Axiom Learning·AP PrecalculusChapter 9: Polar Functions
极函数 — Polar Coordinates, Graphs, and Rates of Change
本章概览
在直角坐标系中，点用 $(x,y)$ 表示；而在极坐标系中，点用距离 $r$（离原点的距离）和角度 $\theta$（方向）来描述，写作 $(r,\theta)$。本章将学习如何在两种坐标系之间转换、如何绘制极坐标函数图像，以及如何计算极函数的变化率。
| Term 术语 | Chinese 中文 | Meaning 含义 |
|---|---|---|
| Polar coordinate system | 极坐标系 | Points described by $(r,\theta)$ |
| Pole | 极点 | Origin of the polar system |
| Polar axis | 极轴 | Positive horizontal ray from the pole |
| $r$ (radius) | 径向坐标/半径 | Directed distance from the pole |
| $\theta$ (angle) | 方向角 | Directed angle from the polar axis |
| Coterminal (polar) | 等价极坐标 | $(r,\theta)=(r,\theta+2\pi k)$ |
| Cardioid | 心形线 | Heart-shaped polar curve |
| Limaçon | 蚶线/利马松曲线 | May have inner loop, dimple, or convex shape |
| Rose curve | 玫瑰线 | Petal-shaped; $n$ or $2n$ petals |
| Modulus | 模 | $|z|=\sqrt{a^2+b^2}$ for complex $z=a+bi$ |
| Argument | 辐角 | Angle $\theta$ in polar form of complex number |
9.1 Trigonometry and Polar Coordinates 极坐标
The Polar Coordinate System 极坐标系
In the polar coordinate system, a point is described by the ordered pair $(r,\theta)$ where:
If $r > 0$: move $r$ units along the terminal side of $\theta$.
If $r < 0$: move $|r|$ units in the opposite direction.
- $r$ is the radius（半径）— directed distance from the origin (pole)
- $\theta$ is the directed angle（有向角）— measured from the positive horizontal ray (polar axis)
If $r > 0$: move $r$ units along the terminal side of $\theta$.
If $r < 0$: move $|r|$ units in the opposite direction.
Multiple Representations of the Same Point 一点多种极坐标
A key feature of polar coordinates: one point has infinitely many representations.
$$\boxed{(r,\theta) = (r,\,\theta+2\pi k)} \quad \text{(same direction, full rotations)}$$
$$\boxed{(r,\theta) = (-r,\,\theta+(2k+1)\pi)} \quad \text{(opposite direction)}$$
where $k$ is any integer.
为什么一点有多种极坐标？
- 旋转整数圈 $2\pi$ 后回到同一位置，所以 $(r,\theta)=(r,\theta+2\pi k)$。
- 如果 $r$ 取负值，则向相反方向走，相当于角度加 $\pi$：$(r,\theta)=(-r,\theta+\pi)$。
- 直角坐标中，每个点只有唯一表示；极坐标中有无穷多种表示——这是极坐标的重要特征！
🤝 一起做 Together T-1 — Multiple Representations
Show that the following three polar coordinates represent the same point: $\left(5,\dfrac{7\pi}{6}\right)$, $\quad \left(-5,\dfrac{\pi}{6}\right)$, $\quad \left(5,-\dfrac{5\pi}{6}\right)$
Start from $\left(5,\dfrac{7\pi}{6}\right)$.
This is the reference point: $r=5$, angle $\dfrac{7\pi}{6}$.
This is the reference point: $r=5$, angle $\dfrac{7\pi}{6}$.
Check $\left(-5,\dfrac{\pi}{6}\right)$.
Using $(-r,\theta) = (r,\theta+\pi)$: $\left(-5,\dfrac{\pi}{6}\right) = \left(5,\,\dfrac{\pi}{6}+\pi\right) = \left(5,\dfrac{7\pi}{6}\right)$ ✓
Using $(-r,\theta) = (r,\theta+\pi)$: $\left(-5,\dfrac{\pi}{6}\right) = \left(5,\,\dfrac{\pi}{6}+\pi\right) = \left(5,\dfrac{7\pi}{6}\right)$ ✓
Check $\left(5,-\dfrac{5\pi}{6}\right)$.
$-\dfrac{5\pi}{6} + 2\pi = -\dfrac{5\pi}{6}+\dfrac{12\pi}{6} = \dfrac{7\pi}{6}$ ✓
So $\left(5,-\dfrac{5\pi}{6}\right)=\left(5,\dfrac{7\pi}{6}\right)$ ✓
$-\dfrac{5\pi}{6} + 2\pi = -\dfrac{5\pi}{6}+\dfrac{12\pi}{6} = \dfrac{7\pi}{6}$ ✓
So $\left(5,-\dfrac{5\pi}{6}\right)=\left(5,\dfrac{7\pi}{6}\right)$ ✓
All three represent the same point. ✓
Polar ↔ Rectangular Conversion 坐标转换
Polar → Rectangular 极坐标 → 直角坐标
$$x = r\cos\theta$$
$$y = r\sin\theta$$
Rectangular → Polar 直角坐标 → 极坐标
$$r = \sqrt{x^2+y^2}$$
$$\tan\theta = \frac{y}{x} \quad (\text{check quadrant!})$$
⚠ 注意象限
从直角坐标转极坐标时，不能只用 $\arctan(y/x)$！必须根据点所在象限修正角度。$\arctan$ 只返回 $(-\pi/2,\pi/2)$ 内的值，可能与实际象限不符。
| Location | Condition | Angle $\theta$ |
|---|---|---|
| Quadrant I | $x>0,\,y>0$ | $\theta = \arctan(y/x)$ |
| Quadrant II | $x<0,\,y>0$ | $\theta = \pi - \arctan(|y/x|)$ |
| Quadrant III | $x<0,\,y<0$ | $\theta = \pi + \arctan(|y/x|)$ |
| Quadrant IV | $x>0,\,y<0$ | $\theta = 2\pi - \arctan(|y/x|)$ |
🤝 一起做 Together T-2 — Polar to Rectangular
Convert $\left(6,\,\dfrac{2\pi}{3}\right)$ to rectangular coordinates.
Apply $x=r\cos\theta$, $y=r\sin\theta$.
$x = 6\cos\dfrac{2\pi}{3} = 6\cdot\left(-\dfrac{1}{2}\right) = -3$
$y = 6\sin\dfrac{2\pi}{3} = 6\cdot\dfrac{\sqrt{3}}{2} = 3\sqrt{3}$
$x = 6\cos\dfrac{2\pi}{3} = 6\cdot\left(-\dfrac{1}{2}\right) = -3$
$y = 6\sin\dfrac{2\pi}{3} = 6\cdot\dfrac{\sqrt{3}}{2} = 3\sqrt{3}$
Rectangular: $(-3,\;3\sqrt{3})$
🤝 一起做 Together T-3 — Rectangular to Polar
Convert $(3,\,3\sqrt{3})$ to polar coordinates with $r>0$ and $0\le\theta<2\pi$.
Find $r$.
$r = \sqrt{3^2+(3\sqrt{3})^2} = \sqrt{9+27} = \sqrt{36} = 6$
$r = \sqrt{3^2+(3\sqrt{3})^2} = \sqrt{9+27} = \sqrt{36} = 6$
Find $\theta$. Point $(3,3\sqrt{3})$ is in Quadrant I.
$\tan\theta = \dfrac{3\sqrt{3}}{3} = \sqrt{3} \Rightarrow \theta = \arctan(\sqrt{3}) = \dfrac{\pi}{3}$
$\tan\theta = \dfrac{3\sqrt{3}}{3} = \sqrt{3} \Rightarrow \theta = \arctan(\sqrt{3}) = \dfrac{\pi}{3}$
Polar: $\left(6,\,\dfrac{\pi}{3}\right)$
Polar Form of a Complex Number 复数的极形式
A complex number $z = a+bi$ can be written in polar form:
$$z = r(\cos\theta + i\sin\theta) = r\,\text{cis}\,\theta$$
where $r = \sqrt{a^2+b^2}$ is the modulus（模）and $\theta$ with $\tan\theta = b/a$ is the argument（辐角）.
极形式的几何意义
在复平面中，$z=a+bi$ 对应点 $(a,b)$。极形式就是用这个点的距离原点（模 $r$）和方向角（辐角 $\theta$）来描述这个点，与极坐标系完全对应。
🤝 一起做 Together T-4 — Complex Number to Polar Form
Write $z = 1 - \sqrt{3}\,i$ in polar form.
Step 1 — Find $r$.
$r = \sqrt{1^2+(-\sqrt{3})^2} = \sqrt{1+3} = 2$
$r = \sqrt{1^2+(-\sqrt{3})^2} = \sqrt{1+3} = 2$
Step 2 — Find $\theta$. Point $(1,-\sqrt{3})$ is in Quadrant IV.
Reference angle: $\arctan\!\left(\dfrac{\sqrt{3}}{1}\right) = \dfrac{\pi}{3}$
QIV: $\theta = 2\pi - \dfrac{\pi}{3} = \dfrac{5\pi}{3}$
Reference angle: $\arctan\!\left(\dfrac{\sqrt{3}}{1}\right) = \dfrac{\pi}{3}$
QIV: $\theta = 2\pi - \dfrac{\pi}{3} = \dfrac{5\pi}{3}$
Step 3 — Write polar form.
$$z = 2\!\left(\cos\frac{5\pi}{3}+i\sin\frac{5\pi}{3}\right) = 2\,\text{cis}\frac{5\pi}{3}$$
$$z = 2\!\left(\cos\frac{5\pi}{3}+i\sin\frac{5\pi}{3}\right) = 2\,\text{cis}\frac{5\pi}{3}$$
$z = 2\!\left(\cos\dfrac{5\pi}{3}+i\sin\dfrac{5\pi}{3}\right)$
📝 练习 E-1 — Coordinate Conversion
(a) Convert $\left(4,\,\dfrac{5\pi}{4}\right)$ to rectangular. (b) Convert $(-2,\,2)$ to polar with $r>0$, $0\le\theta<2\pi$.
(a) $x=4\cos\dfrac{5\pi}{4}=-2\sqrt{2}$, $y=4\sin\dfrac{5\pi}{4}=-2\sqrt{2}$ → $(-2\sqrt{2},-2\sqrt{2})$
(b) $r=\sqrt{4+4}=2\sqrt{2}$; QII: $\theta=\pi-\dfrac{\pi}{4}=\dfrac{3\pi}{4}$ → $\left(2\sqrt{2},\,\dfrac{3\pi}{4}\right)$
(b) $r=\sqrt{4+4}=2\sqrt{2}$; QII: $\theta=\pi-\dfrac{\pi}{4}=\dfrac{3\pi}{4}$ → $\left(2\sqrt{2},\,\dfrac{3\pi}{4}\right)$
📝 练习 E-2 — The Other Direction & Points on the Axes 反向转换与坐标轴上的点
(a) Write $4\left(\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}\right)$ in the form $a+bi$. (b) Write $3\left(\cos\frac{3\pi}{2}+i\sin\frac{3\pi}{2}\right)$ in the form $a+bi$.
(c) Convert $(0,-5)$ and $(-4,0)$ to polar coordinates with $r>0$, $0\le\theta<2\pi$.
(a) $4\left(-\frac12\right)+4\left(\frac{\sqrt3}{2}\right)i=-2+2\sqrt3\,i$. (b) $3(0)+3(-1)i=-3i$.
(c) $(0,-5)$: $x=0$, so $\arctan\frac{y}{x}$ is undefined; the point is on the negative $y$-axis: $\left(5,\frac{3\pi}{2}\right)$. $(-4,0)$: on the negative $x$-axis: $(4,\pi)$.
CED form: $\theta=\arctan\frac{y}{x}$ for $x>0$, and $\theta=\arctan\frac{y}{x}+\pi$ for $x<0$.
(c) $(0,-5)$: $x=0$, so $\arctan\frac{y}{x}$ is undefined; the point is on the negative $y$-axis: $\left(5,\frac{3\pi}{2}\right)$. $(-4,0)$: on the negative $x$-axis: $(4,\pi)$.
CED form: $\theta=\arctan\frac{y}{x}$ for $x>0$, and $\theta=\arctan\frac{y}{x}+\pi$ for $x<0$.
9.2 Polar Function Graphs 极函数图像
What Is a Polar Function? 什么是极函数？
A polar function（极函数）is written as $r = f(\theta)$.
- Input: angle $\theta$; Output: radius $r$
- Each point has coordinates $(r,\theta) = (f(\theta),\,\theta)$
- As $\theta$ changes, the point rotates around the pole while $r$ changes
极函数 vs 直角函数图像的区别
直角函数图像是"左右移动"，而极函数图像是"旋转"：随着 $\theta$ 增大，点绕极点旋转，同时半径 $r$ 随函数变化。极函数可以产生螺旋、花瓣、心形等在直角坐标中很难描述的美丽曲线。
Interactive Polar Curve Tracer · 极坐标曲线描线器
Watch angle $\theta$ rotate while the signed radius $r$ determines the plotted point. 观察“转角度 + 走半径”如何生成曲线。
Complete curve 完整轮廓
Traced path 已描曲线
Angle direction $\theta$
Plotted radius 实际半径
使用描线器时重点观察
先看紫色虚线给出的原始角度 $\theta$，再看 $r$ 的正负。当 $r<0$ 时，橙色半径会转到 $\theta+\pi$ 的方向；因此 rose curve 的花瓣即使出现负半径，也仍然会被画在正确位置。切换三种曲线，比较 periodic（周期型）图像与不断向外扩张的 spiral。
Basic Polar Graphs 基本极坐标图像
Circle centered at pole 以极点为圆心的圆: $r = a$
A circle centered at the pole with radius $|a|$.
A circle centered at the pole with radius $|a|$.
Line through pole 过极点的直线: $\theta = \alpha$
A line through the pole making angle $\alpha$ with the polar axis.
A line through the pole making angle $\alpha$ with the polar axis.
Circles from sine and cosine:
$r = a\cos\theta$ → circle through pole, diameter $|a|$ on $x$-axis, centered at $\left(\dfrac{a}{2},0\right)$
$r = a\sin\theta$ → circle through pole, diameter $|a|$ on $y$-axis, centered at $\left(0,\dfrac{a}{2}\right)$
$r = a\cos\theta$ → circle through pole, diameter $|a|$ on $x$-axis, centered at $\left(\dfrac{a}{2},0\right)$
$r = a\sin\theta$ → circle through pole, diameter $|a|$ on $y$-axis, centered at $\left(0,\dfrac{a}{2}\right)$
为什么 $r=a\cos\theta$ 是圆？
两边乘以 $r$：$r^2 = ar\cos\theta$。用 $r^2=x^2+y^2$ 和 $r\cos\theta=x$ 代入：$x^2+y^2 = ax$，即 $\left(x-\dfrac{a}{2}\right)^2+y^2=\dfrac{a^2}{4}$，这正是圆心 $\left(\dfrac{a}{2},0\right)$、半径 $\dfrac{|a|}{2}$ 的圆。
Common Polar Graph Types 常见极坐标曲线
| Name 名称 | Chinese | General Form | Key Feature 特征 |
|---|---|---|---|
| Circle | 圆 | $r=a$; $r=a\cos\theta$; $r=a\sin\theta$ | Fixed distance or circular equation |
| Line | 直线 | $\theta=\alpha$ | Constant angle through pole |
| Cardioid | 心形线 | $r=a(1\pm\cos\theta)$ or $r=a(1\pm\sin\theta)$ | Heart-shaped; passes through pole |
| Limaçon | 蚶线 | $r=a\pm b\cos\theta$ or $r=a\pm b\sin\theta$ | $a/b$ ratio determines shape (inner loop / dimple / convex) |
| Spiral | 螺线 | $r=a\theta$ | $r$ increases as $\theta$ increases |
| Rose | 玫瑰线 | $r=a\sin(n\theta)$ or $r=a\cos(n\theta)$ | Petal-shaped; $n$ petals if $n$ odd, $2n$ if $n$ even |
玫瑰线花瓣数规律
对于 $r=a\sin(n\theta)$ 或 $r=a\cos(n\theta)$：
- 若 $n$ 为奇数，有 $n$ 片花瓣
- 若 $n$ 为偶数，有 $2n$ 片花瓣
Curve Gallery 常见极坐标曲线图集
Circle 圆
$r=2$
Circle through pole
$r=4\cos\theta$
Cardioid 心形线
$r=2+2\cos\theta$
Inner loop 内环
$r=1+2\cos\theta$
Dimpled 凹陷
$r=3+2\cos\theta$
Convex 凸
$r=4+\cos\theta$
Rose, 3 petals
$r=3\sin3\theta$
Rose, 4 petals
$r=3\cos2\theta$
TI-84 · Polar Graphing 极坐标作图
- MODE → choose POL (and RADIAN).
- Y= → enter $r_1=\dots$ (the X,T,θ,n key types $\theta$).
- WINDOW: θmin $=0$, θmax $=2\pi$ (or the interval you need), θstep $=\pi/24$; then ZOOM → 5: ZSquare so circles look round.
- TRACE shows $\theta$, $r$, and the $(x,y)$ point as the curve is drawn.
Graphing Strategy 绘图策略
- Identify the graph type if possible (circle, cardioid, limaçon, rose, spiral).
- Choose a useful interval for $\theta$, often $[0,2\pi]$.
- Make a table of values for $\theta$ and $r$.
- Plot each polar point $(r,\theta)$, paying attention to negative $r$ values.
- Check whether the curve retraces itself; use symmetry to simplify.
负半径的处理
当 $r<0$ 时，点在 $\theta$ 对应方向的反方向。例如，极坐标 $\left(-3,\dfrac{\pi}{4}\right)$ 与 $\left(3,\dfrac{5\pi}{4}\right)$ 是同一个点。
🤝 一起做 Together T-5 — Describing a Polar Circle
Describe the graph of $r = 6\cos\theta$.
Convert to rectangular form.
Multiply both sides by $r$: $r^2 = 6r\cos\theta$
Substitute $r^2=x^2+y^2$ and $r\cos\theta=x$: $$x^2+y^2 = 6x$$
Multiply both sides by $r$: $r^2 = 6r\cos\theta$
Substitute $r^2=x^2+y^2$ and $r\cos\theta=x$: $$x^2+y^2 = 6x$$
Complete the square.
$(x^2-6x)+y^2=0 \Rightarrow (x-3)^2+y^2=9$
$(x^2-6x)+y^2=0 \Rightarrow (x-3)^2+y^2=9$
Identify.
Circle centered at $(3,0)$ with radius $3$. The circle passes through the origin (pole).
Circle centered at $(3,0)$ with radius $3$. The circle passes through the origin (pole).
Circle centered at $(3,0)$ with radius $3$.
🤝 一起做 Together T-6 — Classifying a Limaçon
Classify the graph of $r = 2 + 4\cos\theta$.
解题思路
$r = a + b\cos\theta$ 的形式是 limaçon（蚶线）。比较 $|a|$ 和 $|b|$ 的大小决定形状：
- $|a| \ge 2|b|$：凸蚶线（convex limaçon）
- $|b| < |a| < 2|b|$：带凹陷的蚶线（dimpled limaçon）
- $|a| = |b|$：心形线（cardioid）
- $|a| < |b|$：有内环的蚶线（limaçon with inner loop）
Identify $a$ and $b$.
$a=2$, $b=4$. Since $|a|=2 < |b|=4$, this is a limaçon with an inner loop（有内环的蚶线）.
$a=2$, $b=4$. Since $|a|=2 < |b|=4$, this is a limaçon with an inner loop（有内环的蚶线）.
Key values.
$\theta=0$: $r=2+4=6$ (max, rightmost point)
$\theta=\pi/2$: $r=2+0=2$
$\theta=\pi$: $r=2-4=-2$, plotted on the opposite side at $(2,0)$: the point of the inner loop farthest from the pole
$\theta=3\pi/2$: $r=2+0=2$
$\theta=0$: $r=2+4=6$ (max, rightmost point)
$\theta=\pi/2$: $r=2+0=2$
$\theta=\pi$: $r=2-4=-2$, plotted on the opposite side at $(2,0)$: the point of the inner loop farthest from the pole
$\theta=3\pi/2$: $r=2+0=2$
Where is the inner loop?
$r<0$ when $2+4\cos\theta<0$, i.e. $\cos\theta<-\tfrac12$: $\;\dfrac{2\pi}{3}<\theta<\dfrac{4\pi}{3}$. The inner loop is traced on this interval; it starts and ends at the pole ($r=0$ at $\theta=\tfrac{2\pi}{3}$ and $\tfrac{4\pi}{3}$).
$r<0$ when $2+4\cos\theta<0$, i.e. $\cos\theta<-\tfrac12$: $\;\dfrac{2\pi}{3}<\theta<\dfrac{4\pi}{3}$. The inner loop is traced on this interval; it starts and ends at the pole ($r=0$ at $\theta=\tfrac{2\pi}{3}$ and $\tfrac{4\pi}{3}$).
Limaçon with inner loop (traced for $\frac{2\pi}{3}<\theta<\frac{4\pi}{3}$); max $r=6$, min $r=-2$.
🤝 一起做 Together T-7 — Spiral
Describe the graph of $r = \theta$ for $\theta\ge 0$.
Interpret the equation.
As $\theta$ increases from $0$, the radius $r=\theta$ increases proportionally. The point rotates around the pole while moving farther away.
As $\theta$ increases from $0$, the radius $r=\theta$ increases proportionally. The point rotates around the pole while moving farther away.
Key values.
$\theta=0$: $r=0$ (at the pole)
$\theta=\pi/2$: $r=\pi/2\approx 1.57$
$\theta=\pi$: $r=\pi\approx 3.14$
$\theta=2\pi$: $r=2\pi\approx 6.28$
$\theta=0$: $r=0$ (at the pole)
$\theta=\pi/2$: $r=\pi/2\approx 1.57$
$\theta=\pi$: $r=\pi\approx 3.14$
$\theta=2\pi$: $r=2\pi\approx 6.28$
Archimedean spiral（阿基米德螺线）: starts at the pole and spirals outward with ever-increasing radius.
📝 练习 E-3 — Identify Polar Graphs
Identify and describe each polar curve: (a) $r=3\sin(2\theta)$ (b) $r=4(1-\sin\theta)$ (c) $r=5$
(a) Rose curve（玫瑰线）: $n=2$ (even) → $2\times2=4$ petals, each with length $3$
(b) Cardioid（心形线）: $a=b=4$; passes through pole; opens downward (due to $-\sin\theta$)
(c) Circle（圆）: centered at pole with radius $5$
(b) Cardioid（心形线）: $a=b=4$; passes through pole; opens downward (due to $-\sin\theta$)
(c) Circle（圆）: centered at pole with radius $5$
Reading the Graph of $r$ vs. $\theta$ 由 $r$–$\theta$ 直角坐标图读极坐标曲线
AP questions often show $r=f(\theta)$ drawn on rectangular axes (horizontal axis $\theta$, vertical axis $r$). Read it like this:
- Where the graph is above the $\theta$-axis ($r>0$), points are plotted in the direction $\theta$; below ($r<0$), they are plotted in the direction $\theta+\pi$.
- Where the graph crosses the $\theta$-axis ($r=0$), the polar curve passes through the pole.
- Distance from the origin is $|r|$: moving away from the $\theta$-axis means moving away from the pole (CED 3.15.A.1–A.2).
- To draw only part of the curve, pick the $\theta$-interval whose endpoints give the desired angles and radii (CED 3.14.A.2).
🤝 一起做 Together T-8 — From the r–θ Graph to the Polar Curve 由直角坐标图到极坐标曲线
The graph of $r=f(\theta)=1+2\cos\theta$ for $0\le\theta\le2\pi$ is shown on rectangular axes (left) and as a polar curve (right).
$r$ vs. $\theta$ (rectangular axes)
polar curve
(a) For which $\theta$ is $r<0$? Where are those points plotted? (b) Which $\theta$-interval traces the inner loop? (c) For which $\theta$ does the curve pass through the pole?
(d) Is the distance from the origin increasing or decreasing on $\left(\frac{2\pi}{3},\pi\right)$? On $\left(\pi,\frac{4\pi}{3}\right)$? On $\left(0,\frac{2\pi}{3}\right)$? Give a reason.
(a) $1+2\cos\theta<0 \Rightarrow \cos\theta<-\tfrac12 \Rightarrow \tfrac{2\pi}{3}<\theta<\tfrac{4\pi}{3}$ (the red part). These points are plotted in the opposite direction, $\theta+\pi$, i.e. on the right side of the pole.
(b) The inner loop is exactly the red part: $\tfrac{2\pi}{3}\le\theta\le\tfrac{4\pi}{3}$.
(c) $r=0$ at $\theta=\tfrac{2\pi}{3}$ and $\theta=\tfrac{4\pi}{3}$ (where the rectangular graph crosses the $\theta$-axis).
(d) On $\left(\frac{2\pi}{3},\pi\right)$: $r$ is negative and decreasing → distance increasing.
On $\left(\pi,\frac{4\pi}{3}\right)$: $r$ is negative and increasing → distance decreasing.
On $\left(0,\frac{2\pi}{3}\right)$: $r$ is positive and decreasing → distance decreasing.
On $\left(\pi,\frac{4\pi}{3}\right)$: $r$ is negative and increasing → distance decreasing.
On $\left(0,\frac{2\pi}{3}\right)$: $r$ is positive and decreasing → distance decreasing.
$r<0$ on $\left(\frac{2\pi}{3},\frac{4\pi}{3}\right)$ (inner loop); through the pole at $\theta=\frac{2\pi}{3},\frac{4\pi}{3}$; distance increasing only on $\left(\frac{2\pi}{3},\pi\right)$ of these three intervals.
📝 练习 E-4 — Petals of a Rose 玫瑰线的花瓣
For $r=3\sin(2\theta)$:
(a) Which $\theta$-interval in $[0,\pi]$ traces the petal in Quadrant I? (b) On $\left[\frac{\pi}{2},\pi\right]$, $r\le0$. In which quadrant is that petal drawn? (c) Which interval traces the petal in Quadrant III?
(d) On $\left(\frac{\pi}{4},\frac{\pi}{2}\right)$, is the point moving toward or away from the origin?
$r$ vs. $\theta$: teal $r\ge0$, red $r\le0$
four petals
(c) $\pi\le\theta\le\frac{3\pi}{2}$ ($r\ge0$ again, angles in QIII). (d) $r$ is positive and decreasing (from 3 to 0) → moving toward the origin.
9.3 Rates of Change in Polar Functions 极函数的变化率
Average Rate of Change 平均变化率
For a polar function $r=f(\theta)$, the average rate of change（平均变化率）of $r$ with respect to $\theta$ on $[\theta_1,\theta_2]$ is:
$$\frac{\Delta r}{\Delta\theta} = \frac{f(\theta_2)-f(\theta_1)}{\theta_2-\theta_1}$$
Units: radius units per radian. This tells how fast $r$ changes per radian of rotation.
Distance from the Pole 与极点的距离
The actual distance from the pole is $|r|$, not $r$.
When $r < 0$, the point is still plotted — on the opposite side — and the distance from the pole is $|r|$.
| Behavior of $r=f(\theta)$ | Distance from pole |
|---|---|
| $r>0$ and increasing | Increasing (moving away) |
| $r>0$ and decreasing | Decreasing (moving closer) |
| $r<0$ and increasing (less negative) | Decreasing (moving closer) |
| $r<0$ and decreasing (more negative) | Increasing (moving away) |
距离 = $|r|$，而不是 $r$
当 $r$ 为负值时，$|r|$ 才是点到极点的真实距离。例如，若 $r$ 从 $-3$ 增大到 $-1$（$r$ 在增大），但 $|r|$ 从 $3$ 减小到 $1$（点实际上在靠近极点）。这是极坐标中最容易混淆的概念之一！
Relative Extrema in Polar Context 极坐标中的相对极值
- A relative maximum（相对最大值）of $r$ often corresponds to a point farthest from the pole.
- A relative minimum（相对最小值）of $r$ often corresponds to a point closest to the pole.
- Exception: when $r$ is negative, always analyze $|r|$ separately.
🤝 一起做 Together T-9 — Behavior of a Cardioid
Describe the behavior of $r = 1+\cos\theta$ over $0\le\theta\le 2\pi$: when is the point moving away from the pole? When is it closest?
Key values of $r$.
$\theta=0$: $r=2$ (maximum, farthest from pole)
$\theta=\pi/2$: $r=1$ (decreasing)
$\theta=\pi$: $r=0$ (at the pole, minimum distance)
$\theta=3\pi/2$: $r=1$ (increasing)
$\theta=2\pi$: $r=2$ (back to maximum)
$\theta=0$: $r=2$ (maximum, farthest from pole)
$\theta=\pi/2$: $r=1$ (decreasing)
$\theta=\pi$: $r=0$ (at the pole, minimum distance)
$\theta=3\pi/2$: $r=1$ (increasing)
$\theta=2\pi$: $r=2$ (back to maximum)
Behavior analysis.
On $[0,\pi]$: $r$ decreases from $2$ to $0$ → point moves closer to pole.
On $[\pi,2\pi]$: $r$ increases from $0$ to $2$ → point moves farther from pole.
On $[0,\pi]$: $r$ decreases from $2$ to $0$ → point moves closer to pole.
On $[\pi,2\pi]$: $r$ increases from $0$ to $2$ → point moves farther from pole.
Farthest at $\theta=0$ ($r=2$); closest (at the pole) at $\theta=\pi$ ($r=0$).
🤝 一起做 Together T-10 — Average Rate of Change on Subintervals
For $r = 4-2\sin\theta$, find the average rate of change on each interval: $\left[0,\dfrac{\pi}{2}\right]$, $\left[\dfrac{\pi}{2},\pi\right]$, $\left[\pi,\dfrac{3\pi}{2}\right]$, $\left[\dfrac{3\pi}{2},2\pi\right]$.
Compute $r$ at key angles.
$r(0) = 4-2(0) = 4$
$r(\pi/2) = 4-2(1) = 2$
$r(\pi) = 4-2(0) = 4$
$r(3\pi/2) = 4-2(-1) = 6$
$r(2\pi) = 4-2(0) = 4$
$r(0) = 4-2(0) = 4$
$r(\pi/2) = 4-2(1) = 2$
$r(\pi) = 4-2(0) = 4$
$r(3\pi/2) = 4-2(-1) = 6$
$r(2\pi) = 4-2(0) = 4$
Compute AROC on each interval.
$\left[0,\dfrac{\pi}{2}\right]$: $\dfrac{2-4}{\pi/2} = \dfrac{-2}{\pi/2} = -\dfrac{4}{\pi}$ (moving closer)
$\left[\dfrac{\pi}{2},\pi\right]$: $\dfrac{4-2}{\pi/2} = \dfrac{4}{\pi}$ (moving farther)
$\left[\pi,\dfrac{3\pi}{2}\right]$: $\dfrac{6-4}{\pi/2} = \dfrac{4}{\pi}$ (moving farther)
$\left[\dfrac{3\pi}{2},2\pi\right]$: $\dfrac{4-6}{\pi/2} = -\dfrac{4}{\pi}$ (moving closer)
$\left[0,\dfrac{\pi}{2}\right]$: $\dfrac{2-4}{\pi/2} = \dfrac{-2}{\pi/2} = -\dfrac{4}{\pi}$ (moving closer)
$\left[\dfrac{\pi}{2},\pi\right]$: $\dfrac{4-2}{\pi/2} = \dfrac{4}{\pi}$ (moving farther)
$\left[\pi,\dfrac{3\pi}{2}\right]$: $\dfrac{6-4}{\pi/2} = \dfrac{4}{\pi}$ (moving farther)
$\left[\dfrac{3\pi}{2},2\pi\right]$: $\dfrac{4-6}{\pi/2} = -\dfrac{4}{\pi}$ (moving closer)
AROCs: $-\dfrac{4}{\pi}$, $\dfrac{4}{\pi}$, $\dfrac{4}{\pi}$, $-\dfrac{4}{\pi}$ (radius units per radian)
📝 练习 E-5 — Polar Distance Analysis
For $r = 3\cos\theta$ on $[0,\pi]$:
(a) When is the point farthest from the pole? (b) Find the AROC of $r$ on $\left[0,\dfrac{\pi}{3}\right]$.
(a) $r(0)=3$ (max), $r(\pi/2)=0$, $r(\pi)=-3$. Distance $|r|$ is max at $\theta=0$ AND $\theta=\pi$ (both give $|r|=3$); farthest from pole at $\theta=0$ and $\theta=\pi$. (These are the same point $(3,0)$: $(-3,\pi)=(3,0)$. The circle $r=3\cos\theta$ is traced once on $[0,\pi]$.)
(b) $r(0)=3$, $r(\pi/3)=3\cos(\pi/3)=3\cdot\tfrac{1}{2}=\tfrac{3}{2}$. AROC $= \dfrac{3/2-3}{\pi/3}=\dfrac{-3/2}{\pi/3}=-\dfrac{9}{2\pi}$
(b) $r(0)=3$, $r(\pi/3)=3\cos(\pi/3)=3\cdot\tfrac{1}{2}=\tfrac{3}{2}$. AROC $= \dfrac{3/2-3}{\pi/3}=\dfrac{-3/2}{\pi/3}=-\dfrac{9}{2\pi}$
📝 练习 E-6 — Rose Curve Analysis
For $r = 2\sin(3\theta)$:
(a) How many petals does the curve have? (b) Find the maximum value of $|r|$ and the angles at which it occurs (on $[0,\pi]$).
(a) $n=3$ (odd) → 3 petals.
(b) $|r|$ is max when $|\sin(3\theta)|=1$, i.e., $3\theta=\pi/2+k\pi$, so $\theta=\pi/6,\,\pi/2,\,5\pi/6$ on $[0,\pi]$. Max $|r|=2$.
(b) $|r|$ is max when $|\sin(3\theta)|=1$, i.e., $3\theta=\pi/2+k\pi$, so $\theta=\pi/6,\,\pi/2,\,5\pi/6$ on $[0,\pi]$. Max $|r|=2$.
Using the Average Rate of Change to Estimate 用平均变化率估计函数值
If $r=f(\theta)$ is known only at some angles, the average rate of change on $[\theta_1,\theta_2]$ gives an estimate inside the interval (CED 3.15.A.5):
$$f(\theta)\approx f(\theta_1)+\frac{f(\theta_2)-f(\theta_1)}{\theta_2-\theta_1}\,(\theta-\theta_1).$$
相当于在两个已知点之间"按平均速率走"，是一种线性估计。
🤝 一起做 Together T-11 — Estimate from a Table 由表格估计
Selected values of a polar function $r=f(\theta)$ are shown.
| $\theta$ | $0$ | $\frac{\pi}{6}$ | $\frac{\pi}{3}$ | $\frac{\pi}{2}$ |
|---|---|---|---|---|
| $f(\theta)$ | 2 | 3.5 | 4.598 | 5 |
(a) Find the average rate of change of $f$ on $\left[\frac{\pi}{6},\frac{\pi}{3}\right]$ and interpret it. (b) Use it to estimate $f\left(\frac{\pi}{4}\right)$. (c) On $\left[0,\frac{\pi}{2}\right]$, is the point moving away from the origin? Explain.
(a) $\dfrac{4.598-3.5}{\pi/3-\pi/6}=\dfrac{1.098}{\pi/6}\approx2.097$: on this interval the radius increases by about 2.097 units per radian.
(b) $f\left(\frac{\pi}{4}\right)\approx3.5+2.097\cdot\frac{\pi}{12}\approx4.049$. (These values come from $f(\theta)=2+3\sin\theta$, whose actual value is $\approx4.121$, so the estimate is close.)
(c) The values are positive and increasing ($2\to3.5\to4.598\to5$), so the distance from the origin is increasing (assuming $f$ is increasing between the table values).
AROC $\approx2.097$ units/radian; $f(\pi/4)\approx4.049$; moving away from the origin.
📝 练习 E-7 — Sign, Direction, and Distance 符号、方向与距离
For $r=2\cos(2\theta)$ on $\left[0,\frac{\pi}{2}\right]$:
(a) On which subinterval is $r>0$? $r<0$? (b) On each subinterval, is the distance from the origin increasing or decreasing? (c) Where is the point at $\theta=\frac{\pi}{2}$ (rectangular coordinates)?
(a) $r>0$ on $\left[0,\frac{\pi}{4}\right)$; $r<0$ on $\left(\frac{\pi}{4},\frac{\pi}{2}\right]$ ($r=0$ at $\theta=\frac{\pi}{4}$: the pole).
(b) On $\left(0,\frac{\pi}{4}\right)$: positive and decreasing → distance decreasing. On $\left(\frac{\pi}{4},\frac{\pi}{2}\right)$: negative and decreasing → distance increasing.
(c) $r=2\cos\pi=-2$: $(x,y)=(-2\cos\frac{\pi}{2},\,-2\sin\frac{\pi}{2})=(0,-2)$, on the negative $y$-axis.
(b) On $\left(0,\frac{\pi}{4}\right)$: positive and decreasing → distance decreasing. On $\left(\frac{\pi}{4},\frac{\pi}{2}\right)$: negative and decreasing → distance increasing.
(c) $r=2\cos\pi=-2$: $(x,y)=(-2\cos\frac{\pi}{2},\,-2\sin\frac{\pi}{2})=(0,-2)$, on the negative $y$-axis.
Chapter 9 Summary 本章总结
| Topic 主题 | Key Formula / Idea 核心公式 |
|---|---|
| Polar point | $(r,\theta)$: $r$ = distance, $\theta$ = angle from polar axis |
| Multiple representations | $(r,\theta)=(r,\theta+2\pi k)=(-r,\theta+\pi)$ |
| Polar → Rectangular | $x=r\cos\theta$, $y=r\sin\theta$ |
| Rectangular → Polar | $r=\sqrt{x^2+y^2}$; $\theta$ from quadrant check |
| Complex polar form | $z=r(\cos\theta+i\sin\theta)$; $r=|z|$ |
| Circle $r=a$ | Circle at pole, radius $|a|$ |
| Circle $r=a\cos\theta$ | Circle through pole, center $\left(\dfrac{a}{2},0\right)$, radius $\dfrac{|a|}{2}$ |
| Cardioid | $r=a(1\pm\cos\theta)$ or $a(1\pm\sin\theta)$; heart-shaped |
| Rose | $r=a\sin(n\theta)$: $n$ odd → $n$ petals; $n$ even → $2n$ petals |
| AROC of polar | $\dfrac{f(\theta_2)-f(\theta_1)}{\theta_2-\theta_1}$ in radius units/radian |
| Distance from pole | Always $|r|$; analyze sign of $r$ carefully |
| Limaçon $a+b\cos\theta$ | $|a|\ge2|b|$ convex; $|b|<|a|<2|b|$ dimpled; $|a|=|b|$ cardioid; $|a|<|b|$ inner loop |
| $r$–$\theta$ graph | Above axis: plot at $\theta$; below: at $\theta+\pi$; crossing: through the pole |
| Part of a curve | Choose the $\theta$-interval whose endpoints give the wanted angles/radii |
| Estimate | $f(\theta)\approx f(\theta_1)+\text{AROC}\cdot(\theta-\theta_1)$ |
考试重点提示
- 极坐标转换时，务必根据象限判断 $\theta$，不能只依赖 $\arctan$。
- 识别极坐标曲线类型：看方程形式→判断是圆、心形线、蚶线还是玫瑰线。
- AROC 计算：单位是"半径/弧度"，注意与普通函数AROC单位区别。
- 距离从极点 = $|r|$，当 $r<0$ 时点在对面，但距离仍是 $|r|$。
- 复数极形式：$r=|z|$ 是模，$\theta$ 是辐角——与极坐标完全对应。
- 看到 $r$ 关于 $\theta$ 的直角坐标图：在 $\theta$ 轴下方的部分画在反方向，与 $\theta$ 轴相交处曲线过极点。
- "正且增 / 负且减 → 远离原点；正且减 / 负且增 → 靠近原点"，答题时要写出 $r$ 的正负和增减作为理由。
