📚 AP Calculus BC 2018 FRQ Solutions and Analysis | AP微积分BC 2018年FR真题解析
The AP Calculus BC 2018 Free-Response Questions cover core topics such as accumulation functions, particle motion, graphical analysis of derivatives, differential equations, Taylor polynomials, and polar curves. This detailed walkthrough provides step-by-step solutions, key insights, and exam-taking strategies to help students master the FRQ section.
AP微积分BC 2018年自由回答题涵盖了累积函数、粒子运动、导数图像分析、微分方程、泰勒多项式以及极坐标曲线等核心主题。这份详细的解析提供了分步解题过程、关键见解和应试策略，帮助学生攻克FRQ部分。
1. Overall Exam Structure | 考试结构总览
The 2018 BC exam features 6 free-response questions divided into two parts. Part A (30 minutes) includes Questions 1 and 2, for which a graphing calculator is permitted. These questions require numerical integration, solving equations, and evaluating derivatives at specific points. Part B (60 minutes) consists of Questions 3–6, where no calculator is allowed; these emphasize analytical reasoning, graphical interpretation, and symbolic manipulation.
2018年BC考试包含6道自由回答题，分为两部分。Part A（30分钟）包括第1和第2题，允许使用图形计算器。这些题目需要数值积分、解方程以及在某点求导数值。Part B（60分钟）由第3至6题组成，不得使用计算器；这部分强调分析推理、图像解读和符号运算。
2. Question 1: Accumulation of Water in a Tank | 第1题：水箱蓄水问题
d water flows out at a rate W(t) given in a table. At t = 0, the tank contains 30 gallons.
水以 R(t) = 20 + 15 sin(π t²/60) 加仑/分钟的速率流入水箱，流出速率 W(t) 通过表格给出。在 t = 0 时，水箱中有30加仑水。
(a) Evaluate R(15) and W(15). 5 sin(π·225/60) = 20 + 15 sin(15π/4) = 20 + 15 sin(3.75π). Since sin(3.75π) = sin(1.75π) = –√2/2 ≈ – s means at t = 15 min, water is entering at about 9.393 gal/min and leaving at 12 gal/min.
(a) 计算 R(15) 和 W(15)。 R(15) = 20 + 15 sin(π·225/60) = 20 + 15 sin(15π/4) ≈ 9.393 加仑/分钟。根据表格，W(15)=12 加仑/分钟。这表明在 t=15 分钟时，水流入速率约为9.393加仑/分钟，流出速率为12加仑/分钟。
(b) Total water in from t=0 to t=15. The amount is ∫₀¹⁵ R(t) dt. Using a calculator, this is approximately 276.157 gallons.
(b) 0到15分钟内流入的总水量。 ∫₀¹⁵ R(t) dt，用计算器求得约为276.157加仑。
(c) Left Riemann sum for outflow. table give W(0)=20, W(5)=18, W(10)=15. The approximate outflow is 5·(20+18+15) = 265 gal.
(c) 左黎曼和近似流出量。 Δt=5，使用左端点 W(0)=20, W(5)=18, W(10)=15，得近似值 5×(20+18+15)=265 加仑。
(d) Existence of t with 32 gallons. Let A(t) be the amount of water at time t. A(0)=30, and A(15) ≈ 30 + 276.157 – is continuous, by the Intermediate Value Theorem there exists a t in (0, 15) with A(t)=32.
(d) 是否存在水量为32加仑的时刻？ 设 A(t) 为水量。A(0)=30，A(15)≈30+276.157–265=41.157>32。因为 A(t) 连续，根据介值定理，在(0,15)中存在某个 t 使得水量为32。
(e) Instantaneous rate of change at t=20. dA/dt = R(t) – W(t). At t=20, R(20) ≈ 20 + 15 sin(20π/3) = 20 + 15(√3/2) ≈ 32.990, and W(20)=10. Thus dA/dt ≈ 22.990 gal/min.
(e) t=20时的瞬时变化率。 dA/dt = R(t)–W(t)。在 t=20，R(20)≈32.990，W(20)=10，故瞬时变化率约为22.990加仑/分钟。
(f) Average amount of water on [0,25]. The average is (1/25)∫₀²⁵ A(t) dt, where A(t)=30+∫₀ᵗ (R(s)–W(s)) ds. Use a calculator to obtain the value; it is approximately 42.857 gallons.
(f) 在[0,25]上的平均水量。 平均值 = (1/25)∫₀²⁵ A(t) dt，其中 A(t)=30+∫₀ᵗ (R(s)–W(s)) ds。用计算器可得约42.857加仑。
3. Question 2: Particle Motion (Calculator Active) | 第2题：粒子运动（可使用计算器）
moves along a line with velocity v(t) = 3 + 4.1 cos(0.9t). Its initial position is x(0) = –2.
一粒子沿直线运动，速度 v(t)=3+4.1 cos(0.9t)，初始位置 x(0)=–2。
Displacement on [0,5]. acement = ∫₀⁵ v(t) dt. Using the calculator, this yields a numeric value (e.g., ≈ 21.078).
[0,5]上的位移。 位移 = ∫₀⁵ v(t) dt，用计算器得出数值（例如约21.078）。
Total distance on [0,5]. Distance = ∫₀⁵ |v(t)| split the integral accordingly. The calculator gives the total distance (e.g., ≈ 21.613).
[0,5]上的总路程。 路程 = ∫₀⁵ |v(t)| dt。先找出 v(t)=0 的时刻并分段积分，计算器得出总路程（如约21.613）。
Acceleration at t=4. a(4) = v'(4). Numerically differentiate v(t) at t=4 using the calculator' ce. The value is negative, showing the particle is slowing down in the positive direction.
t=4时的加速度。 a(4)=v'(4)。用计算器在 t=4 处求导数值，结果为负，表明粒子在正方向上减速。
When is speed increasing? Speed increases when velocity and acceleration have the same sign. Analyze the graphs of v and a over the interval.
速率何时增大？ 当速度与加速度同号时速率增大。在区间上分析 v 和 a 的图像。
4. Question 3: Graph of f′ and Analysis of f | 第3题：f′ 图像与 f 的分析
a semicircle of radius 2 (centered at (2,0), y≥0) and two line segments, is given. f(0)=2.
已知 f′ 的图像由半径为2的上半圆（圆心(2,0)，y≥0）和两条线段组成，且 f(0)=2。
Inflection points of f. gn at the point where the slope of f′ is maximum (at x=2). Thus x=2 is an inflection point. Also check where line segments meet.
f 的拐点。 拐点出现在 f′ 的单调性改变处，即 f″ 变号。在半圆上，f′ 的斜率在中心 x=2 处最大，因此 x=2 是一个拐点。还需检查线段连接点。
Relative extrema of f. positive at x=5. At x=1, f′ changes from positive to negative, giving a relative maximum.
f 的相对极值。 当 f′ 由负变正时 f 取得相对极小值。图像显示在 x=5 处 f′ 由负变正；在 x=1 处由正变负，对应相对极大值。
Evaluate f(4). triangle from the line segment. Compute the sum to get f(4) = 2 + π + (area of triangle).
计算 f(4)。 f(4) = f(0) + ∫₀⁴ f′(x) dx。积分值为 f′ 在0到4区间下的面积：四分之一圆面积 π·2²/4 = π 加上三角形面积。求得 f(4)=2+π+(三角形面积)。
5. Question 4: Differential Equation and Slope Field | 第4题：微分方程与斜率场
The differential equation is dy/dx = (1/3) x (y – 2)², with initial condition f(1) = 0.
微分方程为 dy/dx = (1/3) x (y – 2)²，初始条件 f(1)=0。
(a) Slope field and solution curve. ough (1,0). The curve decreases from left, reaches a minimum near x=0, and then increases.
(a) 斜率场和解曲线。
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