 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ib/paper-1/13d/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

v d t = d s = d r^{2} + r^{2} d θ^{2} .

With the change of radial coordinate

u = lo g r

, the detection functional becomes

P = λ \int \frac{d s}{r} = λ \int d u^{2} + d θ^{2} .

It is therefore

λ

times ordinary path length in the

(u, θ)

-plane. The shortest path is the line segment from

(lo g A, 0)

(lo g B, α)

, so

u (θ) = lo g A + \frac{l o g ( B / A )}{α} θ .

Returning to

r

r (θ) = A exp (\frac{lo g ( B / A )}{α} θ) .

Its minimum detection probability is

P_{m i n} = λ lo g^{2} (B / A) + α^{2} .

The functional depends only on the geometric path and not its parametrization, so tiptoeing and running give the same probability.

For the improved sensor, omit the irrelevant positive factor

λ

and use the Lagrangian

L = \frac{r ˙ ^{2}}{r} + r \dot{θ}^{2} .

The coordinate

θ

is cyclic, so its conjugate momentum is conserved. Equivalently,

h = r \dot{θ}

is constant. Because

L

has no explicit time dependence, the associated conservation of energy gives the constant

E = \frac{r ˙ ^{2}}{r} + r \dot{θ}^{2} .

Multiplying by

r

and using

h = r \dot{θ}

gives the required equation

\overset{r}{˙}^{2} = E r - h^{2} .

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