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ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2023/prelims/mathematics-iv/3/a/iii/solution

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Past exam of the mathematics course of the University of Oxford / 2023 / Prelims / Mathematics IV / 3 / a / iii / Solution by

Codex  0 2026-10-25

For the cardioid,

r^{'} = - a sin θ

. The expression under the square root simplifies to

r^{2} + (r^{'})^{2} = a^{2} (2 + 2 cos θ) = 4 a^{2} cos^{2} (θ /2) .

Since

0 \leq θ \leq θ_{2} \leq π

, the cosine is nonnegative. Therefore

s (θ_{2}) = \int_{0}^{θ_{2}} 2 a cos (θ /2) d θ = 4 a sin (θ_{2} /2) .

The upper half of the cardioid starts at

(2 a, 0)

, passes through

(0, a)

, bends to negative

x

, and ends at the cusp

(0, 0)

when

θ_{2} = π

. Its full upper-half arc length is

4 a

; reflecting gives

8 a

for the full cardioid.

The fundamental theorem of calculus gives

\frac{d s}{d θ _{2}} = 2 a cos (θ_{2} /2) .

This decreases from

2 a

to zero, so the arc length increases fastest at $θ_{2} = 0$ , interpreted as the right-hand rate at the initial endpoint. If only positive endpoints are allowed, the rate has supremum

2 a

θ_{2} ↓ 0

, rather than an attained interior maximum.

Figure 1.
A partial cardioid with $θ_{2} = 2 π /3$ , and the continuation to its cusp
. Details.

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