Axiom Learning·AP PrecalculusChapter 10: Parametric Functions
参数方程 — Planar Motion, Conics, and Implicit Equations
本章概览
普通函数 $y=f(x)$ 只能告诉我们点在哪里，却无法告诉我们点何时到达那里。参数方程引入第三个变量（通常是时间 $t$），将 $x$ 和 $y$ 分别用 $t$ 表示，从而同时描述位置和运动方向。本章还将学习隐式方程和圆锥曲线（抛物线、椭圆、双曲线）。
本章在考纲中的位置 Where this chapter fits
本章对应 CED Unit 4 · Topics 4.1–4.7（Functions Involving Parameters）。Unit 4 是 "Additional Topic Available to Schools"：不在 AP Precalculus 考试范围内（Not assessed on the AP Exam），由学校决定是否讲授。这部分内容为微积分中的隐函数求导以及 BC 的参数方程和平面运动打基础。
| Term 术语 | Chinese 中文 | Meaning 含义 |
|---|---|---|
| Parametric equations | 参数方程 | $x=x(t)$, $y=y(t)$ — express both coordinates via parameter $t$ |
| Parameter | 参数 | The variable $t$ (often time) |
| Planar motion | 平面运动 | Particle moving in the $xy$-plane, position $(x(t),y(t))$ |
| Implicit equation | 隐式方程 | Equation involving both $x$ and $y$, e.g. $x^2+y^2=16$ |
| Explicit equation | 显式方程 | $y$ solved in terms of $x$: $y=f(x)$ |
| Conic section | 圆锥曲线 | Circle, ellipse, parabola, or hyperbola |
| Focus / foci | 焦点 | Fixed point(s) in the definition of conic sections |
| Directrix | 准线 | Fixed line in the definition of a parabola |
| Vertex | 顶点 | Turning point / peak of a conic |
| Major / minor axis | 长轴 / 短轴 | Longest / shortest diameter of an ellipse |
| Asymptote (hyperbola) | 渐近线 | Lines the hyperbola approaches |
10.1 Parametric Functions 参数方程
Parametric Equations 参数方程
Let $x$ and $y$ be functions of the same variable $t$ (the parameter): $$x = x(t), \qquad y = y(t)$$ The curve is the set of ordered pairs $\{(x(t),\,y(t))\}$ as $t$ varies over its domain.
The value of $t$ controls both the location and the direction of motion（运动方向）.
Let $x$ and $y$ be functions of the same variable $t$ (the parameter): $$x = x(t), \qquad y = y(t)$$ The curve is the set of ordered pairs $\{(x(t),\,y(t))\}$ as $t$ varies over its domain.
The value of $t$ controls both the location and the direction of motion（运动方向）.
为什么需要参数方程？
普通函数 $y=f(x)$ 有两大局限：
- 无法描述运动时刻（只知道轨迹，不知道何时到达）
- 无法表示像圆这样的非函数曲线（无法通过竖直线测试）
Graphing Parametric Equations 参数方程作图
- Choose values of $t$ in order (increasing).
- Compute $x(t)$ and $y(t)$ for each $t$.
- Plot ordered pairs $(x(t),\,y(t))$.
- Connect the points in order of increasing $t$.
- Add arrows showing the direction of motion.
🤝 一起做 Together T-1 — Graphing a Parametric Curve
Create a table of values and describe the parametric curve for $(x(t),\,y(t))=(-t^2+t,\;2t-1)$.
Build the table.
| $t$ | $x=-t^2+t$ | $y=2t-1$ | $(x,y)$ |
|---|---|---|---|
| $-1$ | $-2$ | $-3$ | $(-2,-3)$ |
| $0$ | $0$ | $-1$ | $(0,-1)$ |
| $\tfrac{1}{2}$ | $\tfrac{1}{4}$ | $0$ | $(\tfrac{1}{4},0)$ |
| $1$ | $0$ | $1$ | $(0,1)$ |
| $2$ | $-2$ | $3$ | $(-2,3)$ |
Eliminate $t$ to find the shape.
From $y=2t-1$: $t=\dfrac{y+1}{2}$.
Substitute into $x=-t^2+t$: $$x = -\left(\frac{y+1}{2}\right)^2+\frac{y+1}{2} = -\frac{(y+1)^2}{4}+\frac{y+1}{2}$$ This is a horizontal parabola opening to the left, with vertex at $\left(\tfrac{1}{4},0\right)$ (where $t=\tfrac{1}{2}$).
From $y=2t-1$: $t=\dfrac{y+1}{2}$.
Substitute into $x=-t^2+t$: $$x = -\left(\frac{y+1}{2}\right)^2+\frac{y+1}{2} = -\frac{(y+1)^2}{4}+\frac{y+1}{2}$$ This is a horizontal parabola opening to the left, with vertex at $\left(\tfrac{1}{4},0\right)$ (where $t=\tfrac{1}{2}$).
Points labeled by $t$; arrows show the direction of increasing $t$ (shown for $-1\le t\le2$).
Horizontal parabola opening left; as $t$ increases, the curve moves upward (direction of motion: upward ↑).
🤝 一起做 Together T-2 — Table of Values with Restricted Domain
Create a table of values for $(x(t),\,y(t))=\!\left(2t-4,\;\dfrac{t}{2}\right)$, $-2\le t\le 3$.
| $t$ | $x=2t-4$ | $y=t/2$ | $(x,y)$ |
|---|---|---|---|
| $-2$ | $-8$ | $-1$ | $(-8,-1)$ |
| $0$ | $-4$ | $0$ | $(-4,0)$ |
| $1$ | $-2$ | $\tfrac{1}{2}$ | $(-2,\tfrac{1}{2})$ |
| $3$ | $2$ | $\tfrac{3}{2}$ | $(2,\tfrac{3}{2})$ |
Shape. From $x=2t-4$: $t=\dfrac{x+4}{2}$; $y=\dfrac{t}{2}=\dfrac{x+4}{4}$. This is a line segment with slope $\tfrac{1}{4}$, from $(-8,-1)$ to $(2,\tfrac{3}{2})$.
Restricted domain → the graph has a start point ($t=-2$) and an end point ($t=3$).
Line segment, direction: left to right (as $t$ increases, $x$ increases).
10.2 Parametric Functions Modeling Planar Motion 平面运动建模
A parametric function $(x(t),\,y(t))$ models the position of a particle at time $t$.
At time $t=t_0$, the particle is at $\bigl(x(t_0),\;y(t_0)\bigr)$.
At time $t=t_0$, the particle is at $\bigl(x(t_0),\;y(t_0)\bigr)$.
Parametric Particle Motion · 参数粒子运动
One parameter $t$ updates both coordinates at the same time. 同一个 $t$ 同步决定 $x(t)$、$y(t)$、位置与运动方向。
Complete path 完整轨迹
Traveled trail 已走轨迹
Motion direction 运动方向
Current position 当前位置
如何读这个动画？
灰色虚线只告诉我们path（轨迹形状）；绿色轨迹和紫色箭头还告诉我们粒子如何随着 $t$ 增大而运动。特别注意：同一条曲线可以有不同的 parametrization，因此路径相同，不代表起点、方向和速度变化相同。可以暂停后拖动时间轴，逐点检查 $(x(t),y(t))$。
Extrema and Intercepts 极值与截距
- Horizontal extrema（水平极值）: max/min values of $x(t)$ — analyze $x(t)$ separately
- Vertical extrema（竖直极值）: max/min values of $y(t)$ — analyze $y(t)$ separately
- $x$-intercepts: where $y(t)=0$
- $y$-intercepts: where $x(t)=0$
🤝 一起做 Together T-3 — Particle Position
A particle moves so that $x=t^2$, $y=\sqrt{t}$. Graph the path over $0\le t\le 4$ and find the position when $t=10$.
Build table.
| $t$ | $x=t^2$ | $y=\sqrt{t}$ |
|---|---|---|
| $0$ | $0$ | $0$ |
| $1$ | $1$ | $1$ |
| $2$ | $4$ | $\approx1.41$ |
| $4$ | $16$ | $2$ |
Eliminate $t$. $y=\sqrt{t}$ means $t=y^2$, so $x=y^4$ — the path is the curve $x=y^4$ for $y\ge0$.
Path for $0\le t\le4$, from $(0,0)$ to $(16,2)$. (Different scales on the two axes.)
Position at $t=10$. $x(10)=100$, $y(10)=\sqrt{10}\approx3.16$. Position: $(100,\sqrt{10})$.
At $t=10$: position $(100,\sqrt{10})$.
🤝 一起做 Together T-4 — Extrema and Intercepts
For the parametric curve $x=3t^2-2$, $y=2t$, $-2\le t\le2$:
Find: (a) the maximum and minimum values of $x(t)$ and of $y(t)$, with the corresponding points (horizontal and vertical extrema), (b) $x$- and $y$-intercepts.
(a) Extrema.
$x(t)=3t^2-2$: quadratic, minimum at $t=0$: $x(0)=-2$; endpoints $x(\pm2)=10$ (max).
$y(t)=2t$: linear; min at $t=-2$: $y=-4$; max at $t=2$: $y=4$.
Points: leftmost $(-2,0)$ ($t=0$); rightmost $(10,-4)$ and $(10,4)$ ($t=\mp2$); lowest $(10,-4)$; highest $(10,4)$.
$x(t)=3t^2-2$: quadratic, minimum at $t=0$: $x(0)=-2$; endpoints $x(\pm2)=10$ (max).
$y(t)=2t$: linear; min at $t=-2$: $y=-4$; max at $t=2$: $y=4$.
Points: leftmost $(-2,0)$ ($t=0$); rightmost $(10,-4)$ and $(10,4)$ ($t=\mp2$); lowest $(10,-4)$; highest $(10,4)$.
(b) Intercepts.
$x$-intercept: $y(t)=0 \Rightarrow 2t=0 \Rightarrow t=0$. Point: $(x(0),0)=(-2,0)$.
$y$-intercept: $x(t)=0 \Rightarrow 3t^2-2=0 \Rightarrow t=\pm\sqrt{2/3}$.
Points: $\left(0,\pm2\sqrt{2/3}\right)=\left(0,\pm\dfrac{2\sqrt{6}}{3}\right)$.
$x$-intercept: $y(t)=0 \Rightarrow 2t=0 \Rightarrow t=0$. Point: $(x(0),0)=(-2,0)$.
$y$-intercept: $x(t)=0 \Rightarrow 3t^2-2=0 \Rightarrow t=\pm\sqrt{2/3}$.
Points: $\left(0,\pm2\sqrt{2/3}\right)=\left(0,\pm\dfrac{2\sqrt{6}}{3}\right)$.
$x$: min $-2$ ($t=0$), max $10$ ($t=\pm2$); $y$: min $-4$ ($t=-2$), max $4$ ($t=2$).
$x$-int: $(-2,0)$; $y$-int: $\left(0,\pm\dfrac{2\sqrt{6}}{3}\right)$.
$x$-int: $(-2,0)$; $y$-int: $\left(0,\pm\dfrac{2\sqrt{6}}{3}\right)$.
📝 练习 E-1 — Intercepts of Parametric Curve
Find the $x$- and $y$-intercepts of the parametric curve $x=-t^2+t$, $y=2t-1$.
$x$-intercept ($y=0$): $2t-1=0\Rightarrow t=\tfrac{1}{2}$; $x=(-\tfrac{1}{4}+\tfrac{1}{2})=\tfrac{1}{4}$. Point: $\left(\tfrac{1}{4},0\right)$.
$y$-intercept ($x=0$): $-t^2+t=0\Rightarrow t(1-t)=0\Rightarrow t=0$ or $t=1$.
At $t=0$: $(0,-1)$. At $t=1$: $(0,1)$. Two $y$-intercepts: $(0,-1)$ and $(0,1)$.
$y$-intercept ($x=0$): $-t^2+t=0\Rightarrow t(1-t)=0\Rightarrow t=0$ or $t=1$.
At $t=0$: $(0,-1)$. At $t=1$: $(0,1)$. Two $y$-intercepts: $(0,-1)$ and $(0,1)$.
10.3 Parametric Functions and Rates of Change 参数方程的变化率
Direction of Motion 运动方向
Moving Right →
$x(t)$ increasingMoving Left ←
$x(t)$ decreasingMoving Up ↑
$y(t)$ increasingMoving Down ↓
$y(t)$ decreasing分区间判断方向 Direction, interval by interval
把 $x(t)$ 与 $y(t)$ 分开看：先找出每个函数在哪些 $t$ 区间上递增、递减，再组合起来。例如某区间上 $x$ 递减、$y$ 递增 → 粒子向左上运动。注意：① 曲线若两次经过同一点，两次的运动方向可能不同（见 T-8）；② 同一条曲线可以有不同的参数化，起点、方向、快慢都可能不同。
Average Rate of Change 平均变化率
Over an interval $[t_1,t_2]$, the average rates of change of $x$ and of $y$ (each with respect to $t$) are computed independently:
$$\frac{\Delta x}{\Delta t} = \frac{x(t_2)-x(t_1)}{t_2-t_1}, \qquad \frac{\Delta y}{\Delta t} = \frac{y(t_2)-y(t_1)}{t_2-t_1}$$
Their ratio is the slope of the secant line between the points at $t_1$ and $t_2$:
$$\text{slope} = \frac{\Delta y / \Delta t}{\Delta x / \Delta t} = \frac{y(t_2)-y(t_1)}{x(t_2)-x(t_1)} \qquad \bigl(\text{requires } x(t_2)\neq x(t_1)\bigr)$$
🤝 一起做 Together T-5 — Direction and AROC
For $x(t)=3\cos t$, $y(t)=2\sin t$, $0\le t\le 2\pi$: (a) describe the path and the direction of motion, interval by interval; (b) find the average rates of change of $x$ and of $y$ over $\left[0,\frac{\pi}{2}\right]$ and the slope of the secant line between the corresponding points.
Identify the shape.
$\dfrac{x^2}{9}+\dfrac{y^2}{4}=\cos^2 t+\sin^2 t=1$. This is an ellipse with $a=3$, $b=2$.
$\dfrac{x^2}{9}+\dfrac{y^2}{4}=\cos^2 t+\sin^2 t=1$. This is an ellipse with $a=3$, $b=2$.
Direction of motion.
At $t=0$: $(3,0)$ (rightmost point).
At $t=\pi/2$: $(0,2)$ (top). At $t=\pi$: $(-3,0)$. At $t=3\pi/2$: $(0,-2)$.
Interval by interval（分区间）:
Motion: counterclockwise（逆时针）around the ellipse.
At $t=0$: $(3,0)$ (rightmost point).
At $t=\pi/2$: $(0,2)$ (top). At $t=\pi$: $(-3,0)$. At $t=3\pi/2$: $(0,-2)$.
Interval by interval（分区间）:
| $t$-interval | $x(t)=3\cos t$ | $y(t)=2\sin t$ | Motion |
|---|---|---|---|
| $\left(0,\frac{\pi}{2}\right)$ | decreasing | increasing | left & up ↖ |
| $\left(\frac{\pi}{2},\pi\right)$ | decreasing | decreasing | left & down ↙ |
| $\left(\pi,\frac{3\pi}{2}\right)$ | increasing | decreasing | right & down ↘ |
| $\left(\frac{3\pi}{2},2\pi\right)$ | increasing | increasing | right & up ↗ |
(b) AROC over $\left[0,\frac{\pi}{2}\right]$.
$\dfrac{\Delta x}{\Delta t}=\dfrac{0-3}{\pi/2}=-\dfrac{6}{\pi}\approx-1.910$, $\dfrac{\Delta y}{\Delta t}=\dfrac{2-0}{\pi/2}=\dfrac{4}{\pi}\approx1.273$
Slope $=\dfrac{4/\pi}{-6/\pi}=-\dfrac{2}{3}$, the same as $\dfrac{2-0}{0-3}$ for the secant from $(3,0)$ to $(0,2)$.
$\dfrac{\Delta x}{\Delta t}=\dfrac{0-3}{\pi/2}=-\dfrac{6}{\pi}\approx-1.910$, $\dfrac{\Delta y}{\Delta t}=\dfrac{2-0}{\pi/2}=\dfrac{4}{\pi}\approx1.273$
Slope $=\dfrac{4/\pi}{-6/\pi}=-\dfrac{2}{3}$, the same as $\dfrac{2-0}{0-3}$ for the secant from $(3,0)$ to $(0,2)$.
(a) Ellipse $\dfrac{x^2}{9}+\dfrac{y^2}{4}=1$, traversed counterclockwise from $(3,0)$: left & up, left & down, right & down, right & up. (b) $-\dfrac{6}{\pi}$ and $\dfrac{4}{\pi}$; slope $-\dfrac{2}{3}$.
🤝 一起做 Together T-6 — Slope of Secant Line
For $x(t)=t$, $y(t)=1-t^2$, $-1\le t\le 1$, find: (a) AROC of $x(t)$, (b) AROC of $y(t)$, (c) slope of secant from $t=-1$ to $t=1$.
Values at endpoints.
$x(-1)=-1$, $x(1)=1$, $y(-1)=0$, $y(1)=0$.
$x(-1)=-1$, $x(1)=1$, $y(-1)=0$, $y(1)=0$.
(a) AROC of $x$.
$\dfrac{x(1)-x(-1)}{1-(-1)}=\dfrac{1-(-1)}{2}=\dfrac{2}{2}=1$
$\dfrac{x(1)-x(-1)}{1-(-1)}=\dfrac{1-(-1)}{2}=\dfrac{2}{2}=1$
(b) AROC of $y$.
$\dfrac{y(1)-y(-1)}{1-(-1)}=\dfrac{0-0}{2}=0$
$\dfrac{y(1)-y(-1)}{1-(-1)}=\dfrac{0-0}{2}=0$
(c) Slope of secant.
Slope $=\dfrac{y(1)-y(-1)}{x(1)-x(-1)}=\dfrac{0}{2}=0$. (Horizontal secant — makes sense since both points are on the $x$-axis.)
Slope $=\dfrac{y(1)-y(-1)}{x(1)-x(-1)}=\dfrac{0}{2}=0$. (Horizontal secant — makes sense since both points are on the $x$-axis.)
(a) $1$, (b) $0$, (c) slope $= 0$
Analyzing Motion 运动分析
🤝 一起做 Together T-7 — Motion from the Graphs of $x(t)$ and $y(t)$ 由分量图读运动
A particle moves in the plane for $0\le t\le4$. The graphs of its coordinate functions $x(t)$ and $y(t)$ are shown; each is made of line segments.
(a) Find the particle's position at $t=0,1,2,3,4$. (b) On which interval is the particle moving left and down? (c) Find the rightmost and the highest points of the path. (d) When is the particle on the $y$-axis? (e) Find the slope of the secant line between the points at $t=1$ and $t=3$.
(a) Read both graphs at the same $t$.
| $t$ | $0$ | $1$ | $2$ | $3$ | $4$ |
|---|---|---|---|---|---|
| $x(t)$ | $-2$ | $0$ | $2$ | $1$ | $0$ |
| $y(t)$ | $0$ | $2$ | $0$ | $-2$ | $0$ |
| position | $(-2,0)$ | $(0,2)$ | $(2,0)$ | $(1,-2)$ | $(0,0)$ |
(b) Direction. $x$ increases on $(0,2)$ and decreases on $(2,4)$; $y$ increases on $(0,1)$, decreases on $(1,3)$, and increases on $(3,4)$.
Left ($x$ decreasing) and down ($y$ decreasing) at the same time: $\;2<t<3$.
Left ($x$ decreasing) and down ($y$ decreasing) at the same time: $\;2<t<3$.
(c) Extrema. Rightmost = maximum of $x$: $x=2$ at $t=2$ → $(2,0)$. Highest = maximum of $y$: $y=2$ at $t=1$ → $(0,2)$.
(d) $y$-axis. $x(t)=0$ at $t=1$ and $t=4$ → the points $(0,2)$ and $(0,0)$.
(e) Secant slope. $\dfrac{y(3)-y(1)}{x(3)-x(1)}=\dfrac{-2-2}{1-0}=-4$.
The path: starts at $(-2,0)$ when $t=0$ and ends at $(0,0)$ when $t=4$.
(a) $(-2,0),\,(0,2),\,(2,0),\,(1,-2),\,(0,0)$ (b) $2<t<3$ (c) rightmost $(2,0)$, highest $(0,2)$ (d) $t=1$ and $t=4$ (e) $-4$
🤝 一起做 Together T-8 — Same Point, Different Directions 同一点、不同方向
A particle moves along $x(t)=t^2$, $y(t)=t^3-3t$ for $-2\le t\le 2$.
(a) Make a table for $t=-2,-1,0,1,2$ and sketch the path. (b) Show that the particle is at $(3,0)$ at two different times. (c) Describe its direction of motion at each of those times.
(a) Table.
$x=t^2$ decreases for $t<0$ and increases for $t>0$. The cubic $y=t^3-3t$ has turning points at $t=\pm1$: it increases on $(-2,-1)$, decreases on $(-1,1)$, and increases on $(1,2)$.
| $t$ | $-2$ | $-1$ | $0$ | $1$ | $2$ |
|---|---|---|---|---|---|
| $x=t^2$ | $4$ | $1$ | $0$ | $1$ | $4$ |
| $y=t^3-3t$ | $-2$ | $2$ | $0$ | $-2$ | $2$ |
(b) When is the particle at $(3,0)$? $x=3\Rightarrow t^2=3\Rightarrow t=\pm\sqrt3$. $y(\pm\sqrt3)=\pm3\sqrt3\mp3\sqrt3=0$ ✓.
So the particle is at $(3,0)$ when $t=-\sqrt3$ and again when $t=\sqrt3$.
So the particle is at $(3,0)$ when $t=-\sqrt3$ and again when $t=\sqrt3$.
(c) Directions.
$t=-\sqrt3\approx-1.732$ is in $(-2,-1)$: $x$ decreasing, $y$ increasing → moving left and up ↖.
$t=\sqrt3\approx1.732$ is in $(1,2)$: $x$ increasing, $y$ increasing → moving right and up ↗.
$t=-\sqrt3\approx-1.732$ is in $(-2,-1)$: $x$ decreasing, $y$ increasing → moving left and up ↖.
$t=\sqrt3\approx1.732$ is in $(1,2)$: $x$ increasing, $y$ increasing → moving right and up ↗.
The path crosses itself at $(3,0)$. Red: direction at $t=-\sqrt3$; blue: direction at $t=\sqrt3$.
The particle is at $(3,0)$ when $t=-\sqrt3$ (moving left & up) and when $t=\sqrt3$ (moving right & up). The direction at a point depends on when the particle is there.
📝 练习 E-2 — Particle Motion Analysis
A particle moves with $x(t)=t^2-4$, $y(t)=t$, $-3\le t\le3$.
(a) When is the particle moving right? (b) Find its position at $t=2$. (c) Find AROC of $x$ over $[0,2]$.
(a) $x(t)=t^2-4$ is increasing when $t>0$ (for $t\in[0,3]$). Moving right when $t>0$.
(b) $x(2)=0$, $y(2)=2$ → position $(0,2)$.
(c) AROC of $x$: $\dfrac{x(2)-x(0)}{2-0}=\dfrac{0-(-4)}{2}=2$
(b) $x(2)=0$, $y(2)=2$ → position $(0,2)$.
(c) AROC of $x$: $\dfrac{x(2)-x(0)}{2-0}=\dfrac{0-(-4)}{2}=2$
10.4 Parametrically Defined Lines and Circles 直线与圆的参数方程
Line Segment Parametrization 线段的参数方程
A line segment from $(x_0,y_0)$ at $t=0$ to $(x_1,y_1)$ at $t=1$:
$$x(t) = x_0 + (x_1-x_0)\,t, \qquad y(t) = y_0 + (y_1-y_0)\,t, \qquad 0\le t\le 1$$
For the full infinite line, let $t$ range over all reals.
线段参数方程的记忆方式
把它理解为"从起点出发，方向向量乘以 $t$"：方向向量 $= (x_1-x_0,\;y_1-y_0)$；当 $t=0$ 时在起点，$t=1$ 时到终点。$t$ 可以理解为"走了多少比例的路程"。
Initial position + rates 起点 + 速率: if a particle starts at $(x_1,y_1)$ and $x$ changes at a constant rate $a$ and $y$ at a constant rate $b$ (per unit of $t$), then
$$x(t)=x_1+at,\qquad y(t)=y_1+bt$$
The path is a line with slope $\dfrac{b}{a}$ ($a\neq0$). The same segment can be parametrized in many ways: faster, slower, or in the opposite direction.
Circle Parametrization 圆的参数方程
Circle of radius $r$ centered at $(h,k)$:
$$x(t) = h + r\cos t, \qquad y(t) = k + r\sin t, \qquad 0\le t\le 2\pi$$
(Traversed counterclockwise; start at $(h+r,\,k)$.)
Transforming $(\cos t,\sin t)$ 圆周运动的变换
Start from the unit circle: $(x(t),y(t))=(\cos t,\sin t)$, $0\le t\le2\pi$, is one counterclockwise revolution that starts and ends at $(1,0)$. Transformations of it model any circular path:
| Change 改变 | Parametrization | Effect 效果 |
|---|---|---|
| Radius, center | $(h+r\cos t,\;k+r\sin t)$ | radius $r$, center $(h,k)$ |
| Direction | $(\cos t,\;-\sin t)$ | clockwise（顺时针）, still starts at $(1,0)$ |
| Starting point | $(\sin t,\;\cos t)$ | starts at the top $(0,1)$, clockwise |
| Speed / period | $(\cos bt,\;\sin bt)$, $0\le t\le\frac{2\pi}{b}$ | one revolution takes $\frac{2\pi}{b}$ time units |
| Part of the circle | restrict $t$, e.g. $0\le t\le\pi$ | only an arc (here the upper semicircle) |
🤝 一起做 Together T-9 — Parametrizing a Line
Find a parametrization of the line through $A=(-2,3)$ and $B=(3,6)$.
Direction vector.
$(x_1-x_0,\;y_1-y_0) = (3-(-2),\;6-3) = (5,\;3)$
$(x_1-x_0,\;y_1-y_0) = (3-(-2),\;6-3) = (5,\;3)$
Write parametric equations.
$$x(t)=-2+5t, \qquad y(t)=3+3t$$ For the line segment $A$ to $B$: $0\le t\le 1$. For the full line: $t\in\mathbb{R}$.
$$x(t)=-2+5t, \qquad y(t)=3+3t$$ For the line segment $A$ to $B$: $0\le t\le 1$. For the full line: $t\in\mathbb{R}$.
Verify. $t=0$: $(-2,3)=A$ ✓; $t=1$: $(3,6)=B$ ✓
$x(t)=-2+5t$, $y(t)=3+3t$
🤝 一起做 Together T-10 — Parametrizing a Circle
Find parametric equations for the circle $(x-1)^2+(y+2)^2=9$.
Read off center and radius.
Center $(h,k)=(1,-2)$, radius $r=3$.
Center $(h,k)=(1,-2)$, radius $r=3$.
Apply the formula.
$$x(t) = 1+3\cos t, \qquad y(t) = -2+3\sin t, \qquad 0\le t\le 2\pi$$
$$x(t) = 1+3\cos t, \qquad y(t) = -2+3\sin t, \qquad 0\le t\le 2\pi$$
$x(t)=1+3\cos t$, $y(t)=-2+3\sin t$, $0\le t\le 2\pi$
🤝 一起做 Together T-11 — Segment from a Starting Point and Rates 由起点和速率写线段
A drone starts at $(-1,4)$ (units: km). Each minute, its $x$-coordinate increases by $2$ and its $y$-coordinate decreases by $3$. It flies in a straight line for $3$ minutes.
(a) Write a parametrization for $0\le t\le3$. (b) Where does it stop? (c) Give a different parametrization of the same segment, traced from the stopping point back to the start.
(a) Initial position $(-1,4)$; rates $\dfrac{\Delta x}{\Delta t}=2$, $\dfrac{\Delta y}{\Delta t}=-3$:
$$x(t)=-1+2t,\qquad y(t)=4-3t,\qquad 0\le t\le3$$
(b) $t=3$: $(-1+6,\;4-9)=(5,-5)$. The slope of the path is $\dfrac{-3}{2}$.
(c) Start at $(5,-5)$ and reverse both rates: $x(t)=5-2t$, $y(t)=-5+3t$, $0\le t\le3$. Same set of points, opposite direction.
(a) $(-1+2t,\;4-3t)$, $0\le t\le3$ (b) $(5,-5)$ (c) $(5-2t,\;-5+3t)$, $0\le t\le3$ (one of many)
🤝 一起做 Together T-12 — Clockwise Motion with a Given Start and Period 顺时针与周期
A point moves clockwise around the circle $x^2+y^2=9$. It starts at $(0,3)$ and completes one revolution every $4$ seconds.
(a) Write a parametrization for $0\le t\le4$. (b) Find the position at $t=2.5$. (c) Which interval of $t$ traces only the quarter-circle in Quadrant I?
Start and direction. $(\sin\theta,\cos\theta)$ starts at $(0,1)$ and moves clockwise (at $\theta=\frac{\pi}{2}$ it is at $(1,0)$). Multiply by the radius $3$.
Period. One revolution ($\theta$ from $0$ to $2\pi$) takes $4$ seconds, so $\theta=\dfrac{2\pi}{4}t=\dfrac{\pi}{2}t$:
$$x(t)=3\sin\!\left(\frac{\pi}{2}t\right),\qquad y(t)=3\cos\!\left(\frac{\pi}{2}t\right),\qquad 0\le t\le4$$
Check: $t=0$: $(0,3)$ ✓; $t=1$: $(3,0)$ ✓ (clockwise); $t=4$: back to $(0,3)$ ✓.
(b) $\frac{\pi}{2}(2.5)=\frac{5\pi}{4}$: $\;\left(3\sin\frac{5\pi}{4},\;3\cos\frac{5\pi}{4}\right)=\left(-\frac{3\sqrt2}{2},\,-\frac{3\sqrt2}{2}\right)\approx(-2.121,\,-2.121)$.
(c) Quadrant I is traced from $(0,3)$ to $(3,0)$: $0\le t\le1$.
Amber arc: the Quadrant I part, traced for $0\le t\le1$.
(a) $\left(3\sin\frac{\pi t}{2},\;3\cos\frac{\pi t}{2}\right)$, $0\le t\le4$ (b) $\approx(-2.121,\,-2.121)$ (c) $0\le t\le1$
📝 练习 E-3 — Circle Parametrization
Write parametric equations for the circle $x^2+(y-4)^2=25$.
Center $(0,4)$, radius $5$: $x(t)=5\cos t$, $y(t)=4+5\sin t$, $0\le t\le 2\pi$
📝 练习 E-4 — Circular Paths 圆周路径
Write a parametrization for each motion on the circle $(x+2)^2+(y-1)^2=16$.
(a) One counterclockwise revolution starting at $(2,1)$, $0\le t\le2\pi$. (b) One clockwise revolution starting at the top point $(-2,5)$, $0\le t\le2\pi$. (c) Same as (a), but one revolution takes $\pi$ seconds.
(a) $x=-2+4\cos t$, $y=1+4\sin t$, $0\le t\le2\pi$.
(b) $x=-2+4\sin t$, $y=1+4\cos t$, $0\le t\le2\pi$. Check $t=\frac{\pi}{2}$: $(2,1)$, to the right of the top point, so the motion is clockwise ✓.
(c) $x=-2+4\cos 2t$, $y=1+4\sin 2t$, $0\le t\le\pi$.
(b) $x=-2+4\sin t$, $y=1+4\cos t$, $0\le t\le2\pi$. Check $t=\frac{\pi}{2}$: $(2,1)$, to the right of the top point, so the motion is clockwise ✓.
(c) $x=-2+4\cos 2t$, $y=1+4\sin 2t$, $0\le t\le\pi$.
10.5 Implicitly Defined Functions 隐式方程
- Explicit equation（显式方程）: $y$ is isolated → $y=f(x)$
- Implicit equation（隐式方程）: involves both $x$ and $y$, neither necessarily isolated
Examples: $x^2+y^2=16$, $x^3+y^3=1$
隐式方程与显式方程的关系
圆 $x^2+y^2=16$ 是隐式方程。它无法用单一函数 $y=f(x)$ 表示（不满足竖直线测试），但可以拆成两个显式函数：上半圆 $y=\sqrt{16-x^2}$ 和下半圆 $y=-\sqrt{16-x^2}$。
Graphing an Implicit Equation 隐式方程作图
- Choose values of one variable (for example $x$).
- Substitute and solve for the other variable; there may be $0$, $1$, $2$, or more solutions.
- Plot all the solution points and connect them smoothly.
- Each solved branch $y=g(x)$ is a function whose graph is part of the curve.
Rate of Change on Implicit Curves 隐式曲线上的变化率
For two points on an implicit curve that are close together:
- $\dfrac{\Delta y}{\Delta x}>0$: the two variables increase together or decrease together
- $\dfrac{\Delta y}{\Delta x}<0$: as one variable increases, the other decreases
- $\dfrac{\Delta y}{\Delta x}=0$: $y$ does not change → horizontal（水平）; $\dfrac{\Delta x}{\Delta y}=0$: $x$ does not change → vertical（竖直）
🤝 一起做 Together T-13 — Rate of Change on a Circle
Given $x^2+y^2=4$, find the average rate of change of $y$ with respect to $x$: (a) from $(-2,0)$ to $(0,2)$ (b) from $(0,2)$ to $(2,0)$
Verify points are on the circle.
$(-2)^2+0^2=4$ ✓; $0^2+2^2=4$ ✓; $2^2+0^2=4$ ✓
$(-2)^2+0^2=4$ ✓; $0^2+2^2=4$ ✓; $2^2+0^2=4$ ✓
(a) From $(-2,0)$ to $(0,2)$.
AROC $= \dfrac{2-0}{0-(-2)} = \dfrac{2}{2} = 1$ (both $x$ and $y$ increase → positive)
AROC $= \dfrac{2-0}{0-(-2)} = \dfrac{2}{2} = 1$ (both $x$ and $y$ increase → positive)
(b) From $(0,2)$ to $(2,0)$.
AROC $= \dfrac{0-2}{2-0} = \dfrac{-2}{2} = -1$ ($x$ increases, $y$ decreases → negative)
AROC $= \dfrac{0-2}{2-0} = \dfrac{-2}{2} = -1$ ($x$ increases, $y$ decreases → negative)
(a) $1$ (b) $-1$
🤝 一起做 Together T-14 — Graphing $x^2+xy+y^2=3$ 隐式曲线作图
(a) Find the points on $x^2+xy+y^2=3$ with $x=-2,-1,0,1,2$. Are there points with $x>2$?
(b) Solve for $y$ to write two functions whose graphs together make up the curve.
(c) Find $\dfrac{\Delta y}{\Delta x}$ from $(-2,1)$ to $(-1,2)$ and from $(-1,2)$ to $(1,1)$, and interpret the signs.
(d) Use the symmetry of the equation to find the highest and lowest points. Where is the curve vertical, and where is it horizontal?
(a) For fixed $x$, solve a quadratic in $y$.
$y^2+xy+(x^2-3)=0\;\Rightarrow\; y=\dfrac{-x\pm\sqrt{x^2-4(x^2-3)}}{2}=\dfrac{-x\pm\sqrt{12-3x^2}}{2}$
For $|x|>2$, $12-3x^2<0$: no real $y$. The whole curve lies in $-2\le x\le2$.
$y^2+xy+(x^2-3)=0\;\Rightarrow\; y=\dfrac{-x\pm\sqrt{x^2-4(x^2-3)}}{2}=\dfrac{-x\pm\sqrt{12-3x^2}}{2}$
| $x$ | $-2$ | $-1$ | $0$ | $1$ | $2$ |
|---|---|---|---|---|---|
| $\sqrt{12-3x^2}$ | $0$ | $3$ | $\sqrt{12}$ | $3$ | $0$ |
| $y$ | $1$ | $2,\;-1$ | $\pm\sqrt3\approx\pm1.732$ | $1,\;-2$ | $-1$ |
(b) Two functions. $\;y_1=\dfrac{-x+\sqrt{12-3x^2}}{2}$ (upper part) and $y_2=\dfrac{-x-\sqrt{12-3x^2}}{2}$ (lower part), each with domain $[-2,2]$. Each passes the vertical line test; together they form the curve.
(c) Signs. $(-2,1)\to(-1,2)$: $\dfrac{2-1}{-1-(-2)}=1>0$, so $x$ and $y$ increase together.
$(-1,2)\to(1,1)$: $\dfrac{1-2}{1-(-1)}=-\dfrac12<0$, so as $x$ increases, $y$ decreases.
$(-1,2)\to(1,1)$: $\dfrac{1-2}{1-(-1)}=-\dfrac12<0$, so as $x$ increases, $y$ decreases.
(d) Symmetry. Swapping $x$ and $y$ gives the same equation, so the graph is symmetric about $y=x$. The leftmost and rightmost points $(-2,1)$ and $(2,-1)$ (only one $y$-value there) reflect to the lowest point $(1,-2)$ and the highest point $(-1,2)$.
Vertical at $(-2,1)$ and $(2,-1)$ ($x$ momentarily stops changing); horizontal at $(-1,2)$ and $(1,-2)$ ($y$ momentarily stops changing).
Vertical at $(-2,1)$ and $(2,-1)$ ($x$ momentarily stops changing); horizontal at $(-1,2)$ and $(1,-2)$ ($y$ momentarily stops changing).
A tilted ellipse. Teal: $y_1$; purple: $y_2$; dots: the solutions from the table.
(a) $(-2,1)$; $(-1,2),(-1,-1)$; $(0,\pm\sqrt3)$; $(1,1),(1,-2)$; $(2,-1)$; none with $x>2$. (b) $y=\dfrac{-x\pm\sqrt{12-3x^2}}{2}$. (c) $1$ (together); $-\dfrac12$ (opposite). (d) highest $(-1,2)$, lowest $(1,-2)$; vertical at $(\pm2,\mp1)$, horizontal at $(\mp1,\pm2)$.
10.6 Conic Sections 圆锥曲线
圆锥曲线是怎么来的？
圆锥曲线是用一个平面截双圆锥（两个顶点相对的圆锥）得到的四种曲线：圆、椭圆、抛物线、双曲线。截的角度不同，得到的曲线形状也不同。
Parabola 抛物线
Equidistant from focus and directrix.Vertex at peak/trough.
Ellipse 椭圆
Sum of distances to two foci = constant.Has major and minor axes.
Circle 圆
Special ellipse: both foci at center.$x^2+y^2=r^2$
Hyperbola 双曲线
Difference of distances to two foci = constant.Two branches, asymptotes.
Parabola 抛物线
$y+2=\frac12x^2$
Circle 圆
$x^2+y^2=9$
Ellipse 椭圆
$\frac{x^2}{16}+\frac{y^2}{4}=1$
Hyperbola 双曲线
$\frac{x^2}{4}-\frac{y^2}{9}=1$
Parabolas 抛物线
A parabola（抛物线）is the set of all points equidistant from a fixed point (focus（焦点）) and a fixed line (directrix（准线）).
The vertex（顶点）is the midpoint between focus and directrix.
The vertex（顶点）is the midpoint between focus and directrix.
Vertical Parabola 竖向抛物线（opens up or down）:
$$y-k = a(x-h)^2 \qquad \text{Vertex: }(h,k); \quad a>0\text{ opens up}, \;a<0\text{ opens down}$$
Horizontal Parabola 横向抛物线（opens right or left）:
$$x-h = a(y-k)^2 \qquad \text{Vertex: }(h,k); \quad a>0\text{ opens right}, \;a<0\text{ opens left}$$
区分竖向与横向
竖向抛物线：$y$ 是关于 $x$ 的函数 → 可以过竖直线测试。横向抛物线：$x$ 是关于 $y$ 的函数 → 不能过竖直线测试（不是函数）。
🤝 一起做 Together T-15 — Parabola: Vertex and Orientation
State the vertex and orientation of $(x-1)^2 = -4(y+8)$.
Rewrite in standard form.
$(x-1)^2 = -4(y+8)$ matches $a(x-h)^2=(y-k)$ form.
Solve for $y$: $y+8 = -\dfrac{1}{4}(x-1)^2 \Rightarrow y-(-8)= -\dfrac{1}{4}(x-1)^2$
$(x-1)^2 = -4(y+8)$ matches $a(x-h)^2=(y-k)$ form.
Solve for $y$: $y+8 = -\dfrac{1}{4}(x-1)^2 \Rightarrow y-(-8)= -\dfrac{1}{4}(x-1)^2$
Read off parameters.
Vertex: $(h,k) = (1,-8)$; $a=-\dfrac{1}{4}<0$ → opens downward.
Vertex: $(h,k) = (1,-8)$; $a=-\dfrac{1}{4}<0$ → opens downward.
Vertex $(1,-8)$; opens downward.
🤝 一起做 Together T-16 — Writing a Parabola Equation
Write the equation of a parabola with vertex $(-2,4)$ that passes through $(0,0)$ and opens to the right.
Use horizontal parabola form.
$x-h = a(y-k)^2 \Rightarrow x-(-2) = a(y-4)^2 \Rightarrow x+2=a(y-4)^2$
$x-h = a(y-k)^2 \Rightarrow x-(-2) = a(y-4)^2 \Rightarrow x+2=a(y-4)^2$
Plug in $(0,0)$ to find $a$.
$0+2=a(0-4)^2 \Rightarrow 2=16a \Rightarrow a=\dfrac{1}{8}$
Since $a=\dfrac{1}{8}>0$, the parabola opens right ✓
$0+2=a(0-4)^2 \Rightarrow 2=16a \Rightarrow a=\dfrac{1}{8}$
Since $a=\dfrac{1}{8}>0$, the parabola opens right ✓
Final equation.
$$x+2=\frac{1}{8}(y-4)^2$$
$$x+2=\frac{1}{8}(y-4)^2$$
$x+2=\dfrac{1}{8}(y-4)^2$
Ellipses 椭圆
An ellipse（椭圆）is the set of all points whose sum of distances to two fixed foci is constant.
Standard Forms 标准形式（center at origin）:
Horizontal ellipse ($a>b>0$, longer along $x$-axis): $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, \quad c^2=a^2-b^2$$ Vertices: $(\pm a,0)$; Foci: $(\pm c,0)$
Vertical ellipse ($a>b>0$, longer along $y$-axis): $$\frac{x^2}{b^2}+\frac{y^2}{a^2}=1, \quad c^2=a^2-b^2$$ Vertices: $(0,\pm a)$; Foci: $(0,\pm c)$
Centered at $(h,k)$: $$\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1 \quad\text{(horizontal major axis)}$$ $$\frac{(x-h)^2}{b^2}+\frac{(y-k)^2}{a^2}=1 \quad\text{(vertical major axis)}$$
CED notation 考纲写法: the CED writes every ellipse as $\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1$ with $a$ = horizontal radius（水平半径）and $b$ = vertical radius（竖直半径）, whichever is larger. A circle is the case $a=b$.
本页的 $a$ 表示长半轴（semi-major axis）。两种写法都对；但写参数方程 $x=h+(\text{水平半径})\cos t$、$y=k+(\text{竖直半径})\sin t$ 时，一定要看清哪个分母在 $x$ 下面。
本页的 $a$ 表示长半轴（semi-major axis）。两种写法都对；但写参数方程 $x=h+(\text{水平半径})\cos t$、$y=k+(\text{竖直半径})\sin t$ 时，一定要看清哪个分母在 $x$ 下面。
🤝 一起做 Together T-17 — Ellipse Properties
Describe the key properties of $\dfrac{x^2}{9}+\dfrac{y^2}{4}=1$.
Read $a$ and $b$.
$a^2=9\Rightarrow a=3$; $b^2=4\Rightarrow b=2$. Since $a>b$, major axis is horizontal.
$a^2=9\Rightarrow a=3$; $b^2=4\Rightarrow b=2$. Since $a>b$, major axis is horizontal.
Find $c$.
$c^2=a^2-b^2=9-4=5\Rightarrow c=\sqrt{5}$
$c^2=a^2-b^2=9-4=5\Rightarrow c=\sqrt{5}$
List properties.
Center: $(0,0)$; Vertices: $(\pm3,0)$; Co-vertices: $(0,\pm2)$; Foci: $(\pm\sqrt{5},0)$.
Center: $(0,0)$; Vertices: $(\pm3,0)$; Co-vertices: $(0,\pm2)$; Foci: $(\pm\sqrt{5},0)$.
Center $(0,0)$; vertices $(\pm3,0)$; co-vertices $(0,\pm2)$; foci $(\pm\sqrt{5},0)$.
🤝 一起做 Together T-18 — Ellipse Centered at $(h,k)$
Find the center and foci of $\dfrac{(x-1)^2}{4}+\dfrac{(y+7)^2}{17}=1$.
Read center.
Center: $(h,k)=(1,-7)$
Center: $(h,k)=(1,-7)$
Identify $a^2$ and $b^2$.
Denominators: $4$ and $17$. Since $17>4$, $a^2=17$ (vertical major axis), $b^2=4$.
$c^2=17-4=13\Rightarrow c=\sqrt{13}$
Denominators: $4$ and $17$. Since $17>4$, $a^2=17$ (vertical major axis), $b^2=4$.
$c^2=17-4=13\Rightarrow c=\sqrt{13}$
Foci.
Vertical major axis → foci at $(h,k\pm c)=(1,\,-7\pm\sqrt{13})$.
Vertical major axis → foci at $(h,k\pm c)=(1,\,-7\pm\sqrt{13})$.
Center $(1,-7)$; foci $(1,\,-7\pm\sqrt{13})$.
Hyperbolas 双曲线
A hyperbola（双曲线）is the set of all points where the difference of distances to two foci is constant. It has two branches and two asymptotes.
Horizontal Hyperbola 横向双曲线:
$$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1, \quad c^2=a^2+b^2$$
Vertices: $(\pm a,0)$; Foci: $(\pm c,0)$; Asymptotes: $y=\pm\dfrac{b}{a}x$
Vertical Hyperbola 竖向双曲线: $$\frac{y^2}{a^2}-\frac{x^2}{b^2}=1, \quad c^2=a^2+b^2$$ Vertices: $(0,\pm a)$; Foci: $(0,\pm c)$; Asymptotes: $y=\pm\dfrac{a}{b}x$
Centered at $(h,k)$（CED form: $a$ goes with $x$, $b$ goes with $y$）: $$\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1 \;\text{(opens left/right)}$$ $$\frac{(y-k)^2}{b^2}-\frac{(x-h)^2}{a^2}=1 \;\text{(opens up/down)}$$ Asymptotes (both cases): $\;y-k=\pm\dfrac{b}{a}(x-h)$
不论用哪种字母写法：渐近线斜率 $=\pm\dfrac{\sqrt{y\text{ 下面的分母}}}{\sqrt{x\text{ 下面的分母}}}$，并且渐近线经过中心 $(h,k)$。
椭圆 vs 双曲线关键区别
- 椭圆：$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$（加号 +），$c^2=a^2-b^2$（$c<a$）
- 双曲线：$\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$（减号 −），$c^2=a^2+b^2$（$c>a$）
- 椭圆无渐近线；双曲线有两条渐近线 $y=\pm\dfrac{b}{a}x$
🤝 一起做 Together T-19 — Hyperbola Key Features
List the key features of $9x^2-y^2=81$.
Rewrite in standard form.
Divide by $81$: $\dfrac{x^2}{9}-\dfrac{y^2}{81}=1$
Divide by $81$: $\dfrac{x^2}{9}-\dfrac{y^2}{81}=1$
Read $a$ and $b$.
$a^2=9\Rightarrow a=3$; $b^2=81\Rightarrow b=9$. Horizontal hyperbola.
$a^2=9\Rightarrow a=3$; $b^2=81\Rightarrow b=9$. Horizontal hyperbola.
Find $c$ and list features.
$c^2=9+81=90\Rightarrow c=3\sqrt{10}$
Center: $(0,0)$; Vertices: $(\pm3,0)$; Foci: $(\pm3\sqrt{10},0)$; Asymptotes: $y=\pm3x$
$c^2=9+81=90\Rightarrow c=3\sqrt{10}$
Center: $(0,0)$; Vertices: $(\pm3,0)$; Foci: $(\pm3\sqrt{10},0)$; Asymptotes: $y=\pm3x$
Dashed lines: asymptotes $y=\pm3x$. (Different scales on the two axes.)
Horizontal hyperbola; vertices $(\pm3,0)$; foci $(\pm3\sqrt{10},0)$; asymptotes $y=\pm3x$.
General Form → Standard Form 配方化标准式
To identify a conic given as $Ax^2+Cy^2+Dx+Ey+F=0$: group the $x$-terms and the $y$-terms, complete the square in each, then divide so the right side is $1$ (for ellipses and hyperbolas).
Look at $A$ and $C$: same sign → ellipse (circle if $A=C$); opposite signs → hyperbola; one of them $0$ → parabola.
Look at $A$ and $C$: same sign → ellipse (circle if $A=C$); opposite signs → hyperbola; one of them $0$ → parabola.
🤝 一起做 Together T-20 — Completing the Square: Ellipse 配方：椭圆
Write $4x^2+9y^2-16x+54y+61=0$ in standard form. Identify the conic, its center, and its horizontal and vertical radii.
Group and factor. $\;4(x^2-4x)+9(y^2+6y)=-61$
Complete each square (add $4\cdot4=16$ and $9\cdot9=81$ to both sides):
$4(x^2-4x+4)+9(y^2+6y+9)=-61+16+81\;\Rightarrow\;4(x-2)^2+9(y+3)^2=36$
$4(x^2-4x+4)+9(y^2+6y+9)=-61+16+81\;\Rightarrow\;4(x-2)^2+9(y+3)^2=36$
Divide by $36$.
$$\frac{(x-2)^2}{9}+\frac{(y+3)^2}{4}=1$$
Ellipse with center $(2,-3)$, horizontal radius $3$, vertical radius $2$.
$\dfrac{(x-2)^2}{9}+\dfrac{(y+3)^2}{4}=1$: ellipse, center $(2,-3)$, horizontal radius $3$, vertical radius $2$.
📝 练习 E-5 — Conic Identification and Properties
Identify each conic and state its center/vertex and key properties:
(a) $\dfrac{(x-1)^2}{36}-\dfrac{(y+4)^2}{13}=1$ (b) $y-3=2(x+1)^2$
(a) Horizontal hyperbola: center $(1,-4)$; $a=6$, $b=\sqrt{13}$, $c=\sqrt{49}=7$; vertices $(1\pm6,-4)=(-5,-4)$ and $(7,-4)$; foci $(1\pm7,-4)$; asymptotes $y+4=\pm\dfrac{\sqrt{13}}{6}(x-1)$
(b) Vertical parabola: $y-3=2(x+1)^2$; vertex $(-1,3)$; $a=2>0$ opens upward.
(b) Vertical parabola: $y-3=2(x+1)^2$; vertex $(-1,3)$; $a=2>0$ opens upward.
📝 练习 E-6 — Identify by Completing the Square 配方识别
Write in standard form and identify each conic: (a) $y^2-4x^2-4y-8x-4=0$ (b) $x-y^2-6y-5=0$
(a) $(y^2-4y)-4(x^2+2x)=4\;\Rightarrow\;(y^2-4y+4)-4(x^2+2x+1)=4+4-4\;\Rightarrow\;(y-2)^2-4(x+1)^2=4$
$\Rightarrow\;\dfrac{(y-2)^2}{4}-\dfrac{(x+1)^2}{1}=1$: hyperbola opening up and down, center $(-1,2)$, vertices $(-1,0)$ and $(-1,4)$, asymptotes $y-2=\pm2(x+1)$.
(b) $x=y^2+6y+5=(y+3)^2-4\;\Rightarrow\;x+4=(y+3)^2$: horizontal parabola, vertex $(-4,-3)$, opens right.
$\Rightarrow\;\dfrac{(y-2)^2}{4}-\dfrac{(x+1)^2}{1}=1$: hyperbola opening up and down, center $(-1,2)$, vertices $(-1,0)$ and $(-1,4)$, asymptotes $y-2=\pm2(x+1)$.
(b) $x=y^2+6y+5=(y+3)^2-4\;\Rightarrow\;x+4=(y+3)^2$: horizontal parabola, vertex $(-4,-3)$, opens right.
10.7 Parametrization of Functions 函数的参数化
Parametrizing Explicit Functions 显式函数的参数化
If $y=f(x)$, the simplest parametrization is:
$$x(t)=t, \qquad y(t)=f(t)$$
Parametrizing Inverses 反函数的参数化
If $y=f(x)$ is parametrized by $(t,\,f(t))$, then the inverse $x=f^{-1}(y)$ is parametrized by swapping roles:
$$\bigl(f(t),\;t\bigr) \quad \text{(same } t\text{, but } x \text{ and } y \text{ swapped)}$$
Geometrically: reflection over the line $y=x$.
Parametric → Rectangular 消去参数
- Solve one parametric equation for $t$.
- Substitute into the other equation.
- State domain/range restrictions from the interval of $t$.
🤝 一起做 Together T-21 — Parametrizing a Function and Its Inverse
For $f(x)=3x-7$: (a) parametrize $f$, (b) find $f^{-1}(x)$, (c) parametrize $f^{-1}$.
(a) Parametrize $f$.
Let $x(t)=t$, $y(t)=f(t)=3t-7$. So the curve is $(t,\;3t-7)$.
Let $x(t)=t$, $y(t)=f(t)=3t-7$. So the curve is $(t,\;3t-7)$.
(b) Find $f^{-1}$.
$y=3x-7 \Rightarrow x=\dfrac{y+7}{3}$. So $f^{-1}(x)=\dfrac{x+7}{3}$.
$y=3x-7 \Rightarrow x=\dfrac{y+7}{3}$. So $f^{-1}(x)=\dfrac{x+7}{3}$.
(c) Parametrize $f^{-1}$.
Swap $x$ and $y$ roles: $x(t)=3t-7$, $y(t)=t$. Or equivalently, let $x(t)=t$, $y(t)=\dfrac{t+7}{3}$.
Swap $x$ and $y$ roles: $x(t)=3t-7$, $y(t)=t$. Or equivalently, let $x(t)=t$, $y(t)=\dfrac{t+7}{3}$.
(a) $(t,3t-7)$ (b) $f^{-1}(x)=\dfrac{x+7}{3}$ (c) $(3t-7,\,t)$ [or $(t,\dfrac{t+7}{3})$]
🤝 一起做 Together T-22 — Parametric to Rectangular (Ellipse)
Write the parametric equations $x=4\cos t$, $y=3\sin t$, $0\le t\le 2\pi$ in rectangular form.
Isolate $\cos t$ and $\sin t$.
$\cos t = \dfrac{x}{4}$, $\sin t = \dfrac{y}{3}$
$\cos t = \dfrac{x}{4}$, $\sin t = \dfrac{y}{3}$
Apply Pythagorean identity.
$\cos^2 t+\sin^2 t=1 \Rightarrow \dfrac{x^2}{16}+\dfrac{y^2}{9}=1$
$\cos^2 t+\sin^2 t=1 \Rightarrow \dfrac{x^2}{16}+\dfrac{y^2}{9}=1$
$\dfrac{x^2}{16}+\dfrac{y^2}{9}=1$ (ellipse, major axis horizontal, $a=4$, $b=3$)
🤝 一起做 Together T-23 — Parametric to Rectangular (Parabola)
Write $x=2t+1$, $y=4t^2-1$, $-3\le t\le 4$ in rectangular form.
Solve $x=2t+1$ for $t$.
$t = \dfrac{x-1}{2}$
$t = \dfrac{x-1}{2}$
Substitute into $y$.
$y = 4\left(\dfrac{x-1}{2}\right)^2-1 = 4\cdot\dfrac{(x-1)^2}{4}-1 = (x-1)^2-1$
$y = 4\left(\dfrac{x-1}{2}\right)^2-1 = 4\cdot\dfrac{(x-1)^2}{4}-1 = (x-1)^2-1$
State domain and range.
When $t=-3$: $x=2(-3)+1=-5$. When $t=4$: $x=9$. So $-5\le x\le9$.
$y=4t^2-1$ on $-3\le t\le4$: minimum $-1$ at $t=0$, maximum $4(16)-1=63$ at $t=4$. So $-1\le y\le63$.
The graph is the part of the parabola $y=(x-1)^2-1$ from $(-5,35)$ to $(9,63)$, passing through the vertex $(1,-1)$.
When $t=-3$: $x=2(-3)+1=-5$. When $t=4$: $x=9$. So $-5\le x\le9$.
$y=4t^2-1$ on $-3\le t\le4$: minimum $-1$ at $t=0$, maximum $4(16)-1=63$ at $t=4$. So $-1\le y\le63$.
The graph is the part of the parabola $y=(x-1)^2-1$ from $(-5,35)$ to $(9,63)$, passing through the vertex $(1,-1)$.
$y=(x-1)^2-1$ for $-5\le x\le9$ (range $-1\le y\le63$): a piece of a parabola, not a line.
Verifying a Parametrization · Parametrizing Conics 验证参数化 · 圆锥曲线的参数化
Verify 验证: $(x(t),y(t))$ parametrizes an implicitly defined curve if substituting $x(t)$ for $x$ and $y(t)$ for $y$ makes the equation true for every $t$ in the domain.
| Curve 曲线 | Parametrization | Why it works |
|---|---|---|
| $y=f(x)$ (solved for $y$) | $(t,\;f(t))$ | replace $x$ with $t$ |
| Parabola $x-h=a(y-k)^2$ (solved for $x$) | $\bigl(h+a(t-k)^2,\;t\bigr)$ | replace $y$ with $t$ |
| Ellipse $\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$ | $(h+a\cos t,\;k+b\sin t)$, $0\le t\le2\pi$ | $\cos^2t+\sin^2t=1$ |
| Hyperbola $\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$ (left/right) | $(h+a\sec t,\;k+b\tan t)$ | $\sec^2t-\tan^2t=1$ |
| Hyperbola $\frac{(y-k)^2}{b^2}-\frac{(x-h)^2}{a^2}=1$ (up/down) | $(h+a\tan t,\;k+b\sec t)$ | $\sec^2t-\tan^2t=1$ |
🤝 一起做 Together T-24 — Parametrizing the Conics from 10.6 参数化圆锥曲线
Parametrize each curve, then verify by substitution.
(a) the parabola $x+2=\frac{1}{8}(y-4)^2$ (from T-16) (b) the ellipse $\dfrac{(x-1)^2}{4}+\dfrac{(y+7)^2}{17}=1$ (from T-18) (c) the hyperbola $\dfrac{x^2}{9}-\dfrac{y^2}{81}=1$ (from T-19)
(a) Solved for $x$, so let $y=t$.
$x(t)=-2+\frac{1}{8}(t-4)^2$, $y(t)=t$, $t$ any real number.
Verify: $x+2=\frac18(t-4)^2=\frac18(y-4)^2$ ✓
$x(t)=-2+\frac{1}{8}(t-4)^2$, $y(t)=t$, $t$ any real number.
Verify: $x+2=\frac18(t-4)^2=\frac18(y-4)^2$ ✓
(b) Radii. Under $x$: $4$ → horizontal radius $2$. Under $y$: $17$ → vertical radius $\sqrt{17}$.
$x(t)=1+2\cos t$, $y(t)=-7+\sqrt{17}\sin t$, $0\le t\le2\pi$.
Verify: $\dfrac{(2\cos t)^2}{4}+\dfrac{(\sqrt{17}\sin t)^2}{17}=\cos^2t+\sin^2t=1$ ✓
⚠ Not $x=1+\sqrt{17}\cos t$: here the larger radius is the vertical one.
$x(t)=1+2\cos t$, $y(t)=-7+\sqrt{17}\sin t$, $0\le t\le2\pi$.
Verify: $\dfrac{(2\cos t)^2}{4}+\dfrac{(\sqrt{17}\sin t)^2}{17}=\cos^2t+\sin^2t=1$ ✓
⚠ Not $x=1+\sqrt{17}\cos t$: here the larger radius is the vertical one.
(c) Opens left/right, $a=3$, $b=9$:
$x(t)=3\sec t$, $y(t)=9\tan t$, $0\le t\le2\pi$, $t\neq\frac{\pi}{2},\frac{3\pi}{2}$.
Verify: $\dfrac{9\sec^2t}{9}-\dfrac{81\tan^2t}{81}=\sec^2t-\tan^2t=1$ ✓
$x(t)=3\sec t$, $y(t)=9\tan t$, $0\le t\le2\pi$, $t\neq\frac{\pi}{2},\frac{3\pi}{2}$.
Verify: $\dfrac{9\sec^2t}{9}-\dfrac{81\tan^2t}{81}=\sec^2t-\tan^2t=1$ ✓
(a) $\left(-2+\frac18(t-4)^2,\;t\right)$ (b) $(1+2\cos t,\;-7+\sqrt{17}\sin t)$ (c) $(3\sec t,\;9\tan t)$, $t\neq\frac{\pi}{2},\frac{3\pi}{2}$
📝 练习 E-7 — Different Parametrizations of the Same Curve
Find three different parametrizations of $y=2x^2-3$, including one that traces the curve from right to left as $t$ increases.
Option 1: $x(t)=t$, $y(t)=2t^2-3$
Option 2: $x(t)=\dfrac{t}{2}$, $y(t)=2\cdot\dfrac{t^2}{4}-3=\dfrac{t^2}{2}-3$ (or any other substitution $x=at$)
Option 3 (opposite direction): $x(t)=-t$, $y(t)=2t^2-3$; as $t$ increases, $x$ decreases, so the curve is traced from right to left.
Check any option by substitution, e.g. Option 3: $2(-t)^2-3=2t^2-3=y(t)$ ✓
Option 2: $x(t)=\dfrac{t}{2}$, $y(t)=2\cdot\dfrac{t^2}{4}-3=\dfrac{t^2}{2}-3$ (or any other substitution $x=at$)
Option 3 (opposite direction): $x(t)=-t$, $y(t)=2t^2-3$; as $t$ increases, $x$ decreases, so the curve is traced from right to left.
Check any option by substitution, e.g. Option 3: $2(-t)^2-3=2t^2-3=y(t)$ ✓
📝 练习 E-8 — Parametrize and Verify 参数化与验证
(a) Parametrize the hyperbola $\dfrac{(y-2)^2}{4}-(x+1)^2=1$.
(b) Parametrize the ellipse $\dfrac{(x+3)^2}{25}+\dfrac{y^2}{9}=1$, traced counterclockwise from its rightmost point.
(c) Let $f(x)=x^2+1$ for $x\ge0$. Parametrize $f$ and $f^{-1}$.
(d) Does $(x(t),y(t))=(2\cos t,\;\sin t)$ parametrize $x^2+y^2=4$? Explain.
(a) Opens up and down; center $(-1,2)$; $a^2=1$ (under $x$) and $b^2=4$ (under $y$), so $a=1$, $b=2$: $x(t)=-1+\tan t$, $y(t)=2+2\sec t$, $0\le t\le2\pi$, $t\neq\frac{\pi}{2},\frac{3\pi}{2}$.
(b) $x(t)=-3+5\cos t$, $y(t)=3\sin t$, $0\le t\le2\pi$ (starts at $(2,0)$).
(c) $f$: $(t,\;t^2+1)$, $t\ge0$; $f^{-1}$: $(t^2+1,\;t)$, $t\ge0$.
(d) No. $(2\cos t)^2+(\sin t)^2=4\cos^2t+\sin^2t=4-3\sin^2t$, which equals $4$ only when $\sin t=0$, not for every $t$. (It parametrizes the ellipse $\frac{x^2}{4}+y^2=1$.)
(b) $x(t)=-3+5\cos t$, $y(t)=3\sin t$, $0\le t\le2\pi$ (starts at $(2,0)$).
(c) $f$: $(t,\;t^2+1)$, $t\ge0$; $f^{-1}$: $(t^2+1,\;t)$, $t\ge0$.
(d) No. $(2\cos t)^2+(\sin t)^2=4\cos^2t+\sin^2t=4-3\sin^2t$, which equals $4$ only when $\sin t=0$, not for every $t$. (It parametrizes the ellipse $\frac{x^2}{4}+y^2=1$.)
Chapter 10 Summary 本章总结
| Topic 主题 | Key Formula / Idea |
|---|---|
| Parametric definition | $x=x(t)$, $y=y(t)$; $t$ = parameter; curve = $\{(x(t),y(t))\}$ |
| Direction of motion | $x$ increasing → right; $y$ increasing → up |
| $x$-intercept | Solve $y(t)=0$; $y$-intercept: solve $x(t)=0$ |
| Secant slope | $\dfrac{y(t_2)-y(t_1)}{x(t_2)-x(t_1)}$ |
| Line from $(x_0,y_0)$ to $(x_1,y_1)$ | $x=x_0+(x_1-x_0)t$, $y=y_0+(y_1-y_0)t$ |
| Circle center $(h,k)$ radius $r$ | $x=h+r\cos t$, $y=k+r\sin t$ |
| Vertical parabola | $y-k=a(x-h)^2$; opens up ($a>0$) or down ($a<0$) |
| Horizontal parabola | $x-h=a(y-k)^2$; opens right ($a>0$) or left ($a<0$) |
| Ellipse (horizontal) | $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$, $a>b$; $c^2=a^2-b^2$ |
| Hyperbola (horizontal) | $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$; $c^2=a^2+b^2$; asymptotes $y=\pm\dfrac{b}{a}x$ |
| Parametric → rectangular | Solve one eq. for $t$, substitute into other; state domain and range |
| Direction by intervals | Find where $x(t)$ and $y(t)$ increase or decrease separately, then combine |
| Same point, different $t$ | A path can pass through a point twice with different directions |
| Circular motion | $(h+r\cos bt,\;k+r\sin bt)$ counterclockwise, period $\frac{2\pi}{b}$; $(\cos t,-\sin t)$ or $(\sin t,\cos t)$ clockwise |
| Segment from start + rates | $(x_1+at,\;y_1+bt)$; slope $\frac{b}{a}$ |
| Implicit graph | Fix one variable, solve for the other; each branch is a function |
| Hyperbola at $(h,k)$ | Asymptotes $y-k=\pm\frac{b}{a}(x-h)$ (CED form) |
| General form | Complete the square; same signs → ellipse, opposite → hyperbola, one square → parabola |
| Parametrize conics | Ellipse $(h+a\cos t,\;k+b\sin t)$; hyperbola $(h+a\sec t,\;k+b\tan t)$; parabola: replace $y$ (or $x$) with $t$ |
考试重点提示
- 参数方程作图：务必标出运动方向（箭头）以及起点、终点；只画轨迹会丢掉参数方程最重要的信息。（本章属于 Unit 4，不在 AP 考试范围内。）
- 圆锥曲线判断：两个平方项同号是椭圆（圆），异号是双曲线，只有一个平方项是抛物线；一般式要先配方。
- 椭圆的 $c^2=a^2-b^2$（减法，$c<a$）；双曲线的 $c^2=a^2+b^2$（加法，$c>a$）。
- 消去参数时，用三角恒等式 $\cos^2t+\sin^2t=1$ 可以轻松得到椭圆方程。
- 隐式曲线的变化率：关注两点之间 $\Delta y/\Delta x$ 的符号（正/负/零）。
- 写椭圆的参数方程：$\cos t$ 配水平半径，$\sin t$ 配竖直半径，先看清哪个分母在 $x$ 下面。
- 双曲线的参数方程用 $\sec t$、$\tan t$，要排除 $t=\frac{\pi}{2},\frac{3\pi}{2}$。
