbola in standard form is virtually identical to that of an ellipse. One slight hitch lies in the definition: The difference between two numbers is always positive. Let *P* be a point on the hyperbola with coordinates $\left( {x,y} \right).$ Then the definition of the hyperbola gives $\left| {d\left( {P,F_{1}} \right) - d\left( {P,F_{2}} \right)} \right| = \text{constant}.$ To simplify the derivation, assume that *P* is on the right branch of the hyperbola, so the absolute value bars drop. If it is on the left branch, then the subtraction is reversed. The vertex of the right branch has coordinates $\left( {a,0} \right),$ so

双曲线标准方程的推导与椭圆几乎相同。定义中有一点小障碍：两个数的差总是正的。设 *P* 是双曲线上一点，坐标为 $\left( {x,y} \right).$ 则双曲线的定义给出 $\left| {d\left( {P,F_{1}} \right) - d\left( {P,F_{2}} \right)} \right| = \text{constant}.$ 为简化推导，假设 *P* 在双曲线的右支上，从而去掉绝对值符号。若它在左支上，则减法次序相反。右支的顶点坐标为 $\left( {a,0} \right),$ 于是

$$d\left( {P,F_{1}} \right) - d\left( {P,F_{2}} \right) = \left( {c + a} \right) - \left( {c - a} \right) = 2a.$$

This equation is therefore true for any point on the hyperbola. Returning to the coordinates $\left( {x,y} \right)$ for *P*:

因此该方程对双曲线上任意一点都成立。回到 *P* 的坐标 $\left( {x,y} \right)$：

$$\begin{array}{rll} {d\left( {P,F_{1}} \right) - d\left( {P,F_{2}} \right)} & = & {2a} \\ {\sqrt{\left( {x + c} \right)^{2} + y^{2}} - \sqrt{\left( {x - c} \right)^{2} + y^{2}}} & = & {2a.} \end{array}$$

Add the second radical from both sides and square both sides:

将第二项根式移到等式一边，并对两边平方：

$$\begin{array}{rll} \sqrt{\left( {x - c} \right)^{2} + y^{2}} & = & {2a + \sqrt{\left( {x + c} \right)^{2} + y^{2}}} \\ {\left( {x - c} \right)^{2} + y^{2}} & = & {4a^{2} + 4a\sqrt{\left( {x + c} \right)^{2} + y^{2}} + \left( {x + c} \right)^{2} + y^{2}} \\ {x^{2} - 2cx + c^{2} + y^{2}} & = & {4a^{2} + 4a\sqrt{\left( {x + c} \right)^{2} + y^{2}} + x^{2} + 2cx + c^{2} + y^{2}} \\ {−2cx} & = & {4a^{2} + 4a\sqrt{\left( {x + c} \right)^{2} + y^{2}} + 2cx.} \end{array}$$

Now isolate the radical on the right-hand side and square again:

现在将根式单独留在右边，再次平方：

$$\begin{array}{rll} {- 2cx} & = & {4a^{2} + 4a\sqrt{\left( {x + c} \right)^{2} + y^{2}} + 2cx} \\ {4a\sqrt{\left( {x + c} \right)^{2} + y^{2}}} & = & {-4a^{2} - 4cx} \\ \sqrt{\left( {x + c} \right)^{2} + y^{2}} & = & {\text{−}a - \frac{cx}{a}} \\ {\left( {x + c} \right)^{2} + y^{2}} & = & {a^{2} + 2cx + \frac{c^{2}x^{2}}{a^{2}}} \\ {x^{2} + 2cx + c^{2} + y^{2}} & = & {a^{2} + 2cx + \frac{c^{2}x^{2}}{a^{2}}} \\ {x^{2} + c^{2} + y^{2}} & = & {a^{2} + \frac{c^{2}x^{2}}{a^{2}}.} \end{array}$$

Isolate the variables on the left-hand side of the equation and the constants on the right-hand side:

将变量移到等式左边，常数移到右边：

$$\begin{array}{rll} & & \\ {x^{2} - \frac{c^{2}x^{2}}{a^{2}} + y^{2}} & = & {a^{2} - c^{2}} \\ {\frac{\left( {a^{2} - c^{2}} \right)x^{2}}{a^{2}} + y^{2}} & = & {a^{2} - c^{2}.} \end{array}$$

Finally, divide both sides by $a^{2} - c^{2}.$ This gives the equation

最后，两边同除以 $a^{2} - c^{2}.$ 得到方程

$$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{a^{2} - c^{2}} = 1.$$

We now define *b* so that $b^{2} = c^{2} - a^{2}.$ This is possible because $c > a.$ Therefore the equation of the hyperbola becomes

现在定义 *b* 使得 $b^{2} = c^{2} - a^{2}.$ 这是可行的，因为 $c > a.$ 于是双曲线方程变为

$$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1.$$

Finally, if the center of the hyperbola is moved from the origin to the point $\left( {h,k} \right),$ we have the following standard form of a hyperbola.

最后，若将双曲线的中心从原点平移到点 $\left( {h,k} \right),$ 便得到双曲线的如下标准形式。

Equation of a Hyperbola in Standard Form 双曲线的标准方程

Consider the hyperbola with center $\left( {h,k} \right),$ a horizontal major axis, and a vertical minor axis. Then the equation of this hyperbola is

考虑中心为 $\left( {h,k} \right),$ 长轴水平、短轴垂直的双曲线。则该双曲线的方程为

$$\frac{\left( {x - h} \right)^{2}}{a^{2}} - \frac{\left( {y - k} \right)^{2}}{b^{2}} = 1$$ (1.14)

and the foci are located at $\left( {h \pm c,k} \right),$ where $c^{2} = a^{2} + b^{2}.$ The equations of the asymptotes are given by $y = k \pm \frac{b}{a}\left( {x - h} \right).$ The equations of the directrices are

焦点位于 $\left( {h \pm c,k} \right),$ 其中 $c^{2} = a^{2} + b^{2}.$ 渐近线方程为 $y = k \pm \frac{b}{a}\left( {x - h} \right).$ 准线方程为

$$x = h \pm \frac{a^{2}}{\sqrt{a^{2} + b^{2}}} = h \pm \frac{a^{2}}{c}.$$

If the major axis is vertical, then the equation of the hyperbola becomes

若长轴垂直，则双曲线方程变为

$$\frac{\left( {y - k} \right)^{2}}{a^{2}} - \frac{\left( {x - h} \right)^{2}}{b^{2}} = 1$$ (1.15)

and the foci are located at $\left( {h,k \pm c} \right),$ where $c^{2} = a^{2} + b^{2}.$ The equations of the asymptotes are given by $y = k \pm \frac{a}{b}\left( {x - h} \right).$ The equations of the directrices are

焦点位于 $\left( {h,k \pm c} \right),$ 其中 $c^{2} = a^{2} + b^{2}.$ 渐近线方程为 $y = k \pm \frac{a}{b}\left( {x - h} \right).$ 准线方程为

$$y = k \pm \frac{a^{2}}{\sqrt{a^{2} + b^{2}}} = k \pm \frac{a^{2}}{c}.$$

If the major axis (transverse axis) is horizontal, then the hyperbola is called horizontal, and if the major axis is vertical then the hyperbola is called vertical. The equation of a hyperbola is in general form if it is in the form $Ax^{2} + By^{2} + Cx + Dy + E = 0,$ where *A* and *B* have opposite signs. In order to convert the equation from general to standard form, use the method of completing the square.

如果长轴（横轴）是水平的，则称该双曲线为水平双曲线；如果长轴是垂直的，则称其为垂直双曲线。当双曲线方程呈形式 $Ax^{2} + By^{2} + Cx + Dy + E = 0,$ 且 *A* 与 *B* 异号时，称其为一般式。要将方程由一般式化为标准形式，可使用配方法。

Finding the Standard Form of a Hyperbola 求双曲线的标准形式

Put the equation $9x^{2} - 16y^{2} + 36x + 32y - 124 = 0$ into standard form and graph the resulting hyperbola. What are the equations of the asymptotes?

将方程 $9x^{2} - 16y^{2} + 36x + 32y - 124 = 0$ 化为标准形式，并画出所得双曲线。其渐近线方程是什么？

Solution 解

First add 124 to both sides of the equation:

首先，在方程两边同加 124：

$$9x^{2} - 16y^{2} + 36x + 32y = 124.$$

Next group the *x* terms together and the *y* terms together, then factor out the common factors:

接着将 *x* 项与 *y* 项分别归类，再提取公因子：

$$\begin{array}{rll} {\left( {9x^{2} + 36x} \right) - \left( {16y^{2} - 32y} \right)} & = & 124 \\ {9\left( {x^{2} + 4x} \right) - 16\left( {y^{2} - 2y} \right)} & = & 124. \end{array}$$

We need to determine the constant that, when added inside each set of parentheses, results in a perfect square. In the first set of parentheses, take half the coefficient of *x* and square it. This gives $\left( \frac{4}{2} \right)^{2} = 4.$ In the second set of parentheses, take half the coefficient of *y* and square it. This gives $\left( \frac{-2}{2} \right)^{2} = 1.$ Add these inside each pair of parentheses. Since the first set of parentheses has a 9 in front, we are actually adding 36 to the left-hand side. Similarly, we are subtracting 16 from the second set of parentheses. Therefore the equation becomes

我们需要确定一个常数，加进每组括号内后可配成完全平方。在第一组括号中，取 *x* 系数的一半再平方，得到 $\left( \frac{4}{2} \right)^{2} = 4.$ 在第二组括号中，取 *y* 系数的一半再平方，得到 $\left( \frac{-2}{2} \right)^{2} = 1.$ 将这些数分别加进各组括号。由于第一组括号前有系数 9，我们实际上给左边加了 36；同理，第二组括号前有系数 16（带负号），我们给左边减了 16。因此方程变为

$$\begin{array}{l} \\ {9\left( {x^{2} + 4x + 4} \right) - 16\left( {y^{2} - 2y + 1} \right) = 124 + 36 - 16} \\ {9\left( {x^{2} + 4x + 4} \right) - 16\left( {y^{2} - 2y + 1} \right) = 144.} \end{array}$$

Next factor both sets of parentheses and divide by 144:

接着对两组括号分别因式分解，并除以 144：

$$\begin{array}{rll} & & \\ {9\left( {x + 2} \right)^{2} - 16\left( {y - 1} \right)^{2}} & = & 144 \\ {\frac{9\left( {x + 2} \right)^{2}}{144} - \frac{16\left( {y - 1} \right)^{2}}{144}} & = & 1 \\ {\frac{\left( {x + 2} \right)^{2}}{16} - \frac{\left( {y - 1} \right)^{2}}{9}} & = & 1. \end{array}$$

The equation is now in standard form. Comparing this to Equation 1.15 gives $h = -2,$ $k = 1,$ $a = 4,$ and $b = 3.$ This is a horizontal hyperbola with center at $\left( {-2,1} \right)$ and asymptotes given by the equations $y = 1 \pm \frac{3}{4}\left( {x + 2} \right).$ The graph of this hyperbola appears in the following figure.

方程现已化为标准形式。与方程 1.15 比较可得 $h = -2,$ $k = 1,$ $a = 4,$ $b = 3.$ 这是一条水平双曲线，中心在 $\left( {-2,1} \right),$ 渐近线方程为 $y = 1 \pm \frac{3}{4}\left( {x + 2} \right).$ 该双曲线的图形见下图。

Put the equation $4y^{2} - 9x^{2} + 16y + 18x - 29 = 0$ into standard form and graph the resulting hyperbola. What are the equations of the asymptotes?

将方程 $4y^{2} - 9x^{2} + 16y + 18x - 29 = 0$ 化为标准形式，并画出所得双曲线。其渐近线方程是什么？

Hyperbolas also have interesting reflective properties. A ray directed toward one focus of a hyperbola is reflected by a hyperbolic mirror toward the other focus. This concept is illustrated in the following figure.

双曲线还有有趣的反射性质。射向双曲线一个焦点的光线，经双曲镜面反射后会指向另一个焦点。下图说明了这一概念。

This property of the hyperbola has important applications. It is used in radio direction finding (since the difference in signals from two towers is constant along hyperbolas), and in the construction of mirrors inside telescopes (to reflect light coming from the parabolic mirror to the eyepiece). Another interesting fact about hyperbolas is that for a comet entering the solar system, if the speed is great enough to escape the Sun's gravitational pull, then the path that the comet takes as it passes through the solar system is hyperbolic.

双曲线的这一性质有重要应用。它用于无线电定向（因为来自两座塔的信号之差沿双曲线为常数），也用于望远镜内部反射镜的制造（将来自抛物镜面的光反射到目镜）。关于双曲线另一个有趣的事实是：对于进入太阳系的彗星，若其速度大到足以逃脱太阳的引力，则它在穿过太阳系时所走的路径是双曲的。

Eccentricity and Directrix 离心率与准线

An alternative way to describe a conic section involves the directrices, the foci, and a new property called eccentricity. We will see that the value of the eccentricity of a conic section can uniquely define that conic.

描述圆锥曲线的另一种方式涉及准线、焦点以及一个称为离心率的新性质。我们将看到，圆锥曲线的离心率取值可以唯一地确定该圆锥曲线。

The eccentricity *e* of a conic section is defined to be the distance from any point on the conic section to its focus, divided by the perpendicular distance from that point to the nearest directrix. This value is constant for any conic section, and can define the conic section as well:

圆锥曲线的离心率 *e* 定义为：圆锥曲线上任意一点到其焦点的距离，除以该点到最近准线的垂直距离。此值对任何圆锥曲线都是常数，并且也能用来定义该圆锥曲线：

1. If $e = 1,$ the conic is a parabola.

1. 若 $e = 1,$ 该圆锥曲线为抛物线。

2. If $e < 1,$ it is an ellipse.

2. 若 $e < 1,$ 它为椭圆。

3. If $e > 1,$ it is a hyperbola.

3. 若 $e > 1,$ 它为双曲线。

The eccentricity of a circle is zero. The directrix of a conic section is the line that, together with the point known as the focus, serves to define a conic section. Hyperbolas and noncircular ellipses have two foci and two associated directrices. Parabolas have one focus and one directrix.

圆的离心率为零。圆锥曲线的准线是一条直线，它与称为焦点的点一起用来定义圆锥曲线。双曲线和非圆的椭圆有两个焦点和两条相应准线。抛物线有一个焦点和一条准线。

The three conic sections with their directrices appear in the following figure.

三条圆锥曲线及其准线如下图所示。

Recall from the definition of a parabola that the distance from any point on the parabola to the focus is equal to the distance from that same point to the directrix. Therefore, by definition, the eccentricity of a parabola must be 1. The equations of the directrices of a horizontal ellipse are $x = \text{±}\frac{a^{2}}{c}.$ The right vertex of the ellipse is located at $\left( {a,0} \right)$ and the right focus is $\left( {c,0} \right).$ Therefore the distance from the vertex to the focus is $a - c$ and the distance from the vertex to the right directrix is $\frac{a^{2}}{c} - a.$ This gives the eccentricity as

由抛物线的定义回顾可知，抛物线上任意一点到焦点的距离等于该点到准线的距离。因此，按定义，抛物线的离心率必为 1。水平椭圆的准线方程为 $x = \text{±}\frac{a^{2}}{c}.$ 椭圆的右顶点位于 $\left( {a,0} \right),$ 右焦点位于 $\left( {c,0} \right).$ 于是顶点到焦点的距离为 $a - c,$ 顶点到右准线的距离为 $\frac{a^{2}}{c} - a.$ 据此得到离心率

$$e = \frac{a - c}{\frac{a^{2}}{c} - a} = \frac{c\left( {a - c} \right)}{a^{2} - ac} = \frac{c\left( {a - c} \right)}{a\left( {a - c} \right)} = \frac{c}{a}.$$

Since $c < a,$ this step proves that the eccentricity of an ellipse is less than 1. The directrices of a horizontal hyperbola are also located at $x = \text{±}\frac{a^{2}}{c},$ and a similar calculation shows that the eccentricity of a hyperbola is also $e = \frac{c}{a}.$ However in this case we have $c > a,$ so the eccentricity of a hyperbola is greater than 1.

由于 $c < a,$ 这一步证明了椭圆的离心率小于 1。水平双曲线的准线也位于 $x = \text{±}\frac{a^{2}}{c},$ 类似计算表明双曲线的离心率同样为 $e = \frac{c}{a}.$ 但此时有 $c > a,$ 故双曲线的离心率大于 1。

Determining Eccentricity of a Conic Section 确定圆锥曲线的离心率

Determine the eccentricity of the ellipse described by the equation

求由下列方程所描述的椭圆的离心率：

$$\frac{\left( {x - 3} \right)^{2}}{16} + \frac{\left( {y + 2} \right)^{2}}{25} = 1.$$

Solution 解

From the equation we see that $a = 5$ and $b = 4.$ The value of *c* can be calculated using the equation $a^{2} = b^{2} + c^{2}$ for an ellipse. Substituting the values of *a* and *b* and solving for *c* gives $c = 3.$ Therefore the eccentricity of the ellipse is $e = \frac{c}{a} = \frac{3}{5} = 0.6.$

由方程可见 $a = 5,$ $b = 4.$ 对于椭圆，可用关系式 $a^{2} = b^{2} + c^{2}$ 计算 *c* 的值。代入 *a* 与 *b* 的值并解出 *c* 得 $c = 3.$ 因此该椭圆的离心率为 $e = \frac{c}{a} = \frac{3}{5} = 0.6.$

Determine the eccentricity of the hyperbola described by the equation

求由下列方程所描述的双曲线的离心率：

$$\frac{\left( {y - 3} \right)^{2}}{49} - \frac{\left( {x + 2} \right)^{2}}{25} = 1.$$

Polar Equations of Conic Sections 圆锥曲线的极坐标方程

Sometimes it is useful to write or identify the equation of a conic section in polar form. To do this, we need the concept of the focal parameter. The focal parameter of a conic section *p* is defined as the distance from a focus to the nearest directrix. The following table gives the focal parameters for the different types of conics, where *a* is the length of the semi-major axis (i.e., half the length of the major axis), *c* is the distance from the origin to the focus, and *e* is the eccentricity. In the case of a parabola, *a* represents the distance from the vertex to the focus.

有时用极坐标形式写出或识别圆锥曲线的方程很有用。为此，我们需要焦点参数的概念。圆锥曲线的焦点参数 p 定义为从一个焦点到最近准线的距离。下表给出各类圆锥曲线的焦点参数，其中 a 为半长轴（即长轴长度的一半）之长，c 为从原点到焦点的距离，e 为离心率。对于抛物线，a 表示顶点到焦点的距离。

| Conic | *e* | *p* |

| 圆锥曲线 | *e*（离心率） | *p*（焦点参数） |

|-----------|-------------|-------------------------------------------------------------------|

| Ellipse | $0 < e < 1$ | $\frac{a^{2} - c^{2}}{c} = \frac{a\left( {1 - e^{2}} \right)}{e}$ |

| 椭圆 | $0 < e < 1$ | $\frac{a^{2} - c^{2}}{c} = \frac{a\left( {1 - e^{2}} \right)}{e}$ |

| Parabola | $e = 1$ | $2a$ |

| 抛物线 | $e = 1$ | $2a$ |

| Hyperbola | $e > 1$ | $\frac{c^{2} - a^{2}}{c} = \frac{a\left( {e^{2} - 1} \right)}{e}$ |

| 双曲线 | $e > 1$ | $\frac{c^{2} - a^{2}}{c} = \frac{a\left( {e^{2} - 1} \right)}{e}$ |

Table 1.1 Eccentricities and Focal Parameters of the Conic Sections

表 1.1 圆锥曲线的离心率与焦点参数

Using the definitions of the focal parameter and eccentricity of the conic section, we can derive an equation for any conic section in polar coordinates. In particular, we assume that one of the foci of a given conic section lies at the pole. Then using the definition of the various conic sections in terms of distances, it is possible to prove the following theorem.

利用圆锥曲线的焦点参数与离心率的定义，可以推导出任意圆锥曲线在极坐标下的方程。特别地，我们假设给定圆锥曲线的一个焦点位于极点。然后，利用各类圆锥曲线关于距离的定义，即可证明下面的定理。

Polar Equation of Conic Sections 圆锥曲线的极坐标方程

The polar equation of a conic section with focal parameter *p* is given by

具有焦点参数 p 的圆锥曲线的极坐标方程由下式给出

$$r = \frac{ep}{1 \pm e\ \text{cos}\ \theta}\ \text{or}\ r = \frac{ep}{1 \pm e\ \text{sin}\ \theta}.$$

In the equation on the left, the major axis of the conic section is horizontal, and in the equation on the right, the major axis is vertical. To work with a conic section written in polar form, first make the constant term in the denominator equal to 1. This can be done by dividing both the numerator and the denominator of the fraction by the constant that appears in front of the plus or minus in the denominator. Then the coefficient of the sine or cosine in the denominator is the eccentricity. This value identifies the conic. If cosine appears in the denominator, then the conic is horizontal. If sine appears, then the conic is vertical. If both appear then the axes are rotated. The center of the conic is not necessarily at the origin. The center is at the origin only if the conic is a circle (i.e., $e = 0).$

在左边方程中，圆锥曲线的长轴是水平的；在右边方程中，长轴是垂直的。处理极坐标形式的圆锥曲线时，首先令分母中的常数项等于 1。为此，可将分数的分子与分母同除以分母中出现在加减号前的那个常数。于是分母中正弦或余弦的系数就是离心率，据此可判定圆锥曲线的类型。若分母中出现余弦，则曲线为水平的；若出现正弦，则为垂直的；若两者都出现，则坐标轴被旋转。圆锥曲线的中心不一定在原点，只有当它是圆（即 $e = 0)$）时，中心才在原点。

Graphing a Conic Section in Polar Coordinates 在极坐标中绘制圆锥曲线图形

Identify and create a graph of the conic section described by the equation

识别并绘制由下列方程所描述的圆锥曲线图形

$$r = \frac{3}{1 + 2\ \text{cos}\ \theta}.$$

Solution 解答

The constant term in the denominator is 1, so the eccentricity of the conic is 2. This is a hyperbola. The focal parameter *p* can be calculated by using the equation $ep = 3.$ Since $e = 2,$ this gives $p = \frac{3}{2}.$ The cosine function appears in the denominator, so the hyperbola is horizontal. Pick a few values for $\theta$ and create a table of values. Then we can graph the hyperbola (Figure 1.55).

分母中的常数项为 1，故该圆锥曲线的离心率为 2，这是一条双曲线。利用方程 $ep = 3.$ 可求得焦点参数 p；由于 $e = 2,$ 得 $p = \frac{3}{2}.$ 分母中出现余弦函数，因此该双曲线是水平的。选取若干 $\theta$ 值列出数值表，即可画出该双曲线（图 1.55）。

| $\theta$ | $r$ | $\theta$ | $r$ |

|------------------|------------------------------------------|------------------|------------------------------------------|

| 0 | 1 | $\pi$ | −3 |

| $\frac{\pi}{4}$ | $\frac{3}{1 + \sqrt{2}} \approx 1.2426$ | $\frac{5\pi}{4}$ | $\frac{3}{1 - \sqrt{2}} \approx -7.2426$ |

| $\frac{\pi}{2}$ | 3 | $\frac{3\pi}{2}$ | 3 |

| $\frac{3\pi}{4}$ | $\frac{3}{1 - \sqrt{2}} \approx -7.2426$ | $\frac{7\pi}{4}$ | $\frac{3}{1 + \sqrt{2}} \approx 1.2426$ |

Identify and create a graph of the conic section described by the equation

识别并绘制由下列方程所描述的圆锥曲线图形

$$r = \frac{4}{1 - 0.8\ \text{sin}\ \theta}.$$

General Equations of Degree Two 二次一般方程

A general equation of degree two can be written in the form

二次一般方程可以写成如下形式

$$Ax^{2} + Bxy + Cy^{2} + Dx + Ey + F = 0.$$

The graph of an equation of this form is a conic section. If $B \neq 0$ then the coordinate axes are rotated. To identify the conic section, we use the discriminant of the conic section $4AC - B^{2}.$ One of the following cases must be true:

这种形式的方程的图形是圆锥曲线。若 $B \neq 0$，则坐标轴被旋转。为识别圆锥曲线，我们使用其判别式 $4AC - B^{2}.$ 必然出现下列情形之一：

1. $4AC - B^{2} > 0.$ If so, the graph is an ellipse.

1. $4AC - B^{2} > 0.$ 此时图形为椭圆。

2. $4AC - B^{2} = 0.$ If so, the graph is a parabola.

2. $4AC - B^{2} = 0.$ 此时图形为抛物线。

3. $4AC - B^{2} < 0.$ If so, the graph is a hyperbola.

3. $4AC - B^{2} < 0.$ 此时图形为双曲线。

The simplest example of a second-degree equation involving a cross term is $xy = 1.$ This equation can be solved for *y* to obtain $y = \frac{1}{x}.$ The graph of this function is called a *rectangular hyperbola* as shown.

含交叉项的最简单的二次方程例子是 $xy = 1.$ 解出 y 可得 $y = \frac{1}{x}.$ 该函数的图形称为直角双曲线，如下图所示。

The asymptotes of this hyperbola are the *x* and *y* coordinate axes. To determine the angle $\theta$ of rotation of the conic section, we use the formula $\text{cot}\ 2\theta = \frac{A - C}{B}.$ In this case $A = C = 0$ and $B = 1,$ so $\text{cot}\ 2\theta = {{(0 - 0)}\text{/}{1 = 0}}$ and $\theta = 45\text{°}.$ The method for graphing a conic section with rotated axes involves determining the coefficients of the conic in the rotated coordinate system. The new coefficients are labeled $A^{\prime},B^{\prime},C^{\prime},D^{\prime},E^{\prime},\ \text{and}\ F^{\prime},$ and are given by the formulas

该双曲线的渐近线是 x 轴与 y 轴。为确定圆锥曲线的旋转角 $\theta$，使用公式 $\text{cot}\ 2\theta = \frac{A - C}{B}.$ 此处 $A = C = 0$、$B = 1$，故 $\text{cot}\ 2\theta = {{(0 - 0)}\text{/}{1 = 0}}$ 且 $\theta = 45\text{°}.$ 绘制旋转坐标轴下圆锥曲线图形的方法，是先确定曲线在旋转坐标系中的系数。这些新系数记为 $A^{\prime},B^{\prime},C^{\prime},D^{\prime},E^{\prime},\ \text{and}\ F^{\prime},$ 由下列公式给出

$$\begin{array}{rll} A^{\prime} & = & {A\ \text{cos}^{2}\theta + B\ \text{cos}\ \theta\ \text{sin}\ \theta + C\ \text{sin}^{2}\theta} \\ B^{\prime} & = & 0 \\ C^{\prime} & = & {A\ \text{sin}^{2}\theta - B\ \text{sin}\ \theta\ \text{cos}\ \theta + C\ \text{cos}^{2}\theta} \\ D^{\prime} & = & {D\ \text{cos}\ \theta + E\ \text{sin}\ \theta} \\ E^{\prime} & = & {\text{−}D\ \text{sin}\ \theta + E\ \text{cos}\ \theta} \\ F^{\prime} & = & {F.} \end{array}$$

旋转后的各系数由下列公式给出： $$\begin{array}{rll} A^{\prime} & = & {A\ \text{cos}^{2}\theta + B\ \text{cos}\ \theta\ \text{sin}\ \theta + C\ \text{sin}^{2}\theta} \\ B^{\prime} & = & 0 \\ C^{\prime} & = & {A\ \text{sin}^{2}\theta - B\ \text{sin}\ \theta\ \text{cos}\ \theta + C\ \text{cos}^{2}\theta} \\ D^{\prime} & = & {D\ \text{cos}\ \theta + E\ \text{sin}\ \theta} \\ E^{\prime} & = & {\text{−}D\ \text{sin}\ \theta + E\ \text{cos}\ \theta} \\ F^{\prime} & = & {F.} \end{array}$$

The procedure for graphing a rotated conic is the following:

绘制旋转圆锥曲线的步骤如下：

1. Identify the conic section using the discriminant $4AC - B^{2}.$

1. 用判别式 $4AC - B^{2}$ 识别圆锥曲线。

2. Determine $\theta$ using the formula $\text{cot}\ 2\theta = \frac{A - C}{B}.$

2. 用公式 $\text{cot}\ 2\theta = \frac{A - C}{B}$ 确定 $\theta$。

3. Calculate $A^{\prime},B^{\prime},C^{\prime},D^{\prime},E^{\prime},\ \text{and}\ F^{\prime}.$

3. 计算 $A^{\prime},B^{\prime},C^{\prime},D^{\prime},E^{\prime},\ \text{and}\ F^{\prime}.$

4. Rewrite the original equation using $A^{\prime},B^{\prime},C^{\prime},D^{\prime},E^{\prime},\ \text{and}\ F^{\prime}.$

4. 用 $A^{\prime},B^{\prime},C^{\prime},D^{\prime},E^{\prime},\ \text{and}\ F^{\prime}$ 重写原方程。

5. Draw a graph using the rotated equation.

5. 用旋转后的方程作图。

Identifying a Rotated Conic 识别旋转后的圆锥曲线

Identify the conic and calculate the angle of rotation of axes for the curve described by the equation

识别由下列方程所描述的曲线属于哪类圆锥曲线，并计算坐标轴的旋转角

$$13x^{2} - 6\sqrt{3}xy + 7y^{2} - 256 = 0.$$

Solution 解答

In this equation, $A = 13,B = -6\sqrt{3},C = 7,D = 0,E = 0,$ and $F = -256.$ The discriminant of this equation is $4AC - B^{2} = 4(13)(7) - \left( {-6\sqrt{3}} \right)^{2} = 364 - 108 = 256.$ Therefore this conic is an ellipse. To calculate the angle of rotation of the axes, use $\text{cot}\ 2\theta = \frac{A - C}{B}.$ This gives

在此方程中，$A = 13,B = -6\sqrt{3},C = 7,D = 0,E = 0,$ 且 $F = -256.$ 其判别式为 $4AC - B^{2} = 4(13)(7) - \left( {-6\sqrt{3}} \right)^{2} = 364 - 108 = 256.$ 因此该圆锥曲线是椭圆。为计算坐标轴的旋转角，使用 $\text{cot}\ 2\theta = \frac{A - C}{B}.$ 于是得到

$$\begin{array}{cl} {\text{cot}\ 2\theta} & {= \frac{A - C}{B}} \\ & {= \frac{13 - 7}{-6\sqrt{3}}} \\ & {= - \frac{\sqrt{3}}{3}.} \end{array}$$

Therefore $2\theta = 120^{\text{o}}$ and $\theta = 60^{\text{o}},$ which is the angle of the rotation of the axes.

因此 $2\theta = 120^{\text{o}}$、$\theta = 60^{\text{o}}$，这就是坐标轴的旋转角。

To determine the rotated coefficients, use the formulas given above:

为确定旋转后的系数，使用上面给出的公式：

$$\begin{array}{cll} A^{\prime} & = & {A\ \text{cos}^{2}\theta + B\ \text{cos}\ \theta\ \text{sin}\ \theta + C\ \text{sin}^{2}\theta} \\ & = & {13\text{cos}^{2}60 + \left( {-6\sqrt{3}} \right)\ \text{cos}\ 60\ \text{sin}\ 60 + 7\text{sin}^{2}60} \\ & = & {13\left( \frac{1}{2} \right)^{2} - 6\sqrt{3}\left( \frac{1}{2} \right)\left( \frac{\sqrt{3}}{2} \right) + 7\left( \frac{\sqrt{3}}{2} \right)^{2}} \\ & = & 4, \\ B^{\prime} & = & {0,} \\ C^{\prime} & = & {A\ \text{sin}^{2}\theta - B\ \text{sin}\ \theta\ \text{cos}\ \theta + C\ \text{cos}^{2}\theta} \\ & = & {13\text{sin}^{2}60 + \left( {-6\sqrt{3}} \right)\ \text{sin}\ 60\ \text{cos}\ 60 = 7\text{cos}^{2}60} \\ & = & {\left( \frac{\sqrt{3}}{2} \right)^{2} + 6\sqrt{3}\left( \frac{\sqrt{3}}{2} \right)\left( \frac{1}{2} \right) + 7\left( \frac{1}{2} \right)^{2}} \\ & = & 16, \\ D^{\prime} & = & {D\ \text{cos}\ \theta + E\ \text{sin}\ \theta} \\ & = & {(0)\ \text{cos}\ 60 + (0)\ \text{sin}\ 60} \\ & = & 0, \\ E^{\prime} & = & {\text{−}D\ \text{sin}\ \theta + E\ \text{cos}\ \theta} \\ & = & {\text{−}(0)\ \text{sin}\ 60 + (0)\ \text{cos}\ 60} \\ & = & 0, \\ F^{\prime} & = & F \\ & = & -256. \end{array}$$

旋转后的各系数计算如下： $$\begin{array}{cll} A^{\prime} & = & {A\ \text{cos}^{2}\theta + B\ \text{cos}\ \theta\ \text{sin}\ \theta + C\ \text{sin}^{2}\theta} \\ & = & {13\text{cos}^{2}60 + \left( {-6\sqrt{3}} \right)\ \text{cos}\ 60\ \text{sin}\ 60 + 7\text{sin}^{2}60} \\ & = & {13\left( \frac{1}{2} \right)^{2} - 6\sqrt{3}\left( \frac{1}{2} \right)\left( \frac{\sqrt{3}}{2} \right) + 7\left( \frac{\sqrt{3}}{2} \right)^{2}} \\ & = & 4, \\ B^{\prime} & = & {0,} \\ C^{\prime} & = & {A\ \text{sin}^{2}\theta - B\ \text{sin}\ \theta\ \text{cos}\ \theta + C\ \text{cos}^{2}\theta} \\ & = & {13\text{sin}^{2}60 + \left( {-6\sqrt{3}} \right)\ \text{sin}\ 60\ \text{cos}\ 60 = 7\text{cos}^{2}60} \\ & = & {\left( \frac{\sqrt{3}}{2} \right)^{2} + 6\sqrt{3}\left( \frac{\sqrt{3}}{2} \right)\left( \frac{1}{2} \right) + 7\left( \frac{1}{2} \right)^{2}} \\ & = & 16, \\ D^{\prime} & = & {D\ \text{cos}\ \theta + E\ \text{sin}\ \theta} \\ & = & {(0)\ \text{cos}\ 60 + (0)\ \text{sin}\ 60} \\ & = & 0, \\ E^{\prime} & = & {\text{−}D\ \text{sin}\ \theta + E\ \text{cos}\ \theta} \\ & = & {\text{−}(0)\ \text{sin}\ 60 + (0)\ \text{cos}\ 60} \\ & = & 0, \\ F^{\prime} & = & F \\ & = & -256. \end{array}$$

The equation of the conic in the rotated coordinate system becomes

该圆锥曲线在旋转坐标系中的方程变为

$$\begin{array}{rll} & & \\ {4\left( x^{\prime} \right)^{2} + 16\left( y^{\prime} \right)^{2}} & = & 256 \\ {\frac{\left( x^{\prime} \right)^{2}}{64} + \frac{\left( y^{\prime} \right)^{2}}{16}} & = & 1. \end{array}$$

A graph of this conic section appears as follows.

该圆锥曲线的图形如下所示。

Identify the conic and calculate the angle of rotation of axes for the curve described by the equation

识别由下列方程所描述的曲线属于哪类圆锥曲线，并计算坐标轴的旋转角

$$3x^{2} + 5xy - 2y^{2} - 125 = 0.$$

Section 1.5 Exercises 1.5 节习题

For the following exercises, determine the equation of the parabola using the information given.

对于以下习题，根据所给信息确定抛物线的方程。

255.

Focus $\left( {4,0} \right)$ and directrix $x = -4$

焦点 $\left( {4,0} \right)$，准线 $x = -4$

256\.

Focus $\left( {0,-3} \right)$ and directrix $y = 3$

焦点 $\left( {0,-3} \right)$，准线 $y = 3$

257.

Focus $\left( {0,0.5} \right)$ and directrix $y = -0.5$

焦点 $\left( {0,0.5} \right)$，准线 $y = -0.5$

258\.

Focus $\left( {2,\ 3} \right)$ and directrix $x = -2$

焦点 $\left( {2,\ 3} \right)$，准线 $x = -2$

259.

Focus $\left( {0,2} \right)$ and directrix $y = 4$

焦点 $\left( {0,2} \right)$，准线 $y = 4$

260\.

Focus $\left( {-1,4} \right)$ and directrix $x = 5$

焦点 $\left( {-1,4} \right)$，准线 $x = 5$

261.

Focus $\left( {-3,5} \right)$ and directrix $y = 1$

焦点 $\left( {-3,5} \right)$，准线 $y = 1$

262\.

Focus $\left( {\frac{5}{2},-4} \right)$ and directrix $x = \frac{7}{2}$

焦点 $\left( {\frac{5}{2},-4} \right)$，准线 $x = \frac{7}{2}$

For the following exercises, determine the equation of the ellipse using the information given.

对于以下习题，根据所给信息确定椭圆的方程。

263.

Endpoints of major axis at $\left( {4,0} \right),\left( {-4,0} \right)$ and foci located at $\left( {2,0} \right),\left( {-2,0} \right)$

长轴端点位于 $\left( {4,0} \right),\left( {-4,0} \right)$，焦点位于 $\left( {2,0} \right),\left( {-2,0} \right)$

264\.

Endpoints of major axis at $\left( {0,5} \right),\left( {0,-5} \right)$ and foci located at $\left( {0,3} \right),\left( {0,-3} \right)$

长轴端点位于 $\left( {0,5} \right),\left( {0,-5} \right)$，焦点位于 $\left( {0,3} \right),\left( {0,-3} \right)$

265.

Endpoints of minor axis at $\left( {0,2} \right),\left( {0,-2} \right)$ and foci located at $\left( {3,0} \right),\left( {-3,0} \right)$

短轴端点位于 $\left( {0,2} \right),\left( {0,-2} \right)$，焦点位于 $\left( {3,0} \right),\left( {-3,0} \right)$

266\.

Endpoints of major axis at $\left( {-3,3} \right),\left( {7,3} \right)$ and foci located at $\left( {-2,3} \right),\left( {6,3} \right)$

长轴端点位于 $\left( {-3,3} \right),\left( {7,3} \right)$，焦点位于 $\left( {-2,3} \right),\left( {6,3} \right)$

267.

Endpoints of major axis at $\left( {-3,5} \right),\left( {-3,-3} \right)$ and foci located at $\left( {-3,3} \right),\left( {-3,-1} \right)$

长轴端点位于 $\left( {-3,5} \right),\left( {-3,-3} \right)$，焦点位于 $\left( {-3,3} \right),\left( {-3,-1} \right)$

268\.

Endpoints of minor axis at $\left( {0,0} \right),\left( {0,4} \right)$ and foci located at $\left( {5,2} \right),\left( {-5,2} \right)$

短轴端点位于 $\left( {0,0} \right),\left( {0,4} \right)$，焦点位于 $\left( {5,2} \right),\left( {-5,2} \right)$

269.

Foci located at $\left( {2,0} \right),\ \left( {-2,0} \right)$ and eccentricity of $\frac{1}{2}$

焦点位于 $\left( {2,0} \right),\ \left( {-2,0} \right)$，离心率为 $\frac{1}{2}$

270\.

Foci located at $\left( {0,-3} \right),\ \left( {0,3} \right)$ and eccentricity of $\frac{3}{4}$

焦点位于 $\left( {0,-3} \right),\ \left( {0,3} \right)$，离心率为 $\frac{3}{4}$

For the following exercises, determine the equation of the hyperbola using the information given.

对于以下习题，根据所给信息确定双曲线的方程。

271.

Vertices located at $\left( {5,0} \right),\left( {-5,0} \right)$ and foci located at $\left( {6,0} \right),\left( {-6,0} \right)$

顶点位于 $\left( {5,0} \right),\left( {-5,0} \right)$，焦点位于 $\left( {6,0} \right),\left( {-6,0} \right)$

272\.

Vertices located at $\left( {0,2} \right),\left( {0,-2} \right)$ and foci located at $\left( {0,3} \right),\left( {0,-3} \right)$

顶点位于 $\left( {0,2} \right),\left( {0,-2} \right)$，焦点位于 $\left( {0,3} \right),\left( {0,-3} \right)$

273.

Endpoints of the conjugate axis located at $\left( {0,3} \right),\left( {0,-3} \right)$ and foci located $\left( {4,0} \right),\left( {-4,0} \right)$

共轭轴端点位于 $\left( {0,3} \right),\left( {0,-3} \right)$，焦点位于 $\left( {4,0} \right),\left( {-4,0} \right)$

274\.

Vertices located at $\left( {0,1} \right),\left( {6,1} \right)$ and focus located at $\left( {8,1} \right)$

顶点位于 $\left( {0,1} \right),\left( {6,1} \right)$，焦点位于 $\left( {8,1} \right)$

275.

Vertices located at $\left( {-2,0} \right),\left( {-2,-4} \right)$ and focus located at $\left( {-2,-8} \right)$

顶点位于 $\left( {-2,0} \right),\left( {-2,-4} \right)$，焦点位于 $\left( {-2,-8} \right)$

276\.

Endpoints of the conjugate axis located at $\left( {3,2} \right),\left( {3,4} \right)$ and focus located at $\left( {3,7} \right)$

共轭轴端点位于 $\left( {3,2} \right),\left( {3,4} \right)$，焦点位于 $\left( {3,7} \right)$

277.

Foci located at $( - 6,0),(6,0)$ and eccentricity of 3

焦点位于 $( - 6,0),(6,0)$，离心率为 3

278\.

$\left( {0,10} \right),\left( {0,-10} \right)$ and eccentricity of 2.5

$\left( {0,10} \right),\left( {0,-10} \right)$，离心率为 2.5

For the following exercises, consider the following polar equations of conics. Determine the eccentricity and identify the conic.

对于以下习题，考虑下列圆锥曲线的极坐标方程。确定其离心率并判别圆锥曲线的类型。

279.

$r = \frac{-1}{1 + \text{cos}\ \theta}$

280\.

$r = \frac{8}{2 - \text{sin}\ \theta}$

281.

$r = \frac{5}{2 + \text{sin}\ \theta}$

282\.

$r = \frac{5}{-1 + 2\ \text{sin}\ \theta}$

283.

$r = \frac{3}{2 - 6\ \text{sin}\ \theta}$

284\.

$r = \frac{3}{-4 + 3\ \text{sin}\ \theta}$

For the following exercises, find a polar equation of the conic with focus at the origin and eccentricity and directrix as given.

对于以下习题，求以原点为焦点、给定离心率和准线的圆锥曲线的极坐标方程。

285.

$\text{Directrix:}\ x = 4;e = \frac{1}{5}$

286\.

$\text{Directrix:}\ x = -4;e = 5$

287.

$\text{Directrix: y} = 2;e = 2$

288\.

$\text{Directrix: y} = -2;e = \frac{1}{2}$

For the following exercises, sketch the graph of each conic.

对于以下习题，画出每个圆锥曲线的图形。

289.

$r = \frac{1}{1 + \text{sin}\ \theta}$

290\.

$r = \frac{1}{1 - \text{cos}\ \theta}$

291.

$r = \frac{4}{1 + \text{cos}\ \theta}$

292\.

$r = \frac{10}{5 + 4\ \text{sin}\ \theta}$

293.

$r = \frac{15}{3 - 2\ \text{cos}\ \theta}$

294\.

$r = \frac{32}{3 + 5\ \text{sin}\ \theta}$

295.

$r(2 + \text{sin}\ \theta) = 4$

296\.

$r = \frac{3}{2 + 6\ \text{sin}\ \theta}$

297.

$r = \frac{3}{-4 + 2\ \text{sin}\ \theta}$

298.

$\begin{array}{l}{\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1} \\\end{array}$

299.

$\frac{x^{2}}{4} + \frac{y^{2}}{16} = 1$

300\.

$4x^{2} + 9y^{2} = 36$

301.

$25x^{2} - 4y^{2} = 100$

302\.

$\frac{x^{2}}{16} - \frac{y^{2}}{9} = 1$

303.

$x^{2} = 12y$

304\.

$y^{2} = 20x$

305.

$12x = 5y^{2}$

For the following equations, determine which of the conic sections is described.

对于下列方程，判别它们分别描述的是哪一类圆锥曲线。

306\.

$xy = 4$

307.

$x^{2} + 4xy - 2y^{2} - 6 = 0$

308\.

$x^{2} + 2\sqrt{3}xy + 3y^{2} - 6 = 0$

309.

$x^{2} - xy + y^{2} - 2 = 0$

310\.

$34x^{2} - 24xy + 41y^{2} - 25 = 0$

311.

$52x^{2} - 72xy + 73y^{2} + 40x + 30y - 75 = 0$

312\.

The mirror in an automobile headlight has a parabolic cross section, with the lightbulb at the focus. On a schematic, the equation of the parabola is given as $x^{2} = 4y.$ At what coordinates should you place the lightbulb?

汽车前灯中的反射镜具有抛物线形横截面，灯泡位于焦点处。在一张示意图中，抛物线方程给出为 $x^{2} = 4y.$ 灯泡应放置在哪个坐标处？

313.

A satellite dish is shaped like a paraboloid of revolution. The receiver is to be located at the focus. If the dish is 12 feet across at its opening and 4 feet deep at its center, where should the receiver be placed?

卫星天线呈旋转抛物面形状。接收器应放置在焦点处。若天线开口处直径为 12 英尺，中心处深 4 英尺，接收器应放置在何处？

314\.

Consider the satellite dish of the preceding problem. If the dish is 8 feet across at the opening and 2 feet deep, where should we place the receiver?

考虑上一题的卫星天线。若天线开口处直径为 8 英尺，深 2 英尺，接收器应放置在何处？

315.

A searchlight is shaped like a paraboloid of revolution. A light source is located 1 foot from the base along the axis of symmetry. If the opening of the searchlight is 3 feet across, find the depth.

探照灯呈旋转抛物面形状。光源沿对称轴位于距底面 1 英尺处。若探照灯开口直径为 3 英尺，求其深度。

316\.

Whispering galleries are rooms designed with elliptical ceilings. A person standing at one focus can whisper and be heard by a person standing at the other focus because all the sound waves that reach the ceiling are reflected to the other person. If a whispering gallery has a length of 120 feet and the foci are located 30 feet from the center, find the height of the ceiling at the center.

窃窃私语廊是顶部为椭圆形的房间。一个人站在一个焦点处低语，可被站在另一个焦点处的人听到，因为所有到达顶部的声波都会反射到另一人处。若窃窃私语廊长度为 120 英尺，焦点距中心 30 英尺，求中心处顶部的高度。

317.

A person is standing 8 feet from the nearest wall in a whispering gallery. If that person is at one focus and the other focus is 80 feet away, what is the length and the height at the center of the gallery?

在窃窃私语廊中，一个人距最近的墙壁 8 英尺站立。若该人位于一个焦点，另一焦点相距 80 英尺，求画廊的长度及中心处的高度。

For the following exercises, determine the polar equation form of the orbit given the length of the major axis and eccentricity for the orbits of the comets or planets. Distance is given in astronomical units (AU).

对于以下习题，根据彗星或行星轨道的长轴长度与离心率，确定其轨道的极坐标方程形式。距离以天文单位（AU）给出。

318\.

Halley’s Comet: length of major axis = 35.88, eccentricity = 0.967

哈雷彗星：长轴长度 = 35.88，离心率 = 0.967

319.

Hale-Bopp Comet: length of major axis = 525.91, eccentricity = 0.995

海尔-波普彗星：长轴长度 = 525.91，离心率 = 0.995

320\.

Mars: length of major axis = 3.049, eccentricity = 0.0934

火星：长轴长度 = 3.049，离心率 = 0.0934

321.

Jupiter: length of major axis = 10.408, eccentricity = 0.0484

木星：长轴长度 = 10.408，离心率 = 0.0484

Key Terms 关键术语

angular coordinate

角坐标

$\theta$ the angle formed by a line segment connecting the origin to a point in the polar coordinate system with the positive radial (*x*) axis, measured counterclockwise

$\theta$ 是由连接原点与极坐标系中某点的线段，与正径向（*x*）轴之间所成的角，按逆时针方向度量。

cardioid

心形线

a plane curve traced by a point on the perimeter of a circle that is rolling around a fixed circle of the same radius; the equation of a cardioid is $r = a\left( {1 + \text{sin}\ \theta} \right)$ or $r = a\left( {1 + \text{cos}\ \theta} \right)$

一种平面曲线，由在一个固定圆（半径相同）外围滚动的圆周上一点所描绘的轨迹；心形线的方程为 $r = a\left( {1 + \text{sin}\ \theta} \right)$ 或 $r = a\left( {1 + \text{cos}\ \theta} \right)$。

conic section

圆锥曲线

a conic section is any curve formed by the intersection of a plane with a cone of two nappes

圆锥曲线是由一个平面与双叶圆锥相交而形成的任意曲线。

cusp

尖点

a pointed end or part where two curves meet

两条曲线相交处的尖端点或部分。

cycloid

摆线

the curve traced by a point on the rim of a circular wheel as the wheel rolls along a straight line without slippage

当圆轮沿直线无滑动地滚动时，轮缘上一点所描绘的曲线。

directrix

准线

a directrix (plural: directrices) is a line used to construct and define a conic section; a parabola has one directrix; ellipses and hyperbolas have two

准线（复数：directrices）是用于构造和定义圆锥曲线的一条直线；抛物线有一条准线；椭圆和双曲线各有两条准线。

discriminant

判别式

the value $4AC - B^{2},$ which is used to identify a conic when the equation contains a term involving $xy,$ is called a discriminant

当方程中含有涉及 $xy$ 的项时，用于判别圆锥曲线类型的值 $4AC - B^{2},$ 称为判别式。

eccentricity

离心率

the eccentricity is defined as the distance from any point on the conic section to its focus divided by the perpendicular distance from that point to the nearest directrix

离心率定义为圆锥曲线上任意一点到其焦点的距离，除以该点到最近准线的垂直距离。

focal parameter

焦参数

the focal parameter is the distance from a focus of a conic section to the nearest directrix

焦参数是圆锥曲线的一个焦点到最近准线的距离。

focus

焦点

a focus (plural: foci) is a point used to construct and define a conic section; a parabola has one focus; an ellipse and a hyperbola have two

焦点（复数：foci）是用于构造和定义圆锥曲线的一个点；抛物线有一个焦点；椭圆和双曲线各有两个焦点。

general form

一般式

an equation of a conic section written as a general second-degree equation

以一般二次方程形式写出的圆锥曲线方程。

limaçon

蚶线

the graph of the equation $r = a + b\ \text{sin}\ \theta$ or $r = a + b\ \text{cos}\ \theta.$ If $a = b$ then the graph is a cardioid

方程 $r = a + b\ \text{sin}\ \theta$ 或 $r = a + b\ \text{cos}\ \theta$ 的图形。若 $a = b$，则图形为心形线。

major axis

长轴

the major axis of a conic section passes through the vertex in the case of a parabola or through the two vertices in the case of an ellipse or hyperbola; it is also an axis of symmetry of the conic; also called the transverse axis

圆锥曲线的长轴在抛物线情形下通过顶点，在椭圆或双曲线情形下通过两个顶点；它也是该圆锥曲线的对称轴；也称横轴。

minor axis

短轴

the minor axis is perpendicular to the major axis and intersects the major axis at the center of the conic, or at the vertex in the case of the parabola; also called the conjugate axis

短轴垂直于长轴，并在圆锥曲线的中心（或在抛物线情形下的顶点处）与长轴相交；也称共轭轴。

nappe

半锥面

a nappe is one half of a double cone

半锥面是双锥的一半。

orientation

定向

the direction that a point moves on a graph as the parameter increases

参数增大时点在图形上移动的方向。

parameter

参数

an independent variable that both *x* and *y* depend on in a parametric curve; usually represented by the variable *t*

在参数曲线中 *x* 与 *y* 共同依赖的独立变量；通常用变量 *t* 表示。

parameterization of a curve

曲线的参数化

rewriting the equation of a curve defined by a function $y = f(x)$ as parametric equations

将函数 $y = f(x)$ 所定义曲线的方程改写为参数方程。

parametric curve

参数曲线

the graph of the parametric equations $x(t)$ and $y(t)$ over an interval $a \leq t \leq b$ combined with the equations

参数方程 $x(t)$ 与 $y(t)$ 在区间 $a \leq t \leq b$ 上的图形与这些方程共同构成。

parametric equations

参数方程

the equations $x = x(t)$ and $y = y(t)$ that define a parametric curve

定义参数曲线的方程 $x = x(t)$ 与 $y = y(t)$。

polar axis

极轴

the horizontal axis in the polar coordinate system corresponding to $r \geq 0$

极坐标系中对应于 $r \geq 0$ 的水平轴。

polar coordinate system

极坐标系

a system for locating points in the plane. The coordinates are $r,$ the radial coordinate, and $\theta,$ the angular coordinate

平面上确定点位置的一种坐标系。其坐标为 $r$（径向坐标）与 $\theta$（角坐标）。

polar equation

极坐标方程

an equation or function relating the radial coordinate to the angular coordinate in the polar coordinate system

极坐标系中把径向坐标与角坐标联系起来的方程或函数。

pole

极点

the central point of the polar coordinate system, equivalent to the origin of a Cartesian system

极坐标系的中心点，相当于直角坐标系的原点。

radial coordinate

极径

$r$ the coordinate in the polar coordinate system that measures the distance from a point in the plane to the pole

$r$ 是极坐标系中测量平面上一点到极点距离的坐标。

rose

玫瑰线

graph of the polar equation $r = a\ \text{cos}\ 2\theta$ or $r = a\ \text{sin}\ 2\theta$ for a positive constant *a*

极坐标方程 $r = a\ \text{cos}\ 2\theta$ 或 $r = a\ \text{sin}\ 2\theta$（*a* 为正常数）的图形。

space-filling curve

空间填充曲线

a curve that completely occupies a two-dimensional subset of the real plane

完全占据实平面中某个二维子集的曲线。

standard form

标准式

an equation of a conic section showing its properties, such as location of the vertex or lengths of major and minor axes

显示圆锥曲线性质的（如顶点位置或长、短轴长度）的方程形式。

vertex

顶点

a vertex is an extreme point on a conic section; a parabola has one vertex at its turning point. An ellipse has two vertices, one at each end of the major axis; a hyperbola has two vertices, one at the turning point of each branch

顶点是圆锥曲线上的极值点；抛物线在其转向点处有一个顶点。椭圆有两个顶点，分别位于长轴的两端；双曲线有两个顶点，分别位于每个分支的转向点。

Key Equations 关键公式

| | |

（表格分隔行）

|-----------------------------------------------------|---------------------------------------------------------------------------------------------------------------------------------------------------------------|

（表格分隔行）

| Derivative of parametric equations | $\frac{dy}{dx} = \frac{{dy}\text{/}{dt}}{{dx}\text{/}{dt}} = \frac{y^{\prime}(t)}{x^{\prime}(t)}$ |

| 参数方程的导数 | $\frac{dy}{dx} = \frac{{dy}\text{/}{dt}}{{dx}\text{/}{dt}} = \frac{y^{\prime}(t)}{x^{\prime}(t)}$ |

| Second-order derivative of parametric equations | $\frac{d^{2}y}{dx^{2}} = \frac{d}{dx}\left( \frac{dy}{dx} \right) = \frac{\left( {d\text{/}{dt}} \right)\left( {{dy}\text{/}{dx}} \right)}{{dx}\text{/}{dt}}$ |

| 参数方程的二阶导数 | $\frac{d^{2}y}{dx^{2}} = \frac{d}{dx}\left( \frac{dy}{dx} \right) = \frac{\left( {d\text{/}{dt}} \right)\left( {{dy}\text{/}{dx}} \right)}{{dx}\text{/}{dt}}$ |

| Area under a parametric curve | $A = {\int_{a}^{b}{y(t)x^{\prime}(t)\ dt}}$ |

| 参数曲线下的面积 | $A = {\int_{a}^{b}{y(t)x^{\prime}(t)\ dt}}$ |

| Arc length of a parametric curve | $s = {\int_{t_{1}}^{t_{2}}{\sqrt{\left( \frac{dx}{dt} \right)^{2} + \left( \frac{dy}{dt} \right)^{2}}dt}}$ |

| 参数曲线的弧长 | $s = {\int_{t_{1}}^{t_{2}}{\sqrt{\left( \frac{dx}{dt} \right)^{2} + \left( \frac{dy}{dt} \right)^{2}}dt}}$ |

| Surface area generated by a parametric curve | $S = 2\pi{\int_{a}^{b}{y(t)\sqrt{\left( {x^{\prime}(t)} \right)^{2} + \left( {y^{\prime}(t)} \right)^{2}}dt}}$ |

| 参数曲线生成的曲面面积 | $S = 2\pi{\int_{a}^{b}{y(t)\sqrt{\left( {x^{\prime}(t)} \right)^{2} + \left( {y^{\prime}(t)} \right)^{2}}dt}}$ |

| | |

（空行）

|-----------------------------------------------|------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------|

（表格分隔行）

| Area of a region bounded by a polar curve | $A = \frac{1}{2}{\int_{\alpha}^{\beta}{\left\lbrack {f(\theta)} \right\rbrack^{2}d\theta}} = \frac{1}{2}{\int_{\alpha}^{\beta}{r^{2}d\theta}}$ |

| 极坐标曲线所围区域的面积 | $A = \frac{1}{2}{\int_{\alpha}^{\beta}{\left\lbrack {f(\theta)} \right\rbrack^{2}d\theta}} = \frac{1}{2}{\int_{\alpha}^{\beta}{r^{2}d\theta}}$ |

| Arc length of a polar curve | $L = {\int_{\alpha}^{\beta}\sqrt{\left\lbrack {f(\theta)} \right\rbrack^{2} + \left\lbrack {f^{\prime}(\theta)} \right\rbrack^{2}}}d\theta = {\int_{\alpha}^{\beta}{\sqrt{r^{2} + \left( \frac{dr}{d\theta} \right)^{2}}d\theta}}$ |

| 极坐标曲线的弧长 | $L = {\int_{\alpha}^{\beta}\sqrt{\left\lbrack {f(\theta)} \right\rbrack^{2} + \left\lbrack {f^{\prime}(\theta)} \right\rbrack^{2}}}d\theta = {\int_{\alpha}^{\beta}{\sqrt{r^{2} + \left( \frac{dr}{d\theta} \right)^{2}}d\theta}}$ |

Key Concepts 关键概念

1.1 Parametric Equations 1.1 参数方程

Parametric equations provide a convenient way to describe a curve. A parameter can represent time or some other meaningful quantity.
It is often possible to eliminate the parameter in a parameterized curve to obtain a function or relation describing that curve.
There is always more than one way to parameterize a curve.
Parametric equations can describe complicated curves that are difficult or perhaps impossible to describe using rectangular coordinates.

参数方程提供了一种描述曲线的便捷方式。参数可以表示时间或其他有意义的量。
对于参数化曲线，通常可以消去参数，从而得到描述该曲线的函数或关系式。
参数化一条曲线总是不止一种方法。
参数方程能够描述复杂的曲线，这些曲线用直角坐标往往难以描述，甚至无法描述。

1.2 Calculus of Parametric Curves 1.2 参数曲线微积分

The derivative of the parametrically defined curve $x = x(t)$ and $y = y(t)$ can be calculated using the formula $\frac{dy}{dx} = \frac{y^{\prime}(t)}{x^{\prime}(t)}.$ Using the derivative, we can find the equation of a tangent line to a parametric curve.
The area between a parametric curve and the *x*-axis can be determined by using the formula $A = {\int_{t_{1}}^{t_{2}}{y(t)x^{\prime}(t)\ dt}}.$
The arc length of a parametric curve can be calculated by using the formula $s = {\int_{t_{1}}^{t_{2}}{\sqrt{\left( \frac{dx}{dt} \right)^{2} + \left( \frac{dy}{dt} \right)^{2}}dt}}.$
The surface area of a volume of revolution revolved around the *x*-axis is given by $S = 2\pi{\int_{a}^{b}{y(t)\sqrt{\left( {x^{\prime}(t)} \right)^{2} + \left( {y^{\prime}(t)} \right)^{2}}dt}}.$ If the curve is revolved around the *y*-axis, then the formula is $S = 2\pi{\int_{a}^{b}{x(t)\sqrt{\left( {x^{\prime}(t)} \right)^{2} + \left( {y^{\prime}(t)} \right)^{2}}dt}}.$

由参数定义的曲线 $x = x(t)$ 与 $y = y(t)$ 的导数可用公式 $\frac{dy}{dx} = \frac{y^{\prime}(t)}{x^{\prime}(t)}.$ 计算。利用该导数，可以求出参数曲线的切线方程。
参数曲线与 x 轴之间的面积可由公式 $A = {\int_{t_{1}}^{t_{2}}{y(t)x^{\prime}(t)\ dt}}.$ 确定。
参数曲线的弧长可由公式 $s = {\int_{t_{1}}^{t_{2}}{\sqrt{\left( \frac{dx}{dt} \right)^{2} + \left( \frac{dy}{dt} \right)^{2}}dt}}.$ 计算。
绕 x 轴旋转所得旋转体的表面积由 $S = 2\pi{\int_{a}^{b}{y(t)\sqrt{\left( {x^{\prime}(t)} \right)^{2} + \left( {y^{\prime}(t)} \right)^{2}}dt}}.$ 给出。若曲线绕 y 轴旋转，则公式为 $S = 2\pi{\int_{a}^{b}{x(t)\sqrt{\left( {x^{\prime}(t)} \right)^{2} + \left( {y^{\prime}(t)} \right)^{2}}dt}}.$

1.3 Polar Coordinates 1.3 极坐标

The polar coordinate system provides an alternative way to locate points in the plane.
Convert points between rectangular and polar coordinates using the formulas

极坐标系提供了一种在平面上确定点位置的替代方法。
利用以下公式在直角坐标与极坐标之间转换点：

$$x = r\ \text{cos}\ \theta\ \text{and}\ y = r\ \text{sin}\ \theta$$ and $$r = \sqrt{x^{2} + y^{2}}\ \text{and}\ \text{tan}\ \theta = \frac{y}{x}.$$

$$x = r\ \text{cos}\ \theta\ \text{and}\ y = r\ \text{sin}\ \theta$$ 以及 $$r = \sqrt{x^{2} + y^{2}}\ \text{and}\ \text{tan}\ \theta = \frac{y}{x}.$$

To sketch a polar curve from a given polar function, make a table of values and take advantage of periodic properties.
Use the conversion formulas to convert equations between rectangular and polar coordinates.
Identify symmetry in polar curves, which can occur through the pole, the horizontal axis, or the vertical axis.

要根据给定的极坐标函数绘制极坐标曲线，可列出数值表并利用其周期性。
利用转换公式在直角坐标与极坐标之间转换方程。
识别极坐标曲线中的对称性，它可能关于极点、水平轴或竖直轴出现。

1.4 Area and Arc Length in Polar Coordinates 1.4 极坐标下的面积与弧长

The area of a region in polar coordinates defined by the equation $r = f(\theta)$ with $\alpha \leq \theta \leq \beta$ is given by the integral $A = \frac{1}{2}{\int_{\alpha}^{\beta}\left\lbrack {f(\theta)} \right\rbrack}^{2}d\theta.$
To find the area between two curves in the polar coordinate system, first find the points of intersection, then subtract the corresponding areas.
The arc length of a polar curve defined by the equation $r = f(\theta)$ with $\alpha \leq \theta \leq \beta$ is given by the integral $L = {\int_{\alpha}^{\beta}{\sqrt{\left\lbrack {f(\theta)} \right\rbrack^{2} + \left\lbrack {f^{\prime}(\theta)} \right\rbrack^{2}}d\theta = {\int_{\alpha}^{\beta}{\sqrt{r^{2} + \left( \frac{dr}{d\theta} \right)^{2}}d\theta}}}}.$

由方程 $r = f(\theta)$、且 $\alpha \leq \theta \leq \beta$ 定义的极坐标区域的面积，由积分 $A = \frac{1}{2}{\int_{\alpha}^{\beta}\left\lbrack {f(\theta)} \right\rbrack}^{2}d\theta.$ 给出。
要求极坐标下两条曲线之间的面积，先求出交点，再相减对应的面积。
由方程 $r = f(\theta)$、且 $\alpha \leq \theta \leq \beta$ 定义的极坐标曲线的弧长，由积分 $L = {\int_{\alpha}^{\beta}{\sqrt{\left\lbrack {f(\theta)} \right\rbrack^{2} + \left\lbrack {f^{\prime}(\theta)} \right\rbrack^{2}}d\theta = {\int_{\alpha}^{\beta}{\sqrt{r^{2} + \left( \frac{dr}{d\theta} \right)^{2}}d\theta}}}}.$ 给出。

1.5 Conic Sections 1.5 圆锥曲线

The equation of a vertical parabola in standard form with given focus and directrix is $y = \frac{1}{4p}\left( {x - h} \right)^{2} + k$ where *p* is the distance from the vertex to the focus and $\left( {h,k} \right)$ are the coordinates of the vertex.
The equation of a horizontal ellipse in standard form is $\frac{\left( {x - h} \right)^{2}}{a^{2}} + \frac{\left( {y - k} \right)^{2}}{b^{2}} = 1$ where the center has coordinates $\left( {h,k} \right),$ the major axis has length 2*a,* the minor axis has length 2*b*, and the coordinates of the foci are $\left( {h \pm c,k} \right),$ where $c^{2} = a^{2} - b^{2}.$
The equation of a horizontal hyperbola in standard form is $\frac{\left( {x - h} \right)^{2}}{a^{2}} - \frac{\left( {y - k} \right)^{2}}{b^{2}} = 1$ where the center has coordinates $\left( {h,k} \right),$ the vertices are located at $\left( {h \pm a,k} \right),$ and the coordinates of the foci are $\left( {h \pm c,k} \right),$ where $c^{2} = a^{2} + b^{2}.$
The eccentricity of an ellipse is less than 1, the eccentricity of a parabola is equal to 1, and the eccentricity of a hyperbola is greater than 1. The eccentricity of a circle is 0.
The polar equation of a conic section with eccentricity *e* is $r = \frac{ep}{1 \pm e\ \text{cos}\ \theta}$ or $r = \frac{ep}{1 \pm e\ \text{sin}\ \theta},$ where *p* represents the focal parameter.
To identify a conic generated by the equation $Ax^{2} + Bxy + Cy^{2} + Dx + Ey + F = 0,$ first calculate the discriminant $D = 4AC - B^{2}.$ If $D > 0$ then the conic is an ellipse, if $D = 0$ then the conic is a parabola, and if $D < 0$ then the conic is a hyperbola.

给定焦点和准线的竖直抛物线标准形式的方程为 $y = \frac{1}{4p}\left( {x - h} \right)^{2} + k$，其中 *p* 为顶点到焦点的距离，$\left( {h,k} \right)$ 为顶点坐标。
水平椭圆的标准形式的方程为 $\frac{\left( {x - h} \right)^{2}}{a^{2}} + \frac{\left( {y - k} \right)^{2}}{b^{2}} = 1$，其中中心坐标为 $\left( {h,k} \right)$，长轴长度为 2*a*，短轴长度为 2*b*，焦点的坐标为 $\left( {h \pm c,k} \right)$，其中 $c^{2} = a^{2} - b^{2}.$
水平双曲线的标准形式的方程为 $\frac{\left( {x - h} \right)^{2}}{a^{2}} - \frac{\left( {y - k} \right)^{2}}{b^{2}} = 1$，其中中心坐标为 $\left( {h,k} \right)$，顶点位于 $\left( {h \pm a,k} \right)$，焦点的坐标为 $\left( {h \pm c,k} \right)$，其中 $c^{2} = a^{2} + b^{2}.$
椭圆的离心率小于 1，抛物线的离心率等于 1，双曲线的离心率大于 1。圆的离心率为 0。
离心率为 *e* 的圆锥曲线的极坐标方程为 $r = \frac{ep}{1 \pm e\ \text{cos}\ \theta}$ 或 $r = \frac{ep}{1 \pm e\ \text{sin}\ \theta}$，其中 *p* 表示焦参数。
要判别由方程 $Ax^{2} + Bxy + Cy^{2} + Dx + Ey + F = 0$ 生成的圆锥曲线，先计算判别式 $D = 4AC - B^{2}$。若 $D > 0$，则为椭圆；若 $D = 0$，则为抛物线；若 $D < 0$，则为双曲线。

Review Exercises 复习题

*True or False?* Justify your answer with a proof or a counterexample.

*判断正误？* 用证明或反例说明你的答案。

322\.

322.

The rectangular coordinates of the point $\left( {4,\frac{5\pi}{6}} \right)$ are $\left( {2\sqrt{3},-2} \right).$

点 $\left( {4,\frac{5\pi}{6}} \right)$ 的直角坐标为 $\left( {2\sqrt{3},-2} \right)$。

323.

The equations $x = \text{cosh}(3t),$ $y = 2\ \text{sinh}(3t)$ represent a hyperbola.

方程 $x = \text{cosh}(3t),$ $y = 2\ \text{sinh}(3t)$ 表示一条双曲线。

324\.

324.

The arc length of the spiral given by $r = \frac{\theta}{2}$ for $0 \leq \theta \leq 3\pi$ is $\frac{9}{4}\pi^{3}.$

由 $r = \frac{\theta}{2}$（$0 \leq \theta \leq 3\pi$）给出的螺线的弧长为 $\frac{9}{4}\pi^{3}$。

325.

Given $x = f(t)$ and $y = g(t),$ if $\frac{dx}{dy} = \frac{dy}{dx},$ then $f(t) = g(t) + \text{C,}$ where C is a constant.

给定 $x = f(t)$ 与 $y = g(t)$，若 $\frac{dx}{dy} = \frac{dy}{dx}$，则 $f(t) = g(t) + \text{C}$，其中 C 为常数。

For the following exercises, sketch the parametric curve and eliminate the parameter to find the Cartesian equation of the curve.

在以下习题中，画出参数曲线并消去参数，求出该曲线的直角坐标方程。

326\.

326.

$x = 1 + t,$ $y = t^{2} - 1,$ $-1 \leq t \leq 1$

327.

$x = e^{t},$ $y = 1 - e^{3t},$ $0 \leq t \leq 1$

328\.

328.

$x = \text{sin}\ \theta,$ $y = 1 - \text{csc}\ \theta,$ $0 \leq \theta \leq 2\pi$

329.

$x = 4\ \text{cos}\ \phi,$ $y = 1 - \text{sin}\ \phi,$ $0 \leq \phi \leq 2\pi$

For the following exercises, sketch the polar curve and determine what type of symmetry exists, if any.

在以下习题中，画出极坐标曲线并确定其具有何种对称性（若有）。

330\.

330.

$r = 4\ \text{sin}\left( \frac{\theta}{3} \right)$

331.

$r = 5\ \text{cos}\left( {5\theta} \right)$

For the following exercises, find the polar equation for the curve given as a Cartesian equation.

在以下习题中，求以直角坐标方程给出的曲线的极坐标方程。

332\.

332.

$x + y = 5$

333.

$y^{2} = 4 + x^{2}$

For the following exercises, find an equation of the tangent line to the given curve. Graph both the function and its tangent line.

在以下习题中，求给定曲线的切线方程。画出该函数及其切线。

334\.

334.

$x = \text{ln}(t),$ $y = t^{2} - 1,$ $t = 1$

335.

$r = 3 + \text{cos}\left( {2\theta} \right),$ $\theta = \frac{3\pi}{4}$

336\.

336.

Find $\frac{dy}{dx},$ $\frac{dx}{dy},$ and $\frac{d^{2}x}{dy^{2}}$ of $y = \left( {2 + e^{\text{−}t}} \right),$ $x = 1 - \text{sin}(t)$

求 $y = \left( {2 + e^{\text{−}t}} \right),$ $x = 1 - \text{sin}(t)$ 的 $\frac{dy}{dx},$ $\frac{dx}{dy},$ 以及 $\frac{d^{2}x}{dy^{2}}$。

For the following exercises, find the area of the region.

在以下习题中，求区域的面积。

337.

$x = t^{2},$ $y = \text{ln}(t),$ $0 \leq t \leq e$

338\.

338.

$r = 1 - \text{sin}\ \theta$ in the first quadrant

第一象限内的 $r = 1 - \text{sin}\ \theta$

For the following exercises, find the arc length of the curve over the given interval.

在以下习题中，求曲线在给定区间上的弧长。

339.

$x = 3t + 4,$ $y = 9t - 2,$ $0 \leq t \leq 3$

340\.

340.

$r = 6\ \text{cos}\ \theta,$ $0 \leq \theta \leq 2\pi.$ Check your answer by geometry.

$r = 6\ \text{cos}\ \theta,$ $0 \leq \theta \leq 2\pi$。用几何方法验证你的答案。

For the following exercises, find the Cartesian equation describing the given shapes.

在以下习题中，求描述给定图形的直角坐标方程。

341.

A parabola with focus $(2,-5)$ and directrix $x = 6$

焦点为 $(2,-5)$、准线为 $x = 6$ 的抛物线

342\.

342.

An ellipse with a major axis length of 10 and foci at $\left( {-7,2} \right)$ and $\left( {1,2} \right)$

长轴长为 10、焦点在 $\left( {-7,2} \right)$ 与 $\left( {1,2} \right)$ 的椭圆

343.

A hyperbola with vertices at $(–2,3)$ and $(-2,-5)$ and foci at $(-2,-6)$ and $(-2,4)$

顶点在 $(–2,3)$ 与 $(-2,-5)$、焦点在 $(-2,-6)$ 与 $(-2,4)$ 的双曲线

For the following exercises, determine the eccentricity and identify the conic. Sketch the conic.

在以下习题中，求离心率并判别圆锥曲线的类型。画出该圆锥曲线。

344\.

344.

$r = \frac{6}{1 + 3\ \text{cos}(\theta)}$

345.

$r = \frac{4}{3 - 2\ \text{cos}\ \theta}$

346\.

346.

$r = \frac{7}{5 - 5\ \text{cos}\ \theta}$

347.

Determine the Cartesian equation describing the orbit of Pluto, the most eccentric orbit around the Sun. The length of the major axis is 39.26 AU and minor axis is 38.07 AU. What is the eccentricity?

求描述冥王星轨道的直角坐标方程，这是绕太阳运行离心率最大的轨道。长轴长为 39.26 AU，短轴长为 38.07 AU。离心率是多少？

348\.

348.

The C/1980 E1 comet was observed in 1980. Given an eccentricity of 1.057 and a perihelion (point of closest approach to the Sun) of 3.364 AU, find the Cartesian equations describing the comet’s trajectory. Are we guaranteed to see this comet again? (*Hint*: Consider the Sun at point $(0,0).)$

C/1980 E1 彗星于 1980 年被观测到。已知其离心率为 1.057，近日点（离太阳最近的点）为 3.364 AU，求描述该彗星轨迹的直角坐标方程。我们能否确定再次看到这颗彗星？（*提示*：设太阳位于点 $(0,0)$。）