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Chapter 4第四章

Contextual Applications of Differentiation微分的实际应用

AP-Style Practice QuestionsAP 风格练习题

EASYMEDIUMHARD

Topics考点 4.1 - 4.7AB



Name:姓名:Period:班级:
PART ITopics 4.1 - 4.7考点 4.1 - 4.7

Multiple Choice Questions选择题

Show all supporting work on scratch paper. On the AP Exam, Section I is split into a no-calculator and a calculator-allowed part, each question below is labeled accordingly.请将所有辅助步骤写在草稿纸上。AP 考试第一部分分为不可使用计算器和可使用计算器两个部分,每道题已标注对应类型。

Q1EASY4.1 Interpreting Derivative4.1 导数的实际意义No Calculator

$W(t)$ is the weight, in kg, of a calf at age $t$ weeks. The most appropriate units for $W'(t)$ are$W(t)$ 是一头小牛在 $t$ 周龄时的体重(单位:千克)。$W'(t)$ 最合适的单位是

Q2EASY4.2 Motion4.2 运动No Calculator

A particle's position is $s(t)=t^{2}-4t$. The particle is at rest when一个质点的位置函数为 $s(t)=t^{2}-4t$。该质点静止时

Q3MEDIUM4.2 Speeding Up4.2 加速运动No Calculator

A particle moves with velocity $v(t)=t^{2}-4t+3$. The particle is speeding up on一个质点的速度函数为 $v(t)=t^{2}-4t+3$。该质点在哪个区间上速率递增?

Q4EASY4.3 Rates (Context)4.3 变化率(实际情境)No Calculator

Oil is pumped into a tank so that $V(t)=10t-t^{2}/4$ gallons at time $t$ minutes. At $t=4$, oil enters the tank at石油被泵入储罐,使得 $t$ 分钟时罐内油量为 $V(t)=10t-t^{2}/4$ 加仑。在 $t=4$ 时,石油注入储罐的速率为

Q5MEDIUM4.4 Related Rates4.4 相关变化率No Calculator

A spherical balloon is inflated so that its volume increases at $36\pi$ cm³/s. At the instant $r=3$ cm, $\dfrac{dr}{dt}=$一个球形气球被充气,其体积以 $36\pi$ cm³/s 的速率增大。当 $r=3$ cm 时,$\dfrac{dr}{dt}=$

Q6MEDIUM4.4 Related Rates (Shadow)4.4 相关变化率(影子)Calculator

A $6$-ft-tall person walks away from a $15$-ft lamppost at $4$ ft/s. The tip of their shadow moves at一个身高 $6$ 英尺的人以 $4$ 英尺/秒的速率远离一根高 $15$ 英尺的路灯。其影子尖端移动的速率为

Q7MEDIUM4.5 Linear Approximation4.5 线性化近似No Calculator

The linear approximation of $f(x)=\sqrt{x}$ at $x=9$ gives $\sqrt{9.4}\approx$$f(x)=\sqrt{x}$ 在 $x=9$ 处的线性近似给出 $\sqrt{9.4}\approx$

Q8HARD4.5 Error Direction4.5 误差方向No Calculator

Let $f$ be twice-differentiable with $f''<0$ on an open interval around $x=a$. The tangent-line approximation to $f$ near $x=a$ is设 $f$ 在 $x=a$ 的某开区间上二阶可导且 $f''<0$。$f$ 在 $x=a$ 附近的切线近似是

Q9EASY4.7 L'Hôpital4.7 洛必达法则No Calculator

$\displaystyle\lim_{x\to 0}\dfrac{\sin x}{x}=$

Q10MEDIUM4.7 L'Hôpital4.7 洛必达法则No Calculator

$\displaystyle\lim_{x\to 0}\dfrac{e^{2x}-1-2x}{x^{2}}=$

Q11MEDIUM4.7 L'Hôpital (∞/∞)4.7 洛必达法则(∞/∞)No Calculator

$\displaystyle\lim_{x\to\infty}\dfrac{\ln x}{\sqrt{x}}=$

Q12HARD4.4 Related Rates (Angle)4.4 相关变化率(角度)Calculator

A spotlight on the ground is $20$ m from a wall. A $2$-m-tall figure walks from the light toward the wall at $1$ m/s. When the figure is $4$ m from the wall, the length of the shadow on the wall changes at a rate closest to地面上有一盏聚光灯,距墙 $20$ 米。一个身高 $2$ 米的人以 $1$ 米/秒的速率从灯处走向墙壁。当此人距墙 $4$ 米时,墙上投影长度的变化速率最接近

Q13MEDIUM4.3 Rates (Economic)4.3 变化率(经济情境)No Calculator

A company's cost in dollars for producing $x$ items is $C(x)=0.01x^{2}+20x+500$. The marginal cost at $x=100$ is某公司生产 $x$ 件产品的成本(美元)为 $C(x)=0.01x^{2}+20x+500$。在 $x=100$ 时的边际成本为

Q14EASY5.1 MVT5.1 微分中值定理Preview of Unit 5No Calculator

Note: The Mean Value Theorem is formally CED Topic 5.1 (Unit 5). Included here as a preview because tabular-rate FRQs in Unit 4 routinely cite it.注:微分中值定理正式属于 CED 考点 5.1(第五单元)。此处作为预习内容纳入,因为第四单元的表格式变化率自由回答题中常常会引用该定理。

Which hypothesis is required for the Mean Value Theorem on $[a,b]$?微分中值定理在 $[a,b]$ 上成立所需的条件是什么?

Q15MEDIUM5.1 MVT5.1 微分中值定理Preview of Unit 5No Calculator

For $f(x)=x^{2}$ on $[1,4]$, the value $c$ guaranteed by the MVT is对于 $f(x)=x^{2}$ 在 $[1,4]$ 上,由微分中值定理保证存在的 $c$ 值为

Q16HARD4.2 Motion (Graph)4.2 运动(图形分析)No Calculator

A particle's velocity graph is shown on $[0,6]$. On what interval is the particle speeding up?已知质点在 $[0,6]$ 上的速度图像。在哪个区间上质点的速率递增?

$v(t)$
Q17MEDIUM4.5 Linearization Error4.5 线性化误差No Calculator

$L(x)$ is the tangent-line approximation to $f(x)=\sin x$ at $x=0$. Then $L(0.1)=$$L(x)$ 是 $f(x)=\sin x$ 在 $x=0$ 处的切线近似。则 $L(0.1)=$

Q18HARD4.7 L'Hôpital (Twice)4.7 洛必达法则(应用两次)No Calculator

$\displaystyle\lim_{x\to 0}\dfrac{x-\sin x}{x^{3}}=$

PART IIShow All Work展示全部解题过程

Free-Response Questions自由回答题

Contextual FRQs require labeled units, clear statement of what the derivative means in context, and justification using signs of $v$ and $a$ for motion problems. For related-rates, include a diagram, equation relating quantities, and differentiation with respect to time.情境类自由回答题要求标注单位、明确说明导数在情境中的含义,并在运动问题中用 $v$ 和 $a$ 的符号进行论证。对于相关变化率问题,需画图、建立各量之间的方程,并对时间求导。

FRQ 1EASY4.2 Motion4.2 运动No Calculator

A particle moves along the $x$-axis with velocity $v(t)=3t^{2}-12t+9$ for $t\ge 0$ (in m/s).一个质点沿 $x$ 轴运动,速度函数为 $v(t)=3t^{2}-12t+9$(单位:m/s),$t\ge 0$。

(a) Find all times $t\ge 0$ when the particle is at rest.求所有 $t\ge 0$ 时质点静止的时刻。
(b) On which interval(s) is the particle moving right?质点在哪个(些)区间上向右运动?
(c) At $t=2$, is the particle speeding up or slowing down? Justify using the signs of $v$ and $a$.在 $t=2$ 时,质点是在加速还是减速?用 $v$ 和 $a$ 的符号进行论证。
FRQ 2MEDIUM4.4 Related Rates4.4 相关变化率Calculator

A water trough is $10$ ft long with cross-section an isosceles triangle ($2$ ft wide at top, $2$ ft deep). Water fills the trough at $3$ ft³/min.一个水槽长 $10$ 英尺,横截面为等腰三角形(顶部宽 $2$ 英尺,深 $2$ 英尺)。水以 $3$ ft³/min 的速率注入水槽。

(a) Express $V$, the water volume, in terms of the depth $h$. Justify using similar triangles.将水的体积 $V$ 用水深 $h$ 表示,并用相似三角形进行推导。
(b) At the instant $h=1$ ft, find $\dfrac{dh}{dt}$. Include units.在 $h=1$ 英尺的瞬间,求 $\dfrac{dh}{dt}$,并标注单位。
(c) Is the water surface width increasing faster or slower when $h=1$ ft vs $h=1.5$ ft? Justify.当 $h=1$ 英尺与 $h=1.5$ 英尺时,水面宽度的增大速率哪个更快?请说明理由。
FRQ 3MEDIUM4.5 Linearization & Error4.5 线性化与误差No Calculator

Let $f(x)=\sqrt[3]{x}$.设 $f(x)=\sqrt[3]{x}$。

(a) Find the linearization $L(x)$ of $f$ at $x=8$.求 $f$ 在 $x=8$ 处的线性化 $L(x)$。
(b) Use $L$ to estimate $\sqrt[3]{8.6}$.用 $L$ 估算 $\sqrt[3]{8.6}$。
(c) Is the estimate an overestimate or underestimate? Justify using $f''$.该估算是高估还是低估?用 $f''$ 进行论证。
FRQ 4HARD4.3 / 4.4 Tabular Rates4.3 / 4.4 表格变化率Calculator

A tank holds $G(t)$ gallons of water at time $t$ minutes. Selected values of $G$:一个储罐在 $t$ 分钟时储有 $G(t)$ 加仑水。$G$ 的部分取值如下:

$t$ (min)$0$$2$$4$$6$$8$
$G(t)$ (gal)$120$$108$$88$$60$$24$
(a) Estimate $G'(4)$ using a symmetric difference quotient. Include units and interpret.用对称差商估算 $G'(4)$,标注单位并解释其含义。
(b) Must there be a time $t\in(0,8)$ for which $G'(t)=-12$? Justify using the MVT.在 $(0,8)$ 内是否必然存在某时刻 $t$ 使得 $G'(t)=-12$?用微分中值定理进行论证。 (MVT is formally Unit 5, Topic 5.1: see Q14/Q15 preview above.)(微分中值定理正式属于第五单元考点 5.1,参见上方 Q14/Q15 预习内容。)
(c) Use the linearization of $G$ at $t=4$ to estimate $G(4.5)$. Is the estimate likely an overestimate or underestimate, based on the concavity suggested by the table? Justify.用 $G$ 在 $t=4$ 处的线性化估算 $G(4.5)$。根据表格所提示的凹凸性,该估算是高估还是低估?请说明理由。
FRQ 5HARD4.7 L'Hôpital & Reasoning4.7 洛必达法则与推理No Calculator

Evaluate each limit. For each, state the indeterminate form before applying L'Hôpital's Rule, and justify each step.求下列各极限。对每一题,在应用洛必达法则前先说明不定式的类型,并对每一步进行论证。

(a) $\displaystyle\lim_{x\to 0}\dfrac{\tan x-x}{x^{3}}$
(b) $\displaystyle\lim_{x\to \infty}\dfrac{(\ln x)^{2}}{x}$
(c) $\displaystyle\lim_{x\to 0^{+}}x\ln x$
(d) Explain why L'Hôpital's Rule fails to evaluate $\displaystyle\lim_{x\to \infty}\dfrac{x+\sin x}{x}$ even though it is an $\tfrac{\infty}{\infty}$ form, then compute the limit directly.说明为何尽管 $\displaystyle\lim_{x\to \infty}\dfrac{x+\sin x}{x}$ 为 $\tfrac{\infty}{\infty}$ 型,洛必达法则却无法用于求该极限,然后直接计算该极限。