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Unit A5 · Calculus I第A5单元 · 微积分I

Implicit Differentiation and Related Rates隐函数求导与相关变化率

University-Style Practice Problems大学风格练习题

MEDIUM HARD CORE PROOF APPLIED

Sections 1 to 6: implicit differentiation, tangent lines, second derivatives, inverse-function derivatives, and related rates (geometry, motion, mixing)第 1 至 6 节：隐函数求导、切线、二阶导数、反函数导数及相关变化率（几何、运动、混合）CALC I

Name:姓名：Date:日期：
PART I  ·  CORE TECHNIQUES核心技巧Computational fluency · 28 marks计算熟练度 · 28分

Implicit Differentiation and Tangent Lines隐函数求导与切线

Differentiate both sides with respect to $x$, treating $y$ as a function of $x$ at every step. Collect $dy/dx$ terms, isolate, and simplify. Show every chain-rule application explicitly.对两边关于 $x$ 求导，每一步都将 $y$ 视为 $x$ 的函数。整理含 $dy/dx$ 的项，求解并化简。明确写出每次链式法则的应用。

Q1MEDIUM CORE implicit differentiation: find $dy/dx$隐函数求导：求 $dy/dx$ [6 marks]

For each curve, find $\dfrac{dy}{dx}$ by implicit differentiation. Simplify your answer.对每条曲线，用隐函数求导法求 $\dfrac{dy}{dx}$，并化简结果。

(a) $x^{3} + y^{3} = 6xy$ [2]
(b) $x^{2} + 4y^{2} - 2xy = 8$ [2]
(c) $\sin(xy) = x + y$ [2]
Q2MEDIUM CORE tangent line to an implicit curve隐式曲线的切线 [8 marks]

Consider the ellipse $\dfrac{x^{2}}{9} + \dfrac{y^{2}}{4} = 1$.考虑椭圆 $\dfrac{x^{2}}{9} + \dfrac{y^{2}}{4} = 1$。

(a) Use implicit differentiation to find $\dfrac{dy}{dx}$.用隐函数求导法求 $\dfrac{dy}{dx}$。 [2]
(b) Find the equation of the tangent line at the point $\left(\tfrac{3\sqrt{2}}{2},\ \sqrt{2}\right)$. Verify the point lies on the ellipse.求曲线在点 $\left(\tfrac{3\sqrt{2}}{2},\ \sqrt{2}\right)$ 处的切线方程，并验证该点在椭圆上。 [4]
(c) Find the coordinates of the two points on the ellipse where the tangent line is vertical.求椭圆上切线为竖直方向的两点坐标。 [2]
Q3HARD CORE second derivative by implicit differentiation隐函数求导法求二阶导数 [8 marks]

Consider the curve $x^{2} - xy + y^{2} = 3$.考虑曲线 $x^{2} - xy + y^{2} = 3$。

(a) Show that $\dfrac{dy}{dx} = \dfrac{2x - y}{x - 2y}$.证明 $\dfrac{dy}{dx} = \dfrac{2x - y}{x - 2y}$。 [3]
(b) Find $\dfrac{d^{2}y}{dx^{2}}$ in terms of $x$ and $y$. You may use the result of (a) without re-deriving it.用 $x$ 和 $y$ 表示 $\dfrac{d^{2}y}{dx^{2}}$。可直接使用 (a) 的结论，无需重新推导。 [4]
(c) Evaluate $\dfrac{d^{2}y}{dx^{2}}$ at the point $(1, 2)$, and state what this value tells you about the concavity of the curve there.在点 $(1, 2)$ 处计算 $\dfrac{d^{2}y}{dx^{2}}$，并说明该值关于曲线在此处凹凸性的含义。 [1]
Q4MEDIUM CORE tangent lines to the folium of Descartes笛卡尔叶形线的切线 [6 marks]

The folium of Descartes is the curve $x^{3} + y^{3} = 3xy$.笛卡尔叶形线的方程为 $x^{3} + y^{3} = 3xy$。

(a) Find $\dfrac{dy}{dx}$ by implicit differentiation.用隐函数求导法求 $\dfrac{dy}{dx}$。 [2]
(b) Find the equation of the tangent line at the point $({\tfrac{3}{2}},{\tfrac{3}{2}})$. Show first that the point is on the curve.求曲线在点 $({\tfrac{3}{2}},{\tfrac{3}{2}})$ 处的切线方程，先验证该点在曲线上。 [3]
(c) The curve passes through the origin. Explain, using the formula for $dy/dx$, why there is no single well-defined tangent slope at $(0,0)$.曲线过原点。利用 $dy/dx$ 的公式，解释为何在 $(0,0)$ 处没有唯一确定的切线斜率。 [1]
PART II  ·  DEFINITIONS AND PROOF定义与证明Rigorous arguments · 26 marks严格论证 · 26分

Inverse-Function Derivatives and Chain-Rule Justification反函数导数与链式法则的论证

In derivation problems, work from the definition: set up an implicit equation, differentiate both sides with respect to $x$, and solve. State the domain restriction that makes the inverse single-valued. Full marks require a completed argument, not just a quoted result.在推导题中，从定义出发：建立隐式方程，对两边关于 $x$ 求导，再求解。说明使反函数为单值函数的定义域限制。获得满分需要完整的论证过程，不能仅引用结论。

Q5HARD PROOF deriving $\tfrac{d}{dx}\arcsin x$ and $\tfrac{d}{dx}\arctan x$推导 $\tfrac{d}{dx}\arcsin x$ 与 $\tfrac{d}{dx}\arctan x$ [10 marks]

Use implicit differentiation to derive each formula from scratch. State the appropriate domain restriction in each case.用隐函数求导法从头推导每个公式，并在每种情况下说明相应的定义域限制。

(a) Starting from $y = \arcsin x$ (i.e. $\sin y = x$), prove that $\dfrac{d}{dx}\arcsin x = \dfrac{1}{\sqrt{1-x^{2}}}$.从 $y = \arcsin x$（即 $\sin y = x$）出发，证明 $\dfrac{d}{dx}\arcsin x = \dfrac{1}{\sqrt{1-x^{2}}}$。 [5]
(b) Starting from $y = \arctan x$ (i.e. $\tan y = x$), prove that $\dfrac{d}{dx}\arctan x = \dfrac{1}{1+x^{2}}$.从 $y = \arctan x$（即 $\tan y = x$）出发，证明 $\dfrac{d}{dx}\arctan x = \dfrac{1}{1+x^{2}}$。 [5]
Q6HARD PROOF related-rates chain-rule setup: justify the method相关变化率的链式法则建立：论证方法 [8 marks]

A spherical balloon is being inflated so that its volume increases at the constant rate $\dfrac{dV}{dt} = 3\ \text{cm}^{3}/\text{s}$.一个球形气球正在充气，其体积以恒定速率 $\dfrac{dV}{dt} = 3\ \text{cm}^{3}/\text{s}$ 增加。

(a) Write the geometric relation $V = \tfrac{4}{3}\pi r^{3}$ and differentiate both sides with respect to time $t$, justifying each step with the chain rule. Express $\dfrac{dr}{dt}$ in terms of $r$ and $\dfrac{dV}{dt}$.写出几何关系 $V = \tfrac{4}{3}\pi r^{3}$，对两边关于时间 $t$ 求导，用链式法则说明每一步，将 $\dfrac{dr}{dt}$ 用 $r$ 和 $\dfrac{dV}{dt}$ 表示。 [4]
(b) Find $\dfrac{dr}{dt}$ when $r = 5\ \text{cm}$. Give exact and decimal (to 3 s.f.) values and include units.求 $r = 5\ \text{cm}$ 时的 $\dfrac{dr}{dt}$，给出精确值及保留3位有效数字的小数值，并注明单位。 [2]
(c) A student argues: "When $r = 5$ I substitute $r = 5$ into $V = \tfrac{4}{3}\pi r^{3}$ first, then differentiate $V = \tfrac{500\pi}{3}$ with respect to $t$ to get $\tfrac{dV}{dt} = 0$." Identify the error in this reasoning and explain why the correct procedure must differentiate the relation before substituting the snapshot value.一位同学认为："当 $r = 5$ 时，先将 $r = 5$ 代入 $V = \tfrac{4}{3}\pi r^{3}$，再对 $V = \tfrac{500\pi}{3}$ 关于 $t$ 求导，得到 $\tfrac{dV}{dt} = 0$。"指出该推理的错误，并解释为何正确做法必须先对关系式求导，再代入瞬时值。 [2]
Q7HARD PROOF inverse function rule and a composite反函数法则与复合函数 [8 marks]

Let $f$ be a differentiable function with $f(2) = 5$ and $f'(2) = 3$. Let $g = f^{-1}$ be the inverse function, assumed to exist on a neighbourhood of $5$.设 $f$ 为可微函数，且 $f(2) = 5$，$f'(2) = 3$。设 $g = f^{-1}$ 为其反函数，假设其在 $5$ 的某邻域内存在。

(a) Starting from the identity $f(g(x)) = x$, differentiate both sides and derive the inverse-function rule $g'(x) = \dfrac{1}{f'(g(x))}$.从恒等式 $f(g(x)) = x$ 出发，对两边求导，推导反函数法则 $g'(x) = \dfrac{1}{f'(g(x))}$。 [3]
(b) Use the result of (a) to find $g'(5)$.利用 (a) 的结论求 $g'(5)$。 [2]
(c) Let $h(x) = \arctan(f(x))$. Using the chain rule and the given data, find $h'(2)$.设 $h(x) = \arctan(f(x))$。利用链式法则和已知数据，求 $h'(2)$。 [3]
PART III  ·  APPLICATIONS AND SYNTHESIS应用与综合Extended problems · 28 marks综合题 · 28分

Related Rates in Context情境中的相关变化率

In each problem: (i) assign variables with units; (ii) write the geometric or physical relation; (iii) differentiate with respect to $t$; (iv) substitute the snapshot values; (v) interpret the sign of the rate. Do not substitute a time-varying quantity before differentiating.每道题须：(i) 定义带单位的变量；(ii) 写出几何或物理关系式；(iii) 对 $t$ 求导；(iv) 代入瞬时值；(v) 解释变化率的符号。求导之前不得代入随时间变化的量。

Q8MEDIUM APPLIED ladder sliding down a wall梯子沿墙下滑 [8 marks]

A 10-metre ladder leans against a vertical wall. The foot of the ladder slides away from the wall along level ground at a constant rate of $0.5\ \text{m/s}$.一把10米长的梯子靠在竖直墙壁上。梯子底端沿水平地面以恒定速率 $0.5\ \text{m/s}$ 离开墙壁滑动。

(a) Let $x$ be the distance from the foot to the wall and $y$ the height of the top on the wall. Write the relation between $x$ and $y$ and differentiate implicitly with respect to $t$.设 $x$ 为底端到墙的距离，$y$ 为顶端在墙上的高度。写出 $x$ 与 $y$ 之间的关系，并对 $t$ 做隐函数求导。 [2]
(b) Find the rate at which the top of the ladder slides down the wall when the foot is $6\ \text{m}$ from the wall. Include units and interpret the sign.求底端距墙 $6\ \text{m}$ 时，梯子顶端沿墙下滑的速率。注明单位并解释符号的含义。 [3]
(c) Find the rate at which the area of the triangle formed by the ladder, the wall, and the ground is changing at the same instant.求同一时刻由梯子、墙壁和地面围成的三角形面积的变化速率。 [3]
Q9HARD APPLIED shadow length and related rates影长与相关变化率 [10 marks]

A street lamp is mounted at the top of a $6\ \text{m}$ pole. A person $1.8\ \text{m}$ tall walks away from the base of the pole along a straight path at $1.2\ \text{m/s}$.一盏路灯安装在 $6\ \text{m}$ 高的灯柱顶端。一名身高 $1.8\ \text{m}$ 的行人以 $1.2\ \text{m/s}$ 的速度沿直线路径远离灯柱底部行走。

(a) Let $x$ be the distance from the person to the base of the pole, and $s$ the length of the person's shadow. Use similar triangles to write a relation between $x$, $s$, and the lamp height.设 $x$ 为行人到灯柱底部的距离，$s$ 为行人影子的长度。利用相似三角形写出 $x$、$s$ 与灯高之间的关系。 [3]
(b) Differentiate your relation with respect to $t$ to find $\dfrac{ds}{dt}$ in terms of $\dfrac{dx}{dt}$.对关系式关于 $t$ 求导，用 $\dfrac{dx}{dt}$ 表示 $\dfrac{ds}{dt}$。 [3]
(c) Find the rate at which the tip of the shadow moves along the ground (i.e. the rate of change of $x + s$ with respect to $t$).求影子顶端沿地面移动的速率（即 $x + s$ 关于 $t$ 的变化率）。 [2]
(d) Explain why $\dfrac{ds}{dt}$ and $\dfrac{d(x+s)}{dt}$ are both constant in this problem, regardless of where the person currently is.解释为何在本题中，无论行人当前在何处，$\dfrac{ds}{dt}$ 和 $\dfrac{d(x+s)}{dt}$ 均为常数。 [2]
Q10HARD APPLIED water draining from an inverted cone倒置圆锥中的水流出问题 [10 marks]

A tank has the shape of an inverted right circular cone with height $12\ \text{m}$ and base radius $4\ \text{m}$ (the base is at the top). Water drains out of the tank at a rate of $2\ \text{m}^{3}/\text{min}$.一个水箱形状为倒置的直圆锥，高 $12\ \text{m}$，底面半径 $4\ \text{m}$（底面在顶部）。水以 $2\ \text{m}^{3}/\text{min}$ 的速率从水箱中流出。

(a) Let $h$ be the depth of water and $r$ the radius of the water surface. Write the similar-triangle relation between $r$ and $h$, and express the volume of water $V$ purely in terms of $h$.设 $h$ 为水深，$r$ 为水面半径。写出 $r$ 与 $h$ 之间的相似三角形关系，并将水的体积 $V$ 纯粹用 $h$ 表示。 [3]
(b) Differentiate $V$ with respect to $t$ and find an expression for $\dfrac{dh}{dt}$ in terms of $h$ and $\dfrac{dV}{dt}$.将 $V$ 对 $t$ 求导，求用 $h$ 和 $\dfrac{dV}{dt}$ 表示的 $\dfrac{dh}{dt}$ 的表达式。 [3]
(c) Find the rate at which the water level is falling when the depth is $h = 3\ \text{m}$. Give an exact value with units, and comment on how the rate changes as $h$ decreases.求水深 $h = 3\ \text{m}$ 时水位下降的速率，给出带单位的精确值，并说明随 $h$ 减小速率的变化趋势。 [2]
(d) At the instant when $h = 3\ \text{m}$, find the rate at which the radius of the water surface is decreasing.在 $h = 3\ \text{m}$ 的瞬间，求水面半径减小的速率。 [2]