StarWise International · AI Exam Prep

AP Calculus AB
考前 AI 押题冲刺手册7-Day AI Cram Prep Guide

🔥 2026 年最可能考:🔥 Most Likely on 2026:
FRQ1 流入/累积100% ·  FRQ3 f' 图像63% ·  FRQ2 粒子运动63%FRQ1 Rate In/Out100% ·  FRQ3 f' Graph63% ·  FRQ2 Linear Particle63%
基于 2014-2019 + 2021-2023 共 54 道 AP Calc AB FRQ + 179 道 MCQ(4 年:2014/15/16/18) · 2020 无考、2024/25 数据待补Based on 2014-2019 + 2021-2023 · 54 AP Calc AB FRQs + 179 MCQs (4 years: 2014/15/16/18) · 2020 COVID, 2024/25 pending

把冲刺时间花在最可能考的地方 · 每题配解题模板 + 踩分点 + 历年真题 + 官方 Scoring GuidelineFocus your cram time on what matters most — each slot includes templates, pitfalls, past FRQs, and official scoring guidelines.

54历年 FRQ
9 年真题覆盖
179MCQ 真题
6 + 6Mock + 槽位
54Historical FRQs
9 yearsReal exam coverage
179Real MCQs
6 + 6Mock + Slots
适合谁Who this is for
  • 距离 AP Calc AB 考试 ≤ 7 天
  • 基础已过关,想冲 4+ / 5 分
  • 希望把时间花在最可能考的地方
  • ≤ 7 days until AP Calc AB exam
  • Fundamentals covered, aiming for 4+ / 5
  • Want to focus time on the most likely tested topics
本手册包含What's included
  • FRQ 六大槽位 9 年热力图 + 2026 押题概率
  • 每个 FRQ 槽位 6-9 道历年真题 + 官方评分标准
  • 2026 AI 押题模拟卷(6 FRQ + 完整解答)
  • 179 道 MCQ 真题库(按 Unit 1-8 考点筛选)
  • iPad 一键导出练习卷 + 考前 24h Cheat Sheet
  • Six FRQ slots · 9-year heat map + 2026 prediction probability
  • 6-9 historical real problems + official scoring guidelines per FRQ slot
  • 2026 AI predicted mock exam (6 FRQs + complete solutions)
  • 179-problem real MCQ bank (filtered by Unit 1-8 topics)
  • One-click iPad practice sheet export + 24h pre-exam Cheat Sheet

📘 AP Calculus AB · 考试介绍📘 AP Calculus AB · Exam Overview

新生先看:AP Calculus AB 考试结构、时长、计算器政策、1-5 分评分标准一页看懂。New here? A one-glance overview of AP Calculus AB — format, timing, calculator policy, and the 1–5 score scale.

Section I · 选择题(MCQ)Section I · Multiple Choice (MCQ)
总 45 题 · 1 小时 45 分钟 · 占总分 50%45 questions · 1 hr 45 min · 50% of total
Part A · 不可用计算器Part A · No Calculator
30 题 · 60 分钟30 Qs · 60 min
Part B · 必须用计算器Part B · Calculator Required
15 题 · 45 分钟15 Qs · 45 min
Section II · 自由回答(FRQ)Section II · Free Response (FRQ)
总 6 题 · 1 小时 30 分钟 · 占总分 50%6 questions · 1 hr 30 min · 50% of total
Part A · 必须用计算器Part A · Calculator Required
2 题 · 30 分钟2 Qs · 30 min
Part B · 不可用计算器Part B · No Calculator
4 题 · 60 分钟4 Qs · 60 min
计算器:推荐 TI-84 Plus / TI-Nspire CX(非 CAS)。CAS 机型在大学先修课程中允许,但部分州府禁用 — 提前确认。Calculator: TI-84 Plus or TI-Nspire CX (non-CAS) recommended. CAS models are CB-approved but some districts ban them — verify ahead.

得分政策 · 1-5 分标准Score Policy · 1–5 Scale

最终成绩按 Raw Score 折算,从 1 到 5。多数美国大学 4-5 可换学分,部分接受 3。Raw scores convert to a 1–5 scaled score. Most US colleges grant credit for 4–5; some accept 3.

AP 分数AP Score原始分区间Raw Score Range说明Description
572 – 110Extremely Well Qualified(相当于大学 A)Extremely Well Qualified (college A)
460 – 71Well Qualified(A- / B+ / B)Well Qualified (A- / B+ / B)
343 – 59Qualified — 通常视作「通过」Qualified — widely considered passing

满分 ≈ 110 分(45 MCQ × 1.25 + 6 FRQ × 9)。Max ≈ 110 (45 MCQ × 1.25 + 6 FRQ × 9).

Part I · FRQ 押题分析Part I · FRQ Prediction Analysis

8 年槽位热力图8-Year Slot Heatmap

横轴是年份,纵轴是 FRQ 位置。同色块 = 同类考点。X-axis: year. Y-axis: FRQ slot. Same color = same topic archetype.

20152016201720182019202120222023
FRQ1
Calc
流入流出率Rate In/Out流入流出率Rate In/Out流入流出率Rate In/Out流入流出率Rate In/Out流入流出率Rate In/Out表格 / MVT / 近似Table · MVT流入流出率Rate In/Out表格 / MVT / 近似Table · MVT
FRQ2
Calc
面积 / 体积Area · Volumex 轴粒子运动Linear Particle流入流出率Rate In/Outx 轴粒子运动Linear Particlex 轴粒子运动Linear Particlex 轴粒子运动Linear Particle面积 / 体积Area · Volumex 轴粒子运动Linear Particle
FRQ3
No Calc
表格 / MVT / 近似Table · MVTf' 图像分析f' Graphf' 图像分析f' Graphf' 图像分析f' Graphf' 图像分析f' Graph面积 / 体积Area · Volumef' 图像分析f' Graph微分方程Diff. Eq.
FRQ4
No Calc
微分方程Diff. Eq.微分方程Diff. Eq.微分方程Diff. Eq.表格 / MVT / 近似Table · MVT相关变化率Related Ratesf' 图像分析f' Graph相关变化率Related Ratesf' 图像分析f' Graph
FRQ5
No Calc
f' 图像分析f' Graph面积 / 体积Area · Volumex 轴粒子运动Linear Particle函数分析Function Analysis面积 / 体积Area · Volume隐函数求导Implicit Diff微分方程Diff. Eq.函数关系综合Connected Funcs
FRQ6
No Calc
隐函数求导Implicit Diff函数关系综合Connected Funcs函数关系综合Connected Funcs微分方程Diff. Eq.函数关系综合Connected Funcs微分方程Diff. Eq.x 轴粒子运动Linear Particle隐函数求导Implicit Diff
三秒看懂(8 年数据):
  • FRQ1 累积类 100% — 6/8 流入流出率 + 2/8 表格累积
  • FRQ2 63% 粒子运动 / 25% 面积 / 13% 流入
  • FRQ3 f' 图像 63% — No calc,半圆 + 线段,FTC 反推
  • FRQ4 微分方程 38% / 相关变化率 25% / f'图 25% / 表格 13%
  • FRQ5 混槽:面积 25% / 函数分析 / 粒子 / 隐函数 / 连接各 13%
  • FRQ6 连接函数 38% / 隐函数 25% / 微分方程 25% / 粒子 13%
  • AB 不考 参数/极坐标/级数(BC 专属)
Three-second overview (8 years of data):
  • FRQ1 Accumulation class 100% — 6/8 Rate In/Out + 2/8 tabular accumulation
  • FRQ2 63% Particle Motion / 25% Area / 13% Flow-In
  • FRQ3 f' Graph 63% — No calc, semicircle + line segments, FTC reverse
  • FRQ4 Differential Equation 38% / Related Rates 25% / f' Graph 25% / Table 13%
  • FRQ5 Mixed slot: Area 25% / Function Analysis / Particle / Implicit / Connected each 13%
  • FRQ6 Connected Functions 38% / Implicit 25% / Differential Equation 25% / Particle 13%
  • AB does not test parametric/polar/series (BC Only)

2026 每个槽位的押题概率2026 Prediction Probability per Slot

概率 = 该题型在 2015-2023 的出现频率。Probability = historical frequency of this topic on 2015-2023 exams.

槽位Slot最可能Top pick备选Backup
FRQ1 (Calc)流入流出率Rate In/Out 75%表格 / MVT / 近似Table · MVT 25%
FRQ2 (Calc)x 轴粒子运动Linear Particle 62%面积 / 体积Area · Volume 25%
FRQ3 (No Calc)f' 图像分析f' Graph 62%
FRQ4 (No Calc)微分方程Diff. Eq. 38%相关变化率Related Rates 25%  / f' 图像分析f' Graph 25%
FRQ5 (No Calc)面积 / 体积Area · Volume 25%
FRQ6 (No Calc)函数关系综合Connected Funcs 38%隐函数求导Implicit Diff 25%  / 微分方程Diff. Eq. 25%

📱 如何用 iPad / Mac 对着网页做题?

本手册三处都有「导出练习卷」功能 —— MCQ(下方)、FRQ 槽位真题(Part II 每个槽位右上)、2026 模拟卷(Part III 顶部):

  1. MCQ 库:筛选想做的题 → 点 导出当前筛选为练习卷 → 每题下带横线笔记格子 + 末尾答题卡 Ⓐ Ⓑ Ⓒ Ⓓ Ⓔ
  2. FRQ 真题:在 Part II 每个槽位(FRQ1-6),点槽位内的 📱 导出 iPad 练习卷 → 自动导出该槽位 9-11 道历年真题,每题预留 2 页解题笔记空间
  3. 2026 模拟卷:Part III 顶部 📱 导出整套模拟卷(无答案) → 6 道原创 FRQ 打印成一份完整卷,每题 2 页笔记空间

iPad 工作流:点导出按钮 → Safari 分享(⬆️)→「打印」→ 双指撑开预览 → 右下分享 → 存入 GoodNotes / Notability → 用 Apple Pencil 在横线格子上写解题过程。做完回网页点「答案 / 解答」对照。

📱 How to practice on iPad / Mac against this page?

Three places support "Export as practice sheet" — MCQ (below), FRQ past questions (Part II, each slot), 2026 mock (Part III top):

  1. MCQ bank: filter → click Export Current Filter as Practice Sheet → each question has writing lines + bubble sheet Ⓐ Ⓑ Ⓒ Ⓓ Ⓔ
  2. FRQ past: in Part II, click 📱 Export Practice Sheet inside a slot → 9-11 past FRQs with 2 pages of writing space per problem
  3. 2026 Mock: Part III top 📱 Export Full Mock (no answers) → 6 original FRQs as one booklet, 2 pages per problem

iPad workflow: click Export → Safari Share (⬆️) → "Print" → pinch preview open → Share bottom-right → save to GoodNotes / Notability → work with Apple Pencil on the lines. Come back to the web page to check answers.

Part II · FRQ 六大槽位 · 模板 + 真题Part II · FRQ Six Slots · Templates + Past Questions

每个 FRQ 槽位:怎么识别 → 解题模板 → 踩分点 → 历年真题(点击展开)。For each FRQ slot: how to identify → solution template → pitfalls → past FRQs (click to expand).

FRQ1 · 流入流出率 / 累积量(AB 100% 必考)FRQ1 · Rate In/Out · Accumulation (AB 100% Must-Know)

概率 100%100% Likely

考 Unit 6(积分与累积变化)。给你流入率 E(t)(或流出率 L(t)),求累积量、最大/最小时刻。每年都考。Tests Unit 6 (integration and accumulation of change). Given rate in E(t) (or rate out L(t)), find accumulated amount, time of max/min. Tested every year.

🔍 怎么一眼认出🔍 How to Identify

  • 有 rate 函数,单位是「per hour / minute / second」
  • 问题涉及「total amount」「accumulated」「net change」「when max/min」
  • Calculator allowed
  • Rate function given, units are 'per hour / minute / second'
  • Question involves 'total amount', 'accumulated', 'net change', 'when max/min'
  • Calculator allowed

📐 解题模板📐 Solution Template

  1. Step 1:净量 \(= \int_a^b (E(t) - L(t))\,dt\)。若求 \(t=b\) 的总量,加上初始值 \(A(0)\)。
  2. Step 2:rate of change = rate 本身。求 \(E(5)\) 直接代入,\(E'(5)\) 直接求导。calc 允许时别手算积分
  3. Step 3:求「when max」→ 解 \(A'(t)=0\),即 \(E(t)=L(t)\);sign chart 判断。
  4. Step 4:若 rate 是表格 → trapezoidal sum \(\sum \frac{b-a}{2}(L(a)+L(b))\)。
  5. Step 5每个 sub-part 都要写单位(fish, liters, dollars)——没单位直接扣分。
  1. Step 1: Net amount \(= \int_a^b (E(t) - L(t))\,dt\). To find total at \(t=b\), add initial value \(A(0)\).
  2. Step 2: rate of change = rate itself. To find \(E(5)\), substitute directly; \(E'(5)\) requires differentiation. Don't compute integrals by hand when calc is allowed.
  3. Step 3: To find 'when max' → solve \(A'(t)=0\), i.e. \(E(t)=L(t)\); use sign chart to decide.
  4. Step 4: If rate is given as a table → trapezoidal sum \(\sum \frac{b-a}{2}(L(a)+L(b))\).
  5. Step 5: Write units for every sub-part (fish, liters, dollars) — no units means deducted points.

⚠️ 踩分点与易错⚠️ Pitfalls & Common Errors

  • ⚠️ 积分下限漏掉初始量 \(A(0)\)(第 3 问最常见)
  • ⚠️ trapezoidal 系数 \((b-a)/2\) 写成 \((b-a)\)
  • ⚠️ 「rate of change of amount」 vs 「rate」概念混
  • ⚠️ 最后一问 interpret 要写「单位 + 时间区间 + 物理意义」
  • ⚠️ Forgetting initial amount \(A(0)\) in the integral (most common on part 3)
  • ⚠️ Writing trapezoidal coefficient \((b-a)/2\) as \((b-a)\)
  • ⚠️ Confusing 'rate of change of amount' vs 'rate'
  • ⚠️ For the last interpretation part, write 'units + time interval + physical meaning'

历年真题(9 道 · 点击展开)Past FRQs (9 · Click to Expand)

2015 FRQ1 · Rainwater R(t) = 20 sin(t²/35) 流入 drainpipeRainwater R(t) = 20 sin(t²/35) into drainpipe

The rate at which rainwater flows into a drainpipe is modeled by the function \(R\), where \(R(t) = 20\sin\left(\dfrac{t^2}{35}\right)\) cubic feet per hour, \(t\) is measured in hours, and \(0 \le t \le 8\). The pipe is partially blocked, allowing water to drain out the other end of the pipe at a rate modeled by \(D(t) = -0.04t^3 + 0.4t^2 + 0.96t\) cubic feet per hour, for \(0 \le t \le 8\). There are 30 cubic feet of water in the pipe at time \(t = 0\).

  1. (A) How many cubic feet of rainwater flow into the pipe during the 8-hour time interval \(0 \le t \le 8\)?
  2. (B) Is the amount of water in the pipe increasing or decreasing at time \(t = 3\) hours? Give a reason for your answer.
  3. (C) At what time \(t\), \(0 \le t \le 8\), is the amount of water in the pipe at a minimum? Justify your answer.
  4. (D) The pipe can hold 50 cubic feet of water before overflowing. For \(t > 8\), water continues to flow into and out of the pipe at the given rates until the pipe begins to overflow. Write, but do not solve, an equation involving one or more integrals that gives the time \(w\) when the pipe will begin to overflow.
📝 点击展开评分标准(Scoring Guideline)📝 Expand Scoring Guideline 2015 FRQ1 SG
2016 FRQ1 AB/BC 同题AB/BC Shared · 水泵入 W(t) = 2000 e^(-t²/20) + 表格移除 RWater pumped in W(t) = 2000 e^(-t²/20) + tabular removal R

Water is pumped into a tank at a rate modeled by \(W(t) = 2000\,e^{-t^{2}/20}\) liters per hour for \(0 \le t \le 8\), where \(t\) is measured in hours. Water is removed from the tank at a rate modeled by \(R(t)\) liters per hour, where \(R\) is differentiable and decreasing on \(0 \le t \le 8\). Selected values of \(R(t)\) are shown in the table above. At time \(t = 0\), there are 50,000 liters of water in the tank.

t (hours)R(t) (liters per hour)
01340
11190
3950
6740
8700
  1. (A) Estimate \(R'(2)\). Show the work that leads to your answer. Indicate units of measure.
  2. (B) Use a left Riemann sum with the four subintervals indicated by the table to estimate the total amount of water removed from the tank during the 8 hours. Is this an overestimate or an underestimate of the total amount of water removed? Give a reason for your answer.
  3. (C) Use your answer from part (b) to find an estimate of the total amount of water in the tank, to the nearest liter, at the end of 8 hours.
  4. (D) For \(0 \le t \le 8\), is there a time \(t\) when the rate at which water is pumped into the tank is the same as the rate at which water is removed from the tank? Explain why or why not.
📝 点击展开评分标准(Scoring Guideline)📝 Expand Scoring Guideline 2016 FRQ1 SG
2017 FRQ1 AB/BC 同题AB/BC Shared · Tank A(h) cross section + flow

A tank has a height of 10 feet. The area of the horizontal cross section of the tank at height \(h\) feet is given by the function \(A\), where \(A(h)\) is measured in square feet. The function \(A\) is continuous and decreases as \(h\) increases. Selected values for \(A(h)\) are given in the table above.

\(h\) (feet)02510
\(A(h)\) (square feet)50.314.46.52.9
  1. (A) Use a left Riemann sum with the three subintervals indicated by the data in the table to approximate the volume of the tank. Indicate units of measure.
  2. (B) Does the approximation in part (a) overestimate or underestimate the volume of the tank? Explain your reasoning.
  3. (C) The area, in square feet, of the horizontal cross section at height \(h\) feet is modeled by the function \(f\) given by \(f(h) = \dfrac{50.3}{e^{0.2h} + h}\). Based on this model, find the volume of the tank. Indicate units of measure.
  4. (D) Water is pumped into the tank. When the height of the water is 5 feet, the height is increasing at the rate of 0.26 foot per minute. Using the model from part (c), find the rate at which the volume of water is changing with respect to time when the height of the water is 5 feet. Indicate units of measure.
📝 点击展开评分标准(Scoring Guideline)📝 Expand Scoring Guideline 2017 FRQ1 SG
2017 FRQ2 · Grocery store 50lb 香蕉 + 顾客移除 R(t)Grocery store 50lb bananas + customer removal R(t)

When a certain grocery store opens, it has 50 pounds of bananas on a display table. Customers remove bananas from the display table at a rate modeled by \(f(t) = 10 + 0.8t \sin\left(\dfrac{t^3}{100}\right)\) for \(0 < t \le 12\), where \(f(t)\) is measured in pounds per hour and \(t\) is the number of hours after the store opened. After the store has been open for three hours, store employees add bananas to the display table at a rate modeled by \(g(t) = 3 + 2.4\ln(t^2 + 2t)\) for \(3 < t \le 12\), where \(g(t)\) is measured in pounds per hour and \(t\) is the number of hours after the store opened.

  1. (A) How many pounds of bananas are removed from the display table during the first 2 hours the store is open?
  2. (B) Find \(f'(7)\). Using correct units, explain the meaning of \(f'(7)\) in the context of the problem.
  3. (C) Is the number of pounds of bananas on the display table increasing or decreasing at time \(t = 5\)? Give a reason for your answer.
  4. (D) How many pounds of bananas are on the display table at time \(t = 8\)?
📝 点击展开评分标准(Scoring Guideline)📝 Expand Scoring Guideline 2017 FRQ2 SG
2018 FRQ1 AB/BC 同题AB/BC Shared · People enter escalator r(t) = 44(t/100)³(1-t/300)

People enter a line for an escalator at a rate modeled by the function \(r\) given by \[r(t) = \begin{cases} 44\left(\dfrac{t}{100}\right)^3 \left(1 - \dfrac{t}{300}\right)^7 & \text{for } 0 \le t \le 300 \\ 0 & \text{for } t > 300 \end{cases}\] where \(r(t)\) is measured in people per second and \(t\) is measured in seconds. As people get on the escalator, they exit the line at a constant rate of 0.7 person per second. There are 20 people in line at time \(t = 0\).

  1. (A) How many people enter the line for the escalator during the time interval \(0 \le t \le 300\)?
  2. (B) During the time interval \(0 \le t \le 300\), there are always people in line for the escalator. How many people are in line at time \(t = 300\)?
  3. (C) For \(t > 300\), what is the first time \(t\) that there are no people in line for the escalator?
  4. (D) For \(0 \le t \le 300\), at what time \(t\) is the number of people in line a minimum? To the nearest whole number, find the number of people in line at this time. Justify your answer.
📝 点击展开评分标准(Scoring Guideline)📝 Expand Scoring Guideline 2018 FRQ1 SG
2019 FRQ1 · Fish enter E(t) = 20+15sin(πt/6), leave L(t)

Fish enter a lake at a rate modeled by the function \(E\) given by \(E(t) = 20 + 15\sin\left(\dfrac{\pi t}{6}\right)\). Fish leave the lake at a rate modeled by the function \(L\) given by \(L(t) = 4 + 2^{0.1t^{2}}\). Both \(E(t)\) and \(L(t)\) are measured in fish per hour, and \(t\) is measured in hours since midnight \((t = 0)\).

  1. (A) How many fish enter the lake over the 5-hour period from midnight \((t = 0)\) to 5 A.M. \((t = 5)\)? Give your answer to the nearest whole number.
  2. (B) What is the average number of fish that leave the lake per hour over the 5-hour period from midnight \((t = 0)\) to 5 A.M. \((t = 5)\)?
  3. (C) At what time \(t\), for \(0 \le t \le 8\), is the greatest number of fish in the lake? Justify your answer.
  4. (D) Is the rate of change in the number of fish in the lake increasing or decreasing at 5 A.M. \((t = 5)\)? Explain your reasoning.
2021 FRQ1 AB/BC 同题AB/BC Shared · 🆕 Bacteria density f(r) tabular (shared BC)🆕 Bacteria density f(r) tabular (AB/BC shared)

The density of a bacteria population in a circular petri dish at a distance \(r\) centimeters from the center of the dish is given by an increasing, differentiable function \(f\), where \(f(r)\) is measured in milligrams per square centimeter. Values of \(f(r)\) for selected values of \(r\) are given in the table above.

\(r\) (centimeters)0122.54
\(f(r)\) (milligrams per square centimeter)1261018
  1. (A) Use the data in the table to estimate \(f'(2.25)\). Using correct units, interpret the meaning of your answer in the context of this problem.
  2. (B) The total mass, in milligrams, of bacteria in the petri dish is given by the integral expression \(2\pi\displaystyle\int_0^4 r\,f(r)\,dr\). Approximate the value of \(2\pi\displaystyle\int_0^4 r\,f(r)\,dr\) using a right Riemann sum with the four subintervals indicated by the data in the table.
  3. (C) Is the approximation found in part (b) an overestimate or underestimate of the total mass of bacteria in the petri dish? Explain your reasoning.
  4. (D) The density of bacteria in the petri dish, for \(1 \le r \le 4\), is modeled by the function \(g\) defined by \(g(r) = 2 - 16\bigl(\cos(1.57\sqrt{r})\bigr)^3\). For what value of \(k\), \(1 < k < 4\), is \(g(k)\) equal to the average value of \(g(r)\) on the interval \(1 \le r \le 4\)?
📝 点击展开评分标准(Scoring Guideline)📝 Expand Scoring Guideline 2021 FRQ1 SG
2022 FRQ1 AB/BC 同题AB/BC Shared · 🆕 Toll plaza A(t) = 450 sin(0.62t) (shared BC)🆕 Toll plaza A(t) = 450 sin(0.62t) (AB/BC shared)

From 5 A.M. to 10 A.M., the rate at which vehicles arrive at a certain toll plaza is given by \(A(t) = 450\sqrt{\sin(0.62t)}\), where \(t\) is the number of hours after 5 A.M. and \(A(t)\) is measured in vehicles per hour. Traffic is flowing smoothly at 5 A.M. with no vehicles waiting in line.

  1. (A) Write, but do not evaluate, an integral expression that gives the total number of vehicles that arrive at the toll plaza from 6 A.M. \((t = 1)\) to 10 A.M. \((t = 5)\).
  2. (B) Find the average value of the rate, in vehicles per hour, at which vehicles arrive at the toll plaza from 6 A.M. \((t = 1)\) to 10 A.M. \((t = 5)\).
  3. (C) Is the rate at which vehicles arrive at the toll plaza at 6 A.M. \((t = 1)\) increasing or decreasing? Give a reason for your answer.
  4. (D) A line forms whenever \(A(t) \ge 400\). The number of vehicles in line at time \(t\), for \(a \le t \le 4\), is given by \(N(t) = \displaystyle\int_a^t (A(x) - 400)\,dx\), where \(a\) is the time when a line first begins to form. To the nearest whole number, find the greatest number of vehicles in line at the toll plaza in the time interval \(a \le t \le 4\). Justify your answer.
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2023 FRQ1 AB/BC 同题AB/BC Shared · 🆕 Gas pumping rate tabular (shared BC)🆕 Gas pumping rate tabular (AB/BC shared)

A customer at a gas station is pumping gasoline into a gas tank. The rate of flow of gasoline is modeled by a differentiable function \(f\), where \(f(t)\) is measured in gallons per second and \(t\) is measured in seconds since pumping began. Selected values of \(f(t)\) are given in the table.

\(t\) (seconds)06090120135150
\(f(t)\) (gallons per second)00.10.150.10.050
  1. (A) Using correct units, interpret the meaning of \(\displaystyle\int_{60}^{135} f(t)\,dt\) in the context of the problem. Use a right Riemann sum with the three subintervals \([60, 90]\), \([90, 120]\), and \([120, 135]\) to approximate the value of \(\displaystyle\int_{60}^{135} f(t)\,dt\).
  2. (B) Must there exist a value of \(c\), for \(60 < c < 120\), such that \(f'(c) = 0\)? Justify your answer.
  3. (C) The rate of flow of gasoline, in gallons per second, can also be modeled by \(g(t) = \left(\dfrac{t}{500}\right)\cos\!\left(\left(\dfrac{t}{120}\right)^2\right)\) for \(0 \le t \le 150\). Using this model, find the average rate of flow of gasoline over the time interval \(0 \le t \le 150\). Show the setup for your calculations.
  4. (D) Using the model \(g\) defined in part (c), find the value of \(g'(140)\). Interpret the meaning of your answer in the context of the problem.
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FRQ2 · 粒子运动 / 面积体积 / 累积FRQ2 · Particle Motion / Area-Volume / Accumulation

概率 60%60% Likely

AB Q2 最混搭槽位。近 5 年:粒子 60% / 面积 20% / 累积 20%。都是 calc-allowed 的 applied 题。AB Q2 is the most mixed slot. Last 5 years: Particle 60% / Area 20% / Accumulation 20%. All calc-allowed applied problems.

🔍 怎么一眼认出🔍 How to Identify

  • 粒子型:给 v(t);问 position、speed、distance、when turn
  • 面积型:给两条曲线的图;问 area + 体积(绕 x 或 y 轴)
  • Particle type: v(t) given; asks position, speed, distance, when turning
  • Area type: graph of two curves given; asks area + volume (around x or y axis)

📐 解题模板📐 Solution Template

  1. 粒子(x 轴)5 步
  2. Step 1:位置 \(x(T) = x(0) + \int_0^T v(t)\,dt\)。
  3. Step 2:speed \(= |v(t)|\);distance \(= \int |v(t)|\,dt\)。
  4. Step 3:when direction changes → \(v(t) = 0\) 且 \(v\) 变号。
  5. Step 4:acceleration \(a(t) = v'(t)\);speed 增加 ⇔ \(v, a\) 同号。
  6. 面积/体积 5 步
  7. Step 1:找交点 → 解 \(f(x) = g(x)\)。
  8. Step 2:面积 \(= \int_a^b |f - g|\,dx\)。
  9. Step 3:绕 x 轴体积 (washer) \(= \pi \int (R^2 - r^2)\,dx\)。
  10. Step 4:cross-section 体积 \(= \int A(x)\,dx\),\(A(x)\) 为已知截面面积。
  1. Particle (x-axis) 5 steps
  2. Step 1: Position \(x(T) = x(0) + \int_0^T v(t)\,dt\).
  3. Step 2: speed \(= |v(t)|\); distance \(= \int |v(t)|\,dt\).
  4. Step 3: When direction changes → \(v(t) = 0\) and \(v\) changes sign.
  5. Step 4: Acceleration \(a(t) = v'(t)\); speed increasing ⇔ \(v, a\) have same sign.
  6. Area/Volume 5 steps
  7. Step 1: Find intersections → solve \(f(x) = g(x)\).
  8. Step 2: Area \(= \int_a^b |f - g|\,dx\).
  9. Step 3: Volume around x-axis (washer) \(= \pi \int (R^2 - r^2)\,dx\).
  10. Step 4: Cross-section volume \(= \int A(x)\,dx\), where \(A(x)\) is the known cross-section area.

⚠️ 踩分点与易错⚠️ Pitfalls & Common Errors

  • ⚠️ speed 忘加绝对值
  • ⚠️ washer 公式忘 \(\pi\) 或外-内平方混
  • ⚠️ 粒子「when is particle moving right」跟 \(v > 0\) 关联
  • ⚠️ distance traveled 不等于 net displacement
  • ⚠️ Forgetting absolute value for speed
  • ⚠️ Washer formula missing \(\pi\) or confusing outer-inner squares
  • ⚠️ Particle 'when is particle moving right' linked to \(v > 0\)
  • ⚠️ Distance traveled is not equal to net displacement

历年真题(7 道 · 点击展开)Past FRQs (7 · Click to Expand)

2015 FRQ2 · Regions R, S 由 f, g 围成 → 面积 + 体积Regions R, S bounded by f, g → area + volume

Let \(R\) and \(S\) be the two regions enclosed by the graphs of \(f\) and \(g\) shown in the figure above.

  1. (A) Find the sum of the areas of regions \(R\) and \(S\).
  2. (B) Region \(S\) is the base of a solid whose cross sections perpendicular to the \(x\)-axis are squares. Find the volume of the solid.
  3. (C) Let \(h\) be the vertical distance between the graphs of \(f\) and \(g\) in region \(S\). Find the rate at which \(h\) changes with respect to \(x\) when \(x = 1.8\).
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2016 FRQ2 · Particle x 轴 v(t) = 1 + 2sin(t²/2)Particle on x-axis v(t) = 1 + 2sin(t²/2)

For \(t \ge 0\), a particle moves along the \(x\)-axis. The velocity of the particle at time \(t\) is given by \(v(t) = 1 + 2\sin\left(\dfrac{t^2}{2}\right)\). The particle is at position \(x = 2\) at time \(t = 4\).

  1. (A) At time \(t = 4\), is the particle speeding up or slowing down?
  2. (B) Find all times \(t\) in the interval \(0 < t < 3\) when the particle changes direction. Justify your answer.
  3. (C) Find the position of the particle at time \(t = 0\).
  4. (D) Find the total distance the particle travels from time \(t = 0\) to time \(t = 3\).
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2018 FRQ2 · Particle v(t) = 10sin(0.4t²)/(t²-t+3)

A particle moves along the x-axis with velocity given by \(v(t) = \dfrac{10\sin(0.4t^{2})}{t^{2} - t + 3}\) for time \(0 \le t \le 3.5\). The particle is at position \(x = -5\) at time \(t = 0\).

  1. (A) Find the acceleration of the particle at time \(t = 3\).
  2. (B) Find the position of the particle at time \(t = 3\).
  3. (C) Evaluate \(\int_{0}^{3.5} v(t)\,dt\), and evaluate \(\int_{0}^{3.5} |v(t)|\,dt\). Interpret the meaning of each integral in the context of the problem.
  4. (D) A second particle moves along the x-axis with position given by \(x_{2}(t) = t^{2} - t\) for \(0 \le t \le 3.5\). At what time \(t\) are the two particles moving with the same velocity?
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2019 FRQ2 · Particle P 表格 v_P + 表达式分析Particle P tabular v_P + expression analysis

The velocity of a particle, \(P\), moving along the x-axis is given by the differentiable function \(v_{P}\), where \(v_{P}(t)\) is measured in meters per hour and \(t\) is measured in hours. Selected values of \(v_{P}(t)\) are shown in the table above. Particle \(P\) is at the origin at time \(t = 0\).

t (hours)v_P(t) (meters per hour)
  1. (A) Justify why there must be at least one time \(t\), for \(0.3 \le t \le 2.8\), at which \(v_{P}'(t)\), the acceleration of particle \(P\), equals 0 meters per hour per hour.
  2. (B) Use a trapezoidal sum with the three subintervals \([0, 0.3]\), \([0.3, 1.7]\), and \([1.7, 2.8]\) to approximate the value of \(\int_{0}^{2.8} v_{P}(t)\,dt\).
  3. (C) A second particle, \(Q\), also moves along the x-axis so that its velocity for \(0 \le t \le 4\) is given by \(v_{Q}(t) = 45\sqrt{t}\cos(0.063t^{2})\) meters per hour. Find the time interval during which the velocity of particle \(Q\) is at least 60 meters per hour. Find the distance traveled by particle \(Q\) during the interval when the velocity of particle \(Q\) is at least 60 meters per hour.
  4. (D) At time \(t = 0\), particle \(Q\) is at position \(x = -90\). Using the result from part (b) and the function \(v_{Q}\) from part (c), approximate the distance between particles \(P\) and \(Q\) at time \(t = 2.8\).
2021 FRQ2 · 🆕 Particle P v_P(t) 表达式🆕 Particle P v_P(t) expression

A particle, \(P\), is moving along the \(x\)-axis. The velocity of particle \(P\) at time \(t\) is given by \(v_P(t) = \sin\left(t^{1.5}\right)\) for \(0 \le t \le \pi\). At time \(t = 0\), particle \(P\) is at position \(x = 5\). A second particle, \(Q\), also moves along the \(x\)-axis. The velocity of particle \(Q\) at time \(t\) is given by \(v_Q(t) = (t-1.8) \cdot 1.25^t\) for \(0 \le t \le \pi\). At time \(t = 0\), particle \(Q\) is at position \(x = 10\).

  1. (A) Find the positions of particles \(P\) and \(Q\) at time \(t = 1\).
  2. (B) Are particles \(P\) and \(Q\) moving toward each other or away from each other at time \(t = 1\)? Explain your reasoning.
  3. (C) Find the acceleration of particle \(Q\) at time \(t = 1\). Is the speed of particle \(Q\) increasing or decreasing at time \(t = 1\)? Explain your reasoning.
  4. (D) Find the total distance traveled by particle \(P\) over the time interval \(0 \le t \le \pi\).
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2022 FRQ2 · 🆕 f = ln(x+3), g = x⁴+2x³ 面积🆕 f = ln(x+3), g = x⁴+2x³ area

Let \(f\) and \(g\) be the functions defined by \(f(x) = \ln(x + 3)\) and \(g(x) = x^4 + 2x^3\). The graphs of \(f\) and \(g\), shown in the figure above, intersect at \(x = -2\) and \(x = B\), where \(B > 0\).

  1. (A) Find the area of the region enclosed by the graphs of \(f\) and \(g\).
  2. (B) For \(-2 \le x \le B\), let \(h(x)\) be the vertical distance between the graphs of \(f\) and \(g\). Is \(h\) increasing or decreasing at \(x = -0.5\)? Give a reason for your answer.
  3. (C) The region enclosed by the graphs of \(f\) and \(g\) is the base of a solid. Cross sections of the solid taken perpendicular to the \(x\)-axis are squares. Find the volume of the solid.
  4. (D) A vertical line in the \(xy\)-plane travels from left to right along the base of the solid described in part (c). The vertical line is moving at a constant rate of 7 units per second. Find the rate of change of the area of the cross section above the vertical line with respect to time when the vertical line is at position \(x = -0.5\).
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2023 FRQ2 · 🆕 Stephen 泳池来回 v(t) 90s🆕 Stephen swimming pool back and forth v(t) 90s

Stephen swims back and forth along a straight path in a 50-meter-long pool for 90 seconds. Stephen's velocity is modeled by \(v(t) = 2.38 e^{-0.02t} \sin\left(\dfrac{\pi}{56} t\right)\), where \(t\) is measured in seconds and \(v(t)\) is measured in meters per second.

  1. (A) Find all times \(t\) in the interval \(0 < t < 90\) at which Stephen changes direction. Give a reason for your answer.
  2. (B) Find Stephen's acceleration at time \(t = 60\) seconds. Show the setup for your calculations, and indicate units of measure. Is Stephen speeding up or slowing down at time \(t = 60\) seconds? Give a reason for your answer.
  3. (C) Find the distance between Stephen's position at time \(t = 20\) seconds and his position at time \(t = 80\) seconds. Show the setup for your calculations.
  4. (D) Find the total distance Stephen swims over the time interval \(0 \le t \le 90\) seconds. Show the setup for your calculations.
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FRQ3 · f' 图像分析(80% 必考)FRQ3 · f' Graph Analysis (80% Must-Know)

概率 80%80% Likely

Unit 5/6。给 f' 的图(半圆 + 线段组合),分析 f。No calculator。Unit 5/6. Graph of f' given (semicircle + line segment combination), analyze f. No calculator.

🔍 怎么一眼认出🔍 How to Identify

  • 一张 f' 的分段图(半圆 + 线段)
  • 问 f 的 increasing/decreasing、concavity、extrema、inflection
  • 通常要 evaluate f(b) 从 f(a) 已知 → 用 FTC
  • A piecewise graph of f' (semicircle + line segments)
  • Asks increasing/decreasing, concavity, extrema, inflection of f
  • Usually requires evaluating f(b) from known f(a) → use FTC

📐 解题模板📐 Solution Template

  1. Step 1:f 单调递增 ↔ f' > 0。找 f' 正负区间。
  2. Step 2:f 凹 ↔ f'' > 0 ↔ f' 递增。看 f' 图的斜率(上升段 = f'' > 0)。
  3. Step 3:Local max:f' 从 + 变 -。Abs max 还要比较端点。
  4. Step 4:Inflection:f'' 变号 ↔ f' 有局部极值或拐点。
  5. Step 5:\(f(b) = f(a) + \int_a^b f'(t)\,dt\)。积分用几何面积(半圆 = \(\pi r^2/2\),三角形 = \(bh/2\))。
  1. Step 1: f increasing ↔ f' > 0. Find intervals where f' is positive/negative.
  2. Step 2: f concave up ↔ f'' > 0 ↔ f' increasing. Look at slope of f' graph (ascending segment = f'' > 0).
  3. Step 3: Local max: f' changes from + to -. Abs max also requires comparing endpoints.
  4. Step 4: Inflection: f'' changes sign ↔ f' has local extremum or corner.
  5. Step 5: \(f(b) = f(a) + \int_a^b f'(t)\,dt\). Compute integral using geometric area (semicircle = \(\pi r^2/2\), triangle = \(bh/2\)).

⚠️ 踩分点与易错⚠️ Pitfalls & Common Errors

  • ⚠️ 把「f' 图」当 f 图分析(常见低级错)
  • ⚠️ 半圆面积忘 \(1/2\)
  • ⚠️ Abs max 只看 critical point 漏端点
  • ⚠️ 积分下方区域(f' < 0)面积要算负值
  • ⚠️ Analyzing 'f' graph' as if it were the f graph (common low-level error)
  • ⚠️ Forgetting \(1/2\) for semicircle area
  • ⚠️ Abs max only looking at critical points, missing endpoints
  • ⚠️ Area below the axis (f' < 0) should be counted as negative

历年真题(8 道 · 点击展开)Past FRQs (8 · Click to Expand)

2016 FRQ3 AB/BC 同题AB/BC Shared · Piecewise-linear f; g(x) = ∫f → concavity

The figure above shows the graph of the piecewise-linear function \(f\). For \(-4 \le x \le 12\), the function \(g\) is defined by \(g(x) = \int_{2}^{x} f(t)\,dt\).

  1. (A) Does \(g\) have a relative minimum, a relative maximum, or neither at \(x = 10\)? Justify your answer.
  2. (B) Does the graph of \(g\) have a point of inflection at \(x = 4\)? Justify your answer.
  3. (C) Find the absolute minimum value and the absolute maximum value of \(g\) on the interval \(-4 \le x \le 12\). Justify your answers.
  4. (D) For \(-4 \le x \le 12\), find all intervals for which \(g(x) \le 0\).
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2017 FRQ3 AB/BC 同题AB/BC Shared · f 在 [-6,5], f(-2)=7; f' 图 → f 分析f on [-6,5], f(-2)=7; f' graph → analyze f

The function \(f\) is differentiable on the closed interval \([-6, 5]\) and satisfies \(f(-2) = 7\). The graph of \(f'\), the derivative of \(f\), consists of a semicircle and three line segments, as shown in the figure above.

  1. (A) Find the values of \(f(-6)\) and \(f(5)\).
  2. (B) On what intervals is \(f\) increasing? Justify your answer.
  3. (C) Find the absolute minimum value of \(f\) on the closed interval \([-6, 5]\). Justify your answer.
  4. (D) For each of \(f''(-5)\) and \(f''(3)\), find the value or explain why it does not exist.
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2018 FRQ3 AB/BC 同题AB/BC Shared · g = f' piecewise-linear on [-5,7]

The graph of the continuous function \(g\), the derivative of the function \(f\), is shown above. The function \(g\) is piecewise linear for \(-5 \le x < 3\), and \(g(x) = 2(x - 4)^2\) for \(3 \le x \le 6\).

  1. (A) If \(f(1) = 3\), what is the value of \(f(-5)\)?
  2. (B) Evaluate \(\int_{1}^{6} g(x)\,dx\).
  3. (C) For \(-5 < x < 6\), on what open intervals, if any, is the graph of \(f\) both increasing and concave up? Give a reason for your answer.
  4. (D) Find the x-coordinate of each point of inflection of the graph of \(f\). Give a reason for your answer.
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2019 FRQ3 · f 在 [-6,5] 图(两线段 + 半圆)f on [-6,5] graph (two line segments + semicircle)

The continuous function \(f\) is defined on the closed interval \(-6 \le x \le 5\). The figure above shows a portion of the graph of \(f\), consisting of two line segments and a quarter of a circle centered at the point \((5, 3)\). It is known that the point \((3, 3 - \sqrt{5})\) is on the graph of \(f\).

  1. (A) If \(\int_{-6}^{5} f(x)\,dx = 7\), find the value of \(\int_{-6}^{-2} f(x)\,dx\). Show the work that leads to your answer.
  2. (B) Evaluate \(\int_{3}^{5} (2f'(x) + 4)\,dx\).
  3. (C) The function \(g\) is given by \(g(x) = \int_{-2}^{x} f(t)\,dt\). Find the absolute maximum value of \(g\) on the interval \(-2 \le x \le 5\). Justify your answer.
  4. (D) Find \(\displaystyle\lim_{x \to 1} \dfrac{10^{x} - 3f'(x)}{f(x) - \arctan(x)}\).
2015 FRQ5 · f' 图 on [-3,4], twice-diff → inflectionf' graph on [-3,4], twice-diff → inflection

The figure above shows the graph of \(f'\), the derivative of a twice-differentiable function \(f\), on the interval \([-3, 4]\). The graph of \(f'\) has horizontal tangents at \(x = -1\), \(x = 1\), and \(x = 3\). The areas of the regions bounded by the \(x\)-axis and the graph of \(f'\) on the intervals \([-2, 1]\) and \([1, 4]\) are 9 and 12, respectively.

  1. (A) Find all \(x\)-coordinates at which \(f\) has a relative maximum. Give a reason for your answer.
  2. (B) On what open intervals contained in \(-3 < x < 4\) is the graph of \(f\) both concave down and decreasing? Give a reason for your answer.
  3. (C) Find the \(x\)-coordinates of all points of inflection for the graph of \(f\). Give a reason for your answer.
  4. (D) Given that \(f(1) = 3\), write an expression for \(f(x)\) that involves an integral. Find \(f(4)\) and \(f(-2)\).
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2021 FRQ4 AB/BC 同题AB/BC Shared · 🆕 f 在 [-4,6] 图 (shared BC)🆕 f on [-4,6] graph (AB/BC shared)

Let \(f\) be a continuous function defined on the closed interval \(-4 \le x \le 6\). The graph of \(f\), consisting of four line segments, is shown above. Let \(G\) be the function defined by \(G(x) = \displaystyle\int_0^x f(t)\,dt\).

  1. (A) On what open intervals is the graph of \(G\) concave up? Give a reason for your answer.
  2. (B) Let \(P\) be the function defined by \(P(x) = G(x)\cdot f(x)\). Find \(P'(3)\).
  3. (C) Find \(\displaystyle\lim_{x \to 2} \dfrac{G(x)}{x^2 - 2x}\).
  4. (D) Find the average rate of change of \(G\) on the interval \([-4, 2]\). Does the Mean Value Theorem guarantee a value \(c\), \(-4 < c < 2\), for which \(G'(c)\) is equal to this average rate of change? Justify your answer.
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2022 FRQ3 AB/BC 同题AB/BC Shared · 🆕 f' 图 f(4)=3 区间 [0,7] (shared BC)🆕 f' graph, f(4)=3, interval [0,7] (AB/BC shared)

Let \(f\) be a differentiable function with \(f(4) = 3\). On the interval \(0 \le x \le 7\), the graph of \(f'\), the derivative of \(f\), consists of a semicircle and two line segments, as shown in the figure above.

  1. (A) Find \(f(0)\) and \(f(5)\).
  2. (B) Find the \(x\)-coordinates of all points of inflection of the graph of \(f\) for \(0 < x < 7\). Justify your answer.
  3. (C) Let \(g\) be the function defined by \(g(x) = f(x) - x\). On what intervals, if any, is \(g\) decreasing for \(0 \le x \le 7\)? Show the analysis that leads to your answer.
  4. (D) For the function \(g\) defined in part (c), find the absolute minimum value on the interval \(0 \le x \le 7\). Justify your answer.
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2023 FRQ4 AB/BC 同题AB/BC Shared · 🆕 f 在 [-2,8], f(2)=1 (shared BC)🆕 f on [-2,8], f(2)=1 (AB/BC shared)

The function \(f\) is defined on the closed interval \([-2, 8]\) and satisfies \(f(2) = 1\). The graph of \(f'\), the derivative of \(f\), consists of two line segments and a semicircle, as shown in the figure.

  1. (A) Does \(f\) have a relative minimum, a relative maximum, or neither at \(x = 6\)? Give a reason for your answer.
  2. (B) On what open intervals, if any, is the graph of \(f\) concave down? Give a reason for your answer.
  3. (C) Find the value of \(\displaystyle\lim_{x \to 2} \dfrac{6f(x) - 3x}{x^2 - 5x + 6}\), or show that it does not exist. Justify your answer.
  4. (D) Find the absolute minimum value of \(f\) on the closed interval \([-2, 8]\). Justify your answer.
📝 点击展开评分标准(Scoring Guideline)📝 Expand Scoring Guideline 2023 FRQ4 SG

FRQ4 · 微分方程 / 表格 / 相关变化率FRQ4 · Differential Equations / Table / Related Rates

概率 60%60% Likely

Unit 7/4。近 5 年:DE 60%(可分离为主)/ tabular 20% / related rates 20%。Unit 7/4. Last 5 years: DE 60% (mostly separable) / tabular 20% / related rates 20%.

🔍 怎么一眼认出🔍 How to Identify

  • DE: 题目给 dy/dx = f(x, y)
  • Tabular: 给表格 + 问 MVT / Riemann / 解释意义
  • Related rates: 两个量关联(h 和 V、r 和 A),知道一个变化率求另一个
  • DE: problem gives dy/dx = f(x, y)
  • Tabular: table given + asks MVT / Riemann / interpret meaning
  • Related rates: two quantities related (h and V, r and A), one rate known, find the other

📐 解题模板📐 Solution Template

  1. 可分离 DE 5 步
  2. Step 1:分离变量 \(g(y)\,dy = h(x)\,dx\)。
  3. Step 2:积分 \(\int g(y)\,dy = \int h(x)\,dx + C\)。
  4. Step 3:代入初始值求 C。
  5. Step 4:解 y(显式表达式)。
  6. Step 5:写出定义域(常忽略)。
  7. Related Rates 4 步
  8. Step 1:写出关联方程(圆柱体积 \(V = \pi r^2 h\)、Pythagoras \(x^2 + y^2 = z^2\))。
  9. Step 2:两边对 t 求导(链式)。
  10. Step 3:代入已知 rate 和当前值。
  11. Step 4:解未知 rate。
  12. Tabular MVT 3 步
  13. MVT:f 在 \([a,b]\) 连续可导 ⇒ \(\exists c\) 使 \(f'(c) = \dfrac{f(b)-f(a)}{b-a}\)。
  14. Trapezoidal:\(\int_a^b f\,dx \approx \sum \dfrac{x_{i+1}-x_i}{2}(f(x_i)+f(x_{i+1}))\)。
  1. Separable DE 5 steps
  2. Step 1: Separate variables \(g(y)\,dy = h(x)\,dx\).
  3. Step 2: Integrate \(\int g(y)\,dy = \int h(x)\,dx + C\).
  4. Step 3: Substitute initial value to find C.
  5. Step 4: Solve for y (explicit expression).
  6. Step 5: State the domain (often overlooked).
  7. Related Rates 4 steps
  8. Step 1: Write the relational equation (cylinder volume \(V = \pi r^2 h\), Pythagoras \(x^2 + y^2 = z^2\)).
  9. Step 2: Differentiate both sides with respect to t (chain rule).
  10. Step 3: Substitute known rates and current values.
  11. Step 4: Solve for the unknown rate.
  12. Tabular MVT 3 steps
  13. MVT: f continuous and differentiable on \([a,b]\) ⇒ \(\exists c\) such that \(f'(c) = \dfrac{f(b)-f(a)}{b-a}\).
  14. Trapezoidal: \(\int_a^b f\,dx \approx \sum \dfrac{x_{i+1}-x_i}{2}(f(x_i)+f(x_{i+1}))\).

⚠️ 踩分点与易错⚠️ Pitfalls & Common Errors

  • ⚠️ DE 分离时 y, dy 和 x, dx 对调
  • ⚠️ related rates 忘了「对 t 求导」那步
  • ⚠️ MVT 条件没写全
  • ⚠️ Trapezoidal 系数 1/2 漏
  • ⚠️ DE: swapping y, dy and x, dx when separating
  • ⚠️ Related rates: forgetting the 'differentiate with respect to t' step
  • ⚠️ MVT conditions not fully stated
  • ⚠️ Trapezoidal coefficient 1/2 missing

历年真题(10 道 · 点击展开)Past FRQs (10 · Click to Expand)

2015 FRQ3 AB/BC 同题AB/BC Shared · Johanna jogs v(t) 表格 → Riemann + MVT (shared BC)Johanna jogs v(t) table → Riemann + MVT (AB/BC shared)

Johanna jogs along a straight path. For \(0 \le t \le 40\), Johanna's velocity is given by a differentiable function \(v\). Selected values of \(v(t)\), where \(t\) is measured in minutes and \(v(t)\) is measured in meters per minute, are given in the table above.

t (minutes)v(t) (meters per minute)
  1. (A) Use the data in the table to estimate the value of \(v'(16)\).
  2. (B) Using correct units, explain the meaning of the definite integral \(\int_{0}^{40} |v(t)|\,dt\) in the context of the problem. Approximate the value of \(\int_{0}^{40} |v(t)|\,dt\) using a right Riemann sum with the four subintervals indicated in the table.
  3. (C) Bob is riding his bicycle along the same path. For \(0 \le t \le 10\), Bob's velocity is modeled by \(B(t) = t^3 - 6t^2 + 300\), where \(t\) is measured in minutes and \(B(t)\) is measured in meters per minute. Find Bob's acceleration at time \(t = 5\).
  4. (D) Based on the model \(B\) from part (c), find Bob's average velocity during the interval \(0 \le t \le 10\).
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2015 FRQ4 · dy/dx = 2x - y 斜率场 + 粒子解dy/dx = 2x - y slope field + particular solution

Consider the differential equation \(\dfrac{dy}{dx} = 2x - y\).

  1. (A) On the axes provided, sketch a slope field for the given differential equation at the six points indicated.
  2. (B) Find \(\dfrac{d^2 y}{dx^2}\) in terms of \(x\) and \(y\). Determine the concavity of all solution curves for the given differential equation in Quadrant II. Give a reason for your answer.
  3. (C) Let \(y = f(x)\) be the particular solution to the differential equation with the initial condition \(f(2) = 3\).
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2016 FRQ4 AB/BC 同题AB/BC Shared · dy/dx = y²/(x-1) 可分离dy/dx = y²/(x-1) separable

Consider the differential equation \(\dfrac{dy}{dx} = x^{2} - \dfrac{1}{2}y\).

  1. (A) Find \(\dfrac{d^{2}y}{dx^{2}}\) in terms of \(x\) and \(y\).
  2. (B) Let \(y = f(x)\) be the particular solution to the given differential equation whose graph passes through the point \((-2, 8)\). Does the graph of \(f\) have a relative minimum, a relative maximum, or neither at the point \((-2, 8)\)? Justify your answer.
  3. (C) Let \(y = g(x)\) be the particular solution to the given differential equation with \(g(-1) = 2\). Find \(\displaystyle\lim_{x \to -1} \dfrac{g(x) - 2}{3(x + 1)^{2}}\). Show the work that leads to your answer.
  4. (D) Let \(y = h(x)\) be the particular solution to the given differential equation with \(h(0) = 2\). Use Euler's method, starting at \(x = 0\) with two steps of equal size, to approximate \(h(1)\).
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2017 FRQ4 AB/BC 同题AB/BC Shared · Potato 冷却 dH/dt = -(1/4)(H-27)Potato cooling dH/dt = -(1/4)(H-27)

At time \(t = 0\), a boiled potato is taken from a pot on a stove and left to cool in a kitchen. The internal temperature of the potato is 91 degrees Celsius (\(^\circ\)C) at time \(t = 0\), and the internal temperature of the potato is greater than \(27^\circ\)C for all times \(t > 0\). The internal temperature of the potato at time \(t\) minutes can be modeled by the function \(H\) that satisfies the differential equation \(\dfrac{dH}{dt} = -\dfrac{1}{4}(H - 27)\), where \(H(t)\) is measured in degrees Celsius and \(H(0) = 91\).

  1. (A) Write an equation for the line tangent to the graph of \(H\) at \(t = 0\). Use this equation to approximate the internal temperature of the potato at time \(t = 3\).
  2. (B) Use \(\dfrac{d^2 H}{dt^2}\) to determine whether your answer in part (a) is an underestimate or an overestimate of the internal temperature of the potato at time \(t = 3\).
  3. (C) For \(t < 10\), an alternate model for the internal temperature of the potato at time \(t\) minutes is the function \(G\) that satisfies the differential equation \(\dfrac{dG}{dt} = -(G - 27)^{2/3}\), where \(G(t)\) is measured in degrees Celsius and \(G(0) = 91\). Find an expression for \(G(t)\). Based on this model, what is the internal temperature of the potato at time \(t = 3\)?
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2018 FRQ4 AB/BC 同题AB/BC Shared · Tree H(t) 表格 + MVT + RiemannTree H(t) table + MVT + Riemann

The height of a tree at time \(t\) is given by a twice-differentiable function \(H\), where \(H(t)\) is measured in meters and \(t\) is measured in years. Selected values of \(H(t)\) are given in the table above.

t (years)H(t) (meters)
21.5
32
56
711
1015
  1. (A) Use the data in the table to estimate \(H'(6)\). Using correct units, interpret the meaning of \(H'(6)\) in the context of the problem.
  2. (B) Explain why there must be at least one time \(t\), for \(2 < t < 10\), such that \(H'(t) = 2\).
  3. (C) Use a trapezoidal sum with the four subintervals indicated by the data in the table to approximate the average height of the tree over the time interval \(2 \le t \le 10\).
  4. (D) The height of the tree, in meters, can also be modeled by the function \(G\), given by \(G(x) = \dfrac{100x}{1 + x}\), where \(x\) is the diameter of the base of the tree, in meters. When the tree is 50 meters tall, the diameter of the base of the tree is increasing at a rate of 0.03 meter per year. According to this model, what is the rate of change of the height of the tree with respect to time, in meters per year, at the time when the tree is 50 meters tall?
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2019 FRQ4 AB/BC 同题AB/BC Shared · Cylindrical barrel rainwater drains, dh/dt

A cylindrical barrel with a diameter of 2 feet contains collected rainwater, as shown in the figure above. The water drains out through a valve (not shown) at the bottom of the barrel. The rate of change of the height \(h\) of the water in the barrel with respect to time \(t\) is modeled by \(\dfrac{dh}{dt} = -\dfrac{1}{10}\sqrt{h}\), where \(h\) is measured in feet and \(t\) is measured in seconds. (The volume \(V\) of a cylinder with radius \(r\) and height \(h\) is \(V = \pi r^2 h\).)

  1. (A) Find the rate of change of the volume of water in the barrel with respect to time when the height of the water is 4 feet. Indicate units of measure.
  2. (B) When the height of the water is 3 feet, is the rate of change of the height of the water with respect to time increasing or decreasing? Explain your reasoning.
  3. (C) At time \(t = 0\) seconds, the height of the water is 5 feet. Use separation of variables to find an expression for \(h\) in terms of \(t\).
2021 FRQ6 · 🆕 Medication amount m(t) DE

A medication is administered to a patient. The amount, in milligrams, of the medication in the patient at time \(t\) hours is modeled by a function \(y = A(t)\) that satisfies the differential equation \(\dfrac{dy}{dt} = \dfrac{12 - y}{3}\). At time \(t = 0\) hours, there are 0 milligrams of the medication in the patient.

  1. (A) A portion of the slope field for the differential equation \(\dfrac{dy}{dt} = \dfrac{12 - y}{3}\) is given below. Sketch the solution curve through the point \((0, 0)\).
  2. (B) Using correct units, interpret the statement \(\lim\limits_{t \to \infty} A(t) = 12\) in the context of this problem.
  3. (C) Use separation of variables to find \(y = A(t)\), the particular solution to the differential equation \(\dfrac{dy}{dt} = \dfrac{12 - y}{3}\) with initial condition \(A(0) = 0\).
  4. (D) A different procedure is used to administer the medication to a second patient. The amount, in milligrams, of the medication in the second patient at time \(t\) hours is modeled by a function \(y = B(t)\) that satisfies the differential equation \(\dfrac{dy}{dt} = 3 - \dfrac{y}{t+2}\). At time \(t = 1\) hour, there are 2.5 milligrams of the medication in the second patient. Is the rate of change of the amount of medication in the second patient increasing or decreasing at time \(t = 1\)? Give a reason for your answer.
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2022 FRQ4 AB/BC 同题AB/BC Shared · 🆕 Ice sculpture 圆锥 tabular r' + dV/dt (shared BC)🆕 Ice sculpture cone tabular r' + dV/dt (AB/BC shared)

An ice sculpture melts in such a way that it can be modeled as a cone that maintains a conical shape as it decreases in size. The radius of the base of the cone is given by a twice-differentiable function \(r\), where \(r(t)\) is measured in centimeters and \(t\) is measured in days. The table above gives selected values of \(r'(t)\), the rate of change of the radius, over the time interval \(0 \le t \le 12\).

\(t\) (days)0371012
\(r'(t)\) (centimeters per day)\(-6.1\)\(-5.0\)\(-4.4\)\(-3.8\)\(-3.5\)
  1. (A) Approximate \(r''(8.5)\) using the average rate of change of \(r'\) over the interval \(7 \le t \le 10\). Show the computations that lead to your answer, and indicate units of measure.
  2. (B) Is there a time \(t\), \(0 \le t \le 3\), for which \(r'(t) = -6\)? Justify your answer.
  3. (C) Use a right Riemann sum with the four subintervals indicated in the table to approximate the value of \(\displaystyle\int_0^{12} r'(t)\,dt\).
  4. (D) The height of the cone decreases at a rate of 2 centimeters per day. At time \(t = 3\) days, the radius is 100 centimeters and the height is 50 centimeters. Find the rate of change of the volume of the cone with respect to time, in cubic centimeters per day, at time \(t = 3\) days. (The volume \(V\) of a cone with radius \(r\) and height \(h\) is \(V = \tfrac{1}{3}\pi r^2 h\).)
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2022 FRQ5 · 🆕 DE dy/dx = (1/2)sin(πx/2)√(y+7)

Consider the differential equation \(\dfrac{dy}{dx} = \dfrac{1}{2}\sin\left(\dfrac{\pi}{2} x\right)\sqrt{y + 7}\). Let \(y = f(x)\) be the particular solution to the differential equation with the initial condition \(f(1) = 2\). The function \(f\) is defined for all real numbers.

  1. (A) A portion of the slope field for the differential equation is given below. Sketch the solution curve through the point \((1, 2)\).
  2. (B) Write an equation for the line tangent to the solution curve in part (a) at the point \((1, 2)\). Use the equation to approximate \(f(0.8)\).
  3. (C) It is known that \(f''(x) > 0\) for \(-1 \le x \le 1\). Is the approximation found in part (b) an overestimate or an underestimate for \(f(0.8)\)? Give a reason for your answer.
  4. (D) Use separation of variables to find \(y = f(x)\), the particular solution to the differential equation \(\dfrac{dy}{dx} = \dfrac{1}{2}\sin\left(\dfrac{\pi}{2} x\right)\sqrt{y + 7}\) with the initial condition \(f(1) = 2\).
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2023 FRQ3 AB/BC 同题AB/BC Shared · 🆕 Milk warming dM/dt = (1/4)(40-M) (shared BC)🆕 Milk warming dM/dt = (1/4)(40-M) (AB/BC shared)

A bottle of milk is taken out of a refrigerator and placed in a pan of hot water to be warmed. The increasing function \(M\) models the temperature of the milk at time \(t\), where \(M(t)\) is measured in degrees Celsius (\(^\circ\)C) and \(t\) is the number of minutes since the bottle was placed in the pan. \(M\) satisfies the differential equation \(\dfrac{dM}{dt} = \tfrac{1}{4}(40 - M)\). At time \(t = 0\), the temperature of the milk is \(5^\circ\)C. It can be shown that \(M(t) < 40\) for all values of \(t\).

  1. (A) A slope field for the differential equation \(\dfrac{dM}{dt} = \tfrac{1}{4}(40 - M)\) is shown. Sketch the solution curve through the point \((0, 5)\).
  2. (B) Use the line tangent to the graph of \(M\) at \(t = 0\) to approximate \(M(2)\), the temperature of the milk at time \(t = 2\) minutes.
  3. (C) Write an expression for \(\dfrac{d^2 M}{dt^2}\) in terms of \(M\). Use \(\dfrac{d^2 M}{dt^2}\) to determine whether the approximation from part (b) is an underestimate or an overestimate for the actual value of \(M(2)\). Give a reason for your answer.
  4. (D) Use separation of variables to find an expression for \(M(t)\), the particular solution to the differential equation \(\dfrac{dM}{dt} = \tfrac{1}{4}(40 - M)\) with initial condition \(M(0) = 5\).
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FRQ5 · 面积体积 / 函数分析 / 粒子FRQ5 · Area-Volume / Function Analysis / Particle

概率 40%40% Likely

槽位变化多。近 5 年:面积体积 40% / 其它各 20%。要同时准备这几类。Slot varies a lot. Last 5 years: Area-Volume 40% / others each 20%. Must prepare all these categories.

🔍 怎么一眼认出🔍 How to Identify

  • 面积体积型:给 f, g 两条曲线 → 围成区域或绕轴旋转
  • 函数分析型:给 f(x) 显式 → critical points、concavity、extrema
  • 粒子型:给 v(t) → position、distance
  • Area-volume type: given curves f, g → bounded region or rotation around axis
  • Function analysis type: explicit f(x) given → critical points, concavity, extrema
  • Particle type: v(t) given → position, distance

📐 解题模板📐 Solution Template

  1. 面积 \(= \int_a^b (\text{top} - \text{bottom})\,dx\)。找交点 \(f=g\)。
  2. Washer 体积 \(= \pi \int (R^2 - r^2)\,dx\),R = 外半径,r = 内半径。
  3. Cross-section \(= \int A(x)\,dx\),A(x) 为给定形状(正方形 \(A=s^2\)、半圆 \(A=\pi r^2/2\))。
  4. Analyze function:求 \(f'(x)\)、\(f'=0\) 找 critical,二阶导 test;sign chart 判 monotonicity。
  1. Area \(= \int_a^b (\text{top} - \text{bottom})\,dx\). Find intersections \(f=g\).
  2. Washer volume \(= \pi \int (R^2 - r^2)\,dx\), R = outer radius, r = inner radius.
  3. Cross-section \(= \int A(x)\,dx\), A(x) is the given shape (square \(A=s^2\), semicircle \(A=\pi r^2/2\)).
  4. Analyze function: find \(f'(x)\), solve \(f'=0\) for critical points, second derivative test; use sign chart to determine monotonicity.

⚠️ 踩分点与易错⚠️ Pitfalls & Common Errors

  • ⚠️ 面积/体积 bounds 找错(没找到所有交点)
  • ⚠️ Washer 外半径 R 和内半径 r 颠倒
  • ⚠️ Cross-section 误用体积公式(忘了截面形状)
  • ⚠️ 函数分析漏 abs extrema(只看 critical point)
  • ⚠️ Area/volume bounds wrong (didn't find all intersections)
  • ⚠️ Washer outer radius R and inner radius r swapped
  • ⚠️ Cross-section misusing volume formula (forgot cross-section shape)
  • ⚠️ Function analysis missing abs extrema (only checking critical points)

历年真题(9 道 · 点击展开)Past FRQs (9 · Click to Expand)

2016 FRQ5 AB/BC 同题AB/BC Shared · Funnel 10in 圆截面 r = (3+h²)/20 体积Funnel 10in circular cross section r = (3+h²)/20 volume

The inside of a funnel of height 10 inches has circular cross sections, as shown in the figure above. At height \(h\), the radius of the funnel is given by \(r = \dfrac{1}{20}(3 + h^{2})\), where \(0 \le h \le 10\). The units of \(r\) and \(h\) are inches.

  1. (A) Find the average value of the radius of the funnel.
  2. (B) Find the volume of the funnel.
  3. (C) The funnel contains liquid that is draining from the bottom. At the instant when the height of the liquid is 3 inches, the radius of the surface of the liquid is decreasing at a rate of \(\dfrac{1}{5}\) inch per second. At this instant, what is the rate of change of the height of the liquid with respect to time?
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2018 FRQ5 · f(x) = e^x cos(x) 分析f(x) = e^x cos(x) analysis

Let \(f\) be the function defined by \(f(x) = e^{x}\cos(x)\).

  1. (A) Find the average rate of change of \(f\) on the interval \(0 \le x \le \pi\).
  2. (B) What is the slope of the line tangent to the graph of \(f\) at \(x = \dfrac{3\pi}{2}\)?
  3. (C) Find the absolute minimum value of \(f\) on the interval \(0 \le x \le 2\pi\). Justify your answer.
  4. (D) Let \(g\) be a differentiable function such that \(g\left(\dfrac{\pi}{2}\right) = 0\). The graph of \(g'\), the derivative of \(g\), is shown below. Find the value of \(\displaystyle\lim_{x \to \pi/2} \dfrac{f(x)}{g(x)}\) or state that it does not exist. Justify your answer.
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2019 FRQ5 · Region 由 g, h, y-axis 围成 → 面积 + 体积Region bounded by g, h, y-axis → area + volume

Let \(R\) be the region enclosed by the graphs of \(g(x) = -2 + 3\cos\left(\dfrac{\pi}{2}x\right)\) and \(h(x) = 6 - 2(x - 1)^{2}\), the y-axis, and the vertical line \(x = 2\), as shown in the figure above.

  1. (A) Find the area of \(R\).
  2. (B) Region \(R\) is the base of a solid. For the solid, at each \(x\) the cross section perpendicular to the x-axis has area \(A(x) = \dfrac{1}{x + 3}\). Find the volume of the solid.
  3. (C) Write, but do not evaluate, an integral expression that gives the volume of the solid generated when \(R\) is rotated about the horizontal line \(y = 6\).
2017 FRQ5 · 两粒子 P, Q x 轴 x_P = ln(t²-2t+10)Two particles P, Q on x-axis x_P = ln(t²-2t+10)

Two particles move along the \(x\)-axis. For \(0 \le t \le 8\), the position of particle \(P\) at time \(t\) is given by \(x_P(t) = \ln(t^2 - 2t + 10)\), while the velocity of particle \(Q\) at time \(t\) is given by \(v_Q(t) = t^2 - 8t + 15\). Particle \(Q\) is at position \(x = 5\) at time \(t = 0\).

  1. (A) For \(0 \le t \le 8\), when is particle \(P\) moving to the left?
  2. (B) For \(0 \le t \le 8\), find all times \(t\) during which the two particles travel in the same direction.
  3. (C) Find the acceleration of particle \(Q\) at time \(t = 2\). Is the speed of particle \(Q\) increasing, decreasing, or neither at time \(t = 2\)? Explain your reasoning.
  4. (D) Find the position of particle \(Q\) the first time it changes direction.
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2015 FRQ5 · f' 图(见 FRQ3)f' graph (see FRQ3)

The figure above shows the graph of \(f'\), the derivative of a twice-differentiable function \(f\), on the interval \([-3, 4]\). The graph of \(f'\) has horizontal tangents at \(x = -1\), \(x = 1\), and \(x = 3\). The areas of the regions bounded by the \(x\)-axis and the graph of \(f'\) on the intervals \([-2, 1]\) and \([1, 4]\) are 9 and 12, respectively.

  1. (A) Find all \(x\)-coordinates at which \(f\) has a relative maximum. Give a reason for your answer.
  2. (B) On what open intervals contained in \(-3 < x < 4\) is the graph of \(f\) both concave down and decreasing? Give a reason for your answer.
  3. (C) Find the \(x\)-coordinates of all points of inflection for the graph of \(f\). Give a reason for your answer.
  4. (D) Given that \(f(1) = 3\), write an expression for \(f(x)\) that involves an integral. Find \(f(4)\) and \(f(-2)\).
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2021 FRQ3 AB/BC 同题AB/BC Shared · 🆕 Spinning toys y = cx√(4-x²); 面积 + 体积 (shared BC)🆕 Spinning toys y = cx√(4-x²); area + volume (AB/BC shared)

A company designs spinning toys using the family of functions \(y = cx\sqrt{4 - x^2}\), where \(c\) is a positive constant. The figure above shows the region in the first quadrant bounded by the \(x\)-axis and the graph of \(y = cx\sqrt{4 - x^2}\), for some \(c\). Each spinning toy is in the shape of the solid generated when such a region is revolved about the \(x\)-axis. Both \(x\) and \(y\) are measured in inches.

  1. (A) Find the area of the region in the first quadrant bounded by the \(x\)-axis and the graph of \(y = cx\sqrt{4 - x^2}\) for \(c = 6\).
  2. (B) It is known that, for \(y = cx\sqrt{4 - x^2}\), \(\dfrac{dy}{dx} = \dfrac{c(4 - 2x^2)}{\sqrt{4 - x^2}}\). For a particular spinning toy, the radius of the largest cross-sectional circular slice is \(1.2\) inches. What is the value of \(c\) for this spinning toy?
  3. (C) For another spinning toy, the volume is \(2\pi\) cubic inches. What is the value of \(c\) for this spinning toy?
📝 点击展开评分标准(Scoring Guideline)📝 Expand Scoring Guideline 2021 FRQ3 SG
2021 FRQ5 · 🆕 隐函数 y² - 2 = y sin(x), y>0🆕 Implicit y² - 2 = y sin(x), y>0

Consider the function \(y = f(x)\) whose curve is given by the equation \(y^2 - 2 = y\sin x\) for \(y > 0\).

  1. (A) Show that \(\dfrac{dy}{dx} = \dfrac{y\cos x}{2y - \sin x}\).
  2. (B) Write an equation for the line tangent to the curve at the point \(\left(0, \sqrt{2}\right)\).
  3. (C) For \(0 \le x \le \pi\) and \(y > 0\), find the coordinates of the point where the line tangent to the curve is horizontal.
  4. (D) Determine whether \(f\) has a relative minimum, a relative maximum, or neither at the point found in part (c). Justify your answer.
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2022 FRQ6 · 🆕 Particle P x_P(t) = 6 - 4e^(-t)

Particle \(P\) moves along the \(x\)-axis such that, for time \(t > 0\), its position is given by \(x_P(t) = 6 - 4e^{-t}\). Particle \(Q\) moves along the \(y\)-axis such that, for time \(t > 0\), its velocity is given by \(v_Q(t) = \dfrac{1}{t^2}\). At time \(t = 1\), the position of particle \(Q\) is \(y_Q(1) = 2\).

  1. (A) Find \(v_P(t)\), the velocity of particle \(P\) at time \(t\).
  2. (B) Find \(a_Q(t)\), the acceleration of particle \(Q\) at time \(t\). Find all times \(t\), for \(t > 0\), when the speed of particle \(Q\) is decreasing. Justify your answer.
  3. (C) Find \(y_Q(t)\), the position of particle \(Q\) at time \(t\).
  4. (D) As \(t \to \infty\), which particle will eventually be farther from the origin? Give a reason for your answer.
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2023 FRQ5 · 🆕 f, g twice-diff 表格🆕 f, g twice-diff table

The functions \(f\) and \(g\) are twice differentiable. The table shown gives values of the functions and their first derivatives at selected values of \(x\).

\(x\)\(f(x)\)\(f'(x)\)\(g(x)\)\(g'(x)\)
0-7453
2-8-362
4-321-2
73085
10-214-8
  1. (A) Let \(h\) be the function defined by \(h(x) = f(g(x))\). Find \(h'(7)\). Show the work that leads to your answer.
  2. (B) Let \(k\) be a differentiable function such that \(k'(x) = (f(x))^2 \cdot g(x)\). Is the graph of \(k\) concave up or concave down at the point where \(x = 4\)? Give a reason for your answer.
  3. (C) Let \(m\) be the function defined by \(m(x) = 5x^3 + \int_0^x f'(t)\, dt\). Find \(m(2)\). Show the work that leads to your answer.
  4. (D) Is the function \(m\) defined in part (c) increasing, decreasing, or neither at \(x = 2\)? Justify your answer.
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FRQ6 · 函数关系综合 / 隐函数 / DEFRQ6 · Connected Functions / Implicit / DE

概率 60%60% Likely

近 5 年:连接函数(composite/chain rule with tables)60% / 隐函数 20% / DE 20%。Last 5 years: Connected functions (composite/chain rule with tables) 60% / Implicit 20% / DE 20%.

🔍 怎么一眼认出🔍 How to Identify

  • 连接函数型:给多个 f, g, h + 表格值 → 求复合函数的导数、切线、极限
  • 隐函数型:给 F(x,y) = 0,dy/dx 已给 → tangent line、d²y/dx²
  • Connected functions type: multiple f, g, h + table values given → find derivatives, tangent lines, limits of composite functions
  • Implicit type: F(x,y) = 0 given, dy/dx given → tangent line, d²y/dx²

📐 解题模板📐 Solution Template

  1. 连接函数 4 招
  2. Chain rule:\((f \circ g)'(x) = f'(g(x))\cdot g'(x)\)。表格题几乎必考。
  3. Product/Quotient:\((fg)' = f'g + fg'\);\((f/g)' = (f'g - fg')/g^2\)。
  4. Tangent line:\(y = f(a) + f'(a)(x - a)\)。
  5. L'Hôpital:\(\lim \frac{f}{g} = \lim \frac{f'}{g'}\)(0/0 或 ∞/∞)。
  6. 隐函数 3 招
  7. Step 1:两边对 x 求导,y 当 y(x) 处理,\(dy/dx\) 保留符号。
  8. Step 2:代入 (x, y) 求值。
  9. Step 3:\(d^2y/dx^2\) 时对 \(dy/dx\) 再求导,含 \(dy/dx\) 项用已知式代换。
  1. Connected Functions 4 techniques
  2. Chain rule: \((f \circ g)'(x) = f'(g(x))\cdot g'(x)\). Almost always tested with table problems.
  3. Product/Quotient: \((fg)' = f'g + fg'\); \((f/g)' = (f'g - fg')/g^2\).
  4. Tangent line: \(y = f(a) + f'(a)(x - a)\).
  5. L'Hôpital: \(\lim \frac{f}{g} = \lim \frac{f'}{g'}\) (0/0 or ∞/∞).
  6. Implicit 3 techniques
  7. Step 1: Differentiate both sides with respect to x, treat y as y(x), keep \(dy/dx\) symbolic.
  8. Step 2: Substitute (x, y) to evaluate.
  9. Step 3: For \(d^2y/dx^2\), differentiate \(dy/dx\) again; substitute known expressions for any \(dy/dx\) terms.

⚠️ 踩分点与易错⚠️ Pitfalls & Common Errors

  • ⚠️ Chain rule 漏内函数导数(最常见低级错)
  • ⚠️ Product 和 Quotient 混
  • ⚠️ L'Hôpital 没检查 0/0 形式就用
  • ⚠️ 隐函数 d²y/dx² 代换 dy/dx 步骤漏
  • ⚠️ Chain rule missing inner derivative (most common low-level error)
  • ⚠️ Confusing Product and Quotient rules
  • ⚠️ Applying L'Hôpital without checking 0/0 form
  • ⚠️ Implicit d²y/dx²: missing the substitution step for dy/dx

历年真题(6 道 · 点击展开)Past FRQs (6 · Click to Expand)

2015 FRQ6 · y³ - xy = 2 隐函数y³ - xy = 2 implicit

Consider the curve given by the equation \(y^3 - xy = 2\). It can be shown that \(\dfrac{dy}{dx} = \dfrac{y}{3y^2 - x}\).

  1. (A) Write an equation for the line tangent to the curve at the point \((-1, 1)\).
  2. (B) Find the coordinates of all points on the curve at which the line tangent to the curve at that point is vertical.
  3. (C) Evaluate \(\dfrac{d^2 y}{dx^2}\) at the point on the curve where \(x = -1\) and \(y = 1\).
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2016 FRQ6 · f, g 有连续二阶导 + 表格f, g have continuous second derivatives + table

The functions \(f\) and \(g\) have continuous second derivatives. The table above gives values of the functions and their derivatives at selected values of \(x\).

\(x\)\(f(x)\)\(f'(x)\)\(g(x)\)\(g'(x)\)
1-6328
22-2-30
38762
6453-1
  1. (A) Let \(k(x) = f(g(x))\). Write an equation for the line tangent to the graph of \(k\) at \(x = 3\).
  2. (B) Let \(h(x) = \dfrac{g(x)}{f(x)}\). Find \(h'(1)\).
  3. (C) Evaluate \(\int_{1}^{3} f''(2x)\,dx\).
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2017 FRQ6 · f(x) = cos(2x) + e^sinx; g 表格f(x) = cos(2x) + e^sinx; g table

Let \(f\) be the function defined by \(f(x) = \cos(2x) + e^{\sin x}\). Let \(g\) be a differentiable function. The table above gives values of \(g\) and its derivative \(g'\) at selected values of \(x\). Let \(h\) be the function whose graph, consisting of five line segments, is shown in the figure above.

\(x\)\(g(x)\)\(g'(x)\)
-510-3
-45-1
-324
-231
-11-2
00-3
  1. (A) Find the slope of the line tangent to the graph of \(f\) at \(x = \pi\).
  2. (B) Let \(k\) be the function defined by \(k(x) = h(f(x))\). Find \(k'(\pi)\).
  3. (C) Let \(m\) be the function defined by \(m(x) = g(-2x) \cdot h(x)\). Find \(m'(2)\).
  4. (D) Is there a number \(c\) in the closed interval \([-5, -3]\) such that \(g'(c) = -4\)? Justify your answer.
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2018 FRQ6 · dy/dx = (1/3)x(y-2)² 可分离dy/dx = (1/3)x(y-2)² separable

Consider the differential equation \(\dfrac{dy}{dx} = \dfrac{1}{3}x(y - 2)^{2}\).

  1. (A) A slope field for the given differential equation is shown below. Sketch the solution curve that passes through the point with coordinates \((0, 2)\), and sketch the solution curve that passes through the point with coordinates \((1, 0)\).
  2. (B) Let \(y = f(x)\) be the particular solution to the given differential equation with initial condition \(f(1) = 0\). Write an equation for the line tangent to the graph of \(y = f(x)\) at \(x = 1\). Use your equation to approximate \(f(0.7)\).
  3. (C) Find the particular solution \(y = f(x)\) to the given differential equation with initial condition \(f(1) = 0\).
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2019 FRQ6 · f, g, h twice-diff; tangent at x=2

Functions \(f\), \(g\), and \(h\) are twice-differentiable functions with \(g(2) = h(2) = 4\). The line \(y = 4 + \dfrac{2}{3}(x - 2)\) is tangent to both the graph of \(g\) at \(x = 2\) and the graph of \(h\) at \(x = 2\).

  1. (A) Find \(h'(2)\).
  2. (B) Let \(a\) be the function given by \(a(x) = 3x^{3}h(x)\). Write an expression for \(a'(x)\). Find \(a'(2)\).
  3. (C) The function \(h\) satisfies \(h(x) = \dfrac{x^{2} - 4}{1 - (f(x))^{3}}\) for \(x \ne 2\). It is known that \(\displaystyle\lim_{x \to 2} h(x)\) can be evaluated using L'Hospital's Rule. Use \(\displaystyle\lim_{x \to 2} h(x)\) to find \(f(2)\) and \(f'(2)\). Show the work that leads to your answers.
  4. (D) It is known that \(g(x) \le h(x)\) for \(1 < x < 3\). Let \(k\) be a function satisfying \(g(x) \le k(x) \le h(x)\) for \(1 < x < 3\). Is \(k\) continuous at \(x = 2\)? Justify your answer.
2023 FRQ6 · 🆕 隐函数 6xy = 2 + y³🆕 Implicit 6xy = 2 + y³

Consider the curve given by the equation \(6xy = 2 + y^3\).

  1. (A) Show that \(\dfrac{dy}{dx} = \dfrac{2y}{y^2 - 2x}\).
  2. (B) Find the coordinates of a point on the curve at which the line tangent to the curve is horizontal, or explain why no such point exists.
  3. (C) Find the coordinates of a point on the curve at which the line tangent to the curve is vertical, or explain why no such point exists.
  4. (D) A particle is moving along the curve. At the instant when the particle is at the point \(\left(\dfrac{1}{2}, -2\right)\), its horizontal position is increasing at a rate of \(\dfrac{dx}{dt} = \dfrac{2}{3}\) unit per second. What is the value of \(\dfrac{dy}{dt}\), the rate of change of the particle's vertical position, at that instant?
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Part III · 2026 AI 押题模拟卷Part III · 2026 AI Mock Exam

基于 2014–2025 押题概率设计的 6 道原创 FRQ。每题都附完整解答 + 评分细则(严格按 CB 标准 9 分制)。建议卡表做完再看答案。6 original FRQs designed to match 2014-2025 prediction probabilities. Each includes a full solution + 9-point AP rubric. Time yourself before revealing the answer.

做题顺序建议Suggested orderFRQ1-2(calc,共 30 分钟)→ 休息 → FRQ3-6(no calc,共 60 分钟)。总计 1h30m。FRQ1-2 (calc, 30 min) → break → FRQ3-6 (no calc, 60 min). 1h30m total.

Mock FRQ1 · Airport Passenger Flow (Rate In/Out)

流入流出率Rate In/Out

A security checkpoint at Terminal 1 of an international airport is open from 6 AM (\(t = 0\)) to 6 PM (\(t = 12\)). Passengers arrive at the checkpoint at a rate modeled by the differentiable function

\[ E(t) = 150 + 100\sin\!\left(\dfrac{\pi t}{6}\right) \quad (\text{passengers/hour}) \]

Passengers pass through security and leave the waiting area (entering the boarding zone) at a rate

\[ D(t) = 120 + 40\cos\!\left(\dfrac{\pi t}{4}\right) \quad (\text{passengers/hour}) \]

where \(0 \le t \le 12\). At 6 AM when the checkpoint opens, there are already 25 passengers waiting in the waiting area. Let \(Q(t)\) denote the number of passengers in the waiting area at time \(t\); then \(Q(0) = 25\).

(a) Find the value of \(E(3) - D(3)\) (accurate to three decimal places), and explain its meaning (include units).

(b) Find the value of \(\displaystyle\int_0^{12} E(t)\,dt\) (accurate to three decimal places), and explain its meaning in the context of the problem.

(c) Write an integral expression for \(Q(t)\), and find the value of \(Q(12)\) (to the nearest integer).

(d) On \(0 \le t \le 12\), at what time does \(Q(t)\) attain its maximum value? Justify your answer and give the value of \(Q\) at that time (to the nearest integer).

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(a) Value and meaning of \(E(3) - D(3)\)

\(E(3) = 150 + 100\sin(\pi/2) = 150 + 100 \cdot 1 = 250\).

\(D(3) = 120 + 40\cos(3\pi/4) = 120 + 40 \cdot \left(-\dfrac{\sqrt{2}}{2}\right) \approx 120 - 28.284 \approx 91.716\).

\[ E(3) - D(3) \approx \mathbf{158.284} \text{ passengers/hour}. \]

Meaning: At \(t = 3\) (9 AM), the number of passengers in the waiting area is increasing at a rate of approximately \(158.284\) passengers/hour (the arrival rate exceeds the passing rate).

Scoring: 1 pt (value) + 1 pt (meaning with units).

(b) \(\int_0^{12} E(t)\,dt\) and its meaning

Note that \(\displaystyle\int_0^{12} \sin(\pi t / 6)\,dt = \left[-\dfrac{6}{\pi}\cos(\pi t/6)\right]_0^{12} = -\dfrac{6}{\pi}[\cos(2\pi) - \cos 0] = 0\).

Therefore \(\displaystyle\int_0^{12} E(t)\,dt = 150 \cdot 12 + 100 \cdot 0 = \mathbf{1800.000}\).

Meaning: During the 12 hours from 6 AM to 6 PM, a total of 1800 passengers arrive at the waiting area of this security checkpoint.

Scoring: 1 pt (integral value) + 1 pt (meaning).

(c) Expression for \(Q(t)\) and \(Q(12)\)

Since \(Q'(t) = E(t) - D(t)\) and \(Q(0) = 25\), we have

\[ Q(t) = 25 + \int_0^{t} \big[E(s) - D(s)\big]\,ds. \]

Note that \(\displaystyle\int_0^{12} \cos(\pi t / 4)\,dt = \left[\dfrac{4}{\pi}\sin(\pi t/4)\right]_0^{12} = \dfrac{4}{\pi}[\sin(3\pi) - \sin 0] = 0\).

So \(\displaystyle\int_0^{12} D(s)\,ds = 120 \cdot 12 + 40 \cdot 0 = 1440\), and therefore

\[ Q(12) = 25 + 1800 - 1440 = \mathbf{385} \text{ passengers}. \]

Scoring: 1 pt (integral expression) + 1 pt (value of \(Q(12)\)).

(d) Maximum value of \(Q\)

Critical points satisfy \(Q'(t) = E(t) - D(t) = 0\), i.e. \(E(t) = D(t)\). Since \(E\) and \(D\) are continuous periodic functions, solving this equation on \([0, 12]\) with a calculator yields three solutions (in increasing order):

\(t_1 \approx 0.143,\quad t_2 \approx 6.360,\quad t_3 \approx 10.726\).

Sign analysis of \(Q'(t) = E(t) - D(t)\):

  • \(t = 0\): \(E(0) - D(0) = 150 - 160 = -10 < 0\)
  • \(t = 3\): \(\approx +158.3 > 0\) (see (a))
  • \(t = 9\): \(E(9) = 150 + 100\sin(3\pi/2) = 50\), \(D(9) = 120 + 40\cos(9\pi/4) = 120 + 20\sqrt{2} \approx 148.28\), difference \(\approx -98.3 < 0\)
  • \(t = 12\): \(E(12) - D(12) = 150 - 80 = 70 > 0\)

So the sign pattern of \(Q' = E - D\) is \(- , + , - , +\) (sign changes at \(t_1, t_2, t_3\)).

Therefore:

  • \(t = t_1 \approx 0.143\): \(Q'\) changes from \(-\) to \(+\) \(\Rightarrow\) \(Q\) has a local minimum
  • \(t = t_2 \approx 6.360\): \(Q'\) changes from \(+\) to \(-\) \(\Rightarrow\) \(Q\) has a local maximum
  • \(t = t_3 \approx 10.726\): \(Q'\) changes from \(-\) to \(+\) \(\Rightarrow\) \(Q\) has a local minimum

Candidates for the absolute maximum: the local max at \(t_2\) and the endpoints \(t = 0, t = 12\).

Computations (integrals via calculator): \(Q(0) = 25\); \(Q(12) = 385\);

\(Q(t_2) = 25 + \displaystyle\int_0^{6.360}\big[E(s) - D(s)\big]\,ds \approx 25 + 618.2 \approx \mathbf{643}\) passengers.

Conclusion: \(Q\) attains its absolute maximum at \(t \approx \mathbf{6.360}\) (approximately 12:22 PM), with value approximately \(\mathbf{643}\) passengers.

Justification: All critical points and both endpoints have been compared, and \(Q(t_2)\) is the largest.

Scoring: 1 pt (identify candidates: critical points + endpoints) + 1 pt (sign analysis to show \(t_2\) is a local max) + 1 pt (compare the 3 candidates and conclude).

Total (9 points): (a) 2 pts + (b) 2 pts + (c) 2 pts + (d) 3 pts.

Mock FRQ2 · Linear Particle Motion (velocity function)

x 轴粒子运动Linear Particle

A particle moves along the \(x\)-axis with velocity function (in meters/second)

\[ v(t) = t^2 \cos\!\left(\dfrac{t}{2}\right) - 2 \]

for \(0 \le t \le 6\). At \(t = 0\) the particle is located at \(x = 1\) (meters).

(a) Find the acceleration of the particle at \(t = 2\) (accurate to three decimal places). Is the particle speeding up or slowing down at this time? Justify your answer.

(b) Find the speed of the particle at \(t = 4\), and state whether the particle is moving to the left or to the right at that time.

(c) Find the position \(x(5)\) of the particle at \(t = 5\) (accurate to three decimal places).

(d) Find the total distance traveled by the particle from \(t = 0\) to \(t = 6\) (accurate to three decimal places).

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(a) Acceleration at \(t = 2\) and speeding up / slowing down

Acceleration \(a(t) = v'(t)\). By the product rule and chain rule:

\[ v'(t) = 2t\cos\!\left(\tfrac{t}{2}\right) - \tfrac{t^2}{2}\sin\!\left(\tfrac{t}{2}\right). \]

\(v'(2) = 4\cos 1 - 2\sin 1 \approx 4(0.54030) - 2(0.84147) \approx 2.16121 - 1.68294 \approx \mathbf{0.478}\) m/s\(^2\).

\(v(2) = 4\cos 1 - 2 \approx 4(0.54030) - 2 \approx 0.16121 > 0\).

Since \(v(2) > 0\) and \(a(2) > 0\) have the same sign, the particle is speeding up at \(t = 2\).

Scoring: 1 pt (value of \(a(2)\)) + 1 pt (sign comparison and conclusion).

(b) Speed at \(t = 4\) and direction

\(v(4) = 16\cos 2 - 2 \approx 16(-0.41615) - 2 \approx -6.658 - 2 \approx -8.658\) m/s.

Speed \(= |v(4)| \approx \mathbf{8.658}\) m/s.

Since \(v(4) < 0\), the particle is moving to the left (in the negative \(x\)-direction).

Scoring: 1 pt (speed) + 1 pt (direction).

(c) Position at \(t = 5\)

\[ x(5) = x(0) + \int_0^5 v(t)\,dt = 1 + \int_0^5 \left[t^2\cos\!\left(\tfrac{t}{2}\right) - 2\right] dt. \]

Using a calculator: \(\displaystyle\int_0^5 v(t)\,dt \approx -21.691\).

Therefore \(x(5) \approx 1 + (-21.691) = \mathbf{-20.691}\) meters.

Scoring: 1 pt (integral expression) + 1 pt (numerical value).

(d) Total distance traveled on \(t \in [0, 6]\)

\[ \text{Total distance} = \int_0^6 |v(t)|\,dt. \]

Using a calculator: \[ \int_0^6 |v(t)|\,dt \approx \mathbf{51.814} \text{ meters}. \]

Scoring: 1 pt (integral setup with absolute value) + 1 pt (numerical value).

Total (9 points): (a) 2 pts + (b) 2 pts + (c) 2 pts (including expression using \(x(0)\)) + (d) 2 pts + 1 pt (accuracy and units).

Mock FRQ3 · f' Graph Analysis (Line Segments + Semicircle)

f' 图像分析f' Graph

Let \(f\) be continuous on \([-5, 7]\) and differentiable on \((-5, 7)\). The graph of its derivative \(f'\) (not shown) consists of the following three pieces:

  • On \([-5, -1]\): \(f'\) is the line segment through \((-5, 4)\) and \((-1, 0)\) (slope \(-1\), i.e. \(f'(x) = -x - 1\)).
  • On \([-1, 3]\): \(f'\) is the lower semicircle with center \((1, 0)\) and radius \(2\), i.e. \(f'(x) = -\sqrt{4 - (x - 1)^2}\). This piece has vertical tangents at the endpoints \(x = -1, 3\) (the graph of \(f'\) is vertical at the left and right endpoints), so \(f'\) is only continuous at \(x = -1\) and \(x = 3\), with corners at those endpoints (\(f'\) is not differentiable there).
  • On \([3, 7]\): \(f'\) is the line segment through \((3, 0)\) and \((7, 2)\) (slope \(\dfrac{1}{2}\)).

It is given that \(f(-1) = 5\).

(a) On \([-5, 7]\), where is \(f\) increasing? Where is \(f\) decreasing? Give a reason.

(b) Find all \(x\)-values in \((-5, 7)\) at which \(f\) has relative extrema (local max / local min), and classify each.

(c) Find the value of \(f(7)\).

(d) Find all \(x\)-values in \((-5, 7)\) at which \(f\) has points of inflection (i.e. where \(f''\) changes sign), and give a reason.

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(a) Intervals where \(f\) is increasing / decreasing

\(f\) is increasing \(\iff f'(x) > 0\), decreasing \(\iff f'(x) < 0\).

  • \([-5, -1]\): \(f'(x) = -x - 1\). \(f'(-5) = 4 > 0\), \(f'(-1) = 0\); on the interior \(f' > 0\) \(\Rightarrow\) \(f\) is increasing.
  • \([-1, 3]\): the lower semicircle gives \(f'(x) \le 0\), with \(f' = 0\) only at \(x = -1, 3\) \(\Rightarrow\) \(f\) is decreasing.
  • \([3, 7]\): \(f'(x) = \tfrac{1}{2}(x - 3)\). \(f'(3) = 0\), \(f'(7) = 2 > 0\) \(\Rightarrow\) \(f\) is increasing.

Increasing: \([-5, -1] \cup [3, 7]\). Decreasing: \([-1, 3]\).

Reason: Based on the sign of \(f'\) as shown above.

Scoring: 1 pt (correct intervals) + 1 pt (reason referencing the sign of \(f'\)).

(b) Relative extrema

Critical points: \(f'(x) = 0\) at \(x = -1\) and \(x = 3\) (although \(f'\) is not differentiable at these points, \(f\) is differentiable there — \(f'\) takes the value \(0\)).

  • \(x = -1\): \(f' > 0\) to the left and \(f' \le 0\) to the right (lower semicircle) \(\Rightarrow\) \(f\) has a relative maximum (local max).
  • \(x = 3\): \(f' \le 0\) to the left and \(f' > 0\) to the right \(\Rightarrow\) \(f\) has a relative minimum (local min).

Scoring: 1 pt (local max) + 1 pt (local min).

(c) Value of \(f(7)\)

By the Fundamental Theorem of Calculus:

\[ f(7) = f(-1) + \int_{-1}^{7} f'(x)\,dx. \]

Split \(\int_{-1}^7 f'\,dx\) into two pieces (signed area between \(f'\) and the \(x\)-axis):

  • \([-1, 3]\): area of the lower semicircle (with \(f' \le 0\), taken as negative): \(-\tfrac{1}{2}\pi \cdot 2^2 = -2\pi\).
  • \([3, 7]\): triangle with base \(4\) and height \(2\), area \(= \tfrac{1}{2} \cdot 4 \cdot 2 = 4\) (\(f' \ge 0\)).

\[ \int_{-1}^{7} f'\,dx = -2\pi + 4. \]

\[ f(7) = 5 + (-2\pi + 4) = \mathbf{9 - 2\pi}. \]

Scoring: 1 pt (FTC setup) + 1 pt (piecewise area) + 1 pt (final answer).

(d) Points of inflection

Inflection point \(\iff\) \(f''\) changes sign \(\iff\) monotonicity of \(f'\) changes. Analyze the monotonicity of \(f'\) on each piece:

  • \([-5, -1]\): \(f'\) is a line segment with slope \(-1\), so \(f'\) is decreasing \(\Rightarrow\) \(f'' < 0\).
  • \([-1, 1]\) (left half of lower semicircle): \(f'\) decreases from \(0\) to \(-2\) (minimum at \(x=1\)), so \(f'\) is decreasing \(\Rightarrow\) \(f'' < 0\).
  • \([1, 3]\) (right half of lower semicircle): \(f'\) increases from \(-2\) to \(0\), so \(f'\) is increasing \(\Rightarrow\) \(f'' > 0\).
  • \([3, 7]\): \(f'\) is a line segment with slope \(\tfrac{1}{2} > 0\), so \(f'\) is increasing \(\Rightarrow\) \(f'' > 0\).

Check sign changes of \(f''\) at each transition point:

  • \(x = -1\): \(f'' < 0\) on the left and \(f'' < 0\) on the right \(\Rightarrow\) no sign change, not an inflection point.
  • \(x = 1\): \(f'' < 0\) on the left and \(f'' > 0\) on the right \(\Rightarrow\) sign change, inflection point.
  • \(x = 3\): \(f'' > 0\) on the left and \(f'' > 0\) on the right \(\Rightarrow\) no sign change, not an inflection point.

Inflection point: only at \(x = 1\).

Scoring: 1 pt (method: inflection at a change in monotonicity of \(f'\)) + 1 pt (inflection point \(x = 1\) with reason).

Total (9 points): (a) 2 pts + (b) 2 pts + (c) 3 pts + (d) 2 pts.

Mock FRQ4 · Separable DE (Newton's Law of Cooling)

微分方程Differential Eq.

A cup of hot chocolate has an initial temperature of \(90\)°C and is placed in a room at \(20\)°C to cool. By Newton's Law of Cooling, the temperature \(T\) (in °C) satisfies the differential equation

\[ \dfrac{dT}{dt} = -\dfrac{1}{10}(T - 20), \quad t \ge 0, \]

where \(t\) is measured in minutes.

(a) Find the value of \(\dfrac{dT}{dt}\) at \(t = 0\) when \(T(0) = 90\), and explain its physical meaning (include units).

(b) Use separation of variables to solve the differential equation, writing the explicit solution \(T(t)\) (using the initial condition).

(c) Find the value of \(t\) at which \(T(t) = 50\) (express using \(\ln\); no calculator approximation required).

(d) Determine the limiting behavior of \(T(t)\) as \(t \to \infty\), and analyze the concavity (concave up / concave down) of the solution \(T(t)\) on \(t \ge 0\). Give a reason.

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(a) Instantaneous rate of change at \(t = 0\)

\[ \left.\dfrac{dT}{dt}\right|_{t = 0} = -\dfrac{1}{10}(90 - 20) = -\dfrac{70}{10} = \mathbf{-7} \text{ °C/minute}. \]

Meaning: At \(t = 0\) (when the cup is first placed), the chocolate's temperature is decreasing at a rate of \(7\) °C per minute.

Scoring: 1 pt (value) + 1 pt (meaning + units).

(b) Explicit solution

Separating variables: \(\dfrac{dT}{T - 20} = -\dfrac{1}{10}\,dt\).

Integrating both sides: \(\ln|T - 20| = -\dfrac{t}{10} + C_1\).

Since \(T(0) = 90 > 20\), and the solution remains \(T > 20\) up to the asymptote, we have \(T - 20 > 0\) and can drop the absolute value:

\(T - 20 = e^{C_1} e^{-t/10} = C e^{-t/10}\).

Substituting the initial condition \(T(0) = 90\): \(90 - 20 = C \cdot 1 \Rightarrow C = 70\).

\[ \boxed{\;T(t) = 20 + 70 e^{-t/10}.\;} \]

Scoring: 1 pt (separation of variables) + 1 pt (integration) + 1 pt (applying initial condition to obtain the explicit solution).

(c) Time at which \(T(t) = 50\)

Set \(20 + 70 e^{-t/10} = 50\):

\(70 e^{-t/10} = 30 \Rightarrow e^{-t/10} = \dfrac{3}{7}\).

\(-\dfrac{t}{10} = \ln\!\left(\dfrac{3}{7}\right) \Rightarrow t = -10\ln\!\left(\dfrac{3}{7}\right) = \mathbf{10\ln\!\left(\dfrac{7}{3}\right)}\) minutes.

Scoring: 1 pt (solving for \(e^{-t/10}\)) + 1 pt (expression for \(t\)).

(d) Limiting behavior and concavity

Limit: \(\displaystyle\lim_{t \to \infty} T(t) = 20 + 70 \cdot 0 = 20\) °C.

That is, the chocolate's temperature approaches the room temperature \(20\) °C asymptotically (but never reaches \(20\)).

Concavity:

\(T'(t) = -\dfrac{70}{10}e^{-t/10} = -7e^{-t/10}\).

\(T''(t) = -7 \cdot \left(-\dfrac{1}{10}\right)e^{-t/10} = \dfrac{7}{10}e^{-t/10} > 0\) for all \(t \ge 0\).

Therefore \(T\) is concave up on \([0, \infty)\).

Explanation: The temperature is decreasing, but the rate of decrease is itself decreasing (cooling slows down as \(T - 20\) gets smaller), giving the "slowing descent" shape that is concave up.

Scoring: 1 pt (limit) + 1 pt (sign of \(T''\) + concave up conclusion).

Total (9 points): (a) 2 pts + (b) 3 pts + (c) 2 pts + (d) 2 pts.

Mock FRQ5 · Area + Solid of Revolution + Cross-Section Volume

面积 / 体积Area/Volume

Let \(R\) be the bounded region (in the first quadrant) enclosed by the curve \(y = \sqrt{x}\) and the line \(y = \dfrac{x}{2}\).

(a) Find the coordinates of the intersection points of the two curves in the first quadrant, and write an integral expression for the area of the region \(R\). Find its exact value.

(b) The region \(R\) is revolved about the \(x\)-axis. Find the exact volume \(V_1\) of the resulting solid.

(c) The region \(R\) is revolved about the horizontal line \(y = -1\). Find the exact volume \(V_2\) of the resulting solid.

(d) Consider a solid with base \(R\) whose cross-sections perpendicular to the \(x\)-axis are isosceles right triangles with hypotenuse lying in \(R\) (i.e. the length of the hypotenuse equals the vertical width of \(R\) at that \(x\)). Find the exact volume \(V_3\) of this solid.

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(a) Intersection points and area

Set \(\sqrt{x} = \dfrac{x}{2}\): squaring both sides gives \(x = \dfrac{x^2}{4} \Rightarrow x^2 - 4x = 0 \Rightarrow x(x - 4) = 0\), so \(x = 0\) or \(x = 4\).

Intersection points: \((0, 0)\) and \((4, 2)\).

On \([0, 4]\), \(\sqrt{x} \ge x/2\) (check with \(x = 1\): \(1 \ge 0.5\) ✓).

\[ \text{Area} = \int_0^4 \left[\sqrt{x} - \dfrac{x}{2}\right] dx = \left[\dfrac{2}{3}x^{3/2} - \dfrac{x^2}{4}\right]_0^4 = \dfrac{2}{3}(8) - \dfrac{16}{4} = \dfrac{16}{3} - 4 = \mathbf{\dfrac{4}{3}}. \]

Scoring: 1 pt (intersection points) + 1 pt (integral expression with correct integrand) + 1 pt (answer \(\tfrac{4}{3}\)).

(b) Revolve about \(x\)-axis: \(V_1\)

Using the washer method (outer radius \(\sqrt{x}\), inner radius \(x/2\)):

\[ V_1 = \pi \int_0^4 \left[(\sqrt{x})^2 - \left(\dfrac{x}{2}\right)^2\right] dx = \pi \int_0^4 \left[x - \dfrac{x^2}{4}\right] dx. \]

\[ = \pi \left[\dfrac{x^2}{2} - \dfrac{x^3}{12}\right]_0^4 = \pi \left[8 - \dfrac{64}{12}\right] = \pi \left[8 - \dfrac{16}{3}\right] = \pi \cdot \dfrac{24 - 16}{3} = \mathbf{\dfrac{8\pi}{3}}. \]

Scoring: 1 pt (washer setup) + 1 pt (evaluated integral).

(c) Revolve about \(y = -1\): \(V_2\)

Translate the distance of each curve to the line \(y = -1\):

  • Outer radius (distance from the upper curve to \(y = -1\)): \(R_{\text{out}}(x) = \sqrt{x} - (-1) = \sqrt{x} + 1\).
  • Inner radius (distance from the lower curve to \(y = -1\)): \(R_{\text{in}}(x) = \dfrac{x}{2} + 1\).

\[ V_2 = \pi \int_0^4 \left[(\sqrt{x} + 1)^2 - \left(\dfrac{x}{2} + 1\right)^2\right] dx. \]

Expand the integrand:

\((\sqrt{x} + 1)^2 = x + 2\sqrt{x} + 1\).

\(\left(\dfrac{x}{2} + 1\right)^2 = \dfrac{x^2}{4} + x + 1\).

Difference: \((x + 2\sqrt{x} + 1) - \left(\dfrac{x^2}{4} + x + 1\right) = 2\sqrt{x} - \dfrac{x^2}{4}\).

\[ V_2 = \pi \int_0^4 \left[2\sqrt{x} - \dfrac{x^2}{4}\right] dx = \pi \left[\dfrac{4}{3}x^{3/2} - \dfrac{x^3}{12}\right]_0^4. \]

Substituting \(x = 4\): \(\dfrac{4}{3}(8) - \dfrac{64}{12} = \dfrac{32}{3} - \dfrac{16}{3} = \dfrac{16}{3}\).

\[ V_2 = \mathbf{\dfrac{16\pi}{3}}. \]

Scoring: 1 pt (correct outer/inner radii including the \(+1\) shift) + 1 pt (simplified integrand) + 1 pt (answer).

(d) Cross-sections are isosceles right triangles: \(V_3\)

At a given \(x\), the vertical width of \(R\) is \(w(x) = \sqrt{x} - \dfrac{x}{2}\), which is the length of the hypotenuse of the isosceles right triangle.

For an isosceles right triangle with hypotenuse \(h\), each leg has length \(\dfrac{h}{\sqrt{2}}\), so the area \(= \dfrac{1}{2} \cdot \text{leg}^2 = \dfrac{1}{2} \cdot \dfrac{h^2}{2} = \dfrac{h^2}{4}\).

So the cross-sectional area is \(A(x) = \dfrac{w(x)^2}{4} = \dfrac{1}{4}\left(\sqrt{x} - \dfrac{x}{2}\right)^2\).

\[ V_3 = \int_0^4 \dfrac{1}{4}\left(\sqrt{x} - \dfrac{x}{2}\right)^2 dx. \]

Expand: \(\left(\sqrt{x} - \dfrac{x}{2}\right)^2 = x - 2\sqrt{x} \cdot \dfrac{x}{2} + \dfrac{x^2}{4} = x - x^{3/2} + \dfrac{x^2}{4}\).

\[ V_3 = \dfrac{1}{4}\int_0^4 \left[x - x^{3/2} + \dfrac{x^2}{4}\right] dx = \dfrac{1}{4}\left[\dfrac{x^2}{2} - \dfrac{2}{5}x^{5/2} + \dfrac{x^3}{12}\right]_0^4. \]

Substituting \(x = 4\): \(\dfrac{16}{2} - \dfrac{2}{5} \cdot 32 + \dfrac{64}{12} = 8 - \dfrac{64}{5} + \dfrac{16}{3}\).

Common denominator (60): \(\dfrac{480}{60} - \dfrac{768}{60} + \dfrac{320}{60} = \dfrac{480 - 768 + 320}{60} = \dfrac{32}{60} = \dfrac{8}{15}\).

\[ V_3 = \dfrac{1}{4} \cdot \dfrac{8}{15} = \mathbf{\dfrac{2}{15}}. \]

Scoring: 1 pt (cross-sectional area formula \(h^2/4\)) + 1 pt (integral setup) + 1 pt (answer).

Total (9 points): (a) 2 pts + (b) 2 pts + (c) 2 pts + (d) 3 pts.

Mock FRQ6 · Table + Chain/Product Rule + Implicit Differentiation

函数关系综合Connected Funcs

Let \(f\) and \(g\) be differentiable for all real numbers. The table below gives values of \(f, f', g, g'\) at selected points:

\(x\)\(-2\)\(0\)\(1\)\(3\)
\(f(x)\)\(4\)\(1\)\(-2\)\(5\)
\(f'(x)\)\(3\)\(-1\)\(2\)\(6\)
\(g(x)\)\(1\)\(3\)\(0\)\(-2\)
\(g'(x)\)\(-2\)\(4\)\(5\)\(1\)

(a) Let \(h(x) = f(x) \cdot g(x)\). Find \(h'(1)\).

(b) Let \(k(x) = f(g(x))\). Find \(k'(0)\). Show the full differentiation steps.

(c) Let \(m(x) = \dfrac{g(x)}{f(x) + 2}\). Find \(m'(-2)\). Give the simplified exact value.

(d) A curve \(C\) is defined implicitly by \(x^2 + x \cdot y + y^3 = 7\).

  1. Find \(\dfrac{dy}{dx}\) (in terms of \(x\) and \(y\)).
  2. Verify whether the point \((1, \sqrt[3]{5})\) is not on the curve (if it is, find the tangent line there; if not, use the point \((2, 1)\) instead to find the tangent line — first verify that \((2, 1)\) lies on the curve).
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(a) \(h'(1)\) (product rule)

\(h'(x) = f'(x)g(x) + f(x)g'(x)\).

\[ h'(1) = f'(1)g(1) + f(1)g'(1) = (2)(0) + (-2)(5) = 0 - 10 = \mathbf{-10}. \]

Scoring: 1 pt (product rule) + 1 pt (substitution and computation).

(b) \(k'(0)\) (chain rule)

\(k'(x) = f'(g(x)) \cdot g'(x)\).

At \(x = 0\): \(g(0) = 3\), so \(f'(g(0)) = f'(3) = 6\).

\(g'(0) = 4\).

\[ k'(0) = f'(3) \cdot g'(0) = 6 \cdot 4 = \mathbf{24}. \]

Scoring: 1 pt (chain rule expression + table lookup) + 1 pt (answer).

(c) \(m'(-2)\) (quotient rule)

\(m(x) = \dfrac{g(x)}{f(x) + 2}\). By the quotient rule:

\[ m'(x) = \dfrac{g'(x)\big[f(x) + 2\big] - g(x) f'(x)}{\big[f(x) + 2\big]^2}. \]

Substituting \(x = -2\): \(f(-2) = 4\), \(f'(-2) = 3\), \(g(-2) = 1\), \(g'(-2) = -2\).

\[ m'(-2) = \dfrac{(-2)(4 + 2) - (1)(3)}{(4 + 2)^2} = \dfrac{(-2)(6) - 3}{36} = \dfrac{-12 - 3}{36} = \dfrac{-15}{36} = \mathbf{-\dfrac{5}{12}}. \]

Scoring: 1 pt (quotient rule formula + substitution) + 1 pt (simplified answer).

(d) Implicit differentiation and tangent line

(i) Differentiate \(x^2 + xy + y^3 = 7\) implicitly with respect to \(x\) (treating \(y\) as a function of \(x\)):

\[ 2x + \left(y + x\dfrac{dy}{dx}\right) + 3y^2\dfrac{dy}{dx} = 0. \]

\(\dfrac{dy}{dx}\big(x + 3y^2\big) = -(2x + y)\).

\[ \dfrac{dy}{dx} = \mathbf{\dfrac{-(2x + y)}{x + 3y^2}}, \quad x + 3y^2 \ne 0. \]

(ii) First check whether \((1, \sqrt[3]{5})\) lies on the curve: substituting gives \(x^2 + xy + y^3 = 1 + \sqrt[3]{5} + 5 = 6 + \sqrt[3]{5}\).

\(\sqrt[3]{5} \approx 1.710 \ne 1\), so \(6 + \sqrt[3]{5} \approx 7.710 \ne 7\); the point is not on the curve.

Check \((2, 1)\) instead: \(4 + 2 + 1 = 7\) ✓, so \((2, 1)\) lies on the curve.

At \((2, 1)\):

\[ \dfrac{dy}{dx}\bigg|_{(2, 1)} = \dfrac{-(2 \cdot 2 + 1)}{2 + 3 \cdot 1^2} = \dfrac{-5}{5} = \mathbf{-1}. \]

Tangent line: \(y - 1 = -1(x - 2)\), i.e. \(\boxed{\;y = -x + 3.\;}\)

Scoring: (i) 1 pt (implicit differentiation setup) + 1 pt (solving for \(dy/dx\)); (ii) 1 pt (verify \((1, \sqrt[3]{5})\) is not on the curve + verify \((2, 1)\) is on the curve) + 1 pt (slope \(-1\) and tangent line).

Total (9 points): (a) 2 pts + (b) 2 pts + (c) 2 pts + (d) 3 pts ((i) implicit differentiation 2 pts + (ii) tangent line 1 pt).

Part IV · MCQ 真题库(179 题)Part IV · MCQ Past Question Bank (179 Qs)

MCQ 真题(2014 / 2015 / 2016 / 2018)。用下面的筛选按年份 / 分卷 / 考点(Unit)组合。点「答案」查看答案。MCQ past questions (2014 / 2015 / 2016 / 2018). Filter by year / section / unit; click 'Answer' to reveal.

年份:Year:
分卷:Section:
考点(Unit):Unit:
答案:Answer:
📱 iPad / 纸笔做题:📱 iPad / Pen-and-Paper: → Safari 分享 → 打印 → 存为 PDF → 传到 GoodNotes/Notability 用 Apple Pencil 写→ Safari Share → Print → Save as PDF → import to GoodNotes / Notability for Apple Pencil
Total: 179 Qs

Part V · 考前 24 小时 Cheat Sheet (AB)

必背公式卡

极限 / 连续

\(\lim_{x\to a} f(x) = L\) 存在 ⇔ 左右极限相等

连续三条件:\(f(a)\) 有定义、\(\lim_{x\to a} f(x)\) 存在、两者相等

可去间断:极限存在但不等于 \(f(a)\)

L'Hôpital:\(\frac{0}{0}\) 或 \(\frac{\infty}{\infty}\) 时 \(\lim \frac{f}{g} = \lim \frac{f'}{g'}\)

导数基础

\((x^n)' = nx^{n-1}\);\((e^x)' = e^x\);\((\ln x)' = 1/x\)

\((\sin x)' = \cos x\);\((\cos x)' = -\sin x\);\((\tan x)' = \sec^2 x\)

\((\arctan x)' = \frac{1}{1+x^2}\);\((\arcsin x)' = \frac{1}{\sqrt{1-x^2}}\)

乘积:\((fg)' = f'g + fg'\);商:\(\left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2}\)

链式:\((f(g(x)))' = f'(g(x)) \cdot g'(x)\)

隐函数与相关变化率

隐函数求导:两边对 \(x\) 求导,\(y\) 视为 \(y(x)\),解 \(dy/dx\)。

相关变化率 4 步:写方程 → 对 \(t\) 求导(链式)→ 代入已知率 → 解未知率

常用:\(V = \pi r^2 h\)、\(x^2+y^2=z^2\)(梯子)、\(V=\frac{4}{3}\pi r^3\)

MVT / 极值 / 凹凸

MVT:\(f\) 在 \([a,b]\) 连续可导 ⇒ \(\exists c\) 使 \(f'(c) = \dfrac{f(b)-f(a)}{b-a}\)

\(f\) 递增 ⇔ \(f' > 0\);\(f\) 凹 ⇔ \(f'' > 0\)

相对极值:\(f'\) 变号;absolute 还要看端点

拐点:\(f''\) 变号 (不仅 \(f''=0\))

积分基础

\(\int x^n\,dx = \frac{x^{n+1}}{n+1} + C\) (\(n \ne -1\))

\(\int \frac{1}{x}\,dx = \ln|x| + C\);\(\int e^x\,dx = e^x + C\)

\(\int \sin x\,dx = -\cos x\);\(\int \cos x\,dx = \sin x\)

\(\int \sec^2 x\,dx = \tan x\);\(\int \frac{1}{1+x^2}\,dx = \arctan x\)

FTC / 累积

FTC I: \(\frac{d}{dx}\int_a^x f(t)\,dt = f(x)\)

FTC II: \(\int_a^b f(x)\,dx = F(b) - F(a)\)

累积:\(A(t) = A(0) + \int_0^t (E(s) - L(s))\,ds\)

平均值:\(\bar f = \dfrac{1}{b-a}\int_a^b f(x)\,dx\)

数值近似

Trapezoidal: \(\int_a^b f\,dx \approx \sum \dfrac{x_{i+1}-x_i}{2}\big(f(x_i)+f(x_{i+1})\big)\)

Left/Right Riemann: \(\sum f(x_i)\Delta x\) 或 \(\sum f(x_{i+1})\Delta x\)

Midpoint: \(\sum f\!\left(\dfrac{x_i+x_{i+1}}{2}\right)\Delta x\)

Linear approx: \(f(x) \approx f(a) + f'(a)(x-a)\)

微分方程(可分离)

\(\frac{dy}{dx} = g(x)h(y)\) ⇒ 分离 \(\frac{dy}{h(y)} = g(x)\,dx\),两边积分

代初始值求 \(C\),然后解出 \(y\)(显式)

Newton cooling: \(\frac{dT}{dt} = -k(T - T_\infty)\) → \(T = T_\infty + Ce^{-kt}\)

AB 不考:slope field (BC 才考)、Euler、Logistic、Taylor 近似

面积 / 体积

面积:\(A = \int_a^b |f(x) - g(x)|\,dx\)(top − bottom)

Disk: \(V = \pi \int_a^b [f(x)]^2\,dx\)

Washer: \(V = \pi \int_a^b \big([R(x)]^2 - [r(x)]^2\big)dx\)

Cross section: \(V = \int_a^b A(x)\,dx\)(A(x) 为已知截面函数)

考场 5 条纪律

  1. Calc 部分(FRQ1, 2)所有积分 / 求根让计算器做。
  2. 每一问写单位(米, 秒, 人, 升等)。没单位 = 丢点。
  3. "Justify your answer" 必写完整因果句(如"因为 \(f'(c)\) 从正变负,故 \(c\) 为 local max")。
  4. MVT 题必须同时写:连续、可导、结论——缺一条扣分。
  5. AB 不考 参数 / 极坐标 / 级数——不要被 BC 模板误导。

最后 10 条 AB 高频陷阱

  1. 链式法则漏内函数导数 \(g'(x)\)(FRQ6 最常见低级错)
  2. 商法则分子:\(f'g - fg'\),别写成 \(fg' - f'g\)
  3. MVT 三条件缺一条(连续、可导、结论)
  4. Trapezoidal 忘了系数 \(\dfrac{1}{2}\)
  5. L'Hôpital 没检查 \(0/0\) 或 \(\infty/\infty\) 就套用
  6. 可分离 DE 分离变量时 \(dy\) 和 \(dx\) 对调
  7. Absolute max/min 只看 critical point 漏端点
  8. Washer 体积:外半径 \(R\) 减内半径 \(r\)(平方再相减,别先减再平方)
  9. 速度(\(v\))vs 速率(speed,取绝对值)混用
  10. f' 图分析时把 f' 当成 f 本身

AB 不考的 BC 内容(别浪费时间)

  • 参数 / 极坐标 / 向量运动(BC Unit 9)— AB 跳过
  • 级数(BC Unit 10)— 收敛测试、Taylor、Maclaurin、Lagrange 都是 BC 专属
  • Euler's method / 斜率场 / Logistic DE — 大多只在 BC 考

Part V · 24-Hour Pre-Exam Cheat Sheet (AB)

Must-Memorize Formula Cards

Limits / Continuity

\(\lim_{x\to a} f(x) = L\) exists ⇔ left and right limits are equal

Three conditions for continuity: \(f(a)\) is defined, \(\lim_{x\to a} f(x)\) exists, and the two are equal

Removable discontinuity: limit exists but is not equal to \(f(a)\)

L'Hôpital: when \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), \(\lim \frac{f}{g} = \lim \frac{f'}{g'}\)

Derivative Basics

\((x^n)' = nx^{n-1}\); \((e^x)' = e^x\); \((\ln x)' = 1/x\)

\((\sin x)' = \cos x\); \((\cos x)' = -\sin x\); \((\tan x)' = \sec^2 x\)

\((\arctan x)' = \frac{1}{1+x^2}\); \((\arcsin x)' = \frac{1}{\sqrt{1-x^2}}\)

Product: \((fg)' = f'g + fg'\); Quotient: \(\left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2}\)

Chain: \((f(g(x)))' = f'(g(x)) \cdot g'(x)\)

Implicit Diff & Related Rates

Implicit differentiation: differentiate both sides with respect to \(x\), treat \(y\) as \(y(x)\), solve for \(dy/dx\).

Related rates 4 steps: write equation → differentiate with respect to \(t\) (chain rule) → substitute known rates → solve for unknown rate

Common formulas: \(V = \pi r^2 h\), \(x^2+y^2=z^2\) (ladder), \(V=\frac{4}{3}\pi r^3\)

MVT / Extrema / Concavity

MVT: \(f\) is continuous and differentiable on \([a,b]\) ⇒ \(\exists c\) such that \(f'(c) = \dfrac{f(b)-f(a)}{b-a}\)

\(f\) increasing ⇔ \(f' > 0\); \(f\) concave up ⇔ \(f'' > 0\)

Relative extrema: \(f'\) changes sign; absolute extrema also require checking endpoints

Inflection point: \(f''\) changes sign (not only \(f''=0\))

Integral Basics

\(\int x^n\,dx = \frac{x^{n+1}}{n+1} + C\) (\(n \ne -1\))

\(\int \frac{1}{x}\,dx = \ln|x| + C\); \(\int e^x\,dx = e^x + C\)

\(\int \sin x\,dx = -\cos x\); \(\int \cos x\,dx = \sin x\)

\(\int \sec^2 x\,dx = \tan x\); \(\int \frac{1}{1+x^2}\,dx = \arctan x\)

FTC / Accumulation

FTC I: \(\frac{d}{dx}\int_a^x f(t)\,dt = f(x)\)

FTC II: \(\int_a^b f(x)\,dx = F(b) - F(a)\)

Accumulation: \(A(t) = A(0) + \int_0^t (E(s) - L(s))\,ds\)

Average value: \(\bar f = \dfrac{1}{b-a}\int_a^b f(x)\,dx\)

Numerical Approximation

Trapezoidal: \(\int_a^b f\,dx \approx \sum \dfrac{x_{i+1}-x_i}{2}\big(f(x_i)+f(x_{i+1})\big)\)

Left/Right Riemann: \(\sum f(x_i)\Delta x\) or \(\sum f(x_{i+1})\Delta x\)

Midpoint: \(\sum f\!\left(\dfrac{x_i+x_{i+1}}{2}\right)\Delta x\)

Linear approx: \(f(x) \approx f(a) + f'(a)(x-a)\)

Differential Equations (Separable)

\(\frac{dy}{dx} = g(x)h(y)\) ⇒ separate \(\frac{dy}{h(y)} = g(x)\,dx\), integrate both sides

Substitute initial value to find \(C\), then solve for \(y\) (explicit)

Newton cooling: \(\frac{dT}{dt} = -k(T - T_\infty)\) → \(T = T_\infty + Ce^{-kt}\)

AB does not test: slope field (BC only), Euler, Logistic, Taylor approximation

Area / Volume

Area: \(A = \int_a^b |f(x) - g(x)|\,dx\) (top − bottom)

Disk: \(V = \pi \int_a^b [f(x)]^2\,dx\)

Washer: \(V = \pi \int_a^b \big([R(x)]^2 - [r(x)]^2\big)dx\)

Cross section: \(V = \int_a^b A(x)\,dx\) (A(x) is the given cross-section function)

5 Exam-Room Rules

  1. Calc section (FRQ1, 2): let the calculator handle all integrals / root-finding.
  2. Write units for every part (meters, seconds, people, liters, etc.). No units = lost point.
  3. "Justify your answer" must be a complete cause-and-effect sentence (e.g. "Since \(f'(c)\) changes from positive to negative, \(c\) is a local max").
  4. MVT problems must state all three: continuous, differentiable, and conclusion — missing any one loses points.
  5. AB does not test parametric / polar / series — don't be misled by BC templates.

Top 10 AB High-Frequency Traps

  1. Chain rule missing inner derivative \(g'(x)\) (most common low-level error on FRQ6)
  2. Quotient rule numerator: \(f'g - fg'\), don't write as \(fg' - f'g\)
  3. MVT missing one of three conditions (continuous, differentiable, conclusion)
  4. Trapezoidal forgetting coefficient \(\dfrac{1}{2}\)
  5. Applying L'Hôpital without checking \(0/0\) or \(\infty/\infty\) form
  6. Separable DE: swapping \(dy\) and \(dx\) when separating variables
  7. Absolute max/min: only checking critical points, missing endpoints
  8. Washer volume: outer radius \(R\) minus inner radius \(r\) (square first, then subtract, not the other way around)
  9. Velocity (\(v\)) vs speed (absolute value) confusion
  10. Treating the f' graph as if it were the f graph

BC Content AB Does Not Test (Don't Waste Time)

  • Parametric / Polar / Vector motion (BC Unit 9) — AB skips
  • Series (BC Unit 10) — Convergence tests, Taylor, Maclaurin, Lagrange are all BC Only
  • Euler's method / Slope fields / Logistic DE — mostly tested only on BC