 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ia/paper-3/9b/b/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Differentiation gives

r^{'} (t) = 3 sin t cos t (- 2 cos t, sin t, 3 sin t),

and

∣ r^{'} ∣ = 6∣ sin t cos t ∣

. Thus

T = \frac{σ}{2} (- 2 cos t, sin t, 3 sin t), σ = sgn (sin t cos t) .

The plus sign applies on

(0, π /2) \cup (π, 3 π /2)

and the minus sign on

(π /2, π) \cup (3 π /2, 2 π)

On each open quadrant,

N = σ (sin t, \frac{1}{2} cos t, \frac{3}{2} cos t), B = (0, \frac{3}{2}, - \frac{1}{2}) .

The constant binormal proves that the curve is planar; indeed

z = 3 y

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