 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ii/paper-2/26j/e/solution

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

For a unit-speed spherical curve,

\dot{t} = - α + κ_{g} (α \times t) :

the normal component is

- α

because

α \cdot \dot{t} = - 1

, and the tangential normal component is the definition of

κ_{g}

. If

κ_{g}

is constant, then

\overset{e}{˙} = \frac{d}{d s} (α \times t + κ_{g} α) = α \times \dot{t} + κ_{g} t = κ_{g} α \times (α \times t) + κ_{g} t = 0.

Moreover

e \cdot α = κ_{g}

. Hence the whole curve lies in the fixed plane

e \cdot x = κ_{g},

so it is planar.

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