📚 Case Study Practical Exercise in Year 13 Edexcel Maths | Year 13 Edexcel 数学：案例分析实战演练
In Year 13 Edexcel Mathematics, students are often required to integrate pure, mechanics and statistics strands to solve a single extended problem. This article presents a detailed case study that weaves together differential equations, projectile motion with air resistance, and hypothesis testing on experimental data. By working through each stage, you will deepen your understanding of how mathematical techniques interconnect in real-world modelling.
在 Year 13 Edexcel 数学课程中，学生经常需要综合运用纯数学、力学和统计学知识来解决一个扩展问题。本文提供一个详细的案例分析，将微分方程、含空气阻力的抛体运动以及对实验数据的假设检验融合在一起。通过逐步完成每个环节，你会更深入地理解各种数学方法如何在现实世界建模中相互关联。
1. The Scenario – Launching a Projectile from a Fixed Ramp | 场景设定 – 从固定斜面发射抛体
A small ball is launched from a ramp inclined at an angle θ to the horizontal, with initial speed u. A sensor measures the horizontal range R of the ball when it lands on flat ground. The experiment is repeated 40 times, and the results are recorded. We suspect that air resistance cannot be ignored, so we construct a model that includes a drag force proportional to velocity.
一个小球从与水平面成角度 θ 的斜面发射，初速度为 u。传感器测量小球落在水平地面上的水平射程 R。实验重复 40 次并记录结果。我们怀疑空气阻力不能忽略，因此构建一个包含与速度成正比的阻力的模型。
2. Pure Maths – Setting Up the Equations of Motion | 纯数学 – 建立运动方程
We take the drag force as Fdrag = −kv, where k is a positive constant and v is the instantaneous velocity. Resolving horizontally and vertically gives two second-order ODEs. Let x(t) and y(t) be horizontal and vertical displacements.
我们将阻力取为 Fdrag = −kv，其中 k 为正常数，v 为瞬时速度。分别沿水平和竖直方向分解，得到两个二阶常微分方程。设 x(t) 和 y(t) 分别为水平和竖直位移。
From Newton’s second law, horizontally: m d²x/dt² = −k dx/dt. Vertically: m d²y/dt² = −mg − k dy/dt. These can be reduced to first-order linear ODEs in velocity components.
根据牛顿第二定律，水平方向：m d²x/dt² = −k dx/dt。竖直方向：m d²y/dt² = −mg − k dy/dt。这些方程可以化为关于速度分量的一阶线性常微分方程。
3. Solving the Horizontal Motion | 求解水平方向运动
Let vx = dx/dt. Then m dvx/dt = −k vx. This is separable: ∫ (1/vx) dvx = ∫ (−k/m) dt, giving vx = u cos θ e−(k/m)t. Integrating again, x(t) = (mu cos θ / k)(1 − e−(k/m)t).
令 vx = dx/dt，则 m dvx/dt = −k vx。分离变量：∫ (1/vx) dvx = ∫ (−k/m) dt，解得 vx = u cos θ e−(k/m)t。再次积分得到 x(t) = (mu cos θ / k)(1 − e−(k/m)t)。
As t → ∞, the horizontal distance approaches a finite limit because the drag eventually stops horizontal motion.
当 t → ∞ 时，水平距离趋近一个有限极限，因为阻力最终会使水平运动停止。
4. Solving the Vertical Motion | 求解竖直方向运动
Let vy = dy/dt. The ODE is m dvy/dt = −mg − k vy. Rearranging: dvy/dt + (k/m)vy = −g. This is linear; using an integrating factor e(k/m)t yields vy = (u sin θ + mg/k)e−(k/m)t − mg/k.
令 vy = dy/dt，方程为 m dvy/dt = −mg − k vy。整理得 dvy/dt + (k/m)vy = −g。这是一阶线性方程，使用积分因子 e(k/m)t 解得 vy = (u sin θ + mg/k)e−(k/m)t − mg/k。
Integrating again gives y(t) = (m/k)(u sin θ + mg/k)(1 − e−(k/m)t) − (mg/k)t. The time of flight T is found by setting y(T) = 0, which requires numerical solution unless k is very small.
再次积分得到 y(t) = (m/k)(u sin θ + mg/k)(1 − e−(k/m)t) − (mg/k)t。飞行时间 T 通过令 y(T) = 0 求得，除非 k 非常小，否则需要数值求解。
5. Modelling the Range with Air Resistance | 含空气阻力的射程模型
The range Rdrag is x(T). Using the expressions for x(t) and T, we can write Rdrag as a function of u, θ, m, k, g. For the given experiment, u = 12 m s⁻¹, θ = 35°, m = 0.15 kg, g = 9.81 m s⁻². We estimate k from terminal velocity data as 0.05 N s m⁻¹.
射程 Rdrag 为 x(T)。利用 x(t) 和 T 的表达式，可将 Rdrag 写成 u, θ, m, k, g 的函数。实验中 u = 12 m s⁻¹，θ = 35°，m = 0.15 kg，g = 9.81 m s⁻²。我们根据终端速度数据估算 k = 0.05 N s m⁻¹。
Using a numerical solver, T ≈ 1.32 s, and Rdrag ≈ 10.42 m. By contrast, the ideal no-drag range formula Rideal = (u² sin 2θ)/g gives 13.82 m, a 25% overestimate.
使用数值求解，T ≈ 1.32 s，Rdrag ≈ 10.42 m。相比之下，理想无阻力射程公式 Rideal = (u² sin 2θ)/g 给出 13.82 m，高估约 25%。
6. Statistical Analysis – The Experimental Data | 统计分析 – 实验数据
Forty measurements of the horizontal range (in metres) are collected. A summary of the data: sample mean x̄ = 10.36 m, sample standard deviation s = 0.54 m. We wish to test whether the observed data are consistent with the model prediction Rdrag = 10.42 m.
收集到 40 个水平射程的测量值（单位：米）。数据摘要：样本均值 x̄ = 10.36 m，样本标准差 s = 0.54 m。我们希望检验观测数据是否与模型预测 Rdrag = 10.42 m 一致。
The sample appears roughly symmetric, so a t-test for a population mean is appropriate. We test at the 5% significance level.
样本大致对称，因此可以采用总体均值的 t 检验。我们在 5% 显著性水平下进行检验。
7. Hypothesis Test – Two-Tailed t-Test | 假设检验 – 双尾 t 检验
H₀: μ = 10.42 (the true mean range equals the model prediction)
H₁: μ ≠ 10.42 (the true mean range differs from the prediction)
H₀: μ = 10.42（真实平均射程等于模型预测）
H₁: μ ≠ 10.42（真实平均射程不等于预测值）
Test statistic: t = (x̄ − μ₀) / (s/√n) = (10.36 − 10.42) / (0.54/√40) = −0.06 / 0.0854 ≈ −0.702. Degrees of freedom ν = 39. The critical values for a two-tailed 5% test from t39 are ±2.023. Since |t| = 0.702 < 2.023, we do not reject H₀.
检验统计量：t = (x̄ − μ₀) / (s/√n) = (10.36 − 10.42) / (0.54/√40) = −0.06 / 0.0854 ≈ −0.702。自由度 ν = 39。5% 双尾检验的临界值来自 t39 分布，为 ±2.023。由于 |t| = 0.702 < 2.023，我们不拒绝 H₀。
There is insufficient evidence at the 5% level to conclude that the model’s prediction is incorrect. The small discrepancy could be due to random variation.
在 5% 显著性水平下，没有足够证据表明模型预测不正确。微小的差异可能来自随机波动。
8. Confidence Interval for the True Mean Range | 真实平均射程的置信区间
A 95% confidence interval for μ is x̄ ± tcrit × (s/√n) = 10.36 ± 2.023 × 0.0854 = (10.187, 10.533). The model value 10.42 lies comfortably inside this interval, reinforcing the conclusion of the t-test.
μ 的 95% 置信区间为 x̄ ± tcrit × (s/√n) = 10.36 ± 2.023 × 0.0854 = (10.187, 10.533)。模型值 10.42 稳稳落在区间之内，这进一步支持了 t 检验的结论。
If we had obtained a much larger sample size, the interval would narrow, and even a small systematic error could become detectable.
如果样本量更大，区间将变窄，此时即使很小的系统误差也可能被检测出来。
9. Mechanics Extension – Sensitivity to the Drag Coefficient | 力学拓展 – 对阻力系数的敏感性
The value k = 0.05 was an estimate. In reality, k could vary. By solving the ODEs for k = 0.04 and k = 0.06, we find R ≈ 10.89 m and R ≈ 9.98 m respectively. The range is quite sensitive to k, so precise estimation is important.
k = 0.05 只是一个估计值，实际中 k 可能变化。对 k = 0.04 和 k = 0.06 求解常微分方程，分别得到 R ≈ 10.89 m 和 R ≈ 9.98 m。射程对 k 相当敏感，因此精确估计很重要。
This sensitivity analysis illustrates the power of differential equation modelling and the need for accurate parameter measurement in real-world applications.
这种敏感性分析展示了微分方程建模的力量，也体现了在实际应用中精确测量参数的必要性。
10. Reflection and Exam Tips | 反思与考试技巧
When approaching an Edexcel Year 13 case study question, always identify which strands of maths are being tested. You might need to switch from solving an ODE to interpreting statistical output to critiquing a model. Show clear logical steps, state assumptions explicitly, and use correct notation.
在解答 Edexcel Year 13 案例分析题时，要始终判断题目在考查哪些数学分支。你可能需要从求解常微分方程切换到解读统计结果，再到批判模型。展现清晰的逻辑步骤，明确陈述假设，并使用正确的符号。
If a numerical answer doesn’t match expectations, comment on possible sources of error – this demonstrates the higher-order thinking expected at A* level. Always link conclusions back to the original context of the problem.
如果数值答案与预期不符，应评论可能的误差来源——这体现了 A* 水平所需的高阶思维。永远将结论联系回问题的原始背景。
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