Unit E4: Problem-Solving
Using Calculus单元 E4：微积分的应用
The application layer. Where E1, E2, and E3 build the differentiation and integration machinery, E4 deploys it on word problems: optimisation, kinematics, area, volumes of revolution, and related rates. Every section follows the same four-step pattern: read the geometric or physical setup, write the function or relation, differentiate or integrate, and translate the result back into the problem's language.本单元是应用层。E1、E2、E3 搭好了微分与积分的机器，E4 用它们解应用题：最优化、运动学、面积、旋转体体积、相关变化率。每节都遵循同一四步法：读懂几何或物理情境、写出函数或关系式、求导或积分、再把结果翻译回题目语言。
Optimisation最优化 SL 5.8
- Identify the quantity to optimise. "Maximise the volume" or "minimise the surface area". Call this $Q$.
- Write $Q$ as a function of a single variable. If the problem has more than one variable, find a constraint equation that lets you eliminate all but one.
- Differentiate. Compute $\tfrac{dQ}{dx}$.
- Solve $\tfrac{dQ}{dx} = 0$. The critical points are the candidates for the optimum.
- Verify the optimum. Use the second-derivative test ($Q'' < 0$ for max, $Q'' > 0$ for min) or a sign chart of $Q'$. Check the boundary of the feasible region too if it is bounded.
- 识别待优化的量。"使体积最大"或"使表面积最小"。记为 $Q$。
- 把 $Q$ 写成单变量函数。若题目含多变量，找一个约束方程消去其余变量。
- 求导。算 $\tfrac{dQ}{dx}$。
- 解 $\tfrac{dQ}{dx} = 0$。临界点即极值候选。
- 验证。用二阶导数检验（$Q'' < 0$ 为极大、$Q'' > 0$ 为极小）或一阶导符号表。若可行域有界，端点也要检查。
A rectangular sheet of card measures $24 \text{ cm}$ by $24 \text{ cm}$. Equal squares of side $x$ are cut from each corner and the sides folded up to make an open-top box. Find the value of $x$ that maximises the volume of the box, and state the maximum volume.一张 $24 \text{ cm} \times 24 \text{ cm}$ 的方形纸板，四角各裁掉边长为 $x$ 的小正方形，再把四边折起作成无盖纸盒。求使体积最大的 $x$，并写出最大体积。
Step 1. Quantity to optimise. Volume $V$ of the box.
第 1 步：待优化量。盒子的体积 $V$。
Step 2. Express $V$ as a function of $x$. The base is a square of side $24 - 2x$ and the height is $x$, so
第 2 步：把 $V$ 写成 $x$ 的函数。底面是边长 $24 - 2x$ 的正方形，高为 $x$，故
$$ V(x) \;=\; x \,(24 - 2x)^{2}, \qquad 0 < x < 12. $$Step 3. Differentiate. Use the product rule and chain rule:
第 3 步：求导。乘积法则加链式法则：
$$ V'(x) \;=\; (24 - 2x)^{2} + x \cdot 2 (24 - 2x) \cdot (-2) \;=\; (24 - 2x)\bigl[(24 - 2x) - 4x\bigr] \;=\; (24 - 2x)(24 - 6x). $$Step 4. Solve. $V'(x) = 0 \Rightarrow x = 12$ (rejected; gives zero volume) or $x = 4$.
第 4 步：求解。$V'(x) = 0 \Rightarrow x = 12$（舍去，体积为零）或 $x = 4$。
Step 5. Verify. Sign of $V'(x) = (24 - 2x)(24 - 6x)$ near $x = 4$: at $x = 3$, $V'(3) = 18 \cdot 6 = 108 > 0$; at $x = 5$, $V'(5) = 14 \cdot (-6) < 0$. So $V'$ changes from positive to negative at $x = 4$, confirming a local maximum.
第 5 步：验证。$V'(x) = (24 - 2x)(24 - 6x)$ 在 $x = 4$ 附近的符号：$x = 3$ 时 $V'(3) = 18 \cdot 6 = 108 > 0$；$x = 5$ 时 $V'(5) = 14 \cdot (-6) < 0$。符号由正变负，确为局部极大值。
Result. $V(4) = 4 \cdot 16^{2} = 4 \cdot 256 = 1024 \text{ cm}^{3}$. The maximum volume is $1024 \text{ cm}^{3}$, achieved when $x = 4 \text{ cm}$.
结果。$V(4) = 4 \cdot 16^{2} = 4 \cdot 256 = 1024 \text{ cm}^{3}$。最大体积为 $1024 \text{ cm}^{3}$，对应 $x = 4 \text{ cm}$。
A closed cylindrical can must hold $500\pi\;\text{cm}^3$. Find the radius that minimises the total surface area.一封闭圆柱罐必须容纳 $500\pi\;\text{cm}^3$。求使总表面积最小的半径。
Step 1 — Constraint eliminates one variable. $V=\pi r^2 h = 500\pi \Rightarrow h = 500/r^2$.
第 1 步 — 用约束消去一个变量。$V=\pi r^2 h = 500\pi \Rightarrow h = 500/r^2$。
Step 2 — Write S as a function of r alone.
第 2 步 — 把 S 写成只含 r 的函数。
$$ S = 2\pi r^2 + 2\pi r h = 2\pi r^2 + 2\pi r \cdot \frac{500}{r^2} = 2\pi r^2 + \frac{1000\pi}{r} $$Step 3 — Differentiate and solve $S'=0$.
第 3 步 — 求导并解 $S'=0$。
$$ S' = 4\pi r - \frac{1000\pi}{r^2} = 0 \;\Longrightarrow\; r^3 = 250 \;\Longrightarrow\; r = \sqrt[3]{250} \approx 6.30\;\text{cm} $$Step 4 — Verify with the second derivative. $S'' = 4\pi + \tfrac{2000\pi}{r^3} > 0$ always (for $r>0$), so this critical point is a minimum.
第 4 步 — 用二阶导数验证。$S'' = 4\pi + \tfrac{2000\pi}{r^3} > 0$ 恒成立（$r>0$ 时），故该临界点为极小值。
Kinematics by Differentiation运动学：由位置求速度与加速度 SL 5.9
- Displacement is signed: $s(t_{2}) - s(t_{1})$.
- Distance travelled over $[t_{1}, t_{2}]$ is $\int_{t_{1}}^{t_{2}} |v(t)| \, dt$ (which requires splitting where $v$ changes sign).
- Velocity is signed (direction matters); speed is $|v|$.
- At a change of direction, $v(t) = 0$. At a momentary rest, also $v(t) = 0$ but the particle may not change direction (need $v$ sign change to confirm).
- 位移（
displacement）是带符号的：$s(t_{2}) - s(t_{1})$。 - $[t_{1}, t_{2}]$ 上的路程（
distance travelled）是 $\int_{t_{1}}^{t_{2}} |v(t)| \, dt$（需在 $v$ 变号处分段）。 - 速度带符号（方向有意义）；速率是 $|v|$。
- 变向对应 $v(t) = 0$。瞬时静止也是 $v(t) = 0$，但未必变向（需用 $v$ 变号来判断）。
A particle moves along a straight line with position $s(t) = t^{3} - 6 t^{2} + 9 t + 1$ for $t \ge 0$, where $s$ is in metres and $t$ in seconds. Find the times at which the particle is at rest, and determine whether the particle changes direction at each such time.一质点沿直线运动，位置 $s(t) = t^{3} - 6 t^{2} + 9 t + 1$（$t \ge 0$，$s$ 以米计、$t$ 以秒计）。求质点瞬时静止的时刻，并判断每个时刻是否变向。
Velocity. $v(t) = s'(t) = 3 t^{2} - 12 t + 9 = 3 (t - 1)(t - 3)$.
速度。$v(t) = s'(t) = 3 t^{2} - 12 t + 9 = 3 (t - 1)(t - 3)$。
Times at rest. $v(t) = 0 \Rightarrow t = 1$ or $t = 3$.
静止时刻。$v(t) = 0 \Rightarrow t = 1$ 或 $t = 3$。
Direction change check. Sign of $v$:
变向检验。$v$ 的符号：
| $t$ | $0 \le t < 1$ | $1 < t < 3$ | $t > 3$ |
|---|---|---|---|
| $v(t)$ | $+$ | $-$ | $+$ |
Sign changes at both $t = 1$ ($+$ to $-$) and $t = 3$ ($-$ to $+$). The particle changes direction at both times.
$t = 1$（由 $+$ 转 $-$）与 $t = 3$（由 $-$ 转 $+$）两个时刻都变号。质点在两个时刻都变向。
Acceleration. $a(t) = v'(t) = 6t - 12$, so $a(1) = -6 \text{ m/s}^{2}$ and $a(3) = +6 \text{ m/s}^{2}$ at the two turning points.
加速度。$a(t) = v'(t) = 6t - 12$，故 $a(1) = -6 \text{ m/s}^{2}$、$a(3) = +6 \text{ m/s}^{2}$。
A particle has $v(t) = t^2 - 5t + 6$. Is the particle speeding up or slowing down at $t=1$?质点 $v(t) = t^2 - 5t + 6$。在 $t=1$ 时质点是在加速还是减速？
Key rule: a particle speeds up when $v$ and $a$ have the same sign, and slows down when they have opposite signs.
关键规则：$v$ 与 $a$ 同号时质点加速，异号时减速。
Evaluate v(1). $v(1) = 1 - 5 + 6 = 2 > 0$.
求 v(1)。$v(1) = 1 - 5 + 6 = 2 > 0$。
Evaluate a(1). $a(t) = v'(t) = 2t-5$, so $a(1) = 2-5 = -3 < 0$.
求 a(1)。$a(t) = v'(t) = 2t-5$，故 $a(1) = 2-5 = -3 < 0$。
Conclusion. $v(1)>0$ and $a(1)<0$ — opposite signs, so the particle is slowing down at $t=1$ (even though it's still moving forward).
结论。$v(1)>0$、$a(1)<0$——符号相反，故质点在 $t=1$ 时正在减速（尽管仍在向正方向运动）。
Kinematics by Integration运动学：由速度积分求位置 SL 5.9
Two definite-integral applications.
- Displacement from $t_{1}$ to $t_{2}$ is $\displaystyle\int_{t_{1}}^{t_{2}} v(t) \, dt$. Signed.
- Distance travelled from $t_{1}$ to $t_{2}$ is $\displaystyle\int_{t_{1}}^{t_{2}} |v(t)| \, dt$. Always non-negative. Split the integral wherever $v$ changes sign.
定积分的两类应用。
- 位移（$t_{1}$ 到 $t_{2}$）：$\displaystyle\int_{t_{1}}^{t_{2}} v(t) \, dt$。带符号。
- 路程（$t_{1}$ 到 $t_{2}$）：$\displaystyle\int_{t_{1}}^{t_{2}} |v(t)| \, dt$。非负。在 $v$ 变号处分段积分。
A particle has velocity $v(t) = t^{2} - 4 t + 3$ (m/s) for $0 \le t \le 4$. Find (a) the displacement and (b) the total distance travelled over $[0, 4]$.质点速度 $v(t) = t^{2} - 4 t + 3$（m/s），$0 \le t \le 4$。求 (a) $[0, 4]$ 上的位移；(b) 总路程。
Factor $v$. $v(t) = (t - 1)(t - 3)$, so $v$ changes sign at $t = 1$ and $t = 3$.
分解 $v$。$v(t) = (t - 1)(t - 3)$，故 $v$ 在 $t = 1$ 与 $t = 3$ 处变号。
(a) Displacement.
(a) 位移。
$$ \int_{0}^{4} (t^{2} - 4t + 3) \, dt \;=\; \Bigl[ \tfrac{t^{3}}{3} - 2 t^{2} + 3t \Bigr]_{0}^{4} \;=\; \tfrac{64}{3} - 32 + 12 \;=\; \tfrac{64}{3} - 20 \;=\; \tfrac{4}{3} \text{ m}. $$(b) Distance. Split at $t = 1$ and $t = 3$. On $[0, 1]$ and $[3, 4]$, $v \ge 0$; on $[1, 3]$, $v \le 0$.
(b) 路程。在 $t = 1$、$t = 3$ 处分段。$[0, 1]$ 与 $[3, 4]$ 上 $v \ge 0$，$[1, 3]$ 上 $v \le 0$。
$$ \int_{0}^{1} v \, dt \;=\; \tfrac{1}{3} - 2 + 3 \;=\; \tfrac{4}{3}, \qquad \int_{1}^{3} v \, dt \;=\; \Bigl( 9 - 18 + 9 \Bigr) - \Bigl( \tfrac{1}{3} - 2 + 3 \Bigr) \;=\; 0 - \tfrac{4}{3} \;=\; -\tfrac{4}{3}. $$ $$ \int_{3}^{4} v \, dt \;=\; \Bigl( \tfrac{64}{3} - 32 + 12 \Bigr) - \Bigl( 9 - 18 + 9 \Bigr) \;=\; \tfrac{4}{3} - 0 \;=\; \tfrac{4}{3}. $$Total distance $= \tfrac{4}{3} + \bigl|-\tfrac{4}{3}\bigr| + \tfrac{4}{3} = \tfrac{12}{3} = 4 \text{ m}$.
总路程 $= \tfrac{4}{3} + \bigl|-\tfrac{4}{3}\bigr| + \tfrac{4}{3} = \tfrac{12}{3} = 4 \text{ m}$。
Cross-check. Displacement ($\tfrac{4}{3}$ m) does not equal distance ($4$ m), confirming the particle reversed direction on the way.
互相检验。位移（$\tfrac{4}{3}$ m）$\ne$ 路程（$4$ m），可知质点中途变向。
A particle has $a(t) = 6t$, with $v(0) = 2$ and $s(0) = 1$. Find $s(t)$.质点 $a(t) = 6t$，$v(0) = 2$，$s(0) = 1$。求 $s(t)$。
Step 1 — Integrate a(t) to get v(t), using v(0) to find the constant.
第 1 步 — 积分 a(t) 得 v(t)，用 v(0) 求常数。
$$ v(t) = \int 6t\,dt = 3t^2 + C_1, \quad v(0)=2 \Rightarrow C_1=2 \Rightarrow v(t)=3t^2+2 $$Step 2 — Integrate v(t) to get s(t), using s(0) to find the constant.
第 2 步 — 积分 v(t) 得 s(t)，用 s(0) 求常数。
$$ s(t) = \int (3t^2+2)\,dt = t^3+2t+C_2, \quad s(0)=1 \Rightarrow C_2=1 $$ $$ s(t) = t^3 + 2t + 1 $$Two separate integration constants must each be pinned down by their own initial condition — mixing $v(0)$ into the position constant (or vice versa) is a common slip.
两个积分常数必须分别由各自的初始条件确定——把 $v(0)$ 错用到位置常数上（或反过来）是常见失误。
Area Between Curves曲线之间的面积 SL 5.11
If the two curves cross, find the crossing points (solve $f(x) = g(x)$) and split the integral. On each piece, identify which curve is on top and integrate "upper minus lower" for that piece. Sum (or sum the absolute values) to get total geometric area.
Integration with respect to $y$ (HL preview, used in E4.5 for volumes around the $y$-axis): $$ A \;=\; \int_{c}^{d} \bigl[ x_{\text{right}}(y) - x_{\text{left}}(y) \bigr] \, dy. $$ Useful when the curves are easier to write as $x = $ function of $y$, or when the region is bounded by horizontal lines.
若两曲线相交，先求交点（解 $f(x) = g(x)$），再分段积分。每段判断哪条在上，逐段"上减下"。各段相加（或取绝对值再加）得总几何面积。
对 $y$ 积分（HL 预习，E4.5 绕 $y$ 轴体积要用）： $$ A \;=\; \int_{c}^{d} \bigl[ x_{\text{right}}(y) - x_{\text{left}}(y) \bigr] \, dy. $$ 适用于曲线写为 $x = $（$y$ 的函数）更方便、或区域被水平直线界定时。
Find the area enclosed between $y = x^{2}$ and $y = 2x$.求 $y = x^{2}$ 与 $y = 2x$ 所围成的面积。
Intersection points. $x^{2} = 2x \Rightarrow x^{2} - 2x = 0 \Rightarrow x(x - 2) = 0$, so $x = 0$ and $x = 2$.
交点。$x^{2} = 2x \Rightarrow x(x - 2) = 0$，故 $x = 0$ 与 $x = 2$。
Identify the upper curve on $[0, 2]$. At $x = 1$: $y = x^{2} = 1$ vs $y = 2x = 2$, so the line $2x$ is above the parabola $x^{2}$.
判定 $[0, 2]$ 上的上方曲线。$x = 1$：$y = x^{2} = 1$，$y = 2x = 2$，故直线 $2x$ 在抛物线 $x^{2}$ 上方。
Integrate "upper minus lower".
"上减下"积分。
$$ A \;=\; \int_{0}^{2} (2x - x^{2}) \, dx \;=\; \Bigl[ x^{2} - \tfrac{x^{3}}{3} \Bigr]_{0}^{2} \;=\; 4 - \tfrac{8}{3} \;=\; \tfrac{4}{3}. $$Find the total area enclosed between $y = x^3$ and $y = x$ on $[-1, 1]$.求 $y = x^3$ 与 $y = x$ 在 $[-1, 1]$ 上所围的总面积。
Step 1 — Find all intersection points. $x^3=x \Rightarrow x^3-x=0 \Rightarrow x(x-1)(x+1)=0 \Rightarrow x=-1,0,1$.
第 1 步 — 求所有交点。$x^3=x \Rightarrow x(x-1)(x+1)=0 \Rightarrow x=-1,0,1$。
Step 2 — Check which curve is on top on each sub-interval. At $x=-0.5$: $x=-0.5$ and $x^3=-0.125$, so $x < x^3$ on $[-1,0]$ (the cubic is on top there). At $x=0.5$: $x^3=0.125$ and $x=0.5$, so $x^3 < x$ on $[0,1]$ (the line is on top there). The two curves swap which one is "upper" at $x=0$ — this is exactly why the integral must be split there.
第 2 步 — 检查每个子区间哪条曲线在上。$x=-0.5$ 时：$x=-0.5$，$x^3=-0.125$，故 $[-1,0]$ 上 $x < x^3$（此区间三次曲线在上）。$x=0.5$ 时：$x^3=0.125$，$x=0.5$，故 $[0,1]$ 上 $x^3 < x$（此区间直线在上）。两曲线在 $x=0$ 处互换"上下"关系——这正是必须在此分段的原因。
Step 3 — Integrate "upper $-$ lower" separately on each piece.
第 3 步 — 在每一段分别按"上减下"积分。
$$ \int_{-1}^{0}(x^3-x)\,dx = \left[\frac{x^4}{4}-\frac{x^2}{2}\right]_{-1}^{0} = 0 - \left(\frac14-\frac12\right) = \frac14 $$ $$ \int_{0}^{1}(x-x^3)\,dx = \left[\frac{x^2}{2}-\frac{x^4}{4}\right]_{0}^{1} = \left(\frac12-\frac14\right) - 0 = \frac14 $$Step 4 — Sum the two pieces.
第 4 步 — 两段相加。
$$ A = \frac14 + \frac14 = \frac12 $$Had you (incorrectly) integrated $(x - x^3)$ across the whole interval $[-1,1]$ without splitting, the two halves would cancel to give $0$ — a tell-tale sign that a split was needed but skipped, since a geometric area can never be zero.
若（错误地）不分段、直接对整个 $[-1,1]$ 积分 $(x - x^3)$，两半会相互抵消得到 $0$——这正是"该分段却漏分段"的警示信号，因为几何面积永远不可能为零。
Volumes of Revolution旋转体的体积 SL 5.11 · AHL 5.16
Geometric origin (the "disk method"). Slice the solid into thin disks of thickness $dx$ (or $dy$), each disk approximately a cylinder of radius $y$ (or $x$). Volume of one disk $\approx \pi y^{2} \, dx$. Summing over all disks and taking the limit gives the integral.
几何来源（"圆盘法"）。把立体切成厚度 $dx$（或 $dy$）的薄圆盘，每片近似为半径 $y$（或 $x$）的圆柱。单片体积 $\approx \pi y^{2} \, dx$。求和取极限即得积分。
The region bounded by $y = \sqrt{x}$, the $x$-axis, and the lines $x = 0$ and $x = 4$ is rotated $360^{\circ}$ about the $x$-axis. Find the volume of the resulting solid.由 $y = \sqrt{x}$、$x$ 轴及 $x = 0$、$x = 4$ 围成的区域绕 $x$ 轴旋转 $360^{\circ}$，求所得立体的体积。
Apply the formula.
套用公式。
$$ V \;=\; \pi \int_{0}^{4} (\sqrt{x})^{2} \, dx \;=\; \pi \int_{0}^{4} x \, dx \;=\; \pi \Bigl[ \tfrac{x^{2}}{2} \Bigr]_{0}^{4} \;=\; \pi \cdot 8 \;=\; 8 \pi. $$The volume is $8 \pi$ cubic units.
体积为 $8 \pi$ 立方单位。
The region bounded by $y = x^{2}$, the $y$-axis, and the line $y = 4$ is rotated $360^{\circ}$ about the $y$-axis. Find the volume.由 $y = x^{2}$、$y$ 轴及 $y = 4$ 围成的区域绕 $y$ 轴旋转 $360^{\circ}$。求体积。
Express $x$ in terms of $y$. $y = x^{2} \Rightarrow x = \sqrt{y}$ (taking the positive branch, since the region is in the first quadrant).
把 $x$ 表为 $y$ 的函数。$y = x^{2} \Rightarrow x = \sqrt{y}$（区域在第一象限，取正根）。
Identify the $y$-limits. The region runs from $y = 0$ to $y = 4$.
确定 $y$ 的积分限。$y$ 从 $0$ 到 $4$。
Apply the formula.
套用公式。
$$ V \;=\; \pi \int_{0}^{4} (\sqrt{y})^{2} \, dy \;=\; \pi \int_{0}^{4} y \, dy \;=\; \pi \Bigl[ \tfrac{y^{2}}{2} \Bigr]_{0}^{4} \;=\; 8 \pi. $$The volume is $8 \pi$ cubic units. Note that this happens to equal the $x$-axis volume in E4.5a; this coincidence is not general.
体积为 $8 \pi$ 立方单位。本题数值恰巧与 E4.5a 绕 $x$ 轴的体积一致，纯属巧合，并非一般规律。
The region between $y=x^2$ and $y=x$ (for $0\le x\le1$) is rotated about the $x$-axis. Find the volume.$y=x^2$ 与 $y=x$（$0\le x\le1$）之间的区域绕 $x$ 轴旋转，求体积。
Key idea — a washer, not a solid disk. Since the region is bounded above by $y=x$ (outer radius) and below by $y=x^2$ (inner radius), each slice is an annulus (ring), not a filled disk.
关键思想 —— 是圆环，不是实心圆盘。区域上方以 $y=x$ 为界（外半径），下方以 $y=x^2$ 为界（内半径），每片是环形（圆环），而非实心圆盘。
Subtract inner disk from outer disk, inside the integral.
在积分内部，外盘减内盘。
$$ V = \pi\int_0^1 \left[x^2 - (x^2)^2\right]dx = \pi\int_0^1 (x^2 - x^4)\,dx $$Integrate.
积分。
$$ V = \pi\left[\frac{x^3}{3}-\frac{x^5}{5}\right]_0^1 = \pi\left(\frac13-\frac15\right) = \frac{2\pi}{15} $$This "outer squared minus inner squared" pattern generalizes E4.5a's single-curve formula the same way area-between-curves (E4.4) generalizes area-under-one-curve.
这种"外半径平方减内半径平方"的模式，正如两曲线间面积（E4.4）推广单曲线下面积一样，推广了 E4.5a 单曲线的体积公式。
Related Rates相关变化率 HL AHL 5.13
Four-step method.
- Write down the relation connecting the quantities. For instance, $V = \tfrac{4}{3} \pi r^{3}$ for a sphere.
- Differentiate both sides with respect to $t$. Each variable picks up a $\tfrac{d \cdot}{dt}$ factor by the chain rule.
- Substitute the given rates and current values. Plug in numerical values after differentiating, not before. A rate that is given at a specific instant cannot be used until both sides are differentiated.
- Solve for the unknown rate.
四步法。
- 写出关系式。如球：$V = \tfrac{4}{3} \pi r^{3}$。
- 两边对 $t$ 求导。每个变量都按链式法则带上 $\tfrac{d \cdot}{dt}$ 因子。
- 代入给定变化率与当前值。必须在求导之后代入数值，不能之前代。某时刻的瞬时率必须留到两边求导完才能用。
- 解出待求变化率。
Air is pumped into a spherical balloon at a constant rate of $50 \text{ cm}^{3}/\text{s}$. Find the rate at which the radius is increasing at the moment when the radius is $5 \text{ cm}$.向球形气球以恒定速率 $50 \text{ cm}^{3}/\text{s}$ 充气。求气球半径为 $5 \text{ cm}$ 时半径的增长率。
Step 1. Relation. $V = \tfrac{4}{3} \pi r^{3}$.
第 1 步：关系式。$V = \tfrac{4}{3} \pi r^{3}$。
Step 2. Differentiate both sides with respect to $t$.
第 2 步：对 $t$ 求导。
$$ \frac{dV}{dt} \;=\; 4 \pi r^{2} \cdot \frac{dr}{dt}. $$Step 3. Substitute. Given $\tfrac{dV}{dt} = 50$ and $r = 5$:
第 3 步：代入。已知 $\tfrac{dV}{dt} = 50$、$r = 5$：
$$ 50 \;=\; 4 \pi (5)^{2} \cdot \frac{dr}{dt} \;=\; 100 \pi \cdot \frac{dr}{dt}. $$Step 4. Solve.
第 4 步：求解。
$$ \frac{dr}{dt} \;=\; \frac{50}{100 \pi} \;=\; \frac{1}{2 \pi} \;\approx\; 0.159 \text{ cm/s}. $$The radius is increasing at about $0.16 \text{ cm/s}$ when $r = 5 \text{ cm}$.
$r = 5 \text{ cm}$ 时半径约以 $0.16 \text{ cm/s}$ 增长。
A $5\;\text{m}$ ladder leans against a wall. The base slides away from the wall at $0.5\;\text{m/s}$. Find the rate at which the top slides down when the base is $3\;\text{m}$ from the wall.一架 $5\;\text{m}$ 长的梯子靠在墙上。梯脚以 $0.5\;\text{m/s}$ 的速度滑离墙面。求梯脚距墙 $3\;\text{m}$ 时梯顶下滑的速率。
Step 1 — Relation (Pythagoras, since the ladder length is fixed).
第 1 步 — 关系式（毕氏定理，梯长固定）。
$$ x^2 + y^2 = 25 $$Step 2 — Differentiate with respect to t.
第 2 步 — 对 t 求导。
$$ 2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0 $$Step 3 — Find y when x=3 (needed before substituting). $y = \sqrt{25-9} = 4$.
第 3 步 — 求 x=3 时的 y（代入前需要）。$y = \sqrt{25-9} = 4$。
Step 4 — Substitute all known values and solve.
第 4 步 — 代入所有已知值并求解。
$$ 2(3)(0.5) + 2(4)\frac{dy}{dt} = 0 \;\Longrightarrow\; \frac{dy}{dt} = -\frac{3}{8} = -0.375\;\text{m/s} $$The negative sign confirms $y$ (height) is decreasing — the top is sliding down, consistent with the base sliding away. Reporting $0.375$ m/s without the sign, or without stating "downward," loses the interpretation mark.
负号确认 $y$（高度）在减小——梯顶正在下滑，与梯脚滑离一致。若只报 $0.375$ m/s 而不带符号、不说明"向下"，会丢掉解释分。
Exam Strategy and Common Pitfalls考试策略与常见陷阱
- Draw the figure. The geometric setup is often half the marks; a labelled diagram earns the M1 even before any calculus.
- 画图。几何情境往往占一半分数；带标注的图能在动微积分之前就拿到 M1。
- Justify the maximum or minimum. Either second-derivative test or first-derivative sign chart. Stating "this is the max" without justification costs the A1.
- 正式验证极值。二阶导检验或一阶导符号表。光说"此为极大"无验证要丢 A1。
- Read the question for "displacement" vs "distance". The first wants the signed integral; the second wants the absolute-value integral with sign-change splitting.
- 看题分清"位移"与"路程"。前者是带符号积分，后者要带绝对值并在变号处分段。
- Initial conditions fix $C$. When integrating to find $s$ or $v$, write the $+ C$ explicitly and then use the initial value to solve for $C$.
- 初始条件定 $C$。积分求 $s$ 或 $v$ 时显式写 $+ C$，再用初值解出 $C$。
- $\pi$ goes outside the integral. $V = \pi \int y^{2} \, dx$. Forgetting the $\pi$ is the canonical volume-of-revolution slip.
- $\pi$ 写在积分号外。$V = \pi \int y^{2} \, dx$。漏写 $\pi$ 是绕轴体积题的标志性失误。
- For $y$-axis rotation, square the $x$-expression (the radius from the axis), not the $y$-expression. The radius is always perpendicular to the axis of rotation.
- 绕 $y$ 轴时平方的是 $x$ 表达式（到旋转轴的距离），不是 $y$ 表达式。"半径"总是与旋转轴垂直。
- For related rates, differentiate first, substitute second. Never the other way round.
- 相关变化率：先求导，再代值。顺序不可颠倒。
Flashcards闪卡
Unit E4 Practice Quiz单元 E4 练习测验
Readiness Checklist备考清单
Tick each item when you can do it cold, without notes, on your first attempt.
每一条都要"裸做"做对（不看笔记、一次过）才打勾。
- Set up an optimisation problem: identify $Q$, eliminate a variable using a constraint, then differentiate最优化建模：识别 $Q$、用约束消变量、求导
- Verify a maximum or minimum using the second-derivative test or a sign chart用二阶导检验或符号表正式验证极值
- Apply the kinematics chain $s \to v \to a$ by differentiation用求导走完运动学链 $s \to v \to a$
- Distinguish "changes direction" from "instantaneously at rest" using $v$ sign change用 $v$ 变号区分"变向"与"瞬时静止"
- Apply the kinematics chain $a \to v \to s$ by integration, using initial conditions to fix $C$用积分走完链 $a \to v \to s$，用初始条件定 $C$
- Compute displacement and distance, splitting at zeros of $v$ when computing distance算位移与路程，路程要在 $v$ 零点处分段
- Compute area between two curves using "upper minus lower" integration用"上减下"积分算两条曲线之间的面积
- Find crossing points of two curves and split the integration where needed求两曲线交点并在需要时分段积分
- Apply $V = \pi \int y^{2} \, dx$ for rotation about the $x$-axis绕 $x$ 轴旋转用 $V = \pi \int y^{2} \, dx$
- HL Apply $V = \pi \int x^{2} \, dy$ for rotation about the $y$-axis, solving $y = f(x)$ for $x$绕 $y$ 轴用 $V = \pi \int x^{2} \, dy$，先由 $y = f(x)$ 解出 $x$
- HL Solve a related-rates problem: write the relation, differentiate implicitly, then substitute解相关变化率题：写关系式、隐式求导、再代入数值
- HL Recognise that constants substituted before differentiation kill the rate; always differentiate first认清"先代值再求导"会丢掉变化率；务必先求导
IB Paper-Style PracticeIB 试卷风格练习
E4 Practice and Solutions are on the roadmap. They will ship under Practice Questions/Unit_E4_*.html with the bilingual built-in pattern.
E4 配套的 Practice 与 Solutions 已在排期，上线后位于 Practice Questions/Unit_E4_*.html，采用双语内嵌格式。
How to use this guide本指南使用说明
How to use this guide本指南使用说明
E4 is where calculus stops being an exercise in pattern matching and starts being a tool. Each section is organised by application type, not by technique. Pick the section that matches the question you face.E4 是微积分由"模式匹配练习"转为"工具"的阶段。各节按应用类型组织，不按技巧分。遇到题目，找对应小节即可。
Memorise the kinematics chain: position $s(t)$, velocity $v = s'$, acceleration $a = v'$. Memorise the volume of revolution formula $V = \pi \int y^{2} \, dx$. Practise one optimisation problem (a box with maximum volume from a fixed sheet of card) end to end.
背熟运动学链：位置 $s(t)$、速度 $v = s'$、加速度 $a = v'$。背熟旋转体体积公式 $V = \pi \int y^{2} \, dx$。完整做一道最优化题（如固定面积纸板折盒求最大体积）。
For optimisation, justify the maximum with the second-derivative test or a sign chart, not by inspection. For kinematics, master the distinction between displacement, distance, velocity, and speed. For volumes, do both $x$-axis ($\pi \int y^{2} dx$) and $y$-axis ($\pi \int x^{2} dy$) rotations.
最优化：用二阶导数检验或符号表正式验证极值，不要凭直观。运动学：分清位移与路程、速度与速率。体积：掌握绕 $x$ 轴（$\pi \int y^{2} dx$）与绕 $y$ 轴（$\pi \int x^{2} dy$）两种。