📚 AP Calculus BC Free Response Questions Analysis | AP微积分BC自由响应题真题解析
In this article, we work through a comprehensive free response question that mirrors the style and difficulty of the AP Calculus BC exam. The question integrates parametric equations, polar curves, Taylor series, and convergence analysis — all core topics that frequently appear on the BC test. By following the step-by-step solutions, you will strengthen your problem-solving techniques and exam readiness.
本文详细解析一道与 AP 微积分 BC 考试风格及难度高度吻合的综合自由响应题。题目融合了参数方程、极坐标曲线、泰勒级数与收敛性分析 —— 这些均为 BC 考试中频繁出现的核心考点。通过逐步拆解的解题过程，你将强化自己的解题技巧与应试能力。
1. Overview of the Question | 题目概述
The sample question consists of six parts that test a broad range of calculus BC skills. Parts (a) and (b) involve a particle moving along a parametric curve and require velocity vectors and arc length. Part (c) shifts to polar coordinates to find an enclosed area. Parts (d), (e) and (f) focus on Taylor series, approximation, and radius of convergence for a logarithmic function. This multi-topic design is typical of the BC free response section.
这道样题包含六个小问，全面考查微积分 BC 的多项技能。(a)(b) 两问涉及质点沿参数曲线运动的轨迹，要求计算速度向量与路径长度；(c) 问转向极坐标，求封闭区域的面积；(d)(e)(f) 问则聚焦对数函数的泰勒级数展开、近似值与收敛半径。这种多主题综合设计正是 BC 自由响应题的典型特征。
2. Part (a): Velocity Vector from Parametric Equations | 第(a)问：参数方程下的速度向量
A particle moves along a curve defined by x(t) = t² − 2t and y(t) = t³ − 3t for t ≥ 0. To find the velocity vector at t = 2, we first compute the derivatives with respect to t. The velocity vector is v(t) = ⟨x′(t), y′(t)⟩. Here x′(t) = 2t − 2 and y′(t) = 3t² − 3.
一质点沿参数曲线 x(t) = t² − 2t，y(t) = t³ − 3t (t ≥ 0) 运动。为求 t = 2 时的速度向量，我们先对 t 求导。速度向量为 v(t) = ⟨x′(t), y′(t)⟩。此处 x′(t) = 2t − 2，y′(t) = 3t² − 3。
Substituting t = 2 gives x′(2) = 2(2) − 2 = 2, and y′(2) = 3(2)² − 3 = 9. Therefore:
代入 t = 2 得到 x′(2) = 2(2) − 2 = 2，y′(2) = 3(2)² − 3 = 9。因此：
v(2) = ⟨2, 9⟩
The speed at t = 2 is the magnitude of the velocity vector: |v(2)| = √(2² + 9²) = √85. The direction of motion is given by the vector components.
在 t = 2 时的速率是速度向量的模：|v(2)| = √(2² + 9²) = √85。物体的运动方向由向量的分量给出。
3. Part (b): Total Distance Traveled | 第(b)问：运动距离的计算
The total distance traveled from t = 0 to t = 3 is the integral of the speed over that interval. Using the arc length formula for parametric curves:
从 t = 0 到 t = 3 运动的总距离是该区间上速率的积分。使用参数曲线的弧长公式：
Distance = ∫₀³ √( (x′(t))² + (y′(t))² ) dt
We already have x′(t) = 2t − 2 and y′(t) = 3t² − 3. The integrand becomes √((2t − 2)² + (3t² − 3)²). This integral typically requires a calculator for a numerical approximation. Evaluating numerically yields a value around 22.06 units. On the exam, you would set up the integral and then use the stored function to compute the result.
我们已有 x′(t) = 2t − 2，y′(t) = 3t² − 3。被积函数变为 √((2t − 2)² + (3t² − 3)²)。此积分通常需借助计算器进行数值求解，计算结果约为 22.06 单位。在考试中，你需要正确列出积分表达式，然后利用计算器存储功能求出结果。
Remember to show the complete integral expression before using the calculator, as the setup often earns the majority of the points. The final answer should be reported to at least three decimal places.
请记住，在使用计算器前务必写出完整的积分表达式，因为式子本身通常占据多数分值。最终答案应至少保留三位小数。
4. Part (c): Area Enclosed by Polar Curve | 第(c)问：极坐标曲线围成区域的面积
Now consider the polar curve r(θ) = 2 + 2 sin θ for 0 ≤ θ ≤ 2π. This is a cardioid. The area enclosed by a polar curve is given by A = ½ ∫[α,β] r² dθ. Since the curve traces out the full shape over 0 to 2π, we integrate over that interval:
现在考虑极坐标曲线 r(θ) = 2 + 2 sin θ，θ ∈ [0, 2π]。这是一条心形线。极坐标曲线所围面积公式为 A = ½ ∫[α,β] r² dθ。由于曲线在 0 到 2π 内描绘出完整的封闭图形，我们对整个区间积分：
A = (1/2) ∫₀²π (2 + 2 sin θ)² dθ
Expand the integrand: (2 + 2 sin θ)² = 4 + 8 sin θ + 4 sin² θ. Using the identity sin² θ = (1 − cos 2θ)/2, the expression becomes 4 + 8 sin θ + 2(1 − cos 2θ) = 6 + 8 sin θ − 2 cos 2θ.
展开被积函数：(2 + 2 sin θ)² = 4 + 8 sin θ + 4 sin² θ。利用恒等式 sin² θ = (1 − cos 2θ)/2，表达式化为 4 + 8 sin θ + 2(1 − cos 2θ) = 6 + 8 sin θ − 2 cos 2θ。
Now integrate term by term from 0 to 2π. The integral of 6 over that interval is 12π. The integrals of 8 sin θ and −2 cos 2θ over a full period are both zero. Hence:
现在逐项从 0 到 2π 积分。6 的积分为 12π；8 sin θ 和 −2 cos 2θ 在完整周期上的积分为零。因此：
A = (1/2) × (12π) = 6π
The enclosed area of the cardioid r = 2 + 2 sin θ is 6π square units.
心形线 r = 2 + 2 sin θ 所围区域的面积为 6π 平方单位。
5. Part (d): Taylor Series Expansion | 第(d)问：泰勒级数展开
Let f(x) = ln(1 + x). We are to find the first four nonzero terms and the general term of the Taylor series about x = 0 (Maclaurin series). Recall the derivatives of ln(1 + x): f′(x) = (1 + x)⁻¹, f″(x) = −(1 + x)⁻², f‴(x) = 2(1 + x)⁻³, and in general f⁽ⁿ⁾(x) = (−1)ⁿ⁻¹ (n − 1)! (1 + x)⁻ⁿ for n ≥ 1.
设 f(x) = ln(1 + x)。需求其在 x = 0 处（麦克劳林级数）的前四个非零项及通项。回忆导数：f′(x) = (1 + x)⁻¹，f″(x) = −(1 + x)⁻²，f‴(x) = 2(1 + x)⁻³，一般地 f⁽ⁿ⁾(x) = (−1)ⁿ⁻¹ (n − 1)! (1 + x)⁻ⁿ（n ≥ 1）。
Evaluating at x = 0 gives f(0) = 0, f′(0) = 1, f″(0) = −1, f‴(0) = 2, f⁽⁴⁾(0) = −6, etc. The nth term coefficient is f⁽ⁿ⁾(0)/n! = (−1)ⁿ⁻¹ (n − 1)! / n! = (−1)ⁿ⁻¹ / n. Thus, the series is:
在 x = 0 处求值：f(0) = 0，f′(0) = 1，f″(0) = −1，f‴(0) = 2，f⁽⁴⁾(0) = −6 等。第 n 项系数为 f⁽ⁿ⁾(0)/n! = (−1)ⁿ⁻¹ (n − 1)! / n! = (−1)ⁿ⁻¹ / n。因此级数为：
ln(1 + x) = x − x²/2 + x³/3 − x⁴/4 + … + (−1)ⁿ⁻¹ xⁿ / n + …
The first four nonzero terms are x, −x²/2, x³/3, −x⁴/4. The general term is (−1)ⁿ⁻¹ xⁿ / n for n ≥ 1.
前四个非零项为 x、−x²/2、x³/3、−x⁴/4。通项为 (−1)ⁿ⁻¹ xⁿ / n，n ≥ 1。
6. Part (e): Series Approximation and Error Bound | 第(e)问：级数近似与误差估计
We are asked to approximate ln(1.5) using the series, with error less than 0.001. Set x = 0.5 in the series: ln(1.5) = 0.5 − (0.5)²/2 + (0.5)³/3 − (0.5)⁴/4 + … = 0.5 − 0.125 + 0.041666… − 0.015625 + … . This is an alternating series whose terms decrease in magnitude to zero after the first term. For such series, the error after truncating at the nth term is less than the absolute value of the first omitted term.
题目要求用级数近似 ln(1.5)，并使误差小于 0.001。在级数中代入 x = 0.5：ln(1.5) = 0.5 − (0.5)²/2 + (0.5)³/3 − (0.5)⁴/4 + … = 0.5 − 0.125 + 0.041666… − 0.015625 + …。这是一个交错级数，从第二项开始各项绝对值单调递减趋于零。对这类级数，在第 n 项截断后的误差小于被省略的第一项的绝对值。
Compute terms: a₁ = 0.5, a₂ = −0.125, a₃ ≈ 0.04167, a₄ ≈ −0.015625, a₅ ≈ 0.00625, a₆ ≈ −0.0026, a₇ ≈ 0.00116, a₈ ≈ −0.00054. To achieve error < 0.001, we need to include terms until the next omitted term is less than 0.001 in magnitude. The magnitude of a₈ is about 0.00054, so stopping after the 7th term (n=7) gives an error less than a₈ < 0.001.
计算各项：a₁ = 0.5，a₂ = −0.125，a₃ ≈ 0.04167，a₄ ≈ −0.015625，a₅ ≈ 0.00625，a₆ ≈ −0.0026，a₇ ≈ 0.00116，a₈ ≈ −0.00054。要使误差 < 0.001，需要一直加到下一个被省略项的量小于 0.001 为止。a₈ 的绝对值约为 0.00054，因此截断在第 7 项之后（即包含前 7 项），误差将小于 a₈ < 0.001。
Summing the first seven terms: 0.5 − 0.125 = 0.375; +0.04167 = 0.41667; −0.015625 = 0.401045; +0.00625 = 0.407295; −0.0026 = 0.404695; +0.00116 = 0.405855. Thus ln(1.5) ≈ 0.4059 with error < 0.001. (The true value is about 0.405465.)
前七项之和：0.5 − 0.125 = 0.375；+0.04167 = 0.41667；−0.015625 = 0.401045；+0.00625 = 0.407295；−0.0026 = 0.404695；+0.00116 = 0.405855。故 ln(1.5) ≈ 0.4059，误差 < 0.001。（真实值约为 0.405465。）
7. Part (f): Radius of Convergence | 第(f)问：收敛半径的确定
For the series ∑ (−1)ⁿ⁻¹ xⁿ / n, apply the Ratio Test. Let aₙ = (−1)ⁿ⁻¹ xⁿ / n. Then |aₙ₊₁ / aₙ| = |x| · n/(n+1). Taking the limit as n → ∞ gives |x|. The series converges absolutely when |x| < 1, and diverges when |x| > 1. Hence the radius of convergence is R = 1.
对于级数 ∑ (−1)ⁿ⁻¹ xⁿ / n，应用比值判别法。令 aₙ = (−1)ⁿ⁻¹ xⁿ / n，则 |aₙ₊₁ / aₙ| = |x| · n/(n+1)。当 n → ∞ 时极限为 |x|。当 |x| < 1 时级数绝对收敛，当 |x| > 1 时发散。因此收敛半径 R = 1。
We also need to test the endpoints x = 1 and x = −1. At x = 1, the series becomes ∑ (−1)ⁿ⁻¹ / n, which is the alternating harmonic series and converges conditionally. At x = −1, the series is ∑ (−1)ⁿ⁻¹ (−1)ⁿ / n = ∑ (−1)²ⁿ⁻¹ / n = ∑ −1/n, the negative harmonic series, which diverges. The interval of convergence is (−1, 1].
还需检验端点 x = 1 和 x = −1。x = 1 时，级数为 ∑ (−1)ⁿ⁻¹ / n，即交错调和级数，条件收敛。x = −1 时，级数为 ∑ (−1)ⁿ⁻¹ (−1)ⁿ / n = ∑ −1/n，发散的调和级数。故收敛区间为 (−1, 1]。
8. Exam Tips and Summary | 备考建议与总结
Tackling BC free response questions successfully requires fluency in multiple representations: parametric, polar, and series. Always show the setup explicitly before using a calculator for numeric integration or root-finding. For series approximations, identify whether the series is alternating to apply the easy error bound, and remember to check endpoints for the radius of convergence question. Practice integrating these skills in timed conditions to build confidence.
成功应对 BC 自由响应题需要熟练掌握多种数学表达形式：参数方程、极坐标和级数。在使用计算器进行数值积分或求根前，务必清晰写出积分或方程表达式。对于级数近似，要判断是否为交错级数以利用简洁的误差界，并在收敛半径问题中检验端点。在规定时间内综合练习这些技能，有助于增强考试信心。
This sample analysis covered velocity and arc length for parametric curves, polar area integration, Taylor polynomial construction, alternating series error estimation, and convergence analysis — a microcosm of the BC exam’s free response section. Keep reviewing these concepts regularly, and you’ll be well prepared for test day.
本样题解析涵盖了参数方程的速度与弧长、极坐标面积积分、泰勒多项式的构造、交错级数误差估计以及收敛性分析——这正是 BC 考试自由响应部分的缩影。定期复习这些概念，你将为自己的考试做好充分准备。
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