Companion to the University-Style Practice Set配套大学风格练习题集
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
Find $dy/dx$ by implicit differentiation: (a) $x^{3}+y^{3}=6xy$; (b) $x^{2}+4y^{2}-2xy=8$; (c) $\sin(xy)=x+y$.用隐函数求导法求 $dy/dx$：(a) $x^{3}+y^{3}=6xy$；(b) $x^{2}+4y^{2}-2xy=8$；(c) $\sin(xy)=x+y$。
Differentiate both sides with respect to $x$, applying the chain rule to $y^{3}$ and the product rule to $6xy$: (M1)对两边关于 $x$ 求导，对 $y^{3}$ 应用链式法则，对 $6xy$ 应用乘积法则：(M1)
$$ 3x^{2}+3y^{2}\frac{dy}{dx}=6y+6x\frac{dy}{dx}. $$Collect $dy/dx$ terms: $\dfrac{dy}{dx}(3y^{2}-6x)=6y-3x^{2}$. Divide through by $3$: (A1)整理含 $dy/dx$ 的项：$\dfrac{dy}{dx}(3y^{2}-6x)=6y-3x^{2}$，两边除以 $3$：(A1)
$$ \frac{dy}{dx}=\frac{2y-x^{2}}{y^{2}-2x}. $$Differentiate: $2x+8y\dfrac{dy}{dx}-2y-2x\dfrac{dy}{dx}=0$. (M1) Collect: $\dfrac{dy}{dx}(8y-2x)=2y-2x$. (A1)求导：$2x+8y\dfrac{dy}{dx}-2y-2x\dfrac{dy}{dx}=0$。(M1) 整理：$\dfrac{dy}{dx}(8y-2x)=2y-2x$。(A1)
$$ \frac{dy}{dx}=\frac{2(y-x)}{2(4y-x)}=\frac{y-x}{4y-x}. $$Differentiate the left side using the chain rule on $\sin(xy)$, with the product rule on $xy$: (M1)对左侧用链式法则处理 $\sin(xy)$，对 $xy$ 使用乘积法则求导：(M1)
$$ \cos(xy)\!\left(y+x\frac{dy}{dx}\right)=1+\frac{dy}{dx}. $$Expand and collect: $\dfrac{dy}{dx}(x\cos(xy)-1)=1-y\cos(xy)$. (A1)展开并整理：$\dfrac{dy}{dx}(x\cos(xy)-1)=1-y\cos(xy)$。(A1)
$$ \frac{dy}{dx}=\frac{1-y\cos(xy)}{x\cos(xy)-1}. $$For the ellipse $\dfrac{x^{2}}{9}+\dfrac{y^{2}}{4}=1$: (a) find $dy/dx$; (b) the tangent at $\bigl(\tfrac{3\sqrt{2}}{2},\,\sqrt{2}\bigr)$; (c) where the tangent is vertical.对于椭圆 $\dfrac{x^{2}}{9}+\dfrac{y^{2}}{4}=1$：(a) 求 $dy/dx$；(b) 在点 $\bigl(\tfrac{3\sqrt{2}}{2},\,\sqrt{2}\bigr)$ 处的切线；(c) 切线为竖直方向的点。
Differentiate: $\dfrac{2x}{9}+\dfrac{2y}{4}\dfrac{dy}{dx}=0$. (M1) Solving: (A1)求导：$\dfrac{2x}{9}+\dfrac{2y}{4}\dfrac{dy}{dx}=0$。(M1) 求解：(A1)
$$ \frac{dy}{dx}=-\frac{4x}{9y}. $$Verify: $\dfrac{(3\sqrt{2}/2)^{2}}{9}+\dfrac{(\sqrt{2})^{2}}{4}=\dfrac{9/2}{9}+\dfrac{2}{4}=\dfrac{1}{2}+\dfrac{1}{2}=1$. (M1)验证：$\dfrac{(3\sqrt{2}/2)^{2}}{9}+\dfrac{(\sqrt{2})^{2}}{4}=\dfrac{9/2}{9}+\dfrac{2}{4}=\dfrac{1}{2}+\dfrac{1}{2}=1$。(M1)
Slope at the point: $m=-\dfrac{4\cdot\tfrac{3\sqrt{2}}{2}}{9\cdot\sqrt{2}}=-\dfrac{6\sqrt{2}}{9\sqrt{2}}=-\dfrac{2}{3}$. (A1)该点处的斜率：$m=-\dfrac{4\cdot\tfrac{3\sqrt{2}}{2}}{9\cdot\sqrt{2}}=-\dfrac{6\sqrt{2}}{9\sqrt{2}}=-\dfrac{2}{3}$。(A1)
Point-slope form: $y-\sqrt{2}=-\tfrac{2}{3}\!\left(x-\tfrac{3\sqrt{2}}{2}\right)$. (M1)点斜式：$y-\sqrt{2}=-\tfrac{2}{3}\!\left(x-\tfrac{3\sqrt{2}}{2}\right)$。(M1)
Expand: $y=-\tfrac{2}{3}x+\tfrac{2}{3}\cdot\tfrac{3\sqrt{2}}{2}+\sqrt{2}=-\tfrac{2}{3}x+\sqrt{2}+\sqrt{2}$. Hence (A1)展开：$y=-\tfrac{2}{3}x+\tfrac{2}{3}\cdot\tfrac{3\sqrt{2}}{2}+\sqrt{2}=-\tfrac{2}{3}x+\sqrt{2}+\sqrt{2}$，因此 (A1)
$$ y=-\frac{2}{3}x+2\sqrt{2}. $$The tangent is vertical when $dy/dx$ is undefined, i.e. when $9y=0$ so $y=0$. (M1) Substituting into the ellipse: $\dfrac{x^{2}}{9}=1$, giving $x=\pm 3$. The vertical tangents are at $(\pm 3, 0)$. (A1)当 $dy/dx$ 无意义时切线为竖直方向，即 $9y=0$，故 $y=0$。(M1) 代入椭圆方程：$\dfrac{x^{2}}{9}=1$，得 $x=\pm 3$。竖直切线在 $(\pm 3, 0)$。(A1)
For $x^{2}-xy+y^{2}=3$: (a) show $dy/dx=\frac{2x-y}{x-2y}$; (b) find $d^{2}y/dx^{2}$; (c) evaluate it at $(1,2)$ and interpret.对于 $x^{2}-xy+y^{2}=3$：(a) 证明 $dy/dx=\frac{2x-y}{x-2y}$；(b) 求 $d^{2}y/dx^{2}$；(c) 在 $(1,2)$ 处求值并解释。
Differentiate both sides with respect to $x$, applying the product rule to $xy$: (M1)对两边关于 $x$ 求导，对 $xy$ 应用乘积法则：(M1)
$$ 2x - \left(y + x\frac{dy}{dx}\right) + 2y\frac{dy}{dx} = 0. $$Collect $dy/dx$ terms on the left: $\dfrac{dy}{dx}(-x+2y)=y-2x$. (M1) Multiply numerator and denominator by $-1$: (A1)将含 $dy/dx$ 的项移到左侧：$\dfrac{dy}{dx}(-x+2y)=y-2x$。(M1) 分子分母同乘 $-1$：(A1)
$$ \frac{dy}{dx}=\frac{y-2x}{2y-x}=\frac{2x-y}{x-2y}. \qquad\checkmark $$Write $y'=\dfrac{N}{D}$ where $N=2x-y$ and $D=x-2y$. Differentiate each using part (a): (M1)记 $y'=\dfrac{N}{D}$，其中 $N=2x-y$，$D=x-2y$。利用 (a) 对每项求导：(M1)
$$ N'=2-y'=2-\frac{2x-y}{x-2y}=\frac{2(x-2y)-(2x-y)}{x-2y}=\frac{-3y}{x-2y}, $$ $$ D'=1-2y'=1-\frac{2(2x-y)}{x-2y}=\frac{(x-2y)-2(2x-y)}{x-2y}=\frac{-3x}{x-2y}. $$Apply the quotient rule: $y''=\dfrac{N'D-ND'}{D^{2}}$. (M1) The numerator is:应用商式法则：$y''=\dfrac{N'D-ND'}{D^{2}}$。(M1) 分子为：
$$ N'D - ND' = \frac{-3y}{x-2y}\cdot(x-2y) - (2x-y)\cdot\frac{-3x}{x-2y} = -3y+\frac{3x(2x-y)}{x-2y}. $$Combine over the common denominator $(x-2y)$: (A1)通分为公分母 $(x-2y)$：(A1)
$$ = \frac{-3y(x-2y)+3x(2x-y)}{x-2y}=\frac{-3xy+6y^{2}+6x^{2}-3xy}{x-2y}=\frac{6(x^{2}-xy+y^{2})}{x-2y}. $$Since $x^{2}-xy+y^{2}=3$ on the curve, this equals $\dfrac{18}{x-2y}$. Dividing by $D^{2}=(x-2y)^{2}$: (A1)由于曲线上 $x^{2}-xy+y^{2}=3$，上式等于 $\dfrac{18}{x-2y}$，再除以 $D^{2}=(x-2y)^{2}$：(A1)
$$ \frac{d^{2}y}{dx^{2}}=\frac{18}{(x-2y)^{3}}. $$At $(1,2)$: $x-2y=1-4=-3$, so $\dfrac{d^{2}y}{dx^{2}}=\dfrac{18}{(-3)^{3}}=\dfrac{18}{-27}=-\dfrac{2}{3}$. (A1) Since $y''<0$, the curve is concave down at this point.在 $(1,2)$ 处：$x-2y=1-4=-3$，故 $\dfrac{d^{2}y}{dx^{2}}=\dfrac{18}{(-3)^{3}}=\dfrac{18}{-27}=-\dfrac{2}{3}$。(A1) 因为 $y''<0$，曲线在该点处向下凹。
For the folium $x^{3}+y^{3}=3xy$: (a) find $dy/dx$; (b) the tangent at $(\tfrac{3}{2},\tfrac{3}{2})$; (c) why the slope is undefined at $(0,0)$.对于叶形线 $x^{3}+y^{3}=3xy$：(a) 求 $dy/dx$；(b) 在点 $(\tfrac{3}{2},\tfrac{3}{2})$ 处的切线；(c) 解释为何在 $(0,0)$ 处斜率无定义。
Differentiate: $3x^{2}+3y^{2}\dfrac{dy}{dx}=3y+3x\dfrac{dy}{dx}$. (M1) Collect: $\dfrac{dy}{dx}(y^{2}-x)=y-x^{2}$. (A1)求导：$3x^{2}+3y^{2}\dfrac{dy}{dx}=3y+3x\dfrac{dy}{dx}$。(M1) 整理：$\dfrac{dy}{dx}(y^{2}-x)=y-x^{2}$。(A1)
$$ \frac{dy}{dx}=\frac{y-x^{2}}{y^{2}-x}. $$Check: $(\tfrac{3}{2})^{3}+(\tfrac{3}{2})^{3}=2\cdot\tfrac{27}{8}=\tfrac{27}{4}$ and $3\cdot\tfrac{3}{2}\cdot\tfrac{3}{2}=\tfrac{27}{4}$. (M1) The point is on the curve.验证：$(\tfrac{3}{2})^{3}+(\tfrac{3}{2})^{3}=2\cdot\tfrac{27}{8}=\tfrac{27}{4}$，且 $3\cdot\tfrac{3}{2}\cdot\tfrac{3}{2}=\tfrac{27}{4}$。(M1) 该点在曲线上。
Slope: $m=\dfrac{\tfrac{3}{2}-\tfrac{9}{4}}{\tfrac{9}{4}-\tfrac{3}{2}}=\dfrac{-3/4}{3/4}=-1$. (A1)斜率：$m=\dfrac{\tfrac{3}{2}-\tfrac{9}{4}}{\tfrac{9}{4}-\tfrac{3}{2}}=\dfrac{-3/4}{3/4}=-1$。(A1)
Tangent: $y-\tfrac{3}{2}=-(x-\tfrac{3}{2})$, so $y=-x+3$. (A1)切线：$y-\tfrac{3}{2}=-(x-\tfrac{3}{2})$，故 $y=-x+3$。(A1)
At $(0,0)$, the formula gives $\dfrac{0-0}{0-0}=\dfrac{0}{0}$, which is indeterminate. (R1) Geometrically, the folium is a self-intersecting curve at the origin, crossing itself with two distinct tangent directions, so no single slope can be assigned.在 $(0,0)$ 处，公式给出 $\dfrac{0-0}{0-0}=\dfrac{0}{0}$，为不定式。(R1) 从几何上看，叶形线在原点处自交，以两个不同的切线方向穿过该点，因此无法确定唯一的斜率。
Derive (a) $\dfrac{d}{dx}\arcsin x=\dfrac{1}{\sqrt{1-x^{2}}}$ and (b) $\dfrac{d}{dx}\arctan x=\dfrac{1}{1+x^{2}}$ from scratch via implicit differentiation.用隐函数求导法从头推导 (a) $\dfrac{d}{dx}\arcsin x=\dfrac{1}{\sqrt{1-x^{2}}}$ 和 (b) $\dfrac{d}{dx}\arctan x=\dfrac{1}{1+x^{2}}$。
Let $y=\arcsin x$, so $\sin y = x$ with $y\in\bigl[-\tfrac{\pi}{2},\tfrac{\pi}{2}\bigr]$ (the principal branch). (M1)设 $y=\arcsin x$，则 $\sin y = x$，其中 $y\in\bigl[-\tfrac{\pi}{2},\tfrac{\pi}{2}\bigr]$（主值分支）。(M1)
Differentiate both sides with respect to $x$: (M1)对两边关于 $x$ 求导：(M1)
$$ \cos y\,\frac{dy}{dx}=1 \implies \frac{dy}{dx}=\frac{1}{\cos y}. $$On the principal branch, $\cos y\ge 0$, so we take the positive square root. Using $\sin^{2}y+\cos^{2}y=1$ and $\sin y=x$: (R1)在主值分支上，$\cos y\ge 0$，因此取正平方根。利用 $\sin^{2}y+\cos^{2}y=1$ 以及 $\sin y=x$：(R1)
$$ \cos y=\sqrt{1-\sin^{2}y}=\sqrt{1-x^{2}}. $$Therefore (A1·A1):因此 (A1·A1)：
$$ \frac{d}{dx}\arcsin x=\frac{1}{\sqrt{1-x^{2}}}, \quad x\in(-1,1). $$Let $y=\arctan x$, so $\tan y = x$ with $y\in\bigl(-\tfrac{\pi}{2},\tfrac{\pi}{2}\bigr)$. (M1)设 $y=\arctan x$，则 $\tan y = x$，其中 $y\in\bigl(-\tfrac{\pi}{2},\tfrac{\pi}{2}\bigr)$。(M1)
Differentiate both sides with respect to $x$: (M1)对两边关于 $x$ 求导：(M1)
$$ \sec^{2}y\,\frac{dy}{dx}=1 \implies \frac{dy}{dx}=\cos^{2}y. $$Use the Pythagorean identity $1+\tan^{2}y=\sec^{2}y$, so $\cos^{2}y=\dfrac{1}{\sec^{2}y}=\dfrac{1}{1+\tan^{2}y}$. Since $\tan y=x$: (R1·A1)利用勾股恒等式 $1+\tan^{2}y=\sec^{2}y$，故 $\cos^{2}y=\dfrac{1}{\sec^{2}y}=\dfrac{1}{1+\tan^{2}y}$。因为 $\tan y=x$：(R1·A1)
$$ \frac{d}{dx}\arctan x=\frac{1}{1+x^{2}}, \quad x\in\mathbb{R}. $$(A1) The domain restriction is already satisfied since $\sec^2 y > 0$ everywhere on the principal branch.(A1) 由于主值分支上处处有 $\sec^2 y > 0$，定义域限制自然满足。
A spherical balloon inflates at $dV/dt=3\ \text{cm}^{3}/\text{s}$: (a) derive $dr/dt$ in terms of $r$ and $dV/dt$; (b) find $dr/dt$ when $r=5\ \text{cm}$; (c) identify the error in differentiating after substituting $r=5$.球形气球以 $dV/dt=3\ \text{cm}^{3}/\text{s}$ 的速率充气：(a) 用 $r$ 和 $dV/dt$ 表示 $dr/dt$；(b) 求 $r=5\ \text{cm}$ 时的 $dr/dt$；(c) 指出先代入 $r=5$ 再求导的错误所在。
The geometric relation is $V=\dfrac{4}{3}\pi r^{3}$, where both $V$ and $r$ are functions of time $t$. (M1)几何关系为 $V=\dfrac{4}{3}\pi r^{3}$，其中 $V$ 和 $r$ 均为时间 $t$ 的函数。(M1)
Differentiate both sides with respect to $t$, applying the chain rule to $r^{3}$: (M1)对两边关于 $t$ 求导，对 $r^{3}$ 应用链式法则：(M1)
$$ \frac{dV}{dt}=4\pi r^{2}\frac{dr}{dt}. $$(A1) Solving for $dr/dt$: (A1)(A1) 求解 $dr/dt$：(A1)
$$ \frac{dr}{dt}=\frac{1}{4\pi r^{2}}\,\frac{dV}{dt}. $$Substitute $r=5$ and $dV/dt=3$: (M1)代入 $r=5$ 和 $dV/dt=3$：(M1)
$$ \frac{dr}{dt}=\frac{3}{4\pi(25)}=\frac{3}{100\pi}\approx 0.00955\ \text{cm/s}. $$(A1) The positive sign confirms the radius is increasing.(A1) 正号说明半径正在增大。
The student replaces $r$ with the constant $5$ inside $V=\tfrac{4}{3}\pi r^{3}$ before differentiating, producing $V=\tfrac{500\pi}{3}$, a constant. (R1) Differentiating a constant with respect to $t$ yields $dV/dt=0$, which contradicts the given $dV/dt=3$. The flaw is that $r$ is not a constant; it is a time-varying quantity. The chain rule applies to the relation between variables, and only after differentiating may one substitute the snapshot value $r=5$ at a particular instant. (R1)该同学在求导之前将 $r$ 替换为常数 $5$，使 $V=\tfrac{4}{3}\pi r^{3}$ 变为常数 $V=\tfrac{500\pi}{3}$。(R1) 对常数关于 $t$ 求导得 $dV/dt=0$，与已知的 $dV/dt=3$ 矛盾。错误在于：$r$ 并非常数，而是随时间变化的量。链式法则作用于变量之间的关系，只有在求导之后才能代入 $r=5$ 这一特定瞬时值。(R1)
With $f(2)=5$, $f'(2)=3$, and $g=f^{-1}$: (a) derive the inverse-function rule; (b) find $g'(5)$; (c) find $h'(2)$ where $h(x)=\arctan(f(x))$.已知 $f(2)=5$，$f'(2)=3$，且 $g=f^{-1}$：(a) 推导反函数法则；(b) 求 $g'(5)$；(c) 求 $h'(2)$，其中 $h(x)=\arctan(f(x))$。
By definition of the inverse, $f(g(x))=x$ for all $x$ in the domain. (M1)由反函数的定义，对定义域内所有 $x$ 有 $f(g(x))=x$。(M1)
Differentiate both sides with respect to $x$ using the chain rule on the left: (M1)对两边关于 $x$ 求导，对左侧应用链式法则：(M1)
$$ f'(g(x))\cdot g'(x)=1. $$Divide: (A1)两边除以 $f'(g(x))$：(A1)
$$ g'(x)=\frac{1}{f'(g(x))}. $$Since $f(2)=5$ we have $g(5)=2$. (M1) Therefore:因为 $f(2)=5$，所以 $g(5)=2$。(M1) 因此：
$$ g'(5)=\frac{1}{f'(g(5))}=\frac{1}{f'(2)}=\frac{1}{3}. $$(A1)(A1)
Apply the chain rule to $h(x)=\arctan(f(x))$: (M1)对 $h(x)=\arctan(f(x))$ 应用链式法则：(M1)
$$ h'(x)=\frac{1}{1+(f(x))^{2}}\cdot f'(x). $$Substitute $x=2$, using $f(2)=5$ and $f'(2)=3$: (M1)代入 $x=2$，利用 $f(2)=5$ 和 $f'(2)=3$：(M1)
$$ h'(2)=\frac{f'(2)}{1+(f(2))^{2}}=\frac{3}{1+25}=\frac{3}{26}. $$(A1)(A1)
A 10 m ladder leans against a wall; the foot slides out at $0.5\ \text{m/s}$. Find: (a) the differentiated Pythagorean relation; (b) the rate the top slides down when the foot is 6 m out; (c) the rate of change of the triangle's area at the same instant.一把10米长的梯子靠在墙上，底端以 $0.5\ \text{m/s}$ 的速率滑出。求：(a) 对勾股关系式求导的结果；(b) 底端距墙6米时顶端下滑的速率；(c) 同一时刻三角形面积的变化速率。
Let $x$ be the distance from the foot to the wall and $y$ the height of the top; the ladder has constant length 10 m, so (M1)设 $x$ 为底端到墙的距离，$y$ 为顶端的高度；梯子长度恒为10米，因此 (M1)
$$ x^{2}+y^{2}=100. $$Differentiate both sides with respect to $t$: (A1)对两边关于 $t$ 求导：(A1)
$$ 2x\frac{dx}{dt}+2y\frac{dy}{dt}=0 \implies x\frac{dx}{dt}+y\frac{dy}{dt}=0. $$When $x=6$: $y=\sqrt{100-36}=8\ \text{m}$. (M1) Given $dx/dt=0.5\ \text{m/s}$, solve for $dy/dt$: (A1)当 $x=6$ 时：$y=\sqrt{100-36}=8\ \text{m}$。(M1) 已知 $dx/dt=0.5\ \text{m/s}$，求解 $dy/dt$：(A1)
$$ \frac{dy}{dt}=-\frac{x}{y}\,\frac{dx}{dt}=-\frac{6}{8}\cdot 0.5=-\frac{3}{8}\ \text{m/s}. $$(A1) The negative sign confirms the top is sliding downward.(A1) 负号说明顶端正在向下滑动。
The right triangle has area $A=\tfrac{1}{2}xy$. Differentiate with respect to $t$ using the product rule: (M1)直角三角形的面积为 $A=\tfrac{1}{2}xy$，对 $t$ 用乘积法则求导：(M1)
$$ \frac{dA}{dt}=\frac{1}{2}\!\left(y\frac{dx}{dt}+x\frac{dy}{dt}\right). $$Substitute $x=6$, $y=8$, $dx/dt=0.5$, $dy/dt=-3/8$: (M1)代入 $x=6$，$y=8$，$dx/dt=0.5$，$dy/dt=-3/8$：(M1)
$$ \frac{dA}{dt}=\frac{1}{2}\!\left(8\cdot 0.5+6\cdot\!\left(-\frac{3}{8}\right)\right)=\frac{1}{2}\!\left(4-\frac{9}{4}\right)=\frac{1}{2}\cdot\frac{7}{4}=\frac{7}{8}\ \text{m}^{2}/\text{s}. $$(A1) The area is increasing at this instant because the foot moves out faster than the top falls.(A1) 此时面积在增大，因为底端向外移动的速度快于顶端下落的速度。
A 6 m lamp pole casts the shadow of a 1.8 m person walking at 1.2 m/s. Find: (a) the similar-triangle relation; (b) $ds/dt$; (c) the speed of the shadow tip; (d) why both rates are constant.6米高的灯柱将一名身高1.8米、以1.2米/秒行走的行人的影子投在地面。求：(a) 相似三角形关系；(b) $ds/dt$；(c) 影子顶端的移动速度；(d) 为何两个速率均为常数。
Let $x$ be the person's distance from the base of the pole and $s$ the shadow length. The lamp at height $H=6\ \text{m}$, the person's top at height $h=1.8\ \text{m}$, and the tip of the shadow on the ground form two similar right triangles: (M1)设 $x$ 为行人到灯柱底部的距离，$s$ 为影子的长度。灯的高度 $H=6\ \text{m}$，行人头顶高度 $h=1.8\ \text{m}$，影子顶端在地面，构成两个相似直角三角形：(M1)
$$ \frac{H}{x+s}=\frac{h}{s} \implies Hs=h(x+s) \implies 6s=1.8(x+s). $$(M1) Expand and collect: $6s-1.8s=1.8x$, so $4.2s=1.8x$. Dividing: (A1)(M1) 展开整理：$6s-1.8s=1.8x$，故 $4.2s=1.8x$，两边相除：(A1)
$$ s=\frac{1.8}{4.2}x=\frac{3}{7}x. $$Since $s=\tfrac{3}{7}x$ is a constant multiple of $x$, differentiate directly: (M1)由于 $s=\tfrac{3}{7}x$ 是 $x$ 的常数倍，直接求导：(M1)
$$ \frac{ds}{dt}=\frac{3}{7}\,\frac{dx}{dt}=\frac{3}{7}\cdot 1.2=\frac{3.6}{7}=\frac{18}{35}\ \text{m/s}. $$(A1) The shadow grows at approximately $0.514\ \text{m/s}$. (A1)(A1) 影子以约 $0.514\ \text{m/s}$ 的速率增长。(A1)
The tip of the shadow is at position $x+s$ from the base of the pole. (M1)影子顶端距灯柱底部的位置为 $x+s$。(M1)
$$ \frac{d(x+s)}{dt}=\frac{dx}{dt}+\frac{ds}{dt}=1.2+\frac{18}{35}=\frac{42}{35}+\frac{18}{35}=\frac{60}{35}=\frac{12}{7}\ \text{m/s}. $$(A1)(A1)
The similar-triangle relation gives $s=\tfrac{3}{7}x$, which is linear in $x$. (R1) Because $dx/dt$ is constant and $s$ is a constant multiple of $x$, both $ds/dt=\tfrac{3}{7}\,dx/dt$ and $d(x+s)/dt=\tfrac{10}{7}\,dx/dt$ are constant, independent of the person's current position. (R1)相似三角形关系给出 $s=\tfrac{3}{7}x$，这是关于 $x$ 的线性关系。(R1) 因为 $dx/dt$ 为常数，而 $s$ 是 $x$ 的常数倍，所以 $ds/dt=\tfrac{3}{7}\,dx/dt$ 和 $d(x+s)/dt=\tfrac{10}{7}\,dx/dt$ 均为常数，与行人当前位置无关。(R1)
An inverted cone (height 12 m, top radius 4 m) drains at $2\ \text{m}^{3}/\text{min}$. Find: (a) $V$ in terms of $h$ only; (b) $dh/dt$ in terms of $h$; (c) $dh/dt$ when $h=3\ \text{m}$; (d) $dr/dt$ at the same instant.倒置圆锥形水箱（高12米，顶部半径4米）以 $2\ \text{m}^{3}/\text{min}$ 的速率排水。求：(a) 仅用 $h$ 表示的 $V$；(b) 用 $h$ 表示的 $dh/dt$；(c) $h=3\ \text{m}$ 时的 $dh/dt$；(d) 同一时刻的 $dr/dt$。
The full cone has height $H=12\ \text{m}$ and top radius $R=4\ \text{m}$. By similar triangles the water surface radius $r$ and depth $h$ satisfy: (M1)完整圆锥高 $H=12\ \text{m}$，顶部半径 $R=4\ \text{m}$。由相似三角形，水面半径 $r$ 与深度 $h$ 满足：(M1)
$$ \frac{r}{h}=\frac{R}{H}=\frac{4}{12}=\frac{1}{3} \implies r=\frac{h}{3}. $$Substitute into the cone-volume formula $V=\tfrac{1}{3}\pi r^{2}h$: (M1)代入圆锥体积公式 $V=\tfrac{1}{3}\pi r^{2}h$：(M1)
$$ V=\frac{1}{3}\pi\!\left(\frac{h}{3}\right)^{2}\!h=\frac{\pi h^{3}}{27}. $$(A1)(A1)
Differentiate $V=\dfrac{\pi h^{3}}{27}$ with respect to $t$: (M1)对 $V=\dfrac{\pi h^{3}}{27}$ 关于 $t$ 求导：(M1)
$$ \frac{dV}{dt}=\frac{\pi h^{2}}{9}\,\frac{dh}{dt}. $$(A1) Since water drains out, $dV/dt=-2\ \text{m}^{3}/\text{min}$. Solve: (A1)(A1) 由于水在流出，$dV/dt=-2\ \text{m}^{3}/\text{min}$，求解：(A1)
$$ \frac{dh}{dt}=\frac{9}{\pi h^{2}}\,\frac{dV}{dt}=\frac{9}{\pi h^{2}}\cdot(-2)=\frac{-18}{\pi h^{2}}. $$Substitute $h=3$: (M1)代入 $h=3$：(M1)
$$ \frac{dh}{dt}=\frac{-18}{\pi(9)}=-\frac{2}{\pi}\ \text{m/min}\approx -0.637\ \text{m/min}. $$(A1) As $h$ decreases further, $h^{2}$ shrinks, so $|dh/dt|$ increases: the water level falls ever faster as the tank empties.(A1) 随着 $h$ 继续减小，$h^{2}$ 也缩小，因此 $|dh/dt|$ 增大：水箱越空，水位下降越快。
Since $r=\tfrac{h}{3}$, differentiate: (M1)由于 $r=\tfrac{h}{3}$，对其求导：(M1)
$$ \frac{dr}{dt}=\frac{1}{3}\,\frac{dh}{dt}=\frac{1}{3}\cdot\!\left(-\frac{2}{\pi}\right)=-\frac{2}{3\pi}\ \text{m/min}. $$(A1) The water-surface radius also shrinks, as expected.(A1) 水面半径也在减小，符合预期。