AP Calculus AB/BC · 鼎睿学苑
Unit 9: Parametric Equations,
Polar Coordinates & Vector-Valued Functions
单元 9：参数方程、
极坐标与向量值函数
Extend calculus to curves in the plane — parametric & polar derivatives, arc length, vector motion, and polar areas. BC Only.
将微积分扩展到平面曲线 —— 参数方程与极坐标的导数、弧长、向量运动以及极坐标面积。仅 BC。
Topic 9.1
Defining and Differentiating Parametric Equations
参数方程（parametric equations）的定义与求导
Parametric Equations参数方程
A curve is defined by $x = f(t)$ and $y = g(t)$, where $t$ is the parameter. Each value of $t$ gives a point $(x, y)$ on the curve.
一条参数曲线（
parametric curve）由 $x = f(t)$ 与 $y = g(t)$ 定义，其中 $t$ 是参数（parameter）。每一个 $t$ 值对应曲线上的一个点 $(x, y)$。
The slope of the tangent line to a parametric curve is found using the chain rule:
参数曲线切线（tangent line）的斜率（slope）通过链式法则（chain rule）得到：
First Derivative — Parametric一阶导数 —— 参数方程
$$ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} $$
provided $\frac{dx}{dt} \neq 0$
前提是 $\frac{dx}{dt} \neq 0$
Common Exam Trap常见考试陷阱
Students often compute $\frac{dy}{dt}$ or $\frac{dx}{dt}$ alone and forget to divide. The slope of the tangent is the ratio $\frac{dy}{dx}$, not either derivative alone.
学生常常只算出 $\frac{dy}{dt}$ 或 $\frac{dx}{dt}$，却忘了相除。切线斜率是比值 $\frac{dy}{dx}$，而不是其中任何一个导数本身。
Worked Example
例题
Given: x = t² + 1, y = t³ − 3t Find dy/dx at t = 2. dx/dt = 2t → at t=2: dx/dt = 4 dy/dt = 3t² − 3 → at t=2: dy/dt = 9 dy/dx = 9/4
已知：x = t² + 1，y = t³ − 3t 求 t = 2 处的 dy/dx。 dx/dt = 2t → 当 t=2 时：dx/dt = 4 dy/dt = 3t² − 3 → 当 t=2 时：dy/dt = 9 dy/dx = 9/4
Key Exam Concept关键考点
A parametric curve has a horizontal tangent when $\frac{dy}{dt} = 0$ (and $\frac{dx}{dt} \neq 0$), and a vertical tangent when $\frac{dx}{dt} = 0$ (and $\frac{dy}{dt} \neq 0$).
参数曲线在 $\frac{dy}{dt} = 0$（且 $\frac{dx}{dt} \neq 0$）时有水平切线；在 $\frac{dx}{dt} = 0$（且 $\frac{dy}{dt} \neq 0$）时有垂直切线。
Given $x = \sin(t)$, $y = \cos(t)$, what is $\frac{dy}{dx}$?已知 $x = \sin(t)$，$y = \cos(t)$，求 $\frac{dy}{dx}$。
Correct! $\frac{dy}{dt} = -\sin(t)$, $\frac{dx}{dt} = \cos(t)$. So $\frac{dy}{dx} = \frac{-\sin(t)}{\cos(t)} = -\tan(t)$.正确！$\frac{dy}{dt} = -\sin(t)$，$\frac{dx}{dt} = \cos(t)$。所以 $\frac{dy}{dx} = \frac{-\sin(t)}{\cos(t)} = -\tan(t)$。
Find $\frac{dy}{dt} = -\sin(t)$ and $\frac{dx}{dt} = \cos(t)$. Divide them: $\frac{dy}{dx} = \frac{-\sin(t)}{\cos(t)}$.先求 $\frac{dy}{dt} = -\sin(t)$ 与 $\frac{dx}{dt} = \cos(t)$，再相除：$\frac{dy}{dx} = \frac{-\sin(t)}{\cos(t)}$。
Topic 9.2
Second Derivatives of Parametric Equations
参数方程的二阶导数（second derivative）
To find the concavity of a parametric curve, compute the second derivative $\frac{d^2y}{dx^2}$.
要判断参数曲线的凹凸性，需要计算二阶导数 $\frac{d^2y}{dx^2}$。
Second Derivative — Parametric二阶导数 —— 参数方程
$$ \frac{d^2y}{dx^2} = \frac{\frac{d}{dt} \left[ \frac{dy}{dx} \right]}{\frac{dx}{dt}} $$
Biggest Exam Trap in Unit 9单元 9 最大陷阱
Students often compute $\frac{d^2y}{dt^2} \div \frac{d^2x}{dt^2}$. This is WRONG. You must first find $\frac{dy}{dx}$ as a function of $t$, differentiate that with respect to $t$, then divide by $\frac{dx}{dt}$.
学生常常去算 $\frac{d^2y}{dt^2} \div \frac{d^2x}{dt^2}$，这是错误的。正确做法是先把 $\frac{dy}{dx}$ 写成关于 $t$ 的函数，再对它关于 $t$ 求导，最后除以 $\frac{dx}{dt}$。
Step-by-Step Process
分步流程
Step 1: Find $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$
第 1 步：求 $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$
Step 2: Differentiate $\frac{dy}{dx}$ with respect to $t \rightarrow$ get $\frac{d}{dt}\left(\frac{dy}{dx}\right)$
第 2 步：对 $\frac{dy}{dx}$ 关于 $t$ 求导 $\rightarrow$ 得到 $\frac{d}{dt}\left(\frac{dy}{dx}\right)$
Step 3: Divide by $\frac{dx}{dt} \rightarrow \frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}$
第 3 步：再除以 $\frac{dx}{dt}$ $\rightarrow$ $\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}$
Worked Example — Second Derivative例题 —— 二阶导数
Given: x = t², y = t³ Step 1: dy/dx = (3t²)/(2t) = 3t/2 Step 2: d/dt [3t/2] = 3/2 Step 3: d²y/dx² = (3/2) / (2t) = 3/(4t) At t = 1: d²y/dx² = 3/4 > 0 → concave up
已知：x = t²，y = t³ 第 1 步： dy/dx = (3t²)/(2t) = 3t/2 第 2 步： d/dt [3t/2] = 3/2 第 3 步： d²y/dx² = (3/2) / (2t) = 3/(4t) 当 t = 1：d²y/dx² = 3/4 > 0 → 凹向上
Topic 9.3
Finding Arc Lengths of Parametric Curves
参数曲线的弧长（arc length）
Arc Length — Parametric弧长 —— 参数方程
$$ L = \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} dt $$
Key Idea核心思想
The integrand $\sqrt{(x'(t))^2 + (y'(t))^2}$ represents the speed of the particle at time $t$. Integrating speed over time gives the total distance (arc length).
被积函数 $\sqrt{(x'(t))^2 + (y'(t))^2}$ 表示质点在时间 $t$ 处的速率（
speed）。对速率沿时间积分，得到的就是总路程（即弧长）。
Exam Note考试提示
On the AP exam, you'll typically set up the integral and evaluate with a calculator. Focus on writing the correct integrand and limits of integration.
在 AP 考试中，通常你只需要列出积分式并用计算器求值。重点是写对被积函数和积分上下限。
Worked Example — Arc Length例题 —— 弧长
Find the length of the curve x = 3t, y = 4t, for 0 ≤ t ≤ 5. dx/dt = 3, dy/dt = 4 L = ∫₀⁵ √(9 + 16) dt = ∫₀⁵ 5 dt = 25 (This is just a line segment of length 25 — makes sense!)
求曲线 x = 3t、y = 4t 在 0 ≤ t ≤ 5 上的长度。 dx/dt = 3，dy/dt = 4 L = ∫₀⁵ √(9 + 16) dt = ∫₀⁵ 5 dt = 25 （这就是一段长度为 25 的线段 —— 合情合理！）
Practice: Find the arc length of $x = 5\cos(t), \; y = 5\sin(t)$ for $0 \leq t \leq 2\pi$. (Hint: this is a circle of radius 5.)
练习：求 $x = 5\cos(t), \; y = 5\sin(t)$ 在 $0 \leq t \leq 2\pi$ 上的弧长。（提示：这是半径为 5 的圆。）
Topic 9.4
Defining and Differentiating Vector-Valued Functions
向量值函数（vector-valued function）的定义与求导
Vector-Valued Function向量值函数
A function of the form $\mathbf{r}(t) = \langle x(t), y(t) \rangle = x(t)\mathbf{i} + y(t)\mathbf{j}$. It gives the position of a particle in the plane at time $t$.
形如 $\mathbf{r}(t) = \langle x(t), y(t) \rangle = x(t)\mathbf{i} + y(t)\mathbf{j}$ 的函数，给出质点在时间 $t$ 处于平面中的位置（
position）。
Differentiation works component-wise:
求导按分量逐一进行：
Derivative of Vector Function向量值函数的导数
$$ \mathbf{r}'(t) = \langle x'(t), y'(t) \rangle $$
The derivative $\mathbf{r}'(t)$ gives the velocity vector $\mathbf{v}(t)$. Its direction is tangent to the curve at that point.
导数 $\mathbf{r}'(t)$ 即为速度向量（velocity vector） $\mathbf{v}(t)$。它的方向（direction）与曲线在该点处的切线相切。
Connection联系
Parametric equations and vector-valued functions describe the exact same thing — planar motion. The notation is different, but the calculus is identical.
参数方程与向量值函数描述的是同一件事 —— 平面运动（
planar motion）。记号不同，但微积分的内容完全一样。
Topic 9.5
Integrating Vector-Valued Functions
向量值函数的积分（integral）
Integration also works component-wise:
积分同样按分量逐一进行：
Integral of Vector Function向量值函数的积分
$$ \int \mathbf{r}(t) dt = \left\langle \int x(t) dt, \int y(t) dt \right\rangle + \mathbf{C} $$
Initial Value Problems初值问题
Given $\mathbf{v}(t) = \langle x'(t), y'(t) \rangle$ and an initial position $\mathbf{r}(t_0) = \langle x_0, y_0 \rangle$, integrate each component and solve for the constants using the initial conditions.
已知 $\mathbf{v}(t) = \langle x'(t), y'(t) \rangle$ 和初始位置 $\mathbf{r}(t_0) = \langle x_0, y_0 \rangle$，对每个分量积分，再用初始条件解出积分常数。
Worked Example — Vector IVP例题 —— 向量初值问题
Given: r'(t) = ⟨2t, eᵗ⟩, r(0) = ⟨3, 1⟩ Integrate component-wise: x(t) = t² + C₁ y(t) = eᵗ + C₂ Apply initial conditions r(0) = ⟨3, 1⟩: x(0) = 0 + C₁ = 3 → C₁ = 3 y(0) = 1 + C₂ = 1 → C₂ = 0 Solution: r(t) = ⟨t² + 3, eᵗ⟩
已知：r'(t) = ⟨2t, eᵗ⟩，r(0) = ⟨3, 1⟩ 按分量积分： x(t) = t² + C₁ y(t) = eᵗ + C₂ 代入初始条件 r(0) = ⟨3, 1⟩： x(0) = 0 + C₁ = 3 → C₁ = 3 y(0) = 1 + C₂ = 1 → C₂ = 0 解： r(t) = ⟨t² + 3, eᵗ⟩
Topic 9.6
Solving Motion Problems — Parametric & Vector-Valued
求解运动问题 —— 参数方程与向量值函数
Quick Reference — Planar Motion速查表 —— 平面运动
| Quantity量 | Formula公式 |
|---|---|
| Position位置 | $\mathbf{r}(t) = \langle x(t), y(t) \rangle$ |
| Velocity速度 | $\mathbf{v}(t) = \langle x'(t), y'(t) \rangle$ |
| Speed速率 | $|\mathbf{v}(t)| = \sqrt{(x'(t))^2 + (y'(t))^2}$ |
| Acceleration加速度 | $\mathbf{a}(t) = \langle x''(t), y''(t) \rangle$ |
| Displacement位移 | $\int_{a}^{b} \mathbf{v}(t) dt = \left\langle \int_{a}^{b} x'(t) dt, \int_{a}^{b} y'(t) dt \right\rangle$ |
| Total Distance总路程 | $\int_{a}^{b} |\mathbf{v}(t)| dt \quad$ (Arc Length)（即弧长） |
Displacement ≠ Distance位移 ≠ 路程
Displacement = net change in position (a vector). Distance = total path length (always a positive scalar). Integrate the velocity vector for displacement; integrate speed (magnitude of velocity) for distance.
位移是位置的净变化（一个向量）；路程是总路径长度（一个正的标量）。对速度向量积分得到位移；对速率（速度的模，
magnitude）积分得到路程。
Speed Increasing vs Decreasing速率递增还是递减
Speed is increasing when the velocity and acceleration vectors "agree" — specifically, when the dot product $\mathbf{v}(t) \cdot \mathbf{a}(t) > 0$. Speed is decreasing when $\mathbf{v}(t) \cdot \mathbf{a}(t) < 0$.
当速度向量与加速度向量（
acceleration vector）"方向一致"时，速率在增加 —— 具体来说，就是点积 $\mathbf{v}(t) \cdot \mathbf{a}(t) > 0$。当 $\mathbf{v}(t) \cdot \mathbf{a}(t) < 0$ 时，速率在减小。
A particle moves with velocity $\mathbf{v}(t) = \langle 3, 4 \rangle$. What is its speed?质点（
particle）以速度 $\mathbf{v}(t) = \langle 3, 4 \rangle$ 运动，其速率是多少？Correct! Speed = $|\mathbf{v}(t)| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.正确！速率 = $|\mathbf{v}(t)| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$。
Speed is the magnitude of the velocity vector: $\sqrt{3^2 + 4^2} = \sqrt{25} = 5$. Don't just add the components!速率是速度向量的模：$\sqrt{3^2 + 4^2} = \sqrt{25} = 5$。不要直接把两个分量相加！
Practice: A particle has velocity $\mathbf{v}(t) = \langle 2t, 3 \rangle$ and position $\mathbf{r}(0) = \langle 1, -2 \rangle$. Find $x(2)$.
练习：质点速度为 $\mathbf{v}(t) = \langle 2t, 3 \rangle$，位置 $\mathbf{r}(0) = \langle 1, -2 \rangle$。求 $x(2)$。
Topic 9.7
Defining Polar Coordinates & Differentiating in Polar Form
极坐标（polar coordinate）的定义与极坐标求导
Polar Coordinates极坐标
A point is described by $(r, \theta)$ where $r$ = directed distance from origin and $\theta$ = angle from positive x-axis. Convert to rectangular: $x = r\cos(\theta)$, $y = r\sin(\theta)$.
一个点用 $(r, \theta)$ 表示，其中 $r$ 是从极点（
pole）出发的有向距离（极径，radius），$\theta$ 是从极轴（polar axis）量起的角度（极角，angle）。极坐标与直角坐标的转换：$x = r\cos(\theta)$，$y = r\sin(\theta)$。
A polar curve $r = f(\theta)$ is a special case of parametric equations where $\theta$ is the parameter:
极坐标曲线（polar curve）$r = f(\theta)$ 是参数方程的特殊情形，其中 $\theta$ 担任参数：
$x(\theta) = f(\theta)\cos(\theta)$
$y(\theta) = f(\theta)\sin(\theta)$
Derivative in Polar
极坐标下的导数
Slope of Polar Curve极坐标曲线的斜率
$$ \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{f'(\theta)\sin(\theta) + f(\theta)\cos(\theta)}{f'(\theta)\cos(\theta) - f(\theta)\sin(\theta)} $$
Product Rule Required!必须用乘积法则！
Because $x = r\cos(\theta)$ and $y = r\sin(\theta)$, you must use the product rule when differentiating with respect to $\theta$. Remember: $r$ is a function of $\theta$.
因为 $x = r\cos(\theta)$、$y = r\sin(\theta)$，对 $\theta$ 求导时必须使用乘积法则。记住：$r$ 是 $\theta$ 的函数。
Topic 9.8
Area of a Polar Region — Single Curve
极坐标区域的面积 —— 单曲线
Polar Area Formula极坐标面积公式
$$ A = \frac{1}{2} \int_{\alpha}^{\beta} [r(\theta)]^2 d\theta $$
Where It Comes From公式由来
The area of a thin circular sector with radius $r$ and angle $d\theta$ is $\frac{1}{2}r^2 d\theta$. Integrating these sectors from $\theta = \alpha$ to $\theta = \beta$ gives total area.
半径为 $r$、角度为 $d\theta$ 的薄圆扇形的面积是 $\frac{1}{2}r^2 d\theta$。把这些扇形从 $\theta = \alpha$ 到 $\theta = \beta$ 积分，就得到极坐标曲线所围面积。
Common Exam Traps常见考试陷阱
1. Forgetting the $\frac{1}{2}$ in front of the integral.
1. 忘记积分前的 $\frac{1}{2}$。
2. Forgetting to square $r$ — the integrand is $[r(\theta)]^2$.
2. 忘记把 $r$ 平方 —— 被积函数是 $[r(\theta)]^2$。
3. Using wrong limits — sketch the curve first to identify bounds!
3. 积分上下限写错 —— 先画出极坐标曲线再确定边界！
Worked Example — Area of One Petal of $r = \cos(2\theta)$例题 —— 求 $r = \cos(2\theta)$ 一个花瓣的面积
r = cos(2θ) is a rose with 4 petals. One petal: from θ = −π/4 to θ = π/4 (where r ≥ 0). A = ½ ∫−π/4π/4 [cos(2θ)]² dθ Use identity: cos²(u) = (1 + cos(2u))/2 A = ½ ∫−π/4π/4 (1 + cos(4θ))/2 dθ = ¼ [θ + sin(4θ)/4]−π/4π/4 A = ¼ [(π/4 + 0) − (−π/4 + 0)] = ¼ · π/2 = π/8
r = cos(2θ) 是一个 4 瓣玫瑰线。 一个花瓣：θ 从 −π/4 到 π/4（此时 r ≥ 0）。 A = ½ ∫−π/4π/4 [cos(2θ)]² dθ 用恒等式：cos²(u) = (1 + cos(2u))/2 A = ½ ∫−π/4π/4 (1 + cos(4θ))/2 dθ = ¼ [θ + sin(4θ)/4]−π/4π/4 A = ¼ [(π/4 + 0) − (−π/4 + 0)] = ¼ · π/2 = π/8
What is the area enclosed by the polar curve $r = 3$ from $\theta = 0$ to $\theta = 2\pi$?极坐标曲线 $r = 3$ 在 $\theta = 0$ 到 $\theta = 2\pi$ 所围的面积是多少？
Correct! $A = \frac{1}{2} \int_{0}^{2\pi} (3)^2 d\theta = \frac{1}{2}(9)(2\pi) = 9\pi$. Makes sense, as it's a circle of radius $3$ ($\pi r^2 = 9\pi$).正确！$A = \frac{1}{2} \int_{0}^{2\pi} (3)^2 d\theta = \frac{1}{2}(9)(2\pi) = 9\pi$。这正是半径为 $3$ 的圆的面积（$\pi r^2 = 9\pi$）。
$A = \frac{1}{2} \int_{0}^{2\pi} r^2 d\theta = \frac{1}{2} \int_{0}^{2\pi} 9 d\theta = \frac{1}{2}(9)(2\pi) = 9\pi$. Remember the $\frac{1}{2}$ and to square $r$!$A = \frac{1}{2} \int_{0}^{2\pi} r^2 d\theta = \frac{1}{2} \int_{0}^{2\pi} 9 d\theta = \frac{1}{2}(9)(2\pi) = 9\pi$。别忘了那个 $\frac{1}{2}$，也别忘了把 $r$ 平方！
Topic 9.9
Area Between Two Polar Curves
两条极坐标曲线之间的面积
Area Between Polar Curves两极坐标曲线之间的面积
$$ A = \frac{1}{2} \int_{\alpha}^{\beta} \left( [r_{\text{outer}}(\theta)]^2 - [r_{\text{inner}}(\theta)]^2 \right) d\theta $$
Key Steps关键步骤
1. Sketch both curves to identify which is outer vs inner.
1. 画出两条曲线，分辨外曲线与内曲线。
2. Find intersection points by setting $r_1 = r_2$ (these determine $\alpha$ and $\beta$).
2. 解 $r_1 = r_2$ 找出交点（这些点确定 $\alpha$ 与 $\beta$）。
3. Apply the formula: subtract squares inside the integral, not outside.
3. 代公式：把两个平方放到积分内部相减，不要在积分外面相减。
Exam Strategy
How Unit 9 Appears on the AP Exam
单元 9 在 AP 考试中的考法
MCQ — Common Question Styles选择题 —— 常见题型
Find $\frac{dy}{dx}$ for a parametric curve at a specific t-value.
在指定的 t 值处求 $\frac{dy}{dx}$（参数曲线）。
Set up an arc length or polar area integral (frequently without evaluating).
列出弧长或极坐标面积的积分（常常无需求值）。
Determine speed, velocity, or acceleration from vector components.
由向量分量求速率、速度或加速度。
Identify horizontal/vertical tangents on parametric curves.
判断参数曲线上的水平切线或垂直切线。
FRQ — Common Question Styles自由作答题 —— 常见题型
Particle in the plane: Given velocity components, find position, speed, total distance, or acceleration. Heavily involves initial value problems.
平面内的质点：给出速度分量，求位置、速率、总路程或加速度。这类题往往涉及初值问题。
Polar area: Set up and evaluate area integrals, often between two curves or for one petal of a rose.
极坐标面积：列出并计算面积积分，常见的是两条曲线之间的面积，或玫瑰线一个花瓣的面积。
Top Mistakes That Lose Points最常失分的错误
1. Computing $\frac{d^2y}{dx^2}$ by dividing $\frac{d^2y}{dt^2}$ by $\frac{d^2x}{dt^2}$.
1. 用 $\frac{d^2y}{dt^2}$ 除以 $\frac{d^2x}{dt^2}$ 来算 $\frac{d^2y}{dx^2}$。
2. Forgetting the $\frac{1}{2}$ or the square in polar area integrals.
2. 极坐标面积积分里忘掉 $\frac{1}{2}$ 或忘记把 $r$ 平方。
3. Confusing displacement (vector) with distance traveled (scalar integral of speed).
3. 把位移（向量）与路程（速率的标量积分）弄混。
4. Forgetting initial conditions $+C$ in vector IVPs.
4. 在向量初值问题里忘记加初始条件给出的 $+C$。
Review
Flashcards — Click to Flip
闪卡 —— 点击翻面
First derivative $\frac{dy}{dx}$ for parametric curves?参数曲线的一阶导数 $\frac{dy}{dx}$？
$$ \frac{\frac{dy}{dt}}{\frac{dx}{dt}} $$
provided $\frac{dx}{dt} \neq 0$前提 $\frac{dx}{dt} \neq 0$
Second derivative $\frac{d^2y}{dx^2}$ for parametric curves?参数曲线的二阶导数 $\frac{d^2y}{dx^2}$？
$$ \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}} $$
NOT quotient of 2nd derivatives.不是两个二阶导数相除。
Arc length (parametric distance)?参数曲线弧长？
$$ \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} dt $$
Speed of a particle in the plane?平面内质点的速率？
$$ |\mathbf{v}(t)| = \sqrt{(x'(t))^2 + (y'(t))^2} $$
Polar area formula?极坐标面积公式？
$$ A = \frac{1}{2} \int_{\alpha}^{\beta} [r(\theta)]^2 d\theta $$
When is speed increasing?速率何时递增？
When $\mathbf{v}(t) \cdot \mathbf{a}(t) > 0$
(Velocity and acceleration agree)当 $\mathbf{v}(t) \cdot \mathbf{a}(t) > 0$ 时
（速度与加速度方向一致）
Assessment
Unit 9 — Practice Quiz
单元 9 —— 练习测验
1. Given $x = t^2 - 1$, $y = 2t + 3$, find $\frac{dy}{dx}$.1. 已知 $x = t^2 - 1$、$y = 2t + 3$，求 $\frac{dy}{dx}$。
Correct! $\frac{dx}{dt} = 2t$, $\frac{dy}{dt} = 2$. So $\frac{dy}{dx} = \frac{2}{2t} = \frac{1}{t}$.正确！$\frac{dx}{dt} = 2t$，$\frac{dy}{dt} = 2$。所以 $\frac{dy}{dx} = \frac{2}{2t} = \frac{1}{t}$。
$\frac{dx}{dt} = 2t$ and $\frac{dy}{dt} = 2$. Divide $\frac{dy}{dt}$ by $\frac{dx}{dt}$ to get $\frac{2}{2t} = \frac{1}{t}$.$\frac{dx}{dt} = 2t$、$\frac{dy}{dt} = 2$。用 $\frac{dy}{dt}$ 除以 $\frac{dx}{dt}$ 得到 $\frac{2}{2t} = \frac{1}{t}$。
2. To find $\frac{d^2y}{dx^2}$ for a parametric curve, you should:2. 求参数曲线的 $\frac{d^2y}{dx^2}$，应当：
Correct! $\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}(\frac{dy}{dx})}{\frac{dx}{dt}}$. This is heavily tested!正确！$\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}(\frac{dy}{dx})}{\frac{dx}{dt}}$。这一点几乎每年都考！
The correct process is to differentiate $\frac{dy}{dx}$ with respect to $t$, then divide that result by $\frac{dx}{dt}$.正确流程：先对 $\frac{dy}{dx}$ 关于 $t$ 求导，再把结果除以 $\frac{dx}{dt}$。
3. The total distance traveled by a particle with velocity $\mathbf{v}(t) = \langle \cos t, \sin t \rangle$ from $t = 0$ to $t = \pi$ is:3. 质点速度为 $\mathbf{v}(t) = \langle \cos t, \sin t \rangle$，从 $t = 0$ 到 $t = \pi$ 的总路程是：
Correct! Speed = $\sqrt{\cos^2 t + \sin^2 t} = 1$. Distance = $\int_0^{\pi} 1\,dt = \pi$.正确！速率 = $\sqrt{\cos^2 t + \sin^2 t} = 1$。路程 = $\int_0^{\pi} 1\,dt = \pi$。
Speed = $|\mathbf{v}(t)| = \sqrt{\cos^2 t + \sin^2 t} = 1$. Total distance = $\int_0^{\pi} 1\,dt = \pi$. Don't confuse distance with displacement!速率 = $|\mathbf{v}(t)| = \sqrt{\cos^2 t + \sin^2 t} = 1$。总路程 = $\int_0^{\pi} 1\,dt = \pi$。注意不要把路程与位移混淆！
4. The area of one petal of $r = \sin(2\theta)$ is:4. 曲线 $r = \sin(2\theta)$ 一个花瓣的面积是：
Correct! One petal runs from $\theta = 0$ to $\theta = \frac{\pi}{2}$. $A = \frac{1}{2}\int_0^{\pi/2} \sin^2(2\theta)\,d\theta = \frac{1}{2} \cdot \frac{\pi}{4} = \frac{\pi}{8}$.正确！一个花瓣对应 $\theta$ 从 $0$ 到 $\frac{\pi}{2}$。$A = \frac{1}{2}\int_0^{\pi/2} \sin^2(2\theta)\,d\theta = \frac{1}{2} \cdot \frac{\pi}{4} = \frac{\pi}{8}$。
One petal of $r = \sin(2\theta)$ goes from $0$ to $\frac{\pi}{2}$. Using the half-angle identity: $A = \frac{1}{2}\int_0^{\pi/2} \sin^2(2\theta)\,d\theta = \frac{\pi}{8}$.$r = \sin(2\theta)$ 的一个花瓣对应 $\theta \in [0, \frac{\pi}{2}]$。用半角恒等式：$A = \frac{1}{2}\int_0^{\pi/2} \sin^2(2\theta)\,d\theta = \frac{\pi}{8}$。
5. A particle has $\mathbf{v}(t) = \langle 2, -3 \rangle$ and $\mathbf{a}(t) = \langle 1, 4 \rangle$. Is its speed increasing or decreasing?5. 质点的 $\mathbf{v}(t) = \langle 2, -3 \rangle$，$\mathbf{a}(t) = \langle 1, 4 \rangle$。此时其速率在递增还是递减？
Correct! $\mathbf{v} \cdot \mathbf{a} = (2)(1) + (-3)(4) = 2 - 12 = -10 < 0$. Since the dot product is negative, speed is decreasing.正确！$\mathbf{v} \cdot \mathbf{a} = (2)(1) + (-3)(4) = 2 - 12 = -10 < 0$。点积为负，说明速率正在减小。
Check the dot product: $\mathbf{v} \cdot \mathbf{a} = (2)(1) + (-3)(4) = -10 < 0$. When $\mathbf{v} \cdot \mathbf{a} < 0$, velocity and acceleration oppose each other, so speed decreases.先算点积：$\mathbf{v} \cdot \mathbf{a} = (2)(1) + (-3)(4) = -10 < 0$。当 $\mathbf{v} \cdot \mathbf{a} < 0$ 时，速度与加速度方向相反，速率递减。
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- Find $\frac{dy}{dx}$ for a parametric curve using $\frac{dy/dt}{dx/dt}$用 $\frac{dy/dt}{dx/dt}$ 求参数曲线的 $\frac{dy}{dx}$
- Find $\frac{d^2y}{dx^2}$ correctly (not by dividing second derivatives)正确求出 $\frac{d^2y}{dx^2}$（不是用两个二阶导数相除）
- Identify horizontal and vertical tangent lines on parametric curves识别参数曲线上的水平切线与垂直切线
- Set up and evaluate parametric arc length integrals列出并计算参数曲线的弧长积分
- Differentiate and integrate vector-valued functions component-wise按分量对向量值函数求导与积分
- Solve vector initial value problems (position from velocity + initial condition)解向量初值问题（由速度与初始条件求位置）
- Compute speed, displacement, and total distance from velocity components由速度分量求速率、位移与总路程
- Determine when speed is increasing vs. decreasing ($\mathbf{v} \cdot \mathbf{a}$)用 $\mathbf{v} \cdot \mathbf{a}$ 判断速率递增或递减
- Find position at a later time using $\int$ of velocity components用速度分量的 $\int$ 求之后时刻的位置
- Compute acceleration magnitude and direction from vector components由向量分量求加速度的模与方向
- Convert between polar and rectangular coordinates在极坐标与直角坐标之间互相转换
- Sketch a polar curve and identify its symmetry画出极坐标曲线并判断其对称性
- Find $\frac{dy}{dx}$ for a polar curve using the product rule使用乘积法则求极坐标曲线的 $\frac{dy}{dx}$
- Set up and evaluate polar area integrals (single curve and between two curves)列出并计算极坐标面积积分（单曲线与两曲线之间）
- Distinguish displacement (vector) from distance (scalar) on FRQs在自由作答题中区分位移（向量）与路程（标量）
鼎睿学苑 · Dingrui Scholars — AP Calculus AB/BC Unit 9 Notes