IB Math AA HL · Unit E4: Problem-Solving Using Calculus · Solutions

Syllabus SL 5.8, 5.9, 5.11 · AHL 5.13, 5.16考纲 SL 5.8、5.9、5.11 · AHL 5.13、5.16AA HL

Section A · Worked SolutionsA 节 · 详细解析

Q1EASYPaper 1ASL 5.8 Optimisation (Constraint Substitution)[4 marks]

Rectangular pen against a wall; $40$ m of fencing on three sides. Express $A(w)$ and maximise.紧靠墙的矩形围栏；三边用 $40$ m 围栏。把 $A(w)$ 表达出来并求最大值。

Answers:答案： (a) $A(w) = w(40 - 2w) = 40w - 2w^{2}$ · (b) $w = 10\text{ m},\; A_{\max} = 200\text{ m}^{2}$

(a) Eliminate $\ell$ via the fencing constraint M1·A1

Three sides of fencing: $2w + \ell = 40$, so $\ell = 40 - 2w$. Then $$ A(w) \;=\; w \cdot \ell \;=\; w(40 - 2w) \;=\; 40w - 2w^{2},\qquad 0 < w < 20. $$

(b) Stationary point and second-derivative test M1·A1

$A'(w) = 40 - 4w$. Set $A'(w) = 0 \Rightarrow w = 10$. $A''(w) = -4 < 0$, so $w = 10$ is a maximum. Maximum area: $A(10) = 10 \cdot (40 - 20) = 200$ m$^{2}$ (with $\ell = 20$ m).

Why three sides, not four. The single-most missed mark on pen-against-wall problems is forgetting that the wall replaces one side. If you set up $2w + 2\ell = 40$, you halve the fencing budget and lose the entire question. Read once: which sides are fenced? Then mark them on a tiny sketch before writing any equation. Also note the endpoints $w = 0$ and $w = 20$ are excluded (degenerate pens), so the maximum on the open interval is interior, justifying the $f''$ test.

Pattern. For "perimeter $P$ of three sides against a wall", the area as a function of the perpendicular side $w$ is $A(w) = w(P - 2w)$, a downward parabola with vertex at $w = P/4$ and maximum area $P^{2}/8$. With $P = 40$: $w = 10$ and $A_{\max} = 200$. Recognise the family in five seconds; only do the calculus to earn the M1/A1.

(a) 用围栏约束消去 $\ell$ M1·A1

三边围栏：$2w + \ell = 40$，故 $\ell = 40 - 2w$。则 $$ A(w) \;=\; w \cdot \ell \;=\; w(40 - 2w) \;=\; 40w - 2w^{2},\qquad 0 < w < 20。 $$

(b) 驻点与二阶导检验 M1·A1

$A'(w) = 40 - 4w$。令 $A'(w) = 0 \Rightarrow w = 10$。$A''(w) = -4 < 0$，故 $w = 10$ 为最大值。最大面积：$A(10) = 10 \cdot (40 - 20) = 200$ m$^{2}$（$\ell = 20$ m）。

三边、而非四边。紧靠墙的围栏题最常丢的分就是忘了墙替代了一条边。若列成 $2w + 2\ell = 40$，预算被减半，整题作废。读题先问：哪几条边围栏？在小草图上标出后再写方程。其次端点 $w = 0$ 与 $w = 20$ 排除（退化围栏），故开区间上最大值在内部，$f''$ 检验合理。

模板。"周长 $P$、三边贴墙"问题中，面积关于垂直边 $w$ 为 $A(w) = w(P - 2w)$，开口向下，顶点 $w = P/4$、最大面积 $P^{2}/8$。$P = 40$：$w = 10$、$A_{\max} = 200$。五秒识别此族；只为拿 M1/A1 走完导数即可。

Q2MEDIUMPaper 1ASL 5.9 Kinematics (Times at Rest)[5 marks]

$s(t) = t^{3} - 6t^{2} + 9t + 1$. Find rest times, $a(t)$ at $t = 2$, displacement at rest.$s(t) = t^{3} - 6t^{2} + 9t + 1$。求静止时刻、$t = 2$ 处 $a(t)$、静止瞬刻的位移。

Answers:答案： (a) $v(t) = 3t^{2} - 12t + 9$, rest at $t = 1, 3$ · (b) $a(2) = 0$, neither speeding up nor slowing down at that instant · (c) $s(1) = 5,\; s(3) = 1$

(a) Velocity and rest instants M1·A1

$v(t) = s'(t) = 3t^{2} - 12t + 9 = 3(t^{2} - 4t + 3) = 3(t - 1)(t - 3)$. Set $v = 0$: $t = 1$ or $t = 3$.

(b) Acceleration and speeding-up test at $t = 2$ M1·A1

$a(t) = v'(t) = 6t - 12$, so $a(2) = 0$ and $v(2) = 3(1)(-1) = -3 < 0$. With $a = 0$, neither speeding up nor slowing down at that single instant (it is an inflection of $s$). For $t$ slightly less than $2$, $a < 0$ and $v < 0$ so $|v|$ is increasing (speeding up); for $t$ slightly greater than $2$, $a > 0$ and $v < 0$, so $|v|$ is decreasing (slowing down). $t = 2$ is the transition.

(c) Displacement at rest instants A1

$s(1) = 1 - 6 + 9 + 1 = 5$, and $s(3) = 27 - 54 + 27 + 1 = 1$.

Speeding up $\ne$ accelerating. A particle is speeding up when $v$ and $a$ have the same sign (both positive or both negative), and slowing down when they have opposite signs. Most candidates confuse "$a > 0$" with "speeding up": false in general. The cleanest test is to compute $\dfrac{d}{dt}(v^{2}) = 2v \cdot a$, which is positive iff $v$ and $a$ agree in sign. Here at $t = 2$, $v < 0$ and $a = 0$, so $v \cdot a = 0$: the instant is the precise transition. Examiners reward the sign analysis explicitly; "the particle is speeding up because $a > 0$" loses the R1.

Bonus diagnostic. Speed (i.e. $|v|$) is the quantity reported by a car's speedometer; velocity is signed. A common one-line interpretation: a particle speeds up when its kinetic-energy proxy $v^{2}$ is increasing, which by the chain rule happens iff $v$ and $a$ have the same sign. This view also explains the $t = 1$ and $t = 3$ rest instants here: $v$ changes sign through zero, so $|v|$ momentarily touches zero (full stop), then increases on the other side.

(a) 速度与静止时刻 M1·A1

$v(t) = s'(t) = 3t^{2} - 12t + 9 = 3(t - 1)(t - 3)$。令 $v = 0$：$t = 1$ 或 $t = 3$。

(b) $t = 2$ 处加速度与加速/减速判定 M1·A1

$a(t) = v'(t) = 6t - 12$，故 $a(2) = 0$，$v(2) = 3(1)(-1) = -3 < 0$。$a = 0$ 此瞬刻既不加速也不减速（$s$ 的拐点）。$t < 2$ 邻域：$a < 0$、$v < 0$，$|v|$ 增（加速）；$t > 2$ 邻域：$a > 0$、$v < 0$，$|v|$ 减（减速）。$t = 2$ 为过渡。

(c) 静止瞬刻的位移 A1

$s(1) = 1 - 6 + 9 + 1 = 5$，$s(3) = 27 - 54 + 27 + 1 = 1$。

"加速"与"加速度为正"不是一回事。质点加速（即速率增加）当且仅当 $v$ 与 $a$ 同号；减速即两者异号。多数考生把"$a > 0$"等同于"加速"，一般不对。最干净的检验是 $\dfrac{d}{dt}(v^{2}) = 2v \cdot a$：正值即同号。本题 $t = 2$ 处 $v < 0$、$a = 0$，$v \cdot a = 0$：恰是过渡瞬刻。评卷只奖励显式的符号分析；"$a > 0$ 故加速"会丢 R1。

附加诊断。速率（$|v|$）是车速表读数；速度带符号。一句话直觉：质点加速当且仅当动能代理 $v^{2}$ 增加，链式法则即 $v$ 与 $a$ 同号。本题 $t = 1$ 与 $t = 3$ 静止瞬刻也由此解释：$v$ 过零变号，$|v|$ 触零（完全停止），之后再从另一侧增加。

Q3MEDIUMPaper 1ASL 5.9 Displacement vs Distance[6 marks]

$v(t) = t^{2} - 4t + 3$ on $[0, 4]$. Factor, find displacement and total distance.$v(t) = t^{2} - 4t + 3$，$[0, 4]$。因式分解，求位移与总路程。

Answers:答案： (a) $v(t) = (t - 1)(t - 3),\; v = 0$ at $t = 1, 3$ · (b) displacement $= \dfrac{4}{3}$ m · (c) total distance $= 4$ m

(a) Factor and zeros on $[0, 4]$ A1

$v(t) = t^{2} - 4t + 3 = (t - 1)(t - 3)$. So $v = 0$ at $t = 1$ and $t = 3$.

(b) Displacement on $[0, 4]$ M1·A1

$$ \int_{0}^{4} (t^{2} - 4t + 3)\,dt \;=\; \left[\tfrac{t^{3}}{3} - 2t^{2} + 3t\right]_{0}^{4} \;=\; \tfrac{64}{3} - 32 + 12 \;=\; \tfrac{64}{3} - 20 \;=\; \tfrac{4}{3}. $$ So net displacement $= \tfrac{4}{3}$ m.

(c) Total distance with sign-split M1·A1·A1

The factored form $(t - 1)(t - 3)$ shows: $v > 0$ on $[0, 1)$, $v < 0$ on $(1, 3)$, $v > 0$ on $(3, 4]$. Split: $$ \int_{0}^{4} |v|\,dt \;=\; \int_{0}^{1} v\,dt - \int_{1}^{3} v\,dt + \int_{3}^{4} v\,dt. $$ Antiderivative $F(t) = \tfrac{t^{3}}{3} - 2t^{2} + 3t$. Values: $F(0) = 0$, $F(1) = \tfrac{1}{3} - 2 + 3 = \tfrac{4}{3}$, $F(3) = 9 - 18 + 9 = 0$, $F(4) = \tfrac{64}{3} - 32 + 12 = \tfrac{4}{3}$. $$ \int_{0}^{1} v\,dt = \tfrac{4}{3},\quad \int_{1}^{3} v\,dt = 0 - \tfrac{4}{3} = -\tfrac{4}{3},\quad \int_{3}^{4} v\,dt = \tfrac{4}{3} - 0 = \tfrac{4}{3}. $$ Total distance $= \tfrac{4}{3} + \tfrac{4}{3} + \tfrac{4}{3} = 4$ m.

Displacement vs distance: the one-line rule. Displacement $= \int v\,dt$ (sign matters). Distance $= \int |v|\,dt$ (always non-negative). The two coincide only when $v$ does not change sign on the interval. The classic trap: students compute $\int v\,dt$ and quote it as "distance travelled". Here the displacement is $\tfrac{4}{3}$ m but the actual distance walked is $4$ m, three times larger, because the particle goes out, comes back, and goes out again. Always factor $v$ first, locate the sign changes inside $[a, b]$, then split.

(a) 因式分解与 $[0, 4]$ 上的零点 A1

$v(t) = t^{2} - 4t + 3 = (t - 1)(t - 3)$。$v = 0$ 在 $t = 1$ 和 $t = 3$。

(b) $[0, 4]$ 上的位移 M1·A1

$$ \int_{0}^{4} (t^{2} - 4t + 3)\,dt \;=\; \left[\tfrac{t^{3}}{3} - 2t^{2} + 3t\right]_{0}^{4} \;=\; \tfrac{64}{3} - 32 + 12 \;=\; \tfrac{4}{3}。 $$ 故净位移 $= \tfrac{4}{3}$ m。

(c) 分段处理总路程 M1·A1·A1

因式 $(t - 1)(t - 3)$ 表明：$v > 0$ 在 $[0, 1)$，$v < 0$ 在 $(1, 3)$，$v > 0$ 在 $(3, 4]$。分段： $$ \int_{0}^{4} |v|\,dt \;=\; \int_{0}^{1} v\,dt - \int_{1}^{3} v\,dt + \int_{3}^{4} v\,dt。 $$ 原函数 $F(t) = \tfrac{t^{3}}{3} - 2t^{2} + 3t$。值：$F(0) = 0$、$F(1) = \tfrac{4}{3}$、$F(3) = 0$、$F(4) = \tfrac{4}{3}$。三段分别为 $\tfrac{4}{3}$、$-\tfrac{4}{3}$、$\tfrac{4}{3}$。总路程 $= \tfrac{4}{3} + \tfrac{4}{3} + \tfrac{4}{3} = 4$ m。

位移与路程，一句话区分。位移 $= \int v\,dt$（带符号）；路程 $= \int |v|\,dt$（非负）。仅当 $v$ 在区间上不变号两者才相等。经典陷阱：把 $\int v\,dt$ 当作"路程"。本题位移 $\tfrac{4}{3}$ m，实际走过的路程 $4$ m，三倍之差，因为质点出去、回来、又出去。先因式分解 $v$，定位 $[a, b]$ 内符号变化点，再分段。

Q4HARDPaper 1ASL 5.11 Area Between Curves[5 marks]

Region $R$ between $y = x$ and $y = x^{3}$ on $[0, 1]$. Determine which is on top, compute the area exactly.$[0, 1]$ 上由 $y = x$ 与 $y = x^{3}$ 围成的区域 $R$。判定上下次序，精确求面积。

Answers:答案： (a) $y = x$ above $y = x^{3}$ on $(0, 1)$ · (b) Area $= \dfrac{1}{4}$

(a) Top vs bottom on $(0, 1)$ M1·A1

Sample point: at $x = \tfrac{1}{2}$, $y = x$ gives $\tfrac{1}{2}$ and $y = x^{3}$ gives $\tfrac{1}{8}$, so $x > x^{3}$. More structurally, $x - x^{3} = x(1 - x^{2}) = x(1 - x)(1 + x)$, which is positive for $x \in (0, 1)$ (three positive factors). Hence $y = x$ lies above $y = x^{3}$ on $(0, 1)$.

(b) Compute the area M1·A1·A1

$$ \text{Area} \;=\; \int_{0}^{1} \bigl(x - x^{3}\bigr)\,dx \;=\; \left[\tfrac{x^{2}}{2} - \tfrac{x^{4}}{4}\right]_{0}^{1} \;=\; \tfrac{1}{2} - \tfrac{1}{4} \;=\; \tfrac{1}{4}. $$

Always justify which curve is on top. Examiners reward the explicit "$x > x^{3}$ on $(0, 1)$ because $x(1 - x)(1 + x) > 0$" line. Writing $\int (x^{3} - x)\,dx$ by accident gives the negative of the answer, and "area $= -\tfrac{1}{4}$" loses an A1 every time (areas are non-negative). The sample-point check at $x = \tfrac{1}{2}$ is the quickest sanity test, but the sign-of-$(x - x^{3})$ argument earns the R1 for full marks on a "justify" instruction. Recall: between $0$ and $1$, the lower-power monomial is above; outside that interval, the higher-power dominates.

Generalisation. For $f(x) = x^{n}$ on $[0, 1]$, the area between $y = x$ and $y = x^{n}$ ($n \ge 1$) is $\dfrac{1}{2} - \dfrac{1}{n + 1} = \dfrac{n - 1}{2(n + 1)}$. With $n = 3$ that's $\dfrac{2}{8} = \dfrac{1}{4}$, matching the computation. As $n \to \infty$, the area approaches $\dfrac{1}{2}$ (the full triangle under $y = x$) because $x^{n}$ collapses to the $x$-axis on $[0, 1)$. Carrying this formula in your head halves the time on similar questions and lets you cross-check Paper 1 answers in your head.

(a) $(0, 1)$ 上的上下次序 M1·A1

取样点：$x = \tfrac{1}{2}$ 时，$y = x = \tfrac{1}{2}$，$y = x^{3} = \tfrac{1}{8}$，故 $x > x^{3}$。结构上：$x - x^{3} = x(1 - x)(1 + x)$，三正因子之积在 $(0, 1)$ 上为正。故 $y = x$ 在 $(0, 1)$ 上在 $y = x^{3}$ 之上。

(b) 计算面积 M1·A1·A1

$$ \text{Area} \;=\; \int_{0}^{1} \bigl(x - x^{3}\bigr)\,dx \;=\; \left[\tfrac{x^{2}}{2} - \tfrac{x^{4}}{4}\right]_{0}^{1} \;=\; \tfrac{1}{2} - \tfrac{1}{4} \;=\; \tfrac{1}{4}. $$

必须显式论证哪条在上。评卷奖励显式的"在 $(0, 1)$ 上 $x > x^{3}$，因 $x(1 - x)(1 + x) > 0$"。误写 $\int (x^{3} - x)\,dx$ 得负值，"面积 $= -\tfrac{1}{4}$"必扣 A1（面积非负）。$x = \tfrac{1}{2}$ 抽样最快，但"$x - x^{3}$ 符号"论证才得 R1。记：$0$ 到 $1$ 之间，低次幂在上；区间外，高次幂主导。

推广。$[0, 1]$ 上 $y = x$ 与 $y = x^{n}$（$n \ge 1$）所夹面积为 $\dfrac{1}{2} - \dfrac{1}{n + 1} = \dfrac{n - 1}{2(n + 1)}$。$n = 3$ 时即 $\dfrac{2}{8} = \dfrac{1}{4}$，与计算一致。$n \to \infty$ 时面积趋于 $\dfrac{1}{2}$（即 $y = x$ 下的整个三角形），因 $x^{n}$ 在 $[0, 1)$ 上塌缩至 $x$ 轴。这个公式记住可省去半数计算时间，Paper 1 心算核对足矣。

Section B · Worked SolutionsB 节 · 详细解析

Q5HARDPaper 1BSL 5.8 Multi-Part Optimisation (Cylinder)[11 marks]

Closed cylinder with fixed volume $V_{0}$. Minimise $S = 2\pi r^{2} + 2\pi r h$. Show $h = 2r$ at the optimum.固定体积 $V_{0}$ 的闭合圆柱。最小化 $S = 2\pi r^{2} + 2\pi r h$。证明最优时 $h = 2r$。

Answers:答案： (a) $S(r) = 2\pi r^{2} + \dfrac{2V_{0}}{r}$ · (b) $r^{3} = \dfrac{V_{0}}{2\pi}$ · (c) $S''(r) > 0$ · (d) $h = 2r$

(a) $S$ as a function of $r$ M1·A1

Volume constraint: $V_{0} = \pi r^{2} h \Rightarrow h = \dfrac{V_{0}}{\pi r^{2}}$. Substitute: $$ S(r) \;=\; 2\pi r^{2} + 2\pi r \cdot \dfrac{V_{0}}{\pi r^{2}} \;=\; 2\pi r^{2} + \dfrac{2V_{0}}{r},\qquad r > 0. $$

(b) Stationary point M1·A1·A1

$$ \dfrac{dS}{dr} \;=\; 4\pi r - \dfrac{2V_{0}}{r^{2}}. $$ Set $\dfrac{dS}{dr} = 0$: $4\pi r = \dfrac{2V_{0}}{r^{2}} \Rightarrow 4\pi r^{3} = 2V_{0} \Rightarrow r^{3} = \dfrac{V_{0}}{2\pi}$, so $r = \left(\dfrac{V_{0}}{2\pi}\right)^{1/3}$.

(c) Second-derivative test M1·A1

$$ \dfrac{d^{2}S}{dr^{2}} \;=\; 4\pi + \dfrac{4V_{0}}{r^{3}}. $$ For $r > 0$ and $V_{0} > 0$, both terms are strictly positive, so $\dfrac{d^{2}S}{dr^{2}} > 0$. Hence the stationary point is a minimum.

(d) Show $h = 2r$ M1·A1·A1·R1

From the constraint $h = \dfrac{V_{0}}{\pi r^{2}}$, substitute the optimal $r^{3} = \dfrac{V_{0}}{2\pi}$, i.e. $V_{0} = 2\pi r^{3}$: $$ h \;=\; \dfrac{2\pi r^{3}}{\pi r^{2}} \;=\; 2r. \quad \text{AG} $$ The optimal cylinder has height equal to its diameter ($h = 2r$). This minimises the surface area for a given volume.

"Height $=$ diameter" is a deep optimum, not a coincidence. For a closed cylinder of fixed volume, the minimal-surface shape has $h = 2r$, so the cross-section is a square. Tin cans, however, are taller than this optimum (e.g. soup cans have $h \approx 3r$): the discrepancy is not a mistake but a different cost function — the lid and base typically cost more per unit area (thicker metal) than the side, so the manufacturer's true objective weights $2\pi r^{2}$ more heavily, pushing the optimum to a taller, narrower can. The exam answer ($h = 2r$) is correct under uniform material cost; the engineering answer involves weighted areas. This is a beautiful Paper 3 modelling extension if you want to enrich a study guide.

One-line shortcut. For any "minimise surface area subject to fixed volume" cylindrical/box problem, the optimum has $\dfrac{\partial S}{\partial r} = 0$ together with the constraint substitution; the algebraic condition is that the gradient of the objective is parallel to the gradient of the constraint (Lagrange's principle, even though we substitute by hand here). Geometrically, this always reduces to symmetric proportions: cube for an open box, $h = 2r$ for a closed can, equilateral triangle for a fixed-perimeter triangle, etc.

(a) 把 $S$ 写成 $r$ 的函数 M1·A1

体积约束 $V_{0} = \pi r^{2} h \Rightarrow h = \dfrac{V_{0}}{\pi r^{2}}$。代入： $$ S(r) \;=\; 2\pi r^{2} + 2\pi r \cdot \dfrac{V_{0}}{\pi r^{2}} \;=\; 2\pi r^{2} + \dfrac{2V_{0}}{r},\qquad r > 0。 $$

(b) 驻点 M1·A1·A1

$$ \dfrac{dS}{dr} \;=\; 4\pi r - \dfrac{2V_{0}}{r^{2}}。 $$ 令 $\dfrac{dS}{dr} = 0$：$4\pi r^{3} = 2V_{0}$，故 $r^{3} = \dfrac{V_{0}}{2\pi}$，$r = \left(\dfrac{V_{0}}{2\pi}\right)^{1/3}$。

(c) 二阶导检验 M1·A1

$$ \dfrac{d^{2}S}{dr^{2}} \;=\; 4\pi + \dfrac{4V_{0}}{r^{3}}。 $$ $r > 0$ 且 $V_{0} > 0$，两项严格为正，故 $\dfrac{d^{2}S}{dr^{2}} > 0$，驻点为最小值。

(d) 证 $h = 2r$ M1·A1·A1·R1

由约束 $h = \dfrac{V_{0}}{\pi r^{2}}$，代入最优 $V_{0} = 2\pi r^{3}$： $$ h \;=\; \dfrac{2\pi r^{3}}{\pi r^{2}} \;=\; 2r。 \quad \text{AG} $$ 最优圆柱满足"高 $=$ 直径"（$h = 2r$），给定体积下表面积最小。

"高 $=$ 直径"是一项深层最优，非巧合。定体积闭合圆柱的最小表面形状满足 $h = 2r$，即纵截面为正方形。但实际罐头通常比此更高（如汤罐 $h \approx 3r$）：差异不是错误，而是代价函数不同——盖与底通常单位面积成本更高（金属更厚），制造商真实目标对 $2\pi r^{2}$ 权重更大，把最优推向更高更窄。考场答案（$h = 2r$）假设材料单价均匀；工程答案需考虑加权面积。这是 Paper 3 风格的建模延伸，写学习指南时可用。

一句话捷径。所有"定体积、最小化表面积"的柱体/盒子问题，最优满足 $\dfrac{\partial S}{\partial r} = 0$ 加约束代入；代数条件即目标的梯度与约束的梯度平行（拉格朗日原理，虽然这里我们直接代入）。几何上总归约到对称比例：开口盒为立方体、闭合罐 $h = 2r$、定周长三角形为等边，等等。

Paper 2 · Worked Solutions第二卷 · 详细解析

Q6MEDIUMPaper 2SL 5.11 Volume of Revolution ($x$-axis)[7 marks]

$y = \sqrt{x}$, $x$-axis, $x = 4$, rotated about the $x$-axis. Disc integral, exact value, GDC check.$y = \sqrt{x}$、$x$ 轴、$x = 4$ 绕 $x$ 轴旋转。圆盘积分、精确值、GDC 验证。

Answers:答案： (a) $V = \pi \displaystyle\int_{0}^{4} x\,dx$ · (b) $V = 8\pi$ · (c) $8\pi \approx 25.1$

(a) Region and integral set-up M1·A1

Region: under $y = \sqrt{x}$, above $y = 0$, between $x = 0$ and $x = 4$. Rotating about the $x$-axis, the disc formula gives $$ V \;=\; \pi \int_{0}^{4} y^{2}\,dx \;=\; \pi \int_{0}^{4} \bigl(\sqrt{x}\bigr)^{2}\,dx \;=\; \pi \int_{0}^{4} x\,dx. $$

(b) Exact evaluation M1·A1·A1

$$ V \;=\; \pi \left[\tfrac{x^{2}}{2}\right]_{0}^{4} \;=\; \pi \cdot \tfrac{16}{2} \;=\; 8\pi. $$

(c) GDC numerical check M1·A1

Compute $\pi \cdot 8 = 8\pi \approx 25.1327\ldots$ on the GDC. Quote both: $V = 8\pi$ (exact) $\approx 25.1$ (3 sf).

Disc formula sanity: square $y$, never forget $\pi$. The disc-method integrand is $\pi y^{2}$, not $\pi y$ — each thin disc has radius $y$ and area $\pi y^{2}$. With $y = \sqrt{x}$, the squaring conveniently cancels the radical, making this a Paper-1-style problem in disguise. A second pitfall: confusing the axis of rotation. Rotation about the $y$-axis would force $V = \pi \int x^{2}\,dy$ instead, with $x = y^{2}$ and limits in $y$. Always write the integrand after identifying the axis. Third, "exact $+$ decimal" earns the full mark allocation on Paper 2: state both forms unless asked otherwise.

Cross-check by paraboloid formula. The solid generated by rotating $y = \sqrt{x}$ on $[0, k]$ about the $x$-axis is a paraboloid of revolution with volume $V = \tfrac{1}{2}\pi k^{2}$ (half the volume of the circumscribing cylinder $\pi k \cdot k$). At $k = 4$: $V = \tfrac{1}{2}\pi \cdot 16 = 8\pi$. The "half-cylinder" rule is a useful memory device: for any function $y = c x^{p}$ with $p > 0$ rotated about the $x$-axis on $[0, k]$, the volume is $\dfrac{\pi c^{2} k^{2p + 1}}{2p + 1}$, which equals the cylinder volume $\pi (c k^{p})^{2} \cdot k$ divided by $(2p + 1)$.

(a) 区域与积分设置 M1·A1

区域：$y = \sqrt{x}$ 之下、$y = 0$ 之上、$x \in [0, 4]$。绕 $x$ 轴旋转，圆盘公式： $$ V \;=\; \pi \int_{0}^{4} y^{2}\,dx \;=\; \pi \int_{0}^{4} \bigl(\sqrt{x}\bigr)^{2}\,dx \;=\; \pi \int_{0}^{4} x\,dx。 $$

(b) 精确求值 M1·A1·A1

$$ V \;=\; \pi \left[\tfrac{x^{2}}{2}\right]_{0}^{4} \;=\; \pi \cdot 8 \;=\; 8\pi。 $$

(c) GDC 数值核对 M1·A1

GDC 算 $8\pi \approx 25.1327\ldots$。同时写出：$V = 8\pi$（精确）$\approx 25.1$（3 sf）。

圆盘公式核查：$y$ 必平方、$\pi$ 必保留。圆盘积分号内是 $\pi y^{2}$，不是 $\pi y$——薄盘半径 $y$、面积 $\pi y^{2}$。$y = \sqrt{x}$ 时平方恰消根号，本题"伪 Paper 1"。第二陷阱：混淆旋转轴。绕$y$ 轴则 $V = \pi \int x^{2}\,dy$，$x = y^{2}$、积分限以 $y$ 计。先识轴，再写积分。第三，Paper 2 上"精确 + 小数"才得满分：除非另作要求，两形并写。

抛物面公式核对。$y = \sqrt{x}$ 在 $[0, k]$ 上绕 $x$ 轴旋转所得抛物面体积为 $V = \tfrac{1}{2}\pi k^{2}$（其外接圆柱 $\pi k \cdot k$ 的一半）。$k = 4$：$V = \tfrac{1}{2}\pi \cdot 16 = 8\pi$。"圆柱一半"是好记的口诀：任意 $y = c x^{p}$（$p > 0$）在 $[0, k]$ 上绕 $x$ 轴旋转的体积为 $\dfrac{\pi c^{2} k^{2p + 1}}{2p + 1}$，即外接圆柱体积 $\pi (c k^{p})^{2} \cdot k$ 除以 $(2p + 1)$。

Q7HARDPaper 2AHL 5.13 Related Rates (Conical Tank, HL)[10 marks]

Inverted cone, $r = h/2$, inflow $2$ m$^{3}$/min. Find $V(h)$, $dV/dt$ via implicit, $dh/dt$ at $h = 4$, and explain why $dh/dt$ decreases.倒圆锥 $r = h/2$，入流 $2$ m$^{3}$/min。求 $V(h)$、隐式得 $dV/dt$、$h = 4$ 处 $dh/dt$，并解释为何 $dh/dt$ 减小。

Answers:答案： (a) $V = \dfrac{\pi h^{3}}{12}$ · (b) $\dfrac{dV}{dt} = \dfrac{\pi h^{2}}{4}\,\dfrac{dh}{dt}$ · (c) $\dfrac{dh}{dt}\bigg|_{h=4} = \dfrac{1}{2\pi}$ m/min

(a) $V$ as a function of $h$ M1·A1

Substitute $r = h/2$ into $V = \tfrac{1}{3}\pi r^{2} h$: $$ V \;=\; \tfrac{1}{3}\pi \left(\tfrac{h}{2}\right)^{2} h \;=\; \tfrac{1}{3}\pi \cdot \tfrac{h^{3}}{4} \;=\; \dfrac{\pi h^{3}}{12}. $$

(b) Implicit differentiation w.r.t. $t$ M1·A1·A1

Differentiating $V = \dfrac{\pi h^{3}}{12}$ with respect to $t$ via the chain rule: $$ \dfrac{dV}{dt} \;=\; \dfrac{\pi}{12} \cdot 3 h^{2} \cdot \dfrac{dh}{dt} \;=\; \dfrac{\pi h^{2}}{4}\,\dfrac{dh}{dt}. $$

(c) Evaluate at $h = 4$ M1·A1·A1

Plug in $\dfrac{dV}{dt} = 2$ and $h = 4$: $$ 2 \;=\; \dfrac{\pi (4)^{2}}{4}\,\dfrac{dh}{dt} \;=\; 4\pi \cdot \dfrac{dh}{dt} \;\Longrightarrow\; \dfrac{dh}{dt} \;=\; \dfrac{2}{4\pi} \;=\; \dfrac{1}{2\pi} \text{ m/min} \;\approx\; 0.159 \text{ m/min}. $$

(d) Why $\dfrac{dh}{dt}$ decreases R1·R1

From (b), $\dfrac{dh}{dt} = \dfrac{4}{\pi h^{2}} \cdot \dfrac{dV}{dt}$ with $\dfrac{dV}{dt} = 2$ constant; the cross-section area $\pi h^{2}/4$ grows quadratically with $h$. As the cone widens at higher water levels, the same volume increment $dV$ spreads over a larger surface, so the height increment $dh$ is smaller. Quantitatively, $\dfrac{dh}{dt} \propto \dfrac{1}{h^{2}}$.

The cross-section is the rate's denominator. Every related-rates problem with "water poured at constant rate into a container of varying cross-section" reduces to $\dfrac{dV}{dt} = A(h)\,\dfrac{dh}{dt}$, where $A(h)$ is the cross-sectional area at depth $h$. For a cylinder $A$ is constant and $dh/dt$ is constant. For a cone with $r = h/2$, $A(h) = \pi r^{2} = \pi h^{2}/4$ grows like $h^{2}$ and $dh/dt$ falls like $1/h^{2}$ — at $h = 4$ the surface area is sixteen times what it was at $h = 1$. This single insight collapses every "conical tank, hemispheric bowl, trough" problem to one formula; the only work is computing $A(h)$ for the given geometry.

(a) $V$ 作为 $h$ 的函数 M1·A1

$r = h/2$ 代入 $V = \tfrac{1}{3}\pi r^{2} h$： $$ V \;=\; \tfrac{1}{3}\pi \left(\tfrac{h}{2}\right)^{2} h \;=\; \dfrac{\pi h^{3}}{12}。 $$

(b) 对 $t$ 隐式求导 M1·A1·A1

链式法则： $$ \dfrac{dV}{dt} \;=\; \dfrac{\pi}{12} \cdot 3 h^{2} \cdot \dfrac{dh}{dt} \;=\; \dfrac{\pi h^{2}}{4}\,\dfrac{dh}{dt}。 $$

(c) 在 $h = 4$ 处求值 M1·A1·A1

代入 $\dfrac{dV}{dt} = 2$、$h = 4$： $$ 2 \;=\; \dfrac{\pi (4)^{2}}{4}\,\dfrac{dh}{dt} \;=\; 4\pi \cdot \dfrac{dh}{dt} \;\Longrightarrow\; \dfrac{dh}{dt} \;=\; \dfrac{1}{2\pi} \text{ m/min} \;\approx\; 0.159 \text{ m/min}。 $$

(d) 为何 $\dfrac{dh}{dt}$ 减小 R1·R1

由 (b)，$\dfrac{dh}{dt} = \dfrac{4}{\pi h^{2}} \cdot \dfrac{dV}{dt}$，$\dfrac{dV}{dt} = 2$ 恒定；横截面积 $\pi h^{2}/4$ 随 $h$ 二次增长。水位高时锥变宽，同一体积增量 $dV$ 摊在更大表面，高度增量 $dh$ 减小。定量上 $\dfrac{dh}{dt} \propto \dfrac{1}{h^{2}}$。

横截面就是速率的分母。所有"定速率向变截面容器注水"的相关变化率题，化归为 $\dfrac{dV}{dt} = A(h)\,\dfrac{dh}{dt}$，$A(h)$ 为水深 $h$ 处横截面积。圆柱 $A$ 恒，$dh/dt$ 恒；锥（$r = h/2$）$A(h) = \pi h^{2}/4$ 像 $h^{2}$ 增长，$dh/dt$ 像 $1/h^{2}$ 衰减——$h = 4$ 处表面积是 $h = 1$ 处的十六倍。一个洞见即可破"圆锥水池、半球碗、长槽"所有题；唯一工作就是写出几何对应的 $A(h)$。

Paper 3 · Worked Solutions第三卷 · 详细解析

Q8HARDPaper 3AHL 5.16 + AHL 5.13 $y$-axis Revolution & Inflow (HL)[15 marks]

$R$: first quadrant region under $y = 4$, above $y = x^{2}$, right of $y$-axis. Rotate $R$ about $y$-axis. Inflow $3$ cm$^{3}$/s. Find $V(\Omega)$, $V(h)$, $dh/dt$ at $h = 2$, and time $T$ to fill.$R$：第一象限中 $y = 4$ 之下、$y = x^{2}$ 之上、$y$ 轴之右的区域。绕 $y$ 轴旋转。入流 $3$ cm$^{3}$/s。求 $V(\Omega)$、$V(h)$、$h = 2$ 处 $dh/dt$、注满时间 $T$。

Answers:答案： (a) $V(\Omega) = \pi \displaystyle\int_{0}^{4} y\,dy$ · (b) $V(\Omega) = 8\pi$ cm$^{3}$ · (c) $V(h) = \dfrac{\pi h^{2}}{2}$ · (d) $\dfrac{dh}{dt}\bigg|_{h=2} = \dfrac{3}{2\pi}$ cm/s · (e) $T = \dfrac{8\pi}{3}$ s

(a) Set up $y$-axis disc integral M1·A1·A1

The region $R$ is bounded by the curve $y = x^{2}$ (with $x \ge 0$), the line $y = 4$, and the $y$-axis. Rewrite the curve with $x$ as a function of $y$: $x = \sqrt{y}$ for $0 \le y \le 4$. Rotating about the $y$-axis, the disc at height $y$ has radius $x = \sqrt{y}$, so $$ V(\Omega) \;=\; \pi \int_{0}^{4} x^{2}\,dy \;=\; \pi \int_{0}^{4} y\,dy. $$

(b) Evaluate the total volume M1·A1·A1

$$ V(\Omega) \;=\; \pi \left[\tfrac{y^{2}}{2}\right]_{0}^{4} \;=\; \pi \cdot 8 \;=\; 8\pi \text{ cm}^{3}. $$

(c) Show $V(h) = \dfrac{\pi h^{2}}{2}$ M1·A1·A1

Water of depth $h$ fills the bowl from $y = 0$ to $y = h$, using the same disc formula with upper limit $h$: $$ V(h) \;=\; \pi \int_{0}^{h} y\,dy \;=\; \pi \cdot \tfrac{h^{2}}{2} \;=\; \dfrac{\pi h^{2}}{2}. \quad \text{AG} $$

(d) $\dfrac{dh}{dt}$ as a function of $h$ M1·A1·A1

Differentiate $V = \dfrac{\pi h^{2}}{2}$ implicitly with respect to $t$: $$ \dfrac{dV}{dt} \;=\; \pi h \cdot \dfrac{dh}{dt} \;\Longrightarrow\; \dfrac{dh}{dt} \;=\; \dfrac{1}{\pi h}\,\dfrac{dV}{dt} \;=\; \dfrac{3}{\pi h}. $$ At $h = 2$: $\dfrac{dh}{dt} = \dfrac{3}{2\pi}$ cm/s $\approx 0.477$ cm/s.

(e) Time to fill M1·A1·R1

The bowl fills when $h = 4$, i.e. $V = V(\Omega) = 8\pi$ cm$^{3}$. Since inflow is constant: $$ T \;=\; \dfrac{V(\Omega)}{dV/dt} \;=\; \dfrac{8\pi}{3} \text{ s} \;\approx\; 8.38 \text{ s}. $$ Consistency check. Alternatively, integrate $\dfrac{dt}{dh} = \dfrac{1}{dh/dt} = \dfrac{\pi h}{3}$ from $h = 0$ to $h = 4$: $$ T \;=\; \int_{0}^{4} \dfrac{\pi h}{3}\,dh \;=\; \dfrac{\pi}{3} \cdot \tfrac{h^{2}}{2}\bigg|_{0}^{4} \;=\; \dfrac{\pi}{3} \cdot 8 \;=\; \dfrac{8\pi}{3} \text{ s}. \;\checkmark $$ Both routes agree, confirming the model.

Two-route consistency is a Paper 3 hallmark. Whenever a Paper 3 question asks you to compute the same quantity by two different methods, the verification is the point — examiners reward the explicit cross-check (R1). Here, "total volume divided by constant flow rate" and "integrate $dt/dh$" both give $T = 8\pi/3$. A second insight: for any "water-in-bowl" problem with shape $x = g(y)$, the function $V(h) = \pi \int_{0}^{h} [g(y)]^{2}\,dy$ is a cumulative volume; its derivative $V'(h) = \pi [g(h)]^{2}$ is precisely the cross-section area at the water surface. So $\dfrac{dV}{dt} = V'(h)\,\dfrac{dh}{dt}$ is the same "$A(h) \cdot dh/dt$" recipe from Q7, applied with $A(h) = \pi h$ (a circle of radius $\sqrt{h}$). All these problems are one problem in different costumes.

(a) 绕 $y$ 轴的圆盘积分 M1·A1·A1

区域 $R$ 由 $y = x^{2}$（$x \ge 0$）、直线 $y = 4$、$y$ 轴所围。把曲线写为 $x = \sqrt{y}$（$0 \le y \le 4$）。绕 $y$ 轴旋转，高度 $y$ 处的圆盘半径 $x = \sqrt{y}$： $$ V(\Omega) \;=\; \pi \int_{0}^{4} x^{2}\,dy \;=\; \pi \int_{0}^{4} y\,dy。 $$

(b) 计算总体积 M1·A1·A1

$$ V(\Omega) \;=\; \pi \left[\tfrac{y^{2}}{2}\right]_{0}^{4} \;=\; 8\pi \text{ cm}^{3}。 $$

(c) 证 $V(h) = \dfrac{\pi h^{2}}{2}$ M1·A1·A1

水深 $h$ 时水充满 $y \in [0, h]$，圆盘公式上限换为 $h$： $$ V(h) \;=\; \pi \int_{0}^{h} y\,dy \;=\; \dfrac{\pi h^{2}}{2}。 \quad \text{AG} $$

(d) $\dfrac{dh}{dt}$ 作为 $h$ 的函数 M1·A1·A1

对 $V = \dfrac{\pi h^{2}}{2}$ 关于 $t$ 隐式求导： $$ \dfrac{dV}{dt} \;=\; \pi h \cdot \dfrac{dh}{dt} \;\Longrightarrow\; \dfrac{dh}{dt} \;=\; \dfrac{3}{\pi h}。 $$ $h = 2$：$\dfrac{dh}{dt} = \dfrac{3}{2\pi}$ cm/s $\approx 0.477$ cm/s。

(e) 注满时间 M1·A1·R1

碗满时 $h = 4$，$V = V(\Omega) = 8\pi$。恒定入流： $$ T \;=\; \dfrac{V(\Omega)}{dV/dt} \;=\; \dfrac{8\pi}{3} \text{ s} \;\approx\; 8.38 \text{ s}。 $$ 核对。等价地，由 $\dfrac{dt}{dh} = \dfrac{\pi h}{3}$，从 $h = 0$ 积分到 $h = 4$： $$ T \;=\; \int_{0}^{4} \dfrac{\pi h}{3}\,dh \;=\; \dfrac{\pi}{3} \cdot 8 \;=\; \dfrac{8\pi}{3} \text{ s}。 \;\checkmark $$ 两路一致，模型自洽。

双路一致是 Paper 3 标志。Paper 3 让你用两种方法算同一量时，"核对"本身就是题目的目的——评卷专奖显式互验（R1）。本题"总体积 $\div$ 恒流"与"积分 $dt/dh$"都给 $T = 8\pi/3$。其次：任何"水入碗"问题中，若形状 $x = g(y)$，函数 $V(h) = \pi \int_{0}^{h} [g(y)]^{2}\,dy$ 即累积体积，其导数 $V'(h) = \pi [g(h)]^{2}$ 恰为水面处横截面积。故 $\dfrac{dV}{dt} = V'(h)\,\dfrac{dh}{dt}$ 与 Q7 的 "$A(h) \cdot dh/dt$" 公式相同，此处 $A(h) = \pi h$（半径 $\sqrt{h}$ 的圆）。所有这些题，本质上是同一道题穿不同外衣。