Unit B8: Parametric Equations and Polar Coordinates单元 B8：参数方程与极坐标

Identify the curve $x=t^2-1$, $y=t+2$ for $t\in\mathbb{R}$.

Solve the second equation for the parameter: $t=y-2$. Substitute into the first:

$$x=(y-2)^2-1.$$

This is a parabola opening in the positive $x$ direction with vertex at $(-1,2)$. As $t$ increases, $y=t+2$ increases steadily, so the curve is traced upward along the parabola.

判别曲线 $x=t^2-1$、$y=t+2$（$t\in\mathbb{R}$）。

从第二个方程解出参数：$t=y-2$。代入第一个方程：

$$x=(y-2)^2-1.$$

这是一条开口朝正 $x$ 方向、顶点在 $(-1,2)$ 的抛物线。随着 $t$ 增大，$y=t+2$ 稳定增大，所以曲线沿抛物线自下而上被描出。

Eliminate the parameter from $x=3\cos t$, $y=2\sin t$, $0\le t\le 2\pi$.

Isolate the trigonometric functions: $\cos t=x/3$ and $\sin t=y/2$. Apply $\cos^2 t+\sin^2 t=1$:

$$\frac{x^2}{9}+\frac{y^2}{4}=1.$$

This is an ellipse with semi-axes $3$ and $2$, traced once counterclockwise starting at $(3,0)$.

从 $x=3\cos t$、$y=2\sin t$（$0\le t\le 2\pi$）中消去参数。

把三角函数分离出来：$\cos t=x/3$、$\sin t=y/2$。利用 $\cos^2 t+\sin^2 t=1$：

$$\frac{x^2}{9}+\frac{y^2}{4}=1.$$

这是一条半轴为 $3$ 与 $2$ 的椭圆，从 $(3,0)$ 出发逆时针描一次。

Describe the curve $x=\sin t$, $y=\sin^2 t+1$ for $0\le t\le 2\pi$, and contrast it with the full parabola $y=x^2+1$.

Since $x=\sin t$, substituting gives $y=x^2+1$, so every point of the curve lies on that parabola. But $\sin t$ only takes values in $[-1,1]$, so $x$ never leaves $[-1,1]$. The curve is just the arc of the parabola for $-1\le x\le 1$, not the whole thing.

Moreover the point retraces. As $t$ runs $0\to\pi/2$, $x$ climbs from $0$ to $1$; as $t$ runs $\pi/2\to 3\pi/2$, $x$ falls back through $0$ to $-1$; then $x$ returns to $0$. So the arc is swept several times. The lesson: eliminating the parameter recovers the supporting equation, but the parameter interval determines which portion is actually drawn and how often.

描述曲线 $x=\sin t$、$y=\sin^2 t+1$（$0\le t\le 2\pi$），并与完整抛物线 $y=x^2+1$ 作对比。

由于 $x=\sin t$，代入得 $y=x^2+1$，所以曲线上每一点都落在这条抛物线上。但 $\sin t$ 的取值只在 $[-1,1]$ 内，故 $x$ 始终不超出 $[-1,1]$。曲线只是抛物线上 $-1\le x\le 1$ 的那段弧，并非整条。

而且这点会反复经过。当 $t$ 从 $0$ 到 $\pi/2$ 时，$x$ 从 $0$ 升到 $1$；当 $t$ 从 $\pi/2$ 到 $3\pi/2$ 时，$x$ 经过 $0$ 回落到 $-1$；随后 $x$ 又回到 $0$。所以这段弧被扫过多次。教训是：消去参数能还原出支撑方程，但参数区间决定了实际被描出的是哪一部分以及描了几次。

Parametrize the segment from $A=(1,2)$ to $B=(4,-1)$ so that $t\in[0,1]$ traces it once from $A$ to $B$, and find the parameter value at the midpoint.

The standard convex-combination parametrization is

$$x=1+(4-1)t=1+3t,\qquad y=2+(-1-2)t=2-3t,\qquad 0\le t\le 1.$$

At $t=0$ we get $A$ and at $t=1$ we get $B$, with the point moving in a straight line at constant speed. The midpoint corresponds to $t=\tfrac12$, giving $\bigl(1+\tfrac32,\,2-\tfrac32\bigr)=\bigl(\tfrac52,\tfrac12\bigr)$, exactly the average of $A$ and $B$.

对从 $A=(1,2)$ 到 $B=(4,-1)$ 的线段作参数化，使 $t\in[0,1]$ 从 $A$ 到 $B$ 描出一次，并求中点处的参数值。

标准的凸组合参数化是

$$x=1+(4-1)t=1+3t,\qquad y=2+(-1-2)t=2-3t,\qquad 0\le t\le 1.$$

当 $t=0$ 时得到 $A$，$t=1$ 时得到 $B$，点沿直线以匀速运动。中点对应 $t=\tfrac12$，给出 $\bigl(1+\tfrac32,\,2-\tfrac32\bigr)=\bigl(\tfrac52,\tfrac12\bigr)$，恰好是 $A$ 与 $B$ 的平均。

Parametric form earns its keep on curves that have no clean Cartesian equation. The classic example is the cycloid, the path traced by a point on the rim of a wheel of radius $a$ rolling without slipping along the $x$-axis. Let $t$ be the angle through which the wheel has turned.

After turning by $t$, the center of the wheel has moved forward by the arc length it rolled, namely $at$, and sits at height $a$. So the center is at $(at,a)$. The marked rim point starts at the bottom; relative to the center it has rotated clockwise by $t$, placing it at offset $(-a\sin t,-a\cos t)$ from the center. Adding,

$$x=at-a\sin t=a(t-\sin t),\qquad y=a-a\cos t=a(1-\cos t).$$

There is no elementary function $y=F(x)$ for this curve, because near each cusp ($t=2\pi k$) the curve has a vertical tangent and is not locally a graph. Yet the parametric description is effortless, and the calculus of the earlier sections applies directly: this is the same cycloid whose slope we found in Worked Example 2.1 and whose arch length we found in Worked Example 3.2.

对于没有简洁笛卡尔方程的曲线，参数形式才显出价值。经典例子是摆线（cycloid），即半径为 $a$ 的轮子沿 $x$ 轴无滑动滚动时，轮缘上一点描出的路径。设 $t$ 为轮子已转过的角度。

转过 $t$ 之后，轮心向前移动的距离等于它滚过的弧长，即 $at$，且高度为 $a$，所以轮心位于 $(at,a)$。被标记的轮缘点最初在底部；相对于轮心，它已顺时针转过 $t$，因而相对轮心的偏移为 $(-a\sin t,-a\cos t)$。相加得

$$x=at-a\sin t=a(t-\sin t),\qquad y=a-a\cos t=a(1-\cos t).$$

这条曲线没有初等函数 $y=F(x)$，因为在每个尖点附近（$t=2\pi k$）曲线有竖直切线，局部上并非函数图像。然而参数描述却毫不费力，而且前面各节的微积分可直接套用：这正是我们在例题 2.1 中求斜率、在例题 3.2 中求一拱长度的同一条摆线。

For the cycloid $x=t-\sin t$, $y=1-\cos t$, find the slope at $t=\pi/2$.

Differentiate: $dx/dt=1-\cos t$ and $dy/dt=\sin t$. Hence

$$\frac{dy}{dx}=\frac{\sin t}{1-\cos t}.$$

At $t=\pi/2$, $\sin t=1$ and $1-\cos t=1-0=1$, so the slope is $1$. The point is $(\pi/2-1,\,1)$.

对摆线 $x=t-\sin t$、$y=1-\cos t$，求 $t=\pi/2$ 处的斜率。

求导：$dx/dt=1-\cos t$、$dy/dt=\sin t$。于是

$$\frac{dy}{dx}=\frac{\sin t}{1-\cos t}.$$

当 $t=\pi/2$ 时，$\sin t=1$ 且 $1-\cos t=1-0=1$，故斜率为 $1$。该点为 $(\pi/2-1,\,1)$。

For $x=t^2$, $y=t^3-3t$, find $d^2y/dx^2$ and determine where the curve is concave up.

First $dx/dt=2t$ and $dy/dt=3t^2-3$, so

$$\frac{dy}{dx}=\frac{3t^2-3}{2t}=\frac{3}{2}\left(t-\frac{1}{t}\right).$$

Differentiate the slope in $t$: $\dfrac{d}{dt}\left(\dfrac{dy}{dx}\right)=\dfrac{3}{2}\left(1+\dfrac{1}{t^2}\right)$. Divide by $dx/dt=2t$:

$$\frac{d^2y}{dx^2}=\frac{\tfrac{3}{2}\bigl(1+t^{-2}\bigr)}{2t}=\frac{3(t^2+1)}{4t^3}.$$

Since $t^2+1>0$, the sign matches that of $t^3$, so the curve is concave up for $t>0$ and concave down for $t<0$.

对 $x=t^2$、$y=t^3-3t$，求 $d^2y/dx^2$，并判断曲线在何处下凹（凹向上）。

首先 $dx/dt=2t$、$dy/dt=3t^2-3$，所以

$$\frac{dy}{dx}=\frac{3t^2-3}{2t}=\frac{3}{2}\left(t-\frac{1}{t}\right).$$

把斜率对 $t$ 求导：$\dfrac{d}{dt}\left(\dfrac{dy}{dx}\right)=\dfrac{3}{2}\left(1+\dfrac{1}{t^2}\right)$。再除以 $dx/dt=2t$：

$$\frac{d^2y}{dx^2}=\frac{\tfrac{3}{2}\bigl(1+t^{-2}\bigr)}{2t}=\frac{3(t^2+1)}{4t^3}.$$

因为 $t^2+1>0$，符号与 $t^3$ 相同，所以曲线在 $t>0$ 时凹向上、在 $t<0$ 时凹向下。

For $x=t^2-2t$, $y=t^3-3t$, locate all points with horizontal and vertical tangent lines.

Compute the velocities: $dx/dt=2t-2=2(t-1)$ and $dy/dt=3t^2-3=3(t-1)(t+1)$.

Horizontal tangents need $dy/dt=0$ with $dx/dt\neq 0$. Now $dy/dt=0$ at $t=1$ and $t=-1$. At $t=1$ the denominator $dx/dt=0$ too, so that value is suspect and not a clean horizontal tangent; at $t=-1$ we have $dx/dt=2(-2)=-4\neq 0$, giving a genuine horizontal tangent at $(x,y)=(3,2)$.

Vertical tangents need $dx/dt=0$ with $dy/dt\neq 0$. Now $dx/dt=0$ only at $t=1$, where $dy/dt=3(0)(2)=0$ as well. So $t=1$ makes both velocities vanish; the slope is the indeterminate $0/0$ and the point $(x,y)=(-1,-2)$ is a candidate cusp or self-crossing, not a plain vertical tangent. The honest conclusion: one horizontal tangent at $(3,2)$, and a singular point at $(-1,-2)$ that needs a limit of $dy/dx$ to resolve.

对 $x=t^2-2t$、$y=t^3-3t$，找出所有具有水平切线与竖直切线的点。

计算速度分量：$dx/dt=2t-2=2(t-1)$、$dy/dt=3t^2-3=3(t-1)(t+1)$。

水平切线要求 $dy/dt=0$ 且 $dx/dt\neq 0$。现在 $dy/dt=0$ 在 $t=1$ 与 $t=-1$ 处成立。在 $t=1$ 处分母 $dx/dt$ 也为 $0$，所以这个值可疑，不是干净的水平切线；在 $t=-1$ 处 $dx/dt=2(-2)=-4\neq 0$，给出一个真正的水平切线，位于 $(x,y)=(3,2)$。

竖直切线要求 $dx/dt=0$ 且 $dy/dt\neq 0$。现在 $dx/dt=0$ 只在 $t=1$ 处，而此处 $dy/dt=3(0)(2)=0$ 也为零。所以 $t=1$ 使两个速度分量同时消失；斜率是不定式 $0/0$，点 $(x,y)=(-1,-2)$ 是尖点或自交的候选，而非普通的竖直切线。诚实的结论是：一个水平切线在 $(3,2)$，以及一个奇点在 $(-1,-2)$，后者需要用 $dy/dx$ 的极限来判定。

The slope $y'=dy/dx$ is itself a function along the curve, so it has its own $t$-derivative. To get $d^2y/dx^2$ we want the rate of change of the slope with respect to $x$, not with respect to $t$. Apply the same chain-rule idea that produced the first derivative, but to $y'$ in place of $y$:

$$\frac{d^2y}{dx^2}=\frac{d}{dx}\!\left(\frac{dy}{dx}\right)=\frac{\dfrac{d}{dt}\!\left(\dfrac{dy}{dx}\right)}{\dfrac{dx}{dt}}.$$

This makes the structure transparent: every $x$-derivative along the curve is a $t$-derivative divided by $dx/dt$. Writing $\dot x=dx/dt$, $\dot y=dy/dt$ and so on, one differentiation of the quotient $\dot y/\dot x$ gives the explicit form

$$\frac{d^2y}{dx^2}=\frac{\dot x\,\ddot y-\dot y\,\ddot x}{\dot x^{\,3}}.$$

The numerator $\dot x\ddot y-\dot y\ddot x$ is the same cross product that appears in curvature, and the cube of $\dot x$ in the denominator is exactly why a sign change in $\dot x$ flips concavity. This is why Worked Example 2.2 had $t^3$ in the denominator.

斜率 $y'=dy/dx$ 本身就是沿曲线的一个函数，所以它有自己的 $t$ 导数。要得到 $d^2y/dx^2$，我们要的是斜率对 $x$（而非对 $t$）的变化率。把产生一阶导数的同一链式法则思路，作用到 $y'$（取代 $y$）上：

$$\frac{d^2y}{dx^2}=\frac{d}{dx}\!\left(\frac{dy}{dx}\right)=\frac{\dfrac{d}{dt}\!\left(\dfrac{dy}{dx}\right)}{\dfrac{dx}{dt}}.$$

这把结构讲清楚了：沿曲线的每个 $x$ 导数都是一个 $t$ 导数除以 $dx/dt$。记 $\dot x=dx/dt$、$\dot y=dy/dt$ 等，对商 $\dot y/\dot x$ 求一次导，得到显式形式

$$\frac{d^2y}{dx^2}=\frac{\dot x\,\ddot y-\dot y\,\ddot x}{\dot x^{\,3}}.$$

分子 $\dot x\ddot y-\dot y\ddot x$ 与曲率中出现的叉积相同，而分母里 $\dot x$ 的立方正是 $\dot x$ 变号时凹凸性翻转的原因。这也是例题 2.2 分母中出现 $t^3$ 的缘由。

Find the length of $x=r\cos t$, $y=r\sin t$ for $0\le t\le 2\pi$.

Differentiate: $dx/dt=-r\sin t$ and $dy/dt=r\cos t$. The speed is

$$\sqrt{r^2\sin^2 t+r^2\cos^2 t}=\sqrt{r^2}=r.$$

Therefore $L=\int_0^{2\pi} r\,dt=2\pi r$, the circumference of a circle of radius $r$.

求 $x=r\cos t$、$y=r\sin t$（$0\le t\le 2\pi$）的长度。

求导：$dx/dt=-r\sin t$、$dy/dt=r\cos t$。速率为

$$\sqrt{r^2\sin^2 t+r^2\cos^2 t}=\sqrt{r^2}=r.$$

因此 $L=\int_0^{2\pi} r\,dt=2\pi r$，即半径为 $r$ 的圆的周长。

Partition $[\alpha,\beta]$ as $\alpha=t_0 $$L_n=\sum_{i=1}^n\sqrt{(\Delta x_i)^2+(\Delta y_i)^2}=\sum_{i=1}^n\sqrt{\left(\frac{\Delta x_i}{\Delta t_i}\right)^2+\left(\frac{\Delta y_i}{\Delta t_i}\right)^2}\,\Delta t_i.$$

By the Mean Value Theorem applied to $f$ and $g$ on each subinterval, there are points $t_i^\ast$ and $t_i^{\ast\ast}$ with $\Delta x_i/\Delta t_i=f'(t_i^\ast)$ and $\Delta y_i/\Delta t_i=g'(t_i^{\ast\ast})$. As the mesh tends to zero these sample points coalesce, and continuity of $f'$ and $g'$ makes the sum a Riemann sum for $\sqrt{(f')^2+(g')^2}$. Hence $L_n\to\int_\alpha^\beta\sqrt{(f')^2+(g')^2}\,dt$.

把 $[\alpha,\beta]$ 划分为 $\alpha=t_0 $$L_n=\sum_{i=1}^n\sqrt{(\Delta x_i)^2+(\Delta y_i)^2}=\sum_{i=1}^n\sqrt{\left(\frac{\Delta x_i}{\Delta t_i}\right)^2+\left(\frac{\Delta y_i}{\Delta t_i}\right)^2}\,\Delta t_i.$$

对 $f$ 与 $g$ 在每个子区间上分别应用中值定理（Mean Value Theorem），存在点 $t_i^\ast$ 与 $t_i^{\ast\ast}$ 使 $\Delta x_i/\Delta t_i=f'(t_i^\ast)$、$\Delta y_i/\Delta t_i=g'(t_i^{\ast\ast})$。当网格趋于零时这些取样点合并到一起，而 $f'$ 与 $g'$ 的连续性使该和成为 $\sqrt{(f')^2+(g')^2}$ 的黎曼和（Riemann sum）。于是 $L_n\to\int_\alpha^\beta\sqrt{(f')^2+(g')^2}\,dt$。

Find the length of one arch of the cycloid $x=t-\sin t$, $y=1-\cos t$ for $0\le t\le 2\pi$.

The velocities are $dx/dt=1-\cos t$ and $dy/dt=\sin t$, so the squared speed is

$$(1-\cos t)^2+\sin^2 t=1-2\cos t+\cos^2 t+\sin^2 t=2-2\cos t=2(1-\cos t).$$

Use the half-angle identity $1-\cos t=2\sin^2(t/2)$, so the squared speed is $4\sin^2(t/2)$ and the speed is $2\bigl|\sin(t/2)\bigr|$. On $[0,2\pi]$ we have $t/2\in[0,\pi]$, where $\sin(t/2)\ge 0$, so the absolute value drops:

$$L=\int_0^{2\pi}2\sin\!\frac{t}{2}\,dt=\Bigl[-4\cos\frac{t}{2}\Bigr]_0^{2\pi}=-4\cos\pi+4\cos 0=4+4=8.$$

One arch has length exactly $8$, a clean result first found by Christopher Wren. Note how essential the absolute value was: dropping it carelessly on a wider interval would let positive and negative pieces cancel.

求摆线 $x=t-\sin t$、$y=1-\cos t$（$0\le t\le 2\pi$）一拱的长度。

速度分量为 $dx/dt=1-\cos t$、$dy/dt=\sin t$，所以速率的平方为

$$(1-\cos t)^2+\sin^2 t=1-2\cos t+\cos^2 t+\sin^2 t=2-2\cos t=2(1-\cos t).$$

用半角恒等式 $1-\cos t=2\sin^2(t/2)$，于是速率的平方为 $4\sin^2(t/2)$，速率为 $2\bigl|\sin(t/2)\bigr|$。在 $[0,2\pi]$ 上 $t/2\in[0,\pi]$，此时 $\sin(t/2)\ge 0$，所以绝对值可去掉：

$$L=\int_0^{2\pi}2\sin\!\frac{t}{2}\,dt=\Bigl[-4\cos\frac{t}{2}\Bigr]_0^{2\pi}=-4\cos\pi+4\cos 0=4+4=8.$$

一拱的长度恰好是 $8$，这是由 Christopher Wren 最先得出的一个简洁结果。注意绝对值是多么关键：在更宽的区间上草率地丢掉它，会让正负部分相互抵消。

The arc $x=t$, $y=t^2$ for $0\le t\le 1$ is revolved about the $x$-axis. Set up the surface-area integral and reduce it to a single substitution.

For a parametric arc revolved about the $x$-axis the surface element is $2\pi y\,ds$, where $ds=\sqrt{(dx/dt)^2+(dy/dt)^2}\,dt$. Here $dx/dt=1$, $dy/dt=2t$, so $ds=\sqrt{1+4t^2}\,dt$, and $y=t^2$. Thus

$$S=\int_0^1 2\pi\,t^2\sqrt{1+4t^2}\,dt.$$

The integrand is the textbook form $y\,ds$, with $y$ contributing the radius of revolution and $ds$ the slant length. The key point of method: build the integrand as (circumference)$\times$(arc element), $2\pi y\cdot ds$, exactly as in the Cartesian case, but read $ds$ off the parametric speed.

弧 $x=t$、$y=t^2$（$0\le t\le 1$）绕 $x$ 轴旋转。列出表面积积分，并把它化简为一次换元。

对绕 $x$ 轴旋转的参数弧，表面微元为 $2\pi y\,ds$，其中 $ds=\sqrt{(dx/dt)^2+(dy/dt)^2}\,dt$。此处 $dx/dt=1$、$dy/dt=2t$，所以 $ds=\sqrt{1+4t^2}\,dt$，且 $y=t^2$。于是

$$S=\int_0^1 2\pi\,t^2\sqrt{1+4t^2}\,dt.$$

被积函数是课本形式 $y\,ds$，其中 $y$ 提供旋转半径，$ds$ 提供斜长。方法上的关键：把被积函数构造为（周长）$\times$（弧微元），即 $2\pi y\cdot ds$，与笛卡尔情形完全一样，只不过 $ds$ 要从参数速率读出。

For $x=\cos t$, $y=\sin t$ on $0\le t\le 4\pi$, what does the arc length integral compute, and what is the geometric length of the curve?

The speed is $\sqrt{\sin^2 t+\cos^2 t}=1$, so the integral is $\int_0^{4\pi}1\,dt=4\pi$. But the image is the unit circle, whose geometric length is $2\pi$. The parameter interval traces the circle twice, and the integral faithfully reports the total distance traveled, $4\pi$, not the length of the geometric set.

This is the crucial caveat behind every arc-length problem: the formula integrates speed and therefore measures distance traveled. It equals the geometric length only when the curve is traced exactly once with no backtracking. Always confirm the interval traces the curve a single time before quoting the answer as a length.

对 $x=\cos t$、$y=\sin t$（$0\le t\le 4\pi$），弧长积分算的是什么，而曲线的几何长度又是多少？

速率为 $\sqrt{\sin^2 t+\cos^2 t}=1$，所以积分为 $\int_0^{4\pi}1\,dt=4\pi$。但图像是单位圆，其几何长度为 $2\pi$。参数区间把圆描了两次，积分忠实地报出的是总路程 $4\pi$，而非这个几何集合的长度。

这是每个弧长问题背后的关键警告：公式对速率积分，因而度量的是路程。只有当曲线恰被描出一次且无往返时，它才等于几何长度。在把答案当作长度报出之前，务必确认该区间只把曲线描了一次。

Show that $r=2\cos\theta$ is a circle and find its center and radius.

Multiply both sides by $r$ so that the conversions apply: $r^2=2r\cos\theta$. Replace $r^2=x^2+y^2$ and $r\cos\theta=x$:

$$x^2+y^2=2x\quad\Longrightarrow\quad x^2-2x+y^2=0\quad\Longrightarrow\quad (x-1)^2+y^2=1.$$

This is a circle of radius $1$ centered at $(1,0)$, passing through the origin.

证明 $r=2\cos\theta$ 是一个圆，并求出其圆心与半径。

两边同乘以 $r$，以便套用转换公式：$r^2=2r\cos\theta$。用 $r^2=x^2+y^2$ 与 $r\cos\theta=x$ 替换：

$$x^2+y^2=2x\quad\Longrightarrow\quad x^2-2x+y^2=0\quad\Longrightarrow\quad (x-1)^2+y^2=1.$$

这是一个半径为 $1$、圆心在 $(1,0)$ 的圆，并且过原点。

Write the Cartesian point $(-1,1)$ in polar coordinates with $r>0$ and $0\le\theta<2\pi$.

The radius is $r=\sqrt{(-1)^2+1^2}=\sqrt2$. The point lies in the second quadrant ($x<0$, $y>0$), where $\tan\theta=y/x=-1$ gives $\theta=3\pi/4$. Hence the polar coordinates are

$$\left(\sqrt2,\ \tfrac{3\pi}{4}\right).$$

The naive arctangent $\arctan(-1)=-\pi/4$ would place the point in the fourth quadrant, which is wrong. Adding $\pi$ to land in the correct second quadrant gives $3\pi/4$.

把笛卡尔点 $(-1,1)$ 写成满足 $r>0$ 与 $0\le\theta<2\pi$ 的极坐标。

半径为 $r=\sqrt{(-1)^2+1^2}=\sqrt2$。该点位于第二象限（$x<0$、$y>0$），在那里 $\tan\theta=y/x=-1$ 给出 $\theta=3\pi/4$。因此极坐标为

$$\left(\sqrt2,\ \tfrac{3\pi}{4}\right).$$

天真地取反正切 $\arctan(-1)=-\pi/4$ 会把该点放在第四象限，这是错的。加上 $\pi$ 以落在正确的第二象限，得到 $3\pi/4$。

Find the slope of the tangent line to the cardioid $r=1+\sin\theta$ at $\theta=0$.

A polar curve is the parametric curve $x=r\cos\theta$, $y=r\sin\theta$ with parameter $\theta$. The slope is

$$\frac{dy}{dx}=\frac{dy/d\theta}{dx/d\theta}=\frac{\dfrac{dr}{d\theta}\sin\theta+r\cos\theta}{\dfrac{dr}{d\theta}\cos\theta-r\sin\theta}.$$

Here $dr/d\theta=\cos\theta$. At $\theta=0$: $r=1$, $\cos\theta=1$, $\sin\theta=0$. The numerator is $1\cdot 0+1\cdot 1=1$; the denominator is $1\cdot 1-1\cdot 0=1$. So the slope is $1$. The lesson: never differentiate $r$ alone for a slope. You must convert to $x$ and $y$ first, because the tangent direction mixes radial and angular motion.

求心形线（cardioid）$r=1+\sin\theta$ 在 $\theta=0$ 处切线的斜率。

极坐标曲线就是以 $\theta$ 为参数的参数曲线 $x=r\cos\theta$、$y=r\sin\theta$。斜率为

$$\frac{dy}{dx}=\frac{dy/d\theta}{dx/d\theta}=\frac{\dfrac{dr}{d\theta}\sin\theta+r\cos\theta}{\dfrac{dr}{d\theta}\cos\theta-r\sin\theta}.$$

这里 $dr/d\theta=\cos\theta$。在 $\theta=0$ 处：$r=1$、$\cos\theta=1$、$\sin\theta=0$。分子为 $1\cdot 0+1\cdot 1=1$；分母为 $1\cdot 1-1\cdot 0=1$。所以斜率为 $1$。教训是：求斜率时绝不可只对 $r$ 求导。你必须先转换为 $x$ 与 $y$，因为切线方向混合了径向与角向的运动。

Convert the Cartesian line $x=3$ to polar form, and identify the curve $\theta=\pi/4$.

Since $x=r\cos\theta$, the vertical line $x=3$ becomes $r\cos\theta=3$, that is $r=3\sec\theta$. As $\theta$ ranges over $(-\pi/2,\pi/2)$ the radius sweeps the whole vertical line.

The equation $\theta=\pi/4$ fixes the angle and lets $r$ be any real number. Allowing negative $r$, this is the entire straight line through the origin at $45^\circ$, namely $y=x$. If $r$ were restricted to $r\ge 0$, it would be only the ray into the first quadrant.

将笛卡尔直线 $x=3$ 转成极坐标形式，并辨认曲线 $\theta=\pi/4$。

由于 $x=r\cos\theta$，竖直线 $x=3$ 变为 $r\cos\theta=3$，即 $r=3\sec\theta$。当 $\theta$ 在 $(-\pi/2,\pi/2)$ 上变化时，半径扫过整条竖直线。

方程 $\theta=\pi/4$ 固定了角度，而让 $r$ 取任意实数。允许负的 $r$ 时，这是过原点、倾角 $45^\circ$ 的整条直线，即 $y=x$。若把 $r$ 限制为 $r\ge 0$，它就只是伸入第一象限的那条射线。

How many petals does $r=\cos 2\theta$ have, and what is the length of each petal?

The coefficient is $n=2$, which is even, so the rose has $2n=4$ petals. Each petal reaches its tip where $|\cos 2\theta|=1$, that is at $r=\pm 1$, so every petal has length $1$. A single petal lies between consecutive zeros of $r$: $\cos 2\theta=0$ at $2\theta=\pm\pi/2$, giving $\theta\in[-\pi/4,\pi/4]$ for the petal pointing along the positive $x$-axis. The other three petals follow by the symmetry of $\cos 2\theta$, spaced $90^\circ$ apart.

$r=\cos 2\theta$ 有多少片花瓣，每片花瓣的长度是多少？

系数 $n=2$ 为偶数，所以这条玫瑰线有 $2n=4$ 片花瓣。每片花瓣在 $|\cos 2\theta|=1$，即 $r=\pm 1$ 处达到其尖端，所以每片花瓣长 $1$。单片花瓣位于 $r$ 的相邻零点之间：$\cos 2\theta=0$ 在 $2\theta=\pm\pi/2$ 处，为指向正 $x$ 轴的花瓣给出 $\theta\in[-\pi/4,\pi/4]$。其余三片花瓣由 $\cos 2\theta$ 的对称性得出，彼此间隔 $90^\circ$。

A useful shortcut governs the tangent direction when a polar curve passes through the origin. Suppose $r(\theta_0)=0$ for some angle $\theta_0$ and $r'(\theta_0)\neq 0$. Evaluate the general polar slope at this angle:

$$\frac{dy}{dx}=\frac{r'\sin\theta+r\cos\theta}{r'\cos\theta-r\sin\theta}.$$

At $\theta_0$ the terms containing $r$ vanish because $r=0$, leaving

$$\frac{dy}{dx}\bigg|_{\theta_0}=\frac{r'(\theta_0)\sin\theta_0}{r'(\theta_0)\cos\theta_0}=\tan\theta_0.$$

So the tangent line at the pole is simply the ray $\theta=\theta_0$ itself. This is why a rose $r=\cos 2\theta$, which hits the origin at $\theta=\pi/4,3\pi/4,\dots$, has petals separated by tangent lines along exactly those angles. The rule turns the hard-looking question "in what direction does the curve leave the origin?" into "at what angle is $r=0$?".

当极坐标曲线穿过原点时，有一个实用的捷径控制着切线方向。设对某角度 $\theta_0$ 有 $r(\theta_0)=0$ 且 $r'(\theta_0)\neq 0$。在这个角度处计算一般的极坐标斜率：

$$\frac{dy}{dx}=\frac{r'\sin\theta+r\cos\theta}{r'\cos\theta-r\sin\theta}.$$

在 $\theta_0$ 处，含 $r$ 的项因 $r=0$ 而消失，剩下

$$\frac{dy}{dx}\bigg|_{\theta_0}=\frac{r'(\theta_0)\sin\theta_0}{r'(\theta_0)\cos\theta_0}=\tan\theta_0.$$

所以极点处的切线就是射线 $\theta=\theta_0$ 本身。这就是为什么玫瑰线 $r=\cos 2\theta$（它在 $\theta=\pi/4,3\pi/4,\dots$ 处触及原点）的花瓣恰好被沿这些角度的切线分隔。这条规则把看似棘手的问题“曲线从什么方向离开原点？”化为“$r=0$ 发生在什么角度？”。

Find the area of one petal of $r=\cos 2\theta$.

A petal begins and ends where $r=0$, that is $\cos 2\theta=0$. The first petal runs from $2\theta=-\pi/2$ to $2\theta=\pi/2$, so $\theta$ goes from $-\pi/4$ to $\pi/4$. Then

$$A=\frac{1}{2}\int_{-\pi/4}^{\pi/4}\cos^2 2\theta\,d\theta=\frac{1}{2}\int_{-\pi/4}^{\pi/4}\frac{1+\cos 4\theta}{2}\,d\theta.$$ $$=\frac{1}{4}\left[\theta+\frac{\sin 4\theta}{4}\right]_{-\pi/4}^{\pi/4}=\frac{1}{4}\left(\frac{\pi}{2}+0\right)=\frac{\pi}{8}.$$

求 $r=\cos 2\theta$ 一瓣花瓣的面积。

一瓣花瓣在 $r=0$ 处起止，即 $\cos 2\theta=0$。第一瓣从 $2\theta=-\pi/2$ 到 $2\theta=\pi/2$，所以 $\theta$ 从 $-\pi/4$ 到 $\pi/4$。于是

$$A=\frac{1}{2}\int_{-\pi/4}^{\pi/4}\cos^2 2\theta\,d\theta=\frac{1}{2}\int_{-\pi/4}^{\pi/4}\frac{1+\cos 4\theta}{2}\,d\theta.$$ $$=\frac{1}{4}\left[\theta+\frac{\sin 4\theta}{4}\right]_{-\pi/4}^{\pi/4}=\frac{1}{4}\left(\frac{\pi}{2}+0\right)=\frac{\pi}{8}.$$

Find the area inside the circle $r=2\sin\theta$ and outside the circle $r=1$.

The curves meet where $2\sin\theta=1$, that is $\sin\theta=1/2$, giving $\theta=\pi/6$ and $\theta=5\pi/6$. On that interval $2\sin\theta\ge 1$, so $r=2\sin\theta$ is the outer curve and $r=1$ the inner. Then

$$A=\frac{1}{2}\int_{\pi/6}^{5\pi/6}\bigl((2\sin\theta)^2-1^2\bigr)\,d\theta=\frac{1}{2}\int_{\pi/6}^{5\pi/6}\bigl(4\sin^2\theta-1\bigr)\,d\theta.$$

Use $4\sin^2\theta=2(1-\cos 2\theta)=2-2\cos 2\theta$, so the integrand is $1-2\cos 2\theta$:

$$A=\frac{1}{2}\Bigl[\theta-\sin 2\theta\Bigr]_{\pi/6}^{5\pi/6}=\frac{1}{2}\left(\frac{2\pi}{3}-\bigl(\sin\tfrac{5\pi}{3}-\sin\tfrac{\pi}{3}\bigr)\right)=\frac{1}{2}\left(\frac{2\pi}{3}+\sqrt3\right)=\frac{\pi}{3}+\frac{\sqrt3}{2}.$$

求圆 $r=2\sin\theta$ 内且圆 $r=1$ 外的面积。

两曲线相交于 $2\sin\theta=1$，即 $\sin\theta=1/2$，得到 $\theta=\pi/6$ 与 $\theta=5\pi/6$。在该区间上 $2\sin\theta\ge 1$，所以 $r=2\sin\theta$ 是外曲线、$r=1$ 是内曲线。于是

$$A=\frac{1}{2}\int_{\pi/6}^{5\pi/6}\bigl((2\sin\theta)^2-1^2\bigr)\,d\theta=\frac{1}{2}\int_{\pi/6}^{5\pi/6}\bigl(4\sin^2\theta-1\bigr)\,d\theta.$$

利用 $4\sin^2\theta=2(1-\cos 2\theta)=2-2\cos 2\theta$，所以被积函数为 $1-2\cos 2\theta$：

$$A=\frac{1}{2}\Bigl[\theta-\sin 2\theta\Bigr]_{\pi/6}^{5\pi/6}=\frac{1}{2}\left(\frac{2\pi}{3}-\bigl(\sin\tfrac{5\pi}{3}-\sin\tfrac{\pi}{3}\bigr)\right)=\frac{1}{2}\left(\frac{2\pi}{3}+\sqrt3\right)=\frac{\pi}{3}+\frac{\sqrt3}{2}.$$

Find the area enclosed by the cardioid $r=1+\cos\theta$.

The whole cardioid is traced once as $\theta$ runs from $0$ to $2\pi$, and $r\ge 0$ throughout, so

$$A=\frac{1}{2}\int_0^{2\pi}(1+\cos\theta)^2\,d\theta=\frac{1}{2}\int_0^{2\pi}\bigl(1+2\cos\theta+\cos^2\theta\bigr)\,d\theta.$$

Replace $\cos^2\theta=\tfrac12(1+\cos 2\theta)$, so the integrand becomes $\tfrac32+2\cos\theta+\tfrac12\cos 2\theta$. Over a full period $\int_0^{2\pi}\cos\theta\,d\theta=0$ and $\int_0^{2\pi}\cos 2\theta\,d\theta=0$, leaving only the constant:

$$A=\frac{1}{2}\int_0^{2\pi}\frac{3}{2}\,d\theta=\frac{1}{2}\cdot\frac{3}{2}\cdot 2\pi=\frac{3\pi}{2}.$$

求心形线（cardioid）$r=1+\cos\theta$ 所围的面积。

整条心形线在 $\theta$ 从 $0$ 到 $2\pi$ 时被描出一次，且始终 $r\ge 0$，所以

$$A=\frac{1}{2}\int_0^{2\pi}(1+\cos\theta)^2\,d\theta=\frac{1}{2}\int_0^{2\pi}\bigl(1+2\cos\theta+\cos^2\theta\bigr)\,d\theta.$$

用 $\cos^2\theta=\tfrac12(1+\cos 2\theta)$ 替换，于是被积函数变为 $\tfrac32+2\cos\theta+\tfrac12\cos 2\theta$。在一个完整周期上 $\int_0^{2\pi}\cos\theta\,d\theta=0$ 且 $\int_0^{2\pi}\cos 2\theta\,d\theta=0$，只剩下常数项：

$$A=\frac{1}{2}\int_0^{2\pi}\frac{3}{2}\,d\theta=\frac{1}{2}\cdot\frac{3}{2}\cdot 2\pi=\frac{3\pi}{2}.$$

Partition $[\alpha,\beta]$ into subintervals $[\theta_{i-1},\theta_i]$ of width $\Delta\theta_i$. On the $i$th piece pick a sample angle $\theta_i^\ast$ and approximate the swept region by a true circular sector of radius $f(\theta_i^\ast)$ subtending angle $\Delta\theta_i$. A sector is the fraction $\Delta\theta_i/(2\pi)$ of a full disk of radius $f(\theta_i^\ast)$, so its area is

$$\frac{\Delta\theta_i}{2\pi}\cdot\pi\bigl(f(\theta_i^\ast)\bigr)^2=\frac{1}{2}\bigl(f(\theta_i^\ast)\bigr)^2\,\Delta\theta_i.$$

Summing over all pieces gives $\sum \tfrac12 f(\theta_i^\ast)^2\,\Delta\theta_i$, a Riemann sum for the continuous function $\tfrac12 f(\theta)^2$. Because $f$ is continuous, as the mesh tends to zero this converges to the integral, establishing

$$A=\frac{1}{2}\int_\alpha^\beta f(\theta)^2\,d\theta.$$

The contrast with Cartesian area is the geometry of the element: there the building block is a thin rectangle of area (height)$\times dx$; here it is a thin wedge whose area grows like $r^2$, not $r$, which is exactly why $r$ is squared.

把 $[\alpha,\beta]$ 划分为宽度为 $\Delta\theta_i$ 的子区间 $[\theta_{i-1},\theta_i]$。在第 $i$ 段上取一个样本角 $\theta_i^\ast$，用半径为 $f(\theta_i^\ast)$、张开角为 $\Delta\theta_i$ 的真正圆扇形来近似被扫过的区域。一个圆扇形是半径为 $f(\theta_i^\ast)$ 的整个圆盘的 $\Delta\theta_i/(2\pi)$ 这一部分，所以其面积为

$$\frac{\Delta\theta_i}{2\pi}\cdot\pi\bigl(f(\theta_i^\ast)\bigr)^2=\frac{1}{2}\bigl(f(\theta_i^\ast)\bigr)^2\,\Delta\theta_i.$$

对所有段求和得到 $\sum \tfrac12 f(\theta_i^\ast)^2\,\Delta\theta_i$，这是连续函数 $\tfrac12 f(\theta)^2$ 的一个黎曼和（Riemann sum）。因为 $f$ 连续，当划分粗细趋于零时，它收敛于积分，从而建立

$$A=\frac{1}{2}\int_\alpha^\beta f(\theta)^2\,d\theta.$$

与笛卡尔面积的对比在于微元的几何形状：那里的基本块是面积为（高）$\times dx$ 的细矩形；而这里是一个细楔形，其面积按 $r^2$（而非 $r$）增长，这正是 $r$ 被平方的原因。

Find the length of the spiral $r=e^{\theta}$ from $\theta=0$ to $\theta=2\pi$.

Here $dr/d\theta=e^{\theta}=r$, so $r^2+(dr/d\theta)^2=e^{2\theta}+e^{2\theta}=2e^{2\theta}$. Then

$$L=\int_0^{2\pi}\sqrt{2e^{2\theta}}\,d\theta=\sqrt2\int_0^{2\pi}e^{\theta}\,d\theta=\sqrt2\,\bigl(e^{2\pi}-1\bigr).$$

求螺线 $r=e^{\theta}$ 从 $\theta=0$ 到 $\theta=2\pi$ 的长度。

这里 $dr/d\theta=e^{\theta}=r$，所以 $r^2+(dr/d\theta)^2=e^{2\theta}+e^{2\theta}=2e^{2\theta}$。于是

$$L=\int_0^{2\pi}\sqrt{2e^{2\theta}}\,d\theta=\sqrt2\int_0^{2\pi}e^{\theta}\,d\theta=\sqrt2\,\bigl(e^{2\pi}-1\bigr).$$

Find the total length of the cardioid $r=1+\cos\theta$.

Here $dr/d\theta=-\sin\theta$, so

$$r^2+\left(\frac{dr}{d\theta}\right)^2=(1+\cos\theta)^2+\sin^2\theta=1+2\cos\theta+\cos^2\theta+\sin^2\theta=2+2\cos\theta=2(1+\cos\theta).$$

Apply the half-angle identity $1+\cos\theta=2\cos^2(\theta/2)$, so the quantity under the root is $4\cos^2(\theta/2)$ and the integrand is $2\bigl|\cos(\theta/2)\bigr|$. On $[0,\pi]$, $\theta/2\in[0,\pi/2]$ where $\cos(\theta/2)\ge 0$. By symmetry the upper half ($0$ to $\pi$) is half the perimeter, so

$$L=2\int_0^{\pi}2\cos\frac{\theta}{2}\,d\theta=2\Bigl[4\sin\frac{\theta}{2}\Bigr]_0^{\pi}=2\cdot 4=8.$$

The cardioid has perimeter $8$. Working over $[0,\pi]$ and doubling sidesteps the sign change of $\cos(\theta/2)$ that occurs past $\theta=\pi$.

求心形线（cardioid）$r=1+\cos\theta$ 的总长度。

这里 $dr/d\theta=-\sin\theta$，所以

$$r^2+\left(\frac{dr}{d\theta}\right)^2=(1+\cos\theta)^2+\sin^2\theta=1+2\cos\theta+\cos^2\theta+\sin^2\theta=2+2\cos\theta=2(1+\cos\theta).$$

运用半角恒等式 $1+\cos\theta=2\cos^2(\theta/2)$，于是根号下的量为 $4\cos^2(\theta/2)$，被积函数为 $2\bigl|\cos(\theta/2)\bigr|$。在 $[0,\pi]$ 上，$\theta/2\in[0,\pi/2]$，此处 $\cos(\theta/2)\ge 0$。由对称性，上半部分（$0$ 到 $\pi$）是周长的一半，所以

$$L=2\int_0^{\pi}2\cos\frac{\theta}{2}\,d\theta=2\Bigl[4\sin\frac{\theta}{2}\Bigr]_0^{\pi}=2\cdot 4=8.$$

这条心形线的周长为 $8$。在 $[0,\pi]$ 上积分再翻倍，可以避开 $\theta=\pi$ 之后 $\cos(\theta/2)$ 出现的变号问题。

Set up the length of the Archimedean spiral $r=\theta$ from $\theta=0$ to $\theta=\pi$, and evaluate it.

Here $dr/d\theta=1$, so $r^2+(dr/d\theta)^2=\theta^2+1$ and

$$L=\int_0^{\pi}\sqrt{\theta^2+1}\,d\theta.$$

This is a standard $\sqrt{u^2+1}$ integral with antiderivative $\tfrac12\bigl(\theta\sqrt{\theta^2+1}+\ln(\theta+\sqrt{\theta^2+1})\bigr)$. Evaluating from $0$ to $\pi$,

$$L=\frac{1}{2}\Bigl(\pi\sqrt{\pi^2+1}+\ln\bigl(\pi+\sqrt{\pi^2+1}\bigr)\Bigr).$$

The point of method is that the $r^2$ term, here $\theta^2$, dominates the integrand for large $\theta$, which matches the geometric fact that a spiral's outer turns are much longer than its inner ones.

列出阿基米德螺线 $r=\theta$ 从 $\theta=0$ 到 $\theta=\pi$ 的长度，并求出其值。

这里 $dr/d\theta=1$，所以 $r^2+(dr/d\theta)^2=\theta^2+1$，且

$$L=\int_0^{\pi}\sqrt{\theta^2+1}\,d\theta.$$

这是一个标准的 $\sqrt{u^2+1}$ 积分，其原函数为 $\tfrac12\bigl(\theta\sqrt{\theta^2+1}+\ln(\theta+\sqrt{\theta^2+1})\bigr)$。从 $0$ 到 $\pi$ 求值，

$$L=\frac{1}{2}\Bigl(\pi\sqrt{\pi^2+1}+\ln\bigl(\pi+\sqrt{\pi^2+1}\bigr)\Bigr).$$

这一方法的要点在于：$r^2$ 项（这里是 $\theta^2$）在 $\theta$ 较大时主导被积函数，这与几何事实相符，即螺线外圈比内圈长得多。

Start from the velocity components above and form the sum of their squares. Abbreviate $r'=dr/d\theta$:

$$\left(\frac{dx}{d\theta}\right)^2=r'^2\cos^2\theta-2rr'\cos\theta\sin\theta+r^2\sin^2\theta,$$ $$\left(\frac{dy}{d\theta}\right)^2=r'^2\sin^2\theta+2rr'\sin\theta\cos\theta+r^2\cos^2\theta.$$

Adding, the middle terms $\mp 2rr'\sin\theta\cos\theta$ cancel. Grouping the rest and applying $\cos^2\theta+\sin^2\theta=1$ twice,

$$\left(\frac{dx}{d\theta}\right)^2+\left(\frac{dy}{d\theta}\right)^2=r'^2+r^2.$$

Substituting this into the parametric arc length integral $\int\sqrt{(dx/d\theta)^2+(dy/d\theta)^2}\,d\theta$ gives the polar formula.

从上面的速度分量出发，构造它们的平方和。把 $dr/d\theta$ 简记为 $r'$：

$$\left(\frac{dx}{d\theta}\right)^2=r'^2\cos^2\theta-2rr'\cos\theta\sin\theta+r^2\sin^2\theta,$$ $$\left(\frac{dy}{d\theta}\right)^2=r'^2\sin^2\theta+2rr'\sin\theta\cos\theta+r^2\cos^2\theta.$$

相加，中间项 $\mp 2rr'\sin\theta\cos\theta$ 相消。把其余各项归类，并两次运用 $\cos^2\theta+\sin^2\theta=1$，

$$\left(\frac{dx}{d\theta}\right)^2+\left(\frac{dy}{d\theta}\right)^2=r'^2+r^2.$$

把它代入参数弧长积分 $\int\sqrt{(dx/d\theta)^2+(dy/d\theta)^2}\,d\theta$，便得到极坐标公式。

Identify the conic $r=\dfrac{6}{2+\cos\theta}$ and state its eccentricity.

Put the denominator in the standard form $1+e\cos\theta$ by dividing numerator and denominator by $2$:

$$r=\frac{3}{1+\tfrac12\cos\theta}.$$

Comparing with $r=ed/(1+e\cos\theta)$, the eccentricity is $e=\tfrac12$. Since $0

辨识圆锥曲线 $r=\dfrac{6}{2+\cos\theta}$ 并指出其离心率。

把分子分母同除以 $2$，使分母化为标准形式 $1+e\cos\theta$：

$$r=\frac{3}{1+\tfrac12\cos\theta}.$$

与 $r=ed/(1+e\cos\theta)$ 对照，离心率为 $e=\tfrac12$。由于 $0

Find the focus of the parabola $y^2=12x$.

Compare with the standard form $y^2=4px$: matching gives $4p=12$, so $p=3$. For $y^2=4px$ the focus lies at $(p,0)$, hence the focus is at $(3,0)$ and the directrix is the line $x=-3$.

求抛物线 $y^2=12x$ 的焦点。

与标准形式 $y^2=4px$ 对照：匹配得 $4p=12$，故 $p=3$。对 $y^2=4px$，焦点位于 $(p,0)$，因此焦点在 $(3,0)$，而准线是直线 $x=-3$。

Find the foci and eccentricity of $\dfrac{x^2}{25}+\dfrac{y^2}{9}=1$.

Here $a^2=25$ and $b^2=9$, so $a=5$ and $b=3$ with the major axis along $x$. For an ellipse the focal distance satisfies $c^2=a^2-b^2=25-9=16$, so $c=4$ and the foci are $(\pm 4,0)$. The eccentricity is

$$e=\frac{c}{a}=\frac{4}{5}.$$

Since $0

求 $\dfrac{x^2}{25}+\dfrac{y^2}{9}=1$ 的焦点与离心率。

此处 $a^2=25$、$b^2=9$，故 $a=5$、$b=3$，长轴沿 $x$ 轴。对椭圆，焦距满足 $c^2=a^2-b^2=25-9=16$，所以 $c=4$，焦点为 $(\pm 4,0)$。离心率为

$$e=\frac{c}{a}=\frac{4}{5}.$$

由于 $0

Place a focus at the origin and the directrix as the vertical line $x=d$ to the right. The focus-directrix definition of a conic says every point $P$ satisfies

$$\frac{\text{distance to focus}}{\text{distance to directrix}}=e,$$

a constant, the eccentricity. In polar coordinates the distance from $P=(r,\theta)$ to the focus at the origin is simply $r$. Its $x$-coordinate is $r\cos\theta$, so its horizontal distance to the directrix $x=d$ is $d-r\cos\theta$. The definition reads

$$\frac{r}{\,d-r\cos\theta\,}=e\quad\Longrightarrow\quad r=e\,(d-r\cos\theta)=ed-er\cos\theta.$$

Collect the $r$ terms: $r+er\cos\theta=ed$, that is $r(1+e\cos\theta)=ed$, hence

$$r=\frac{ed}{1+e\cos\theta}.$$

The single parameter $e$ controls the shape. When $e=0$ the equation is $r=$ constant, a circle. When $01$ the denominator changes sign, producing the two branches of a hyperbola. Putting the directrix on a different side or using $\sin\theta$ merely rotates this picture.

把焦点放在原点，准线取为右侧的竖直线 $x=d$。圆锥曲线的焦点与准线的定义说，每一点 $P$ 都满足

$$\frac{\text{distance to focus}}{\text{distance to directrix}}=e,$$

这是个常数，即离心率。在极坐标下，$P=(r,\theta)$ 到原点处焦点的距离就是 $r$。其 $x$ 坐标为 $r\cos\theta$，所以它到准线 $x=d$ 的水平距离为 $d-r\cos\theta$。定义写作

$$\frac{r}{\,d-r\cos\theta\,}=e\quad\Longrightarrow\quad r=e\,(d-r\cos\theta)=ed-er\cos\theta.$$

收集 $r$ 项：$r+er\cos\theta=ed$，即 $r(1+e\cos\theta)=ed$，于是

$$r=\frac{ed}{1+e\cos\theta}.$$

单一参数 $e$ 控制着形状。当 $e=0$ 时方程为 $r=$ 常数，是圆。当 $01$ 时分母变号，产生双曲线的两支。把准线放到另一侧，或改用 $\sin\theta$，只是把这幅图旋转。

Identify the curve $4x^2+9y^2-16x+18y-11=0$.

Group and complete the square in each variable. For $x$: $4(x^2-4x)=4(x-2)^2-16$. For $y$: $9(y^2+2y)=9(y+1)^2-9$. Substitute:

$$4(x-2)^2-16+9(y+1)^2-9-11=0\ \Longrightarrow\ 4(x-2)^2+9(y+1)^2=36.$$

Divide by $36$ to reach standard form:

$$\frac{(x-2)^2}{9}+\frac{(y+1)^2}{4}=1.$$

This is an ellipse centered at $(2,-1)$ with $a=3$ along $x$ and $b=2$ along $y$. Because both squared terms are positive with different coefficients, the equation is an ellipse; equal coefficients would give a circle, and opposite signs a hyperbola.

辨识曲线 $4x^2+9y^2-16x+18y-11=0$。

对每个变量分组并配方。对 $x$：$4(x^2-4x)=4(x-2)^2-16$。对 $y$：$9(y^2+2y)=9(y+1)^2-9$。代入：

$$4(x-2)^2-16+9(y+1)^2-9-11=0\ \Longrightarrow\ 4(x-2)^2+9(y+1)^2=36.$$

除以 $36$ 以达到标准形式：

$$\frac{(x-2)^2}{9}+\frac{(y+1)^2}{4}=1.$$

这是一条以 $(2,-1)$ 为中心的椭圆，沿 $x$ 轴 $a=3$、沿 $y$ 轴 $b=2$。因为两个平方项都为正且系数不同，故该方程是椭圆；系数相等会给出圆，而符号相反则给出双曲线。