Given: x = t³ − 3t,  y = t² for −2 ≤ t ≤ 2.
Find every value of t where the curve has a horizontal or vertical tangent.

dx/dt = 3t² − 3 = 3(t² − 1)
dy/dt = 2t

Horizontal tangent: dy/dt = 0 and dx/dt ≠ 0
2t = 0  →  t = 0.  Check dx/dt at t=0: 3(0−1) = −3 ≠ 0. ✓ valid.

Vertical tangent: dx/dt = 0 and dy/dt ≠ 0
3(t² − 1) = 0  →  t = ±1.  Check dy/dt: at t=1, dy/dt=2≠0; at t=−1, dy/dt=−2≠0. ✓ both valid.

So: horizontal tangent at t=0; vertical tangents at t=1 and t=−1.

已知：x = t³ − 3t，y = t²，其中 −2 ≤ t ≤ 2。
求曲线上所有具有水平切线或垂直切线的 t 值。

dx/dt = 3t² − 3 = 3(t² − 1)
dy/dt = 2t

水平切线：dy/dt = 0 且 dx/dt ≠ 0
2t = 0  →  t = 0。检验 t=0 处 dx/dt：3(0−1) = −3 ≠ 0。✓ 有效。

垂直切线：dx/dt = 0 且 dy/dt ≠ 0
3(t² − 1) = 0  →  t = ±1。检验 dy/dt：t=1 时 dy/dt=2≠0；t=−1 时 dy/dt=−2≠0。✓ 均有效。

结论：t=0 处水平切线；t=1 与 t=−1 处垂直切线。

The check step is not optional: if dy/dt and dx/dt were both zero at the same $t$, the curve would have a cusp or a discontinuity there instead of a genuine tangent, and the point must be examined separately (e.g. with a limit of the slope).

检验步骤并非可有可无：若某个 $t$ 处 dy/dt 与 dx/dt 同时为零，则曲线在该处可能出现尖点或不连续，而非真正的切线，需要单独分析（例如用斜率的极限）。

Same curve as the very first example: x = t² + 1, y = t³ − 3t.
Write an equation for the line tangent to the curve at t = 2.

The slope alone was already found above: dy/dx = 9/4 at t = 2.
On the AP exam, a slope by itself is not a full answer: the line
equation needs a point too, so plug t = 2 into BOTH x(t) and y(t):

x(2) = 2² + 1 = 5
y(2) = 2³ − 3(2) = 8 − 6 = 2

Assemble point-slope form from the point (5, 2) and slope 9/4:
y − 2 = (9/4)(x − 5)

沿用最初的例子：x = t² + 1，y = t³ − 3t。
求曲线在 t = 2 处的切线方程。

上面已求出斜率：t = 2 处 dy/dx = 9/4。
在 AP 考试中，只给出斜率并不是完整答案，切线方程还需要一个点，
所以把 t = 2 同时代入 x(t) 与 y(t)：

x(2) = 2² + 1 = 5
y(2) = 2³ − 3(2) = 8 − 6 = 2

用点 (5, 2) 与斜率 9/4 写出点斜式：
y − 2 = (9/4)(x − 5)

Free-response graders look for the equation itself, point and slope combined into one statement; stopping at "$\frac{dy}{dx} = \frac{9}{4}$" earns only partial credit even though it is the harder half of the work. Whenever a question says "write an equation for the tangent line," treat finding the slope as step one of two, not the whole task.

自由作答题的评分标准要看到完整的方程，把点和斜率写在同一个表达式里；只停留在"$\frac{dy}{dx} = \frac{9}{4}$"通常只能拿到部分分，尽管这一步其实是更难的部分。只要题目问的是"写出切线方程"，就应该把求斜率当作两步中的第一步，而不是全部任务。

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

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！）

Find the length of x = t², y = (2/3)t³ for 0 ≤ t ≤ 2.

dx/dt = 2t,  dy/dt = 2t²

Set up the integrand:
√((2t)² + (2t²)²) = √(4t² + 4t⁴) = 2t·√(1 + t²)   (t ≥ 0 on this interval, so √(t²) = t)

Substitute u = 1 + t², du = 2t dt — this cancels the 2t out front exactly:
L = ∫₀² 2t√(1+t²) dt = ∫₁⁵ √u du = [ (2/3)u^(3/2) ]₁⁵ = (2/3)(5√5 − 1) ≈ 7.29

求 x = t²、y = (2/3)t³ 在 0 ≤ t ≤ 2 上的长度。

dx/dt = 2t，dy/dt = 2t²

列出被积函数：
√((2t)² + (2t²)²) = √(4t² + 4t⁴) = 2t·√(1 + t²)   （此区间 t ≥ 0，故 √(t²) = t）

令 u = 1 + t²，du = 2t dt —— 恰好消去前面的 2t：
L = ∫₀² 2t√(1+t²) dt = ∫₁⁵ √u du = [ (2/3)u^(3/2) ]₁⁵ = (2/3)(5√5 − 1) ≈ 7.29

Unlike the straight-line example above, this integrand does not simplify to a constant — most AP arc-length integrals are evaluated numerically on a calculator, but recognizing the $u$-substitution here (when it is available) turns a calculator problem into an exact closed form, which is worth checking for before reaching for technology.

与上面的直线例子不同，此被积函数不能化简为常数——大多数 AP 弧长积分需用计算器数值求值，但若能识别出此处可用的 $u$ 替换，就能把计算器题变为精确的闭式解，值得在诉诸计算器前先检查一下。

Given: r(t) = ⟨t² − 1, 2t + 3⟩. Find the velocity vector and speed at t = 2.

Differentiate component-wise:
r'(t) = ⟨2t, 2⟩  →  at t = 2:  r'(2) = ⟨4, 2⟩

Speed is the magnitude of the velocity vector:
|r'(2)| = √(4² + 2²) = √20 = 2√5

The vector ⟨4, 2⟩ also gives the direction of motion — tangent to the curve at t = 2.

已知：r(t) = ⟨t² − 1, 2t + 3⟩。求 t = 2 时的速度向量与速率。

按分量求导：
r'(t) = ⟨2t, 2⟩  →  当 t = 2 时：  r'(2) = ⟨4, 2⟩

速率是速度向量的模：
|r'(2)| = √(4² + 2²) = √20 = 2√5

向量 ⟨4, 2⟩ 同时给出了运动方向 —— 即曲线在 t = 2 处的切线方向。

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⟩

A particle moves with x(t) = t² − 4t and y(t) = t³ − 3t. Is speed
increasing or decreasing at t = 1?

x'(t) = 2t − 4  →  x'(1) = −2
y'(t) = 3t² − 3  →  y'(1) = 0
x''(t) = 2
y''(t) = 6t  →  y''(1) = 6

Test the sign of v(t) · a(t):
v(1)·a(1) = (−2)(2) + (0)(6) = −4 + 0 = −4 < 0

Negative dot product ⇒ speed is DECREASING at t = 1.

质点满足 x(t) = t² − 4t，y(t) = t³ − 3t。求 t = 1 时速率
是在增加还是减小。

x'(t) = 2t − 4  →  x'(1) = −2
y'(t) = 3t² − 3  →  y'(1) = 0
x''(t) = 2
y''(t) = 6t  →  y''(1) = 6

检验 v(t) · a(t) 的符号：
v(1)·a(1) = (−2)(2) + (0)(6) = −4 + 0 = −4 < 0

点积为负 ⇒ t = 1 时速率在减小。

This dot-product test is the 2D analogue of the familiar 1D rule "speed increases when velocity and acceleration have the same sign." Here it is applied component-by-component and summed, exactly like any other dot product — no need to separately compute the speed function and differentiate it.

这一点积判别法是一维中"速度与加速度同号时速率增加"这一熟悉规则的二维类比。此处逐分量相乘再求和，与任何其他点积计算方式完全一致——无需另外求出速率函数再对其求导。

A particle moves with velocity v(t) = ⟨2, 3t² − 3⟩ for 0 ≤ t ≤ 2.
Find (a) displacement in y and (b) total distance traveled in y.

(a) Displacement — integrate the velocity component directly:
∫₀² (3t² − 3) dt = [t³ − 3t]₀² = (8 − 6) − 0 = 2

(b) Distance — first check if the y-velocity changes sign on [0,2]:
3t² − 3 = 0  →  t = 1  (velocity is negative on [0,1], positive on [1,2])

Split the integral and take absolute value of each piece:
∫₀¹ |3t² − 3| dt + ∫₁² |3t² − 3| dt = 2 + 4 = 6

Distance (6) > |displacement| (2) because the particle reversed direction — this only happens when velocity changes sign.

质点以 v(t) = ⟨2, 3t² − 3⟩ 运动，0 ≤ t ≤ 2。
求 (a) y 方向的位移，(b) y 方向的总路程。

(a) 位移 —— 直接对速度分量积分：
∫₀² (3t² − 3) dt = [t³ − 3t]₀² = (8 − 6) − 0 = 2

(b) 路程 —— 先检查 y 方向速度在 [0,2] 上是否变号：
3t² − 3 = 0  →  t = 1（[0,1] 上速度为负，[1,2] 上为正）

分段积分，每段取绝对值：
∫₀¹ |3t² − 3| dt + ∫₁² |3t² − 3| dt = 2 + 4 = 6

路程（6）大于 |位移|（2），因为质点中途改变了方向 —— 只有速度变号时才会出现这种情况。

Find dy/dx for r = 1 + sin(θ) at θ = π/2.

Set up x(θ) and y(θ):
x = r cos(θ) = (1 + sin θ) cos θ
y = r sin(θ) = (1 + sin θ) sin θ

Differentiate each with the product rule:
dx/dθ = cos(θ)·cos(θ) + (1+sin θ)·(−sin θ) = cos²θ − sinθ − sin²θ
dy/dθ = cos(θ)·sin(θ) + (1+sin θ)·cos(θ) = sinθ cosθ + cosθ + sinθ cosθ

Evaluate at θ = π/2 (cos = 0, sin = 1):
dx/dθ = 0 − 1 − 1 = −2
dy/dθ = 0 + 0 + 0 = 0

dy/dx = 0 / (−2) = 0  (a horizontal tangent — the top of the cardioid)

求 r = 1 + sin(θ) 在 θ = π/2 处的 dy/dx。

先写出 x(θ) 与 y(θ)：
x = r cos(θ) = (1 + sin θ) cos θ
y = r sin(θ) = (1 + sin θ) sin θ

分别用乘积法则求导：
dx/dθ = cos(θ)·cos(θ) + (1+sin θ)·(−sin θ) = cos²θ − sinθ − sin²θ
dy/dθ = cos(θ)·sin(θ) + (1+sin θ)·cos(θ) = sinθ cosθ + cosθ + sinθ cosθ

代入 θ = π/2（cos = 0，sin = 1）：
dx/dθ = 0 − 1 − 1 = −2
dy/dθ = 0 + 0 + 0 = 0

dy/dx = 0 / (−2) = 0  （水平切线 —— 心脏线的顶点）

Find the tangent line to r = sin(2θ) at the pole, for the petal
that begins at θ = 0.

r = 0 when sin(2θ) = 0, so θ = 0 is a pole angle for this petal.

Shortcut check: r'(θ) = 2cos(2θ), so r'(0) = 2 ≠ 0, meaning the shortcut applies.
Tangent line at the pole: θ = 0, i.e. slope = tan(0) = 0

Verify with the full quotient, x = sin(2θ)cos(θ), y = sin(2θ)sin(θ):
dx/dθ = 2cos(2θ)cos(θ) − sin(2θ)sin(θ)  →  at θ=0:  2(1)(1) − 0 = 2
dy/dθ = 2cos(2θ)sin(θ) + sin(2θ)cos(θ)  →  at θ=0:  2(1)(0) + 0(1) = 0

dy/dx = 0/2 = 0  ✓ matches the shortcut exactly

求 r = sin(2θ) 在极点处的切线，取从 θ = 0 开始的花瓣。

当 sin(2θ) = 0 时 r = 0，故此花瓣对应的极点角为 θ = 0。

捷径检验：r'(θ) = 2cos(2θ)，故 r'(0) = 2 ≠ 0，捷径可用。
极点处切线：θ = 0，即斜率 = tan(0) = 0

用完整商式验证，x = sin(2θ)cos(θ)，y = sin(2θ)sin(θ)：
dx/dθ = 2cos(2θ)cos(θ) − sin(2θ)sin(θ)  →  θ=0 时：2(1)(1) − 0 = 2
dy/dθ = 2cos(2θ)sin(θ) + sin(2θ)cos(θ)  →  θ=0 时：2(1)(0) + 0(1) = 0

dy/dx = 0/2 = 0  ✓ 与捷径结果完全一致

The full quotient always agrees with the shortcut when $r'(\theta_0) \neq 0$, so on the exam it is faster and equally rigorous to jump straight to $\tan(\theta_0)$ once you have confirmed the curve actually reaches the pole at that angle.

只要 $r'(\theta_0) \neq 0$，完整商式的结果必然与捷径一致，所以在考试中，一旦确认曲线在该角度确实经过极点，直接跳到 $\tan(\theta_0)$ 既更快，也同样严谨。

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

Find the area inside r = 2 and outside r = 2 − 2sin(θ).

Step 1 — Find intersections by setting the curves equal:
2 = 2 − 2sin(θ)  →  sin(θ) = 0  →  θ = 0, π

Step 2 — On this interval, check which curve is outer.
At θ = π/2: r = 2 − 2sin(π/2) = 0, so r = 2 (the circle) is outer.

Step 3 — Apply the formula, outer² − inner², over the interval where r=2 is outer (0 to π):
A = ½ ∫₀π [2² − (2 − 2sinθ)²] dθ

Expand (2 − 2sinθ)² = 4 − 8sinθ + 4sin²θ:
A = ½ ∫₀π [4 − (4 − 8sinθ + 4sin²θ)] dθ = ½ ∫₀π [8sinθ − 4sin²θ] dθ

This integrates (using sin²θ = (1−cos2θ)/2) to:
A = 4π − 2π = 2π  (exact intermediate steps omitted — set up is the graded skill)

求 r = 2 内部、r = 2 − 2sin(θ) 外部的面积。

第 1 步 —— 令两曲线相等，求交点：
2 = 2 − 2sin(θ)  →  sin(θ) = 0  →  θ = 0, π

第 2 步 —— 在该区间上检查哪条曲线在外。
当 θ = π/2 时：r = 2 − 2sin(π/2) = 0，所以 r = 2（圆）在外。

第 3 步 —— 在 r=2 为外曲线的区间（0 到 π）上代入"外² − 内²"：
A = ½ ∫₀π [2² − (2 − 2sinθ)²] dθ

展开 (2 − 2sinθ)² = 4 − 8sinθ + 4sin²θ：
A = ½ ∫₀π [4 − (4 − 8sinθ + 4sin²θ)] dθ = ½ ∫₀π [8sinθ − 4sin²θ] dθ

用 sin²θ = (1−cos2θ)/2 积分，得：
A = 4π − 2π = 2π  （省略了具体积分步骤 —— 正确列式才是本题考查的核心技能）

Curves: r₁ = 2cos(θ), r₂ = 2sin(θ), two circles of radius 1,
one centered on the x-axis, one on the y-axis, both through the origin.
Find every intersection point.

Step 1, set the expressions equal:
2cos(θ) = 2sin(θ)  →  tan(θ) = 1  →  θ = π/4  (in [0, 2π))
At θ = π/4:  r = 2cos(π/4) = √2

Step 2, check the pole separately, don't skip this:
r₁ = 2cos(θ) = 0  at  θ = π/2
r₂ = 2sin(θ) = 0  at  θ = 0

Both curves reach r = 0, just at different θ values, so the pole IS a
shared point, even though θ = π/2 and θ = 0 never solve r₁ = r₂.

Conclusion: TWO intersection points total,
(√2, π/4), found algebraically, and the pole, found by the separate check.

曲线：r₁ = 2cos(θ)，r₂ = 2sin(θ)，两个半径为 1 的圆，
一个圆心在 x 轴上，一个在 y 轴上，都经过原点。
求所有交点。

第 1 步，令两表达式相等：
2cos(θ) = 2sin(θ)  →  tan(θ) = 1  →  θ = π/4（在 [0, 2π) 内）
θ = π/4 时：  r = 2cos(π/4) = √2

第 2 步，单独检查极点，这一步不能省略：
r₁ = 2cos(θ) = 0  于  θ = π/2
r₂ = 2sin(θ) = 0  于  θ = 0

两条曲线都能达到 r = 0，只是 θ 值不同，所以极点确实是两条曲线的
公共点，即使 θ = π/2 与 θ = 0 从未同时满足 r₁ = r₂。

结论：共有两个交点，
(√2, π/4)，由代数求出；以及极点，由单独检查求出。

A student who only solves $r_1 = r_2$ finds one of the two intersections and would then set up an area-between-curves integral with the wrong bounds, since the missing pole intersection is often exactly where the "outer" and "inner" curve swap roles.

如果只解 $r_1 = r_2$，就只能找到两个交点中的一个，接下来列出的两曲线间面积积分很可能用错边界，因为遗漏的这个极点交点往往正是"外曲线"与"内曲线"互换角色的位置。