 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2015/ia/paper-3/4a/b/solution

Codex  0 Created 2026-09-24 Updated 2026-10-06

Differentiate the unit tangent vector with

d / d s = v^{- 1} d / d u

and

v^{'} = (x^{'} \cdot x^{''}) / v

. This gives

\frac{d t}{d s} = \frac{x ^{''}}{v ^{2}} - \frac{x ^{'} ( x ^{'} \cdot x ^{''} )}{v ^{4}} .

κ = ∣ d t / d s ∣

. Squaring the displayed expression and simplifying gives

κ^{2} = \frac{v ^{2} ∣ x ^{''} ∣ ^{2} - ( x ^{'} \cdot x ^{''} ) ^{2}}{v ^{6}} = \frac{∣ x ^{'} \times x ^{''} ∣ ^{2}}{v ^{6}}, κ = \frac{∣ x ^{'} \times x ^{''} ∣}{∣ x ^{'} ∣ ^{3}} .

The last equality is the cross product norm identity. Equivalently, the hinted vector triple product rewrites

d t / d s = - x^{'} \times (x^{'} \times x^{''}) / v^{4}

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