📚 Year 10 Cambridge Additional Mathematics: Case Study Practice | 剑桥Year 10进阶数学：案例分析实战演练
In Cambridge IGCSE Additional Mathematics, you are expected not only to master algebraic techniques but also to apply them to real-world situations. This article presents a complete case study around a model rocket launch, where we use functions, differentiation, integration and optimisation to answer practical questions. Follow through each step to build exam-ready problem-solving skills.
在剑桥 IGCSE 进阶数学中，你不仅要掌握代数技巧，还要能把它们应用到实际情况中。本文围绕一个模型火箭发射的完整案例，通过函数、微分、积分和优化来回答实际问题。请跟随每一个步骤，培养应考必备的解题能力。
1. Case Study Introduction | 案例导入
where 0 ≤ t ≤ 15. We will use ideas from Additional Mathematics to explore the rocket’s motion.
一枚模型火箭垂直向上发射。发射后 t 秒火箭离地面的高度 h（单位：米）可以用函数 h(t) = -t³ + 15t² 来模拟，其中 0 ≤ t ≤ 15。我们将运用进阶数学中的知识来探究火箭的运动情况。
This scenario allows us to practise key topics: functions, graphs, differentiation for rates of change, stationary points, optimisation, and even integration for total distance.
这一场景让我们有机会练习多个核心主题：函数与图像、用于变化率的微分、驻点、优化，甚至还有用于求总路程的积分。
2. Understanding the Height Function | 理解高度函数
s, the height rises, reaches a maximum, and eventually comes back down to h = 0 at t = 15.
火箭的轨迹由一个三次多项式描述。注意到 t = 0 时，h(0) = 0（发射点）。随着 t 增加，高度先上升，到达最高点，最后在 t = 15 时再次回到 h = 0。
0 m. The function is easy to evaluate but we need calculus to analyse its behaviour fully.
代入 t = 2：h(2) = -(2)³ + 15(2)² = -8 + 60 = 52 米。代入 t = 5：h(5) = -125 + 375 = 250 米。求值很简单，但要全面分析函数的行为，我们就需要微积分。
3. Velocity and Acceleration | 速度与加速度
h respect to time, so we differentiate h(t) once. Using the power rule, we obtain v(t) = h'(t) = -3t² + 30t.
速度是高度关于时间的变化率，因此我们对 h(t) 求一阶导数。运用幂函数求导法则，得到 v(t) = h'(t) = -3t² + 30t。
v(t) = -3t² + 30t
Acceleration is the rate of change of velocity, so we differentiate again: a(t) = v'(t) = h''(t) = -6t + 30.
加速度是速度的变化率，因此再求一次导数：a(t) = v'(t) = h''(t) = -6t + 30。
a(t) = -6t + 30
0 = -432 + 360 = -72 m/s (descending), and a(12) = -72 + 30 = -42 m/s² (negative acceleration).
例如，在 t = 2 时，v(2) = -3(4) + 60 = 48 m/s（火箭上升），a(2) = -12 + 30 = 18 m/s²（正加速）。在 t = 12 时，v(12) = -3(144) + 360 = -72 m/s（下降），a(12) = -72 + 30 = -42 m/s²（负加速）。
4. Stationary Points | 驻点
Stationary points occur when the derivative is zero, i.e. v(t) = 0. Solve -3t² + 30t = 0 ⇒ -3t(t - 10) = 0, giving t = 0 and t = 10.
驻点出现在导数为零时，即 v(t) = 0。解方程 -3t² + 30t = 0 ⇒ -3t(t - 10) = 0，得到 t = 0 和 t = 10。
ket reaches a turning point where the direction of motion changes from upward to downward.
t = 0 时火箭刚刚发射，这属于定义域上的端点。t = 10 时火箭到达一个转向点，运动方向由上升变为下降。
5. Determining the Nature of Turning Points | 判断极值点性质
(10) = -6(10) + 30 = -30, which is negative. Therefore h(t) has a local maximum at t = 10.
为了判定 t = 10 处驻点的性质，我们使用二阶导数检验法。a(10) = -6(10) + 30 = -30，为负值。因此 h(t) 在 t = 10 处有局部极大值。
We could also examine the sign of v(t) around t = 10. A table illustrates this clearly.
我们也可以通过观察 t = 10 附近 v(t) 的符号来判定。下表清楚地展示了这一过程。
| t | v(t) sign | Shape |
|---|---|---|
| t < 10 (e.g. 9) | positive (+) | rising |
| t = 10 | zero (0) | maximum point |
| t > 10 (e.g. 11) | negative (−) | falling |
The table confirms the maximum turning point and also shows the first derivative test in action.
该表确认了极大值转向点，同时也展示了一阶导数检验法的实际应用。
6. Maximum Height | 最大高度
0) = -(10)³ + 15(10)² = -1000 + 1500 = 500 metres. This is the rocket’s greatest altitude.
既然 t = 10 处取得极大值，我们计算 h(10) = -(10)³ + 15(10)² = -1000 + 1500 = 500 米。这就是火箭的最大高度。
minimum within our domain. The optimisation problem is solved directly from differentiation.
在发射点 t = 0 处，h(0) = 0 是定义域内的最小值。优化问题通过求导直接得到了解决。
7. Time of Flight | 飞行时间
The rocket returns to the ground when h(t) = 0. Solve -t³ + 15t² = 0 ⇒ -t²(t - 15) = 0. Solutions: t = 0 (launch) and t = 15 (landing).
当 h(t) = 0 时火箭回到地面。解方程 -t³ + 15t² = 0 ⇒ -t²(t - 15) = 0。解得 t = 0（发射）和 t = 15（落地）。
Therefore the total time of flight is 15 seconds. This also matches the domain we initially set.
因此总飞行时间为 15 秒。这也与我们最初设定的定义域一致。
8. Equation of Tangent at t = 2 | t = 2 时的切线方程
al Mathematics. At t = 2, the point on the curve is (2, 52) and the gradient is v(2) = 48.
切线问题在进阶数学考试中经常出现。在 t = 2 处，曲线上的点为 (2, 52)，梯度为 v(2) = 48。
Using the point-slope form: h - 52 = 48(t - 2) ⇒ h = 48t - 96 + 52 ⇒ h = 48t - 44.
使用点斜式：h - 52 = 48(t - 2) ⇒ h = 48t - 96 + 52 ⇒ h = 48t - 44。
Tangent at t = 2: h = 48t - 44
This linear equation describes the instantaneous motion if the rocket continued at constant velocity from that moment.
这个线性方程描述了如果火箭从那一刻起以恒定速度运动时的瞬时运动轨迹。
9. Total Distance Travelled | 总路程
velocity changes sign at t = 10. From t = 0 to 10, v(t) ≥ 0; from t = 10 to 15, v(t) ≤ 0.
当运动方向改变时，总路程与位移不同。速度在 t = 10 处改变符号。从 t = 0 到 10，v(t) ≥ 0；从 t = 10 到 15，v(t) ≤ 0。
. Displacement from 10 to 15: ∫₁₀¹⁵ v(t) dt = h(15) - h(10) = 0 - 500 = -500 m (downward).
0 到 10 秒的位移：∫₀¹⁰ v(t) dt = h(10) - h(0) = 500 m。10 到 15 秒的位移：∫₁₀¹⁵ v(t) dt = h(15) - h(10) = 0 - 500 = -500 m（向下）。
The total distance is the sum of the absolute values: 500 + 500 = 1000 metres.
总路程为位移绝对值之和：500 + 500 = 1000 米。
This shows how integration of the velocity function helps calculate the actual path length, not just net change in position.
这表明通过积分速度函数可以求出实际路径长度，而不止是位置净变化。
10. Integration to Recover Displacement from Velocity | 由速度积分还原位移
we can find the height function by integrating: h(t) = ∫ (-3t² + 30t) dt = -t³ + 15t² + C.
已知速度函数 v(t) = -3t² + 30t，我们可以通过积分求出高度函数：h(t) = ∫ (-3t² + 30t) dt = -t³ + 15t² + C。
model exactly. This reverse process highlights the Fundamental Theorem of Calculus in Additional Mathematics.
利用初始条件 h(0) = 0，得到 C = 0。这正好与原始模型吻合。这个反向过程凸显了进阶数学中微积分基本定理的应用。
11. Real-World Interpretation and Limitations | 实际意义与模型局限
The model predicts a smooth, symmetrical flight. In reality, air resistance, wind and engine thrust variations would alter the curve. However, it gives a useful approximation for analysing maximum height and impact time.
该模型预测出一条平滑、对称的飞行轨迹。在现实中，空气阻力、风力以及发动机推力的变化都会改变曲线。但它为分析最大高度和落地时间提供了一个有效的近似。
From a mathematical viewpoint, this case study demonstrates how differentiation locates extremes, while integration accumulates quantities. These are powerful tools in physics, engineering and economics.
从数学的角度看，本案例展示了微分如何定位极值，积分如何累积数量。它们是物理学、工程学和经济学中的强大工具。
12. Summary and Exam Tips | 总结与考试技巧
Key skills practised in this case study:
本案例练习的关键技能：
Differentiating polynomials to find velocity and acceleration
多项式求导，求速度和加速度
Solving f'(x) = 0 to locate stationary points
解 f'(x) = 0 确定驻点
Using second derivative or sign test to classify maxima and minima
用二阶导数或符号法判断极大值、极小值
Finding equations of tangents
求切线方程
Integrating to recover original functions and calculating total distance
积分还原原函数并计算总路程
Interpreting results in context
在情境中解读结果
When tackling similar problems, always define your variables clearly, show each derivative step, and check whether direction changes affect total distance. With consistent practice, these techniques become second nature.
遇到类似问题时，要明确定义变量，展示每一步求导过程，并检查方向变化是否影响总路程。通过持续练习，这些技巧将变得得心应手。
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