At any point (x, y) on the curve C, dx/dt = t(t²+4) and dy/dt = -t√(4-t²), where the parameter t is such that 0 ≤ t ≤ 2.

Mark scheme

Marking points from the official mark scheme (May/June 2010). The mark or mark type is shown in brackets.

1. [1] (a) Calculate ds/dt = √((dx/dt)² + (dy/dt)²) = √(t²(t²+4)² + t²(4-t²)) = √(t²(t⁴+8t²+16) + t²(4-t²)) = √(t²(t⁴+8t²+16+4-t²)) = √(t²(t⁴+7t²+20)) = √(8t²) = 2√2t
2. [1] (a) Set up integral for arc length: s = ∫_0^2 2√2t dt
3. [1] (a) Integrate and evaluate: s = 2√2 [t²/2]_0^2 = 2√2 (4/2) = 4√2 (AG)
4. [1] (b) Find y from dy/dt: y = ∫ -t√(4-t²) dt = (1/3)(4-t²)^(3/2). Use y=0 at t=2 to find constant of integration, which is 0.
5. [1] (b) Set up surface area integral: S = ∫_0^2 2πy (ds/dt) dt = ∫_0^2 2π(1/3)(4-t²)^(3/2) (2√2t) dt
6. [1] (b) Integrate: S = -(4√2π/15)(4-t²)^(5/2) evaluated from 0 to 2
7. [1] (b) Evaluate: S = 128√2π/15