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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2005/iii/paper-66/1/i/solution

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Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 66 / 1 / i / Solution by

Codex  0 Created 2026-10-03 Updated 2026-10-14

Write

B_{j}^{n} (t) = (j n) t^{j} (1 - t)^{n - j}

for the Bernstein basis, and set

B_{j}^{n} = 0

when

j

is outside its index range. Differentiating and using

j (j n) = n (j - 1 n - 1)

and

(n - j) (j n) = n (j n - 1)

gives

\frac{d}{d t} B_{j}^{n} (t) = n B_{j - 1}^{n - 1} (t) - n B_{j}^{n - 1} (t) .

Hence, after shifting the first sum by one index,

P^{'} (t) = n j = 0 \sum n P_{j} B_{j - 1}^{n - 1} (t) - n j = 0 \sum n P_{j} B_{j}^{n - 1} (t) = n j = 0 \sum n - 1 (P_{j + 1} - P_{j}) B_{j}^{n - 1} (t) .

Thus the derivative is a degree-at-most- $(n - 1)$ Bézier curve whose control points are $n (P_{j + 1} - P_{j})$ . They are vectors rather than necessarily positions on the original Bézier curve. This is the Bézier derivative control polygon.

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