Answer only one of the following two alternatives. OR A curve C has parametric equations x = 1 − 3t², y = t(1 – 3t²), for 0 ≤ t ≤ 1/√3. Show that (dx/dt)² + (dy/dt)² = (1+9t²)². Hence find (i) the arc length of C, (ii) the surface area generated when C is rotated through 2π radians about the x-axis. Use the fact that t = y/x to find a cartesian equation of C. Hence show that the polar equation of C is r = sec θ(1 – 3 tan²θ), and state the domain of θ. Find the area of the region enclosed between C and the initial line.

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2016). The mark or mark type is shown in brackets.

1. [M1] dx/dt = -6t, dy/dt = (1-3t²) + t(-6t) = 1-3t²-6t² = 1-9t².
2. [A1] (dx/dt)² + (dy/dt)² = (-6t)² + (1-9t²)² = 36t² + 1 - 18t² + 81t⁴ = 1 + 18t² + 81t⁴ = (1+9t²)². (AG)
3. [M1A1] (i) Arc length L = ∫₀^(1/√3) √((dx/dt)² + (dy/dt)²) dt = ∫₀^(1/√3) (1+9t²) dt = [t + 3t³]₀^(1/√3) = (1/√3) + 3(1/√3)³ = 1/√3 + 3(1/(3√3)) = 1/√3 + 1/√3 = 2/√3.
4. [M1] (ii) Surface Area S = ∫₀^(1/√3) 2πy √((dx/dt)² + (dy/dt)²) dt = ∫₀^(1/√3) 2π t(1-3t²)(1+9t²) dt.
5. [M1A1] = 2π ∫₀^(1/√3) (t+6t³-27t⁵) dt = 2π [1/2 t² + 3/2 t⁴ - 9/2 t⁶]₀^(1/√3). = 2π [1/2(1/3) + 3/2(1/9) - 9/2(1/27)] = 2π [1/6 + 1/6 - 1/6] = π/3.
6. [M1] (Cartesian equation) Substitute t=y/x into x=1-3t²: x = 1 - 3(y/x)² => x² = x - 3y² => x² - x + 3y² = 0.
7. [M1A1] (Polar equation) Use x=r cos θ, y=r sin θ. From x = 1-3t², t=y/x=tan θ. So x = 1-3tan²θ. Substitute x=r cos θ: r cos θ = 1-3tan²θ => r = sec θ (1-3tan²θ). (Shown)
8. [B1] Domain of θ: Since t = tan θ and 0 ≤ t ≤ 1/√3, 0 ≤ tan θ ≤ 1/√3. Thus 0 ≤ θ ≤ π/6.
9. [M1] Area = ∫₀^(π/6) 1/2 r² dθ = 1/2 ∫₀^(π/6) sec² θ (1-3tan²θ)² dθ.
10. [A1A1] = 1/2 ∫₀^(π/6) sec² θ (1 - 6tan²θ + 9tan⁴θ) dθ. Let u = tan θ, du = sec² θ dθ. When θ=0, u=0. When θ=π/6, u=1/√3. = 1/2 ∫₀^(1/√3) (1 - 6u² + 9u⁴) du = 1/2 [u - 2u³ + 9/5 u⁵]₀^(1/√3). = 1/2 [1/√3 - 2(1/√3)³ + 9/5 (1/√3)⁵] = 1/2 [1/√3 - 2/(3√3) + 9/(5 * 9√3)] = 1/2 [1/√3 - 2/(3√3) + 1/(5√3)] = 1/2 [ (15 - 10 + 3)/(15√3) ] = 1/2 [8/(15√3)] = 4/(15√3) = 4√3/45.