📚 Year 10 Edexcel Further Mathematics: Cross-Disciplinary Integrated Problem Training | Edexcel 十年级进阶数学：跨学科综合题型训练
In Edexcel Year 10 Further Mathematics, mastering the ability to apply mathematical concepts across different disciplines is essential for top exam performance. This article provides a series of integrated problem-solving exercises that blend algebra, calculus, vectors, trigonometry, and sequences with real-world contexts from physics, economics, biology, and engineering. Working through these cross-curricular challenges will deepen your understanding, sharpen your analytical skills, and build confidence for both structured questions and open-ended investigations.
在 Edexcel 十年级进阶数学中，掌握跨学科应用数学概念的能力是取得优异成绩的关键。本文提供一系列综合解题训练，将代数、微积分、向量、三角学及数列与物理学、经济学、生物学和工程学的真实情境相结合。通过完成这些跨领域挑战，你将深化理解，提升分析能力，并为结构化题型和开放式探究建立信心。
1. Kinematics and Quadratic Functions | 运动学与二次函数
n by h = 25t − 4.9t² (taking g = 9.8 m/s²). Determine the maximum height reached and the time at which it occurs. Also find the time when the particle returns to the ground.
问题：一质点从地面以 25 m/s 的初速度竖直向上抛出。经过 t 秒后，其高度 h 米由 h = 25t − 4.9t² 给出（取 g = 9.8 m/s²）。求达到的最大高度及相应的时间，并求质点返回地面的时间。
g parabola. Using the vertex formula t = −b/(2a) with a = −4.9 and b = 25 gives t = −25/(2 × −4.9) = 25/9.8 ≈ 2.551 seconds. Substituting back: hmax = 25(2.551) − 4.9(2.551)² ≈ 31.89 metres.
解题思路：高度函数是关于 t 的二次函数，h(t) = −4.9t² + 25t，开口向下，顶点即最大值。利用顶点公式 t = −b/(2a)，其中 a = −4.9，b = 25，得 t = −25/(2 × −4.9) = 25/9.8 ≈ 2.551 秒。代入得 hmax ≈ 25 × 2.551 − 4.9 × (2.551)² ≈ 31.89 米。
For the return to the ground, set h = 0: −4.9t² + 25t = 0 ⇒ metry of parabolic motion under constant acceleration and can be verified by factoring or using the quadratic formula.
返回地面时，令 h = 0：−4.9t² + 25t = 0 ⇒ t(−4.9t + 25) = 0。非零解为 t = 25/4.9 ≈ 5.102 秒，恰好是达到最高点时间的两倍。这一结果简洁地验证了匀加速抛物线运动的对称性，可通过因式分解或求根公式加以验证。
2. Optimisation in Engineering Design | 工程设计中的最优化
Problem: An open-top box is to be made from a rectangular sheet of cardboard measuring 30 cm by 20 cm. Squares of side length x cm are cut from each corner, and the sides are folded up. Find the value of x that maximises the volume of the box, and determine the maximum volume.
问题：用一块 30 cm × 20 cm 的矩形纸板制作一个无盖盒子。从每个角剪去边长为 x cm 的正方形，然后将边折起。求使盒子体积最大的 x 值，并计算最大体积。
ifferentiate: dV/dx = 12x² − 200x + 600. Set the derivative to zero: 12x² − 200x + 600 = 0 ⇒ 3x² − 50x + 150 = 0. Use the quadratic formula x = [50 ± √50² − 4×3×150] / (2×3) = [50 ± √2500 − 1800 ts. The root that fits the physical constraint 0 < x < 10 is x = (50 − 26.46)/6 ≈ 3.92 cm.
用 x 表示体积 V：长 = 30 − 2x，宽 = 20 − 2x，高 = x。因此 V = x(30 − 2x)(20 − 2x) = 4x³ − 100x² + 600x。求导：dV/dx = 12x² − 200x + 600。令导数为零：12x² − 200x + 600 = 0 ⇒ 3x² − 50x + 150 = 0。使用求根公式 x = [50 ± √50² − 4×3×150]/(2×3) = [50 ± √700]/6，√700 ≈ 26.46，得到两个根。符合物理限制 0 < x < 10 的解为 x = (50 − 26.46)/6 ≈ 3.92 cm。
h = 30 − 7.84 = 22.16 cm, width = 20 − 7.84 = 12.16 cm, height = 3.92 cm. Maximum volume Vmax emonstrates the power of calculus in solving real-world engineering design problems while also illustrating how to interpret and reject extraneous mathematical solutions.
将 x ≈ 3.92 代入：长 = 30 − 7.84 = 22.16 cm，宽 = 20 − 7.84 = 12.16 cm，高 = 3.92 cm。最大体积 Vmax ≈ 3.92 × 22.16 × 12.16 ≈ 1056 cm³。此题展示了微积分在解决实际工程设计问题中的强大能力，同时也说明了如何解释并舍去不符合实际情况的数学解。
3. Cost, Revenue and Profit: Calculus in Economics | 成本、收益与利润：经济学中的微积分
Problem: A company produces and sells a gadget. The total cost of producing q units is C
Published by TutorHao | Year 10 进阶数学 Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图