Year 11 SQA Further Mathematics: Case Study Practical Drill | 案例分析实战演练

📚 Year 11 SQA Further Mathematics: Case Study Practical Drill | 案例分析实战演练

This article presents a series of real-world case studies designed to sharpen your problem-solving skills for the SQA Further Mathematics course. Each scenario integrates key concepts from calculus, algebra, trigonometry, and vectors, showing how theoretical tools are applied in practical situations. By working through these examples step by step, you will develop the confidence to tackle unfamiliar problems in the exam.

本文通过一系列真实情境的案例研究，帮助你提升SQA进阶数学的解题能力。每个情景都融合了微积分、代数、三角学和向量等核心概念，展示理论工具在实际问题中的应用。通过逐步解析这些例题，你将建立起应对考试中陌生题型的自信心。

1. Optimisation: Maximising Area of a Rectangular Enclosure | 优化：最大化矩形围栏面积

e width of the field perpendicular to the river be x metres, and the length parallel to the river be y metres. The total fencing gives y + 2x = 100, so y = 100 − 2x. The area A = x y = x(100 − 2x).

一位农民有100米长的围栏，想靠一条直河岸围一个矩形场地，因此河岸一侧无需围栏。设垂直河岸的宽度为x米，平行河岸的长度为y米。总围栏用量为y + 2x = 100，故y = 100 − 2x。面积A = x y = x(100 − 2x)。

To maximise the area, we differentiate A with respect to x and set the derivative to zero.

为求最大面积，我们对A关于x求导，并令导数为零。

A = 100x − 2x²,   dA/dx = 100 − 4x = 0 ⇒ x = 25

5 × 50 = 1250 m². We verify it is a maximum by the second derivative: d²A/dx² = −4 < 0.

于是y = 100 − 2(25) = 50。最大面积为A = 25 × 50 = 1250平方米。由二阶导数d²A/dx² = −4 < 0可确认其为极大值。

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2. Kinematics: Velocity and Acceleration from Displacement | 运动学：由位移求速度和加速度

− 6t² + 9t metres. Find the times when the object is at rest, and determine the acceleration at those instants.

一物体沿直线运动，t秒时相对于固定点O的位移为s(t) = t³ − 6t² + 9t米。求物体静止的时刻，并确定这些时刻的加速度。

Velocity v(t) = ds/dt = 3t² − 12t + 9. The object is at rest when v(t) = 0.

速度v(t) = ds/dt = 3t² − 12t + 9。当v(t) = 0时物体静止。

3t² − 12t + 9 = 0 ⇒ t² − 4t + 3 = 0 ⇒ (t − 1)(t − 3) = 0, so t = 1 s and t = 3 s

a(t) = dv/dt = 6t − 12. At t = 1 s, a = −6 m s⁻² (decelerating); at t = 3 s, a = 6 m s⁻².

加速度a(t) = dv/dt = 6t − 12。t = 1秒时，a = −6 m s⁻²（减速）；t = 3秒时，a = 6 m s⁻²（加速）。

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3. Exponential Decay: Half-life and Remaining Mass | 指数衰减：半衰期与剩余质量

e M₀ is the initial mass and k is a positive constant. Its half-life is 8 years. Find the value of k, and determine after how many years only 10% of the original mass remains.

某放射性物质按模型M(t) = M₀ e⁻ᵏᵗ衰变，M₀为初始质量，k为正常数。其半衰期为8年。求k的值，并计算经过多少年后仅剩原质量的10%。

Half-life means M/M₀ = ½ when t = 8. So ½ = e⁻⁸ᵏ. Taking natural logs: ln(½) = −8k ⇒ k = −ln(½)/8 = ln(2)/8 ≈ 0.08664.

半衰期意指t = 8时M/M₀ = ½。因此½ = e⁻⁸ᵏ。取自然对数：ln(½) = −8k ⇒ k = −ln(½)/8 = ln(2)/8 ≈ 0.08664。

For 10% remaining, we set 0.1 = e⁻ᵏᵗ ⇒ ln(0.1) = −kt ⇒ t = −ln(0.1)/k = ln(10)/k ≈ 2.3026/0.08664 ≈ 26.6 years.

对于剩余10%，设0.1 = e⁻ᵏᵗ ⇒ ln(0.1) = −kt ⇒ t = −ln(0.1)/k = ln(10)/k ≈ 2.3026/0.08664 ≈ 26.6年。

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4. Vectors: Resultant Velocity of a Boat in a Current | 向量：船在水流中的合速度

A boat sails with a speed of 5 m s⁻¹ in a direction 30° north of east relative to the water. The water current flows at 2 m s⁻¹ due east. Find the true speed and bearing of the boat.

一艘船相对于水以5 m s⁻¹的速度沿东偏北30°方向航行。水流以2 m s⁻¹的速度向正东流动。求船的实际速度大小与方位角。

Let i and j be unit vectors east and north. Boat velocity relative to water: v_b/w = 5 cos30° i + 5 sin30° j = (5√3/2) i + (2.5) j. Current velocity: v_c = 2 i. True velocity v = v_b/w + v_c = (5√3/2 + 2) i + 2.5 j.

设i和j分别为向东和向北的单位向量。船对水速度：v_b/w = 5 cos30° i + 5 sin30° j = (5√3/2) i + 2.5 j。水流速度：v_c = 2 i。实际速度v = v_b/w + v_c = (5√3/2 + 2) i + 2.5 j。

Speed = |v| = √[ (5√3/2 + 2)² + (2.5)² ] ≈ √[ (4.33+2)² + 6.25 ] = √[ (6.33)² + 6.25 ] ≈ √46.1 ≈ 6.79 m s⁻¹

Direction: tan θ = 2.5 / (5√3/2+2), so θ ≈ tan⁻¹(2.5/6.33) ≈ 21.6° north of east. Bearing ≈ 068.4°.

方向：tan θ = 2.5 / (5√3/2+2)，故θ ≈ tan⁻¹(2.5/6.33) ≈ 21.6°，即东偏北。方位角约068.4°。

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5. Trigonometry: Height of a Mountain from Two Angles of Elevation | 三角学：由两个仰角求山高

Two observation points, A and B, are 500 m apart on horizontal ground. From A, the angle of elevation to the summit is 28°, and from B it is 42°. Points A and B lie in the same vertical plane as the peak. Find the height of the mountain.

两个观测点A和B在同一水平面上，相距500米。在A点测得山顶仰角为28°，在B点测得仰角为42°。A、B与山峰位于同一竖直平面内。求山的高度。

B is closer to the peak). From right triangles: tan 28° = h / x, tan 42° = h / (500 − x).

设从山顶到地面的垂足为C。设AC = x，则BC = 500 − x（假设B更靠近山峰）。由直角三角形得：tan 28° = h / x，tan 42° = h / (500 − x)。

from both: x tan 28° = (500 − x) tan 42°. Solve for x: x (tan 28° + tan 42°) = 500 tan 42° ⇒ / (0.5317+0.9004) = 450.2 / 1.4321 ≈ 314.4 m. Then h = x tan 28° ≈ 314.4×0.5317 ≈ 167.2 m.

令两式h相等：x tan 28° = (500 − x) tan 42°。解x：x (tan 28° + tan 42°) = 500 tan 42° ⇒ x = 500 tan 42° / (tan 28° + tan 42°)。代入近似值：tan 28° ≈ 0.5317，tan 42° ≈ 0.9004。得x ≈ 450.2 / 1.4321 ≈ 314.4 m，h ≈ 314.4×0.5317 ≈ 167.2米。

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6. Polynomial Functions: Stationary Points and Curve Sketching | 多项式函数：驻点与曲线草图

nd the coordinates of any stationary points, determine their nature, and sketch the curve.

已知f(x) = x³ − 3x² + 4，求全部驻点坐标，判断其性质，并绘制曲线草图。

Differentiate: f′(x) = 3x² − 6x = 3x(x − 2). Set f′(x)=0 ⇒ x=0, x=2. The stationary points are at (0,4) and (2,0). Second derivative: f″(x)=6x − 6.

求导：f′(x) = 3x² − 6x = 3x(x − 2)。令f′(x)=0得x=0, x=2。驻点为(0,4)和(2,0)。二阶导数：f″(x) = 6x − 6。

At x=0, f″(0)=−6 < 0 ⇒ local maximum. At x=2, f″(2)=6 > 0 ⇒ crosses the y-axis at (0,4), and we find x-intercepts by solving f(x)=0. x³ − 3x² + 4 = 0 ⇒ sing) and x=2 (touching). Sketch: rising from −∞, max at (0,4), min at (2,0), then rising.

在x=0处，f″(0)=−6 < 0 ⇒ 局部极大值。x=2处，f″(2)=6 > 0 ⇒ 局部极小值。曲线交y轴于(0,4)，令f(x)=0求x轴截距：x³ − 3x² + 4 = 0 ⇒ (x+1)(x−2)² = 0，得x = −1（穿过）和x=2（触及）。草图：自−∞上升，极大点(0,4)，极小点(2,0)，然后上升。

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7. Integral Calculus: Area Between Two Curves | 积分学：两曲线间的面积

Find the area enclosed by the curves y = x² and y = 2x − x².

求由曲线y = x²与y = 2x − x²所围成区域的面积。

First find points of intersection: x² = 2x − x² ⇒ 2x² − 2x = 0 ⇒ = 2x − x² for 0 < x < 1. The area A = ∫₀¹ [ (2x − x²) − x² ] dx = ∫₀¹ (2x − 2x²) dx.

先求交点：x² = 2x − x² ⇒ 2x² − 2x = 0 ⇒ 2x(x − 1) = 0，得x = 0与x = 1。在区间(0,1)内上方曲线为y = 2x − x²。面积A = ∫₀¹ [ (2x − x²) − x² ] dx = ∫₀¹ (2x − 2x²) dx。

A = [ x² − (2/3)x³ ]₀¹ = 1 − 2/3 = 1/3 square units

The enclosed area is ⅓ unit².

所围面积为⅓平方单位。

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8. Differential Equations: Newton's Law of Cooling | 微分方程：牛顿冷却定律

A cup of coffee at 80°C is placed in a room at 20°C. According to Newton' = −k(T − 20). After 5 minutes, the coffee is 60°C. Find the temperature after 15 minutes.

一杯80°C的咖啡置于20°C的房间中。根据牛顿冷却定律，冷却速率与温差成正比：dT/dt = −k(T − 20)。5分钟后，咖啡温度为60°C。求15分钟后的温度。

Separate variables: ∫ dT/(T−20) = ∫ −k dt ⇒ ln|T−20| = −kt + C. At t=0, T=80 ⇒ C = ln(60). So T−20 = 60 e⁻ᵏᵗ. At t=5, T=60 ⇒ 40 = 60 e⁻⁵ᵏ ⇒ e⁻⁵ᵏ = 40/60 = 2/3 ⇒ −5k = ln(2/3) ⇒ k = −⅕ ln(2/3) = ⅕ ln(1.5).

分离变量：∫ dT/(T−20) = ∫ −k dt ⇒ ln|T−20| = −kt + C。t=0时T=80，得C = ln(60)。故T−20 = 60 e⁻ᵏᵗ。t=5时T=60 ⇒ 40 = 60 e⁻⁵ᵏ ⇒ e⁻⁵ᵏ = 40/60 = 2/3 ⇒ −5k = ln(2/3) ⇒ k = −⅕ ln(2/3) = ⅕ ln(1.5)。

nutes: T−20 = 60 e⁻¹⁵ᵏ = 60 (e⁻⁵ᵏ)³ = 60 (2/3)³ = 60 × (8/27) ≈ 17.78°C. Hence T ≈ 37.8°C.

15分钟后：T−20 = 60 e⁻¹⁵ᵏ = 60 (e⁻⁵ᵏ)³ = 60 (2/3)³ = 60 × (8/27) ≈ 17.78°C。故T ≈ 37.8°C。

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9. Sequences and Series: Loan Repayment with Geometric Sum | 序列与级数：等比数列求和与贷款偿还

A borrower takes out a loan of £10,000 at an annual compound interest rate of 5%. Repayments of £R are made at the end of each year. The loan is to be cleared after 4 years. Formulate the outstanding balance as a geometric series and find R.

借款人贷款£10,000，年复利5%，在每年末偿还£R。贷款需在4年后还清。将未偿还余额表示为等比数列，并求R。

first term 1, ratio 1.05, 4 terms. Sum = 1 × (1.05⁴ − 1) / (1.05 − 1) = (1.05⁴ − 1)/0.05.

4年后未偿还余额：10000 × 1.05⁴ − R(1.05³ + 1.05² + 1.05 + 1) = 0。括号内为首项1、公比1.05、项数4的等比数列。和 = 1 × (1.05⁴ − 1) / (1.05 − 1) = (1.05⁴ − 1)/0.05。

10000 × 1.215506 − R × (1.215506 − 1)/0.05 = 0
12155.06 − R × 4.31012 = 0 ⇒ R = 12155.06 / 4.31012 ≈ 2820.09

The annual repayment is approximately £2820.09.

每年还款额约为£2820.09。

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10. Parametric Equations: Tangent and Normal Lines | 参数方程：切线与法线

1, y = t³ − 3t. Find the equation of the tangent and the normal at the point where t = 2.

一条曲线由参数方程x = t² + 1，y = t³ − 3t定义。求t = 2处的切线与法线方程。

adient dy/dx = (dy/dt)/(dx/dt) = (3t² − 3)/(2t). At t=2, m = (3×4−3)/(4) = (12−3)/4 = 9/4.

t=2时：x=5, y=2。导数：dx/dt = 2t，dy/dt = 3t² − 3。斜率dy/dx = (3t²−3)/(2t)。t=2时，m = (12−3)/4 = 9/4。

Tangent line: y − 2 = (9/4)(x − 5) ⇒ 4y − 8 = 9x − 45 ⇒ 9x − 4y = 37.

切线方程：y − 2 = (9/4)(x − 5) ⇒ 9x − 4y = 37。

Normal gradient = −4/9. Normal line: y − 2 = (−4/9)(x − 5) ⇒ 9y − 18 = −4x + 20 ⇒ 4x + 9y = 38.

法线斜率 = −4/9。法线方程：y − 2 = (−4/9)(x − 5) ⇒ 4x + 9y = 38。

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