= Solution
In <polar coordinates>,
$$
v\,dt=ds=\sqrt{dr^2+r^2d\theta^2}.
$$
With the change of radial coordinate $u=\log r$, the detection functional becomes
$$
P=\lambda\int\frac{ds}{r}
=\lambda\int\sqrt{du^2+d\theta^2}.
$$
It is therefore $\lambda$ times ordinary <path length> in the $(u,\theta)$-plane. The shortest path is the line segment from $(\log A,0)$ to $(\log B,\alpha)$, so
$$
u(\theta)=\log A+\frac{\log(B/A)}{\alpha}\theta.
$$
Returning to $r$ gives the <logarithmic spiral>
$$
\boxed{r(\theta)=A\exp\!\left(\frac{\log(B/A)}{\alpha}\theta\right)}.
$$
Its minimum detection probability is
$$
P_{\min}=\lambda\sqrt{\log^2(B/A)+\alpha^2}.
$$
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 $\lambda$ and use the <Lagrangian>
$$
L=\frac{\dot r^2}{r}+r\dot\theta^2.
$$
The coordinate $\theta$ is cyclic, so its <conjugate momentum> is conserved. Equivalently,
$$
\boxed{h=r\dot\theta}
$$
is constant. Because $L$ has no explicit time dependence, the associated <conservation of energy> gives the constant
$$
\boxed{E=\frac{\dot r^2}{r}+r\dot\theta^2}.
$$
Multiplying by $r$ and using $h=r\dot\theta$ gives the required equation
$$
\boxed{\dot r^2=Er-h^2}.
$$
Back to article page
