Pre-U CCEA Maths Case Study: Roller Coaster Design | Pre-U CCEA 数学案例分析：过山车设计实战

📚 Pre-U CCEA Maths Case Study: Roller Coaster Design | Pre-U CCEA 数学案例分析：过山车设计实战

In Pre-U CCEA Mathematics, case studies bridge the gap between abstract techniques and real-world problem-solving. This article presents a step-by-step investigation into designing a smooth roller coaster track using calculus, mechanics, and optimisation. By working through the modelling, differentiation, energy methods, and safety constraints, you will see how fundamental mathematical principles combine to create a thrilling yet secure ride. Each section pairs an English explanation with its Chinese counterpart to reinforce understanding across both languages.

在 Pre-U CCEA 数学课程中，案例分析是连接抽象技巧与现实问题解决的桥梁。本文逐步研究如何使用微积分、力学和最优化方法设计一条平滑的过山车轨道。通过建模、微分、能量方法和安全约束的推演，你将看到基础数学原理如何结合起来，打造出既刺激又安全的乘坐体验。每节均以英文解说与中文对照，帮助双语理解。

1. Problem Scenario | 问题情境

A theme park requires a new roller coaster segment that starts from a platform at height 10 m, descends smoothly, rises to a peak exactly at a horizontal distance of 50 m, and returns to ground level at 100 m. The track must be a single differentiable curve, the ride must feel exciting but safe, and the maximum speed must not exceed 20 m s⁻¹. Our task is to determine the track profile, compute the motion, and verify the safety criteria.

某主题公园需要一段新的过山车轨道：从高度 10 m 的平台出发，平顺下降，在水平距离 50 m 处恰好达到一个高峰，并在 100 m 处回到地面。轨道必须是单一可微曲线，乘坐体验要既刺激又安全，且最大速度不得超过 20 m s⁻¹。我们的任务是确定轨道轮廓，计算运动过程，并验证安全标准。

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2. Mathematical Modelling – Track Equation | 数学建模 – 轨道方程

We model the track as a cubic polynomial y(x) = ax³ + bx² + cx + d, where x is the horizontal displacement in metres and y is the height above the ground. A cubic allows enough flexibility to meet four boundary conditions while keeping differentiability. The chosen conditions are: at x = 0, y = 10 and the track is horizontal, so y'(0) = 0; at x = 100, y = 0; and at x = 50, there is a stationary point that we intend to be a maximum.

我们将轨道建模为三次多项式 y(x) = ax³ + bx² + cx + d，其中 x 为水平位移（米），y 为离地高度。三次函数具有足够的灵活性，可以在可微的前提下满足四个边界条件。选定的条件为：在 x = 0 处，y = 10 且轨道水平，因此 y'(0) = 0；在 x = 100 处，y = 0；在 x = 50 处有一个驻点，我们期望它是极大值。

These conditions translate into the following equations quickly. From y(0) = 10 we get d = 10. From y'(0) = 0, since y'(x) = 3ax² + 2bx + c, we obtain c = 0. Substituting x = 100, y = 0 yields:

这些条件迅速转化为下列方程。由 y(0) = 10 得 d = 10。由 y'(0) = 0，因为 y'(x) = 3ax² + 2bx + c，得到 c = 0。代入 x = 100, y = 0 给出：

1000000a + 10000b + 0 + 10 = 0 → 1000000a + 10000b = -10.

For the stationary point at x = 50 we set y'(50) = 0:

对于 x = 50 处的驻点，令 y'(50) = 0：

3a(50)² + 2b(50) = 0 → 7500a + 100b = 0.

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3. Solving for Coefficients | 求解系数

From 7500a + 100b = 0 we can express b = -75a. Substituting into the first linear equation: 1000000a + 10000(-75a) = -10 → 1000000a – 750000a = 250000a = -10, therefore a = -10 ÷ 250000 = -0.00004 = -4 × 10⁻⁵. Then b = -75 × (-4 × 10⁻⁵) = 0.003 = 3 × 10⁻³.

由 7500a + 100b = 0 可得 b = -75a。代入第一个线性方程：1000000a + 10000(-75a) = -10 → 1000000a – 750000a = 250000a = -10，因此 a = -10 ÷ 250000 = -0.00004 = -4 × 10⁻⁵。然后 b = -75 × (-4 × 10⁻⁵) = 0.003 = 3 × 10⁻³。

Hence the track profile is y(x) = -4×10⁻⁵ x³ + 3×10⁻³ x² + 10, with domain 0 ≤ x ≤ 100. Before proceeding, we check the height at the intended peak: y(50) = -4×10⁻⁵ × 125000 + 3×10⁻³ × 2500 + 10 = -5 + 7.5 + 10 = 12.5 m, confirming a peak higher than the start.

因此轨道轮廓为 y(x) = -4×10⁻⁵ x³ + 3×10⁻³ x² + 10，定义域 0 ≤ x ≤ 100。继续之前，我们检查预定峰点高度：y(50) = -4×10⁻⁵ × 125000 + 3×10⁻³ × 2500 + 10 = -5 + 7.5 + 10 = 12.5 m，确认有一个高于起点的峰点。

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4. Verifying the Maximum | 验证极大值

A stationary point could be a minimum or point of inflection. We compute the second derivative y”(x) = 6ax + 2b = -2.4×10⁻⁴ x + 0.006. At x = 50: y”(50) = -2.4×10⁻⁴ × 50 + 0.006 = -0.012 + 0.006 = -0.006 < 0. The negative second derivative confirms a local maximum, as required for the crest of the ride.

驻点有可能是极小值或拐点。我们计算二阶导数 y”(x) = 6ax + 2b = -2.4×10⁻⁴ x + 0.006。在 x = 50 处：y”(50) = -2.4×10⁻⁴ × 50 + 0.006 = -0.012 + 0.006 = -0.006 < 0。二阶导数为负确认了局部极大值，符合过山车顶峰的要求。

Additionally, we check the endpoints: at x = 0, y” = 0.006 > 0, indicating the curve is concave up at the start, which matches a dip before rising; at x = 100, y” = -0.024 + 0.006 = -0.018 < 0, so the track ends concave down as it approaches the ground. The cubic provides an elegant shape for a single hill.

另外，我们检查端点：在 x = 0 处，y” = 0.006 > 0，表明曲线起始处上凹，这与先下降后上升的走势吻合；在 x = 100 处，y” = -0.024 + 0.006 = -0.018 < 0，因此接近地面时轨道下凹。这条三次曲线为单峰设计了优美的形状。

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5. Velocity and Energy Conservation | 速度与能量守恒

Assume a car of mass m starts from rest at the top of the platform (x = 0, y = 10) and rolls freely under gravity, ignoring friction and air resistance. Taking gravitational acceleration g = 9.8 m s⁻², the mechanical energy is conserved. With initial speed v₀ = 2 m s⁻¹ as a gentle push, the total initial energy per unit mass is E = ½ v₀² + g·10 = 2 + 98 = 100 J kg⁻¹ (approx). At any position x, speed v satisfies ½ v² + g·y(x) = E.

假设一节车厢质量为 m，从平台顶端（x = 0, y = 10）处静止出发，仅在重力作用下自由滑行，忽略摩擦和空气阻力。取重力加速度 g = 9.8 m s⁻²，机械能守恒。若初始以轻微推动获得初速 v₀ = 2 m s⁻¹，则初始单位质量总能量为 E = ½ v₀² + g·10 = 2 + 98 = 100 J kg⁻¹（近似）。在任意位置 x，速率 v 满足 ½ v² + g·y(x) = E。

v(x) = √[2(E – g·y(x))] = √[200 – 19.6 y(x)].

We evaluate v at key points. At the start, y = 10, v = √(200 – 196) = √4 = 2 m s⁻¹, consistent. At the bottom of the initial dip (x ≈ 22 m, check by solving y'(x) = 0? Actually we can compute the lowest point before the peak: set y'(x)=0 gives x=50 (max) and x=0 (min?), but we know at x=0 it’s a local minimum? y”(0)>0, so x=0 is a local minimum, but we have a descent then ascent. Let’s find the minimum between 0 and 50. y’ = 3ax²+2bx = x(3ax+2b). Setting y’=0 gives x=0 or x = -2b/(3a) = -0.006 / (-0.00012) = 50. So only x=0 and x=50 are stationary. Wait, check: -2b/(3a) = -0.006/( -0.00012 ) = 50. So the only stationary points are at x=0 and x=50. At x=0 y”>0, so it’s a local minimum. That means the track descends from the start! But we set start as platform height 10, and it’s horizontal, yet it is a local minimum? That would mean the track goes upward immediately? Let’s check slope just right of x=0: for very small x>0, y’ ≈ 2bx (since a term negligible), b=0.003 >0, so y’ >0, meaning height increases? That contradicts a descent. Let’s recalculate coefficients: a = -4e-5, b=0.003. At x=0, y’=0. For x small positive, y’ = 3a x² + 2b x ≈ 2b x = 0.006x >0, so y increases! That means the track goes up from the start, but we wanted it to descend first to then rise to a peak. Our condition y'(0)=0 and then a peak at 50 with d=10 implies that the track must go down and back up, but with only two stationary points, the function would either be monotonic change or have a min and max. Since at x=0 we have y”>0 (local min), the function must decrease left of 0? But domain starts at 0, so to the right of 0 the function increases. So the track height would increase from 10 m to 12.5 m over 50 m, then decrease to 0 at 100 m. That means the track is entirely above 10 m until after the peak? This is not a descent—it’s an upward climb then fall. For a roller coaster, you might want a drop first. However, we could reinterpret: the starting platform is at x=0, height=10, and the track goes uphill to a peak at 50, then down to ground. That’s a valid design (a launched coaster or one that climbs first). The problem statement said “descends smoothly, rises to a peak”, but we can adjust by stating that the coaster climbs from the station to a high point and then drops. That’s fine; we can just label it as an upward launch. To avoid confusion, I’ll adjust the problem description to “starts at ground level? No, we already have y(0)=10. So let’s change the scenario: the coaster begins at height 10 m with a small push, climbs to 12.5 m, then drops to ground. This is plausible. The earlier wording “descends smoothly” can be corrected to “climbs to a peak”. In the problem scenario I wrote “descends smoothly, rises to a peak” but that is not consistent with the math. I should correct the problem description in section 1 to match the solved model. I’ll edit section 1: “starts from a platform at height 10 m, climbs smoothly, reaches a peak at 50 m horizontal distance, and then returns to ground level at 100 m.” That way it’s consistent. Then the energy conservation: at start y=10, as it climbs to 12.5 m, it slows down, then accelerates down. With v₀=2, at the peak v will be lower. Let’s recompute E. I used E=100. That’s fine. At peak y=12.5, v = √(200 – 19.6×12.5) = √(200 – 245) = √(-45) → impossible. Oh no! 19.6*12.5 = 245. 200-245 = -45, so v undefined. That means the car cannot reach the peak with initial speed 2 m/s from height 10, because 100 J/kg is less than required potential energy g*12.5 = 122.5 J/kg. So the coaster would not make it over. We need to give it a higher initial speed or start from a higher point. Let’s adjust the initial speed to be higher, say v₀ = 10 m s⁻¹. Then E = 0.5*100 + 98 = 50+98=148 J/kg. Then at peak, v = √(2*148 – 2*9.8*12.5) = √(296 – 245) = √51 ≈ 7.14 m/s. That works. At bottom (x=0? Actually the lowest point on the track is at the end where y=0, so max speed at y=0: v = √(2*148 – 0) = √296 ≈ 17.2 m/s, under 20 m/s. So that satisfies speed constraint. At start (y=10) v=10. Good. So I will set v₀ = 10 m s⁻¹ and recalc energy. I’ll rewrite section 5 accordingly. The math must be consistent. So I’ll edit the scenario to have starting speed 10 m/s. Then E = ½*100 + 9.8*10 = 50 + 98 = 148 J/kg. Then v(x) = √(2*148 – 19.6 y) = √(296 – 19.6 y). At top y=12.5: v = √(296 – 245) = √51 ≈ 7.14 m/s. At bottom y=0: v = √296 ≈ 17.2 m/s. Perfect. Then we can compute accelerations and g-forces. So I’ll rewrite section 5 with these numbers. I’ll also adjust the problem context in section 1 to say “The car receives an initial speed of 10 m s⁻¹ at the station (perhaps from a launch mechanism).” So it’s launched. So overall: launched coaster climbs to 12.5 m, then drops. I’ll revise the text accordingly. Since the output must be consistent, I’ll write the final version correctly.

Now back to section 5: I’ll explain energy conservation, compute E=148 J/kg. Use v₀=10. Then formula. Then evaluate speeds. Also mention max speed at x=100 (y=0) is 17.2 <20. Checks out. I'll do that. Proceed to section 6: Acceleration components. Need tangential and normal acceleration. We'll compute derivatives: v_x = dx/dt, but we may use parameterization. For a track y(x), the car moves along path. We can find curvature and normal acceleration aₙ = v²/ρ, where ρ is radius of curvature. Tangential acceleration a_t = dv/dt = g sinθ? Actually without friction, the only force along track is component of gravity. a_t = -g sinφ, where φ is slope angle, but we can also compute using energy. But we can focus on normal acceleration for g-force. So I'll compute slope, curvature, and then aₙ. Use formula: ρ = |(1 + (y')²)^(3/2) / y''|. That requires y' and y'' at points. Then aₙ = v²/ρ. Express g-force as (aₙ + g cosθ?) Actually the rider feels normal force N = m(g cosθ + v²/ρ). We can compute "g-force" as N/(mg) = cosθ + v²/(gρ). So I'll calculate that at key points, especially at the peak and bottom. This is good for case study. Section 7: radius of curvature. Section 8: g-force. Section 9: safety constraints. Section 10: conclusion. Will need to ensure all formulas Unicode. For slope, y' = dy/dx, then tanθ = y', so sinθ = y'/√(1+y'²), cosθ = 1/√(1+y'²). I'll compute at peak (x=50, y'=0) then ρ = |(1+0)^1.5 / y''| = 1/|y''| = 1/0.006 ≈ 166.67 m. At peak v≈7.14, so aₙ = 51 / 166.67 ≈ 0.306 m/s², small. Then g-force = cos0 + aₙ/g = 1 + 0.306/9.8 ≈ 1.03g, very mild. At bottom (x=100, y=0), slope: y'(100) = 3a(10000) + 2b(100) = 3*(-4e-5)*10000 + 0.006*100 = -1.2 + 0.6 = -0.6. So tanθ = -0.6, angle downwards. Then cosθ = 1/√(1+0.36) = 1/√1.36 ≈ 0.857. y''(100) = -0.018. So ρ = (1+0.36)^(1.5) / 0.018 = 1.36^1.5 / 0.018. 1.36^1.5 = 1.36 * √1.36 ≈ 1.36 * 1.166 = 1.586. Then ρ ≈ 1.586/0.018 ≈ 88.1 m. v at bottom = 17.2, so aₙ = 296/88.1 ≈ 3.36 m/s². Then g-force = cosθ + aₙ/g = 0.857 + 0.343 ≈ 1.20g. Still safe. But we might need to ensure negative g (airtime) isn't too high. At the crest, g-force <1, could give airtime if aₙ less than g? At crest, aₙ centripetal is upward? Actually normal acceleration is towards centre of curvature, which at peak is downward? y''<0, curvature centre is below track, so aₙ direction is downward. The passenger feels a reduced normal force. N/m = g - aₙ if N upward. So g-force = 1 - aₙ/g = 1 - 0.031 = 0.969g. So slightly less than 1g, mild airtime. Good. All consistent. I'll incorporate these calculations. Thus, I need to rewrite the case study with accurate numbers. Will output full corrected version. Now craft the final article. Ensure that each section has English-Chinese pairs. I'll make each section has a few pairs. Now produce final output.TITLE: Pre-U CCEA Maths Case Study: Roller Coaster Design | Pre-U CCEA 数学案例分析：过山车设计实战

📚 Pre-U CCEA Maths Case Study: Roller Coaster Design | Pre-U CCEA 数学案例分析：过山车设计实战

Case studies in Pre-U CCEA Mathematics challenge you to integrate calculus, mechanics, and modelling into one coherent investigation. This article walks through the design of a roller coaster segment from initial specification to safety verification. Every mathematical step is presented first in English, immediately followed by a Chinese translation, so you can deepen both subject knowledge and bilingual fluency.

Pre-U CCEA 数学中的案例分析要求你将微积分、力学和建模融会贯通，形成连贯的研究过程。本文从初始规格到安全验证，逐步完成一段过山车轨道的设计。每个数学步骤都先以英文呈现，紧接着给出中文翻译，帮助你在深化专业知识的同时提升双语能力。

1. Problem Scenario | 问题情境

A theme park plans a launched roller coaster segment. The car leaves the station at a height of 10 m with an initial speed of 10 m s⁻¹. The track must climb smoothly, reach a single peak exactly 50 m horizontally from the start, and return to ground level at 100 m. The total mass of the car with passengers is 600 kg, but many quantities will be calculated per unit mass. Safety rules demand that the maximum speed stays below 20 m s⁻¹ and the g-force experienced by riders does not exceed 4g. Our task is to propose a differentiable track curve, predict the motion, and assess the safety limits.

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