In geometry, a radiodrome is the pursuit curve followed by a point that is pursuing another linearly-moving point. The term is derived from the Greek words "ῥάδιος" (easier) and "δρόμος" (running). The classic (and best-known) form of a radiodrome is known as the "dog curve"; this is the path a dog follows when it swims across a stream with a current after food it has spotted on the other side. Because the dog drifts downwards with the current, it will have to change its heading; it will also have to swim further than if it had computed the optimal heading. This case was described by Pierre Bouguer in 1732.

A radiodrome may alternatively be described as the path a dog follows when chasing a hare, assuming that the hare runs in a straight line at a constant velocity. It is illustrated by the following figure:

The path of a dog chasing a hare running along a vertical straight line at a constant speed. The dog runs towards the momentary position of the hare, and will have to be changing his heading continuously. The speed of the dog is 20% faster than the speed of the hare.

## Mathematical analysis

Introduce a coordinate system with origin at the position of the dog at time zero and with y-axis in the direction the hare is running with the constant speed ${\displaystyle V_{t}}$. The position of the hare at time zero is (Ax, Ay) and at time t it is

${\displaystyle (T_{x}\ ,\ T_{y})\ =\ (A_{x}\ ,\ A_{y}+V_{t}t)}$

(1)

The dog runs with the constant speed ${\displaystyle V_{d}}$ towards the instantaneous position of the hare.

The differential equation corresponding to the movement of the dog, (x(t), y(t)), is consequently

${\displaystyle {\dot {x}}=V_{d}\ {\frac {T_{x}-x}{\sqrt {(T_{x}-x)^{2}+(T_{y}-y)^{2}}}}}$

(2)

${\displaystyle {\dot {y}}=V_{d}\ {\frac {T_{y}-y}{\sqrt {(T_{x}-x)^{2}+(T_{y}-y)^{2}}}}}$

(3)

It is possible to obtain a closed-form analytic expression y=f(x) for the motion of the dog, From (2) and (3) it follows that

${\displaystyle y'(x)={\frac {T_{y}-y}{T_{x}-x}}}$

(4)

Multiplying both sides with ${\displaystyle T_{x}-x}$ and taking the derivative with respect to x using that

${\displaystyle {\frac {dT_{y}}{dx}}\ =\ {\frac {dT_{y}}{dt}}\ {\frac {dt}{dx}}\ =\ {\frac {V_{t}}{V_{d}}}\ {\sqrt {{y'}^{2}+1}}}$

(5)

one gets

${\displaystyle y''={\frac {V_{t}\ {\sqrt {1+{y'}^{2}}}}{V_{d}(A_{x}-x)}}}$

(6)

or

${\displaystyle {\frac {y''}{\sqrt {1+{y'}^{2}}}}={\frac {V_{t}}{V_{d}(A_{x}-x)}}}$

(7)

From this relation it follows that

${\displaystyle \sinh ^{-1}(y')=B-{\frac {V_{t}}{V_{d}}}\ \ln(A_{x}-x)}$

(8)

where B is the constant of integration determined by the initial value of y' at time zero, y' (0)= sinh(B − (Vt /Vd) lnAx), i.e.,

${\displaystyle B={\frac {V_{t}}{V_{d}}}\ \ln(A_{x})+\ln \left(y'(0)+{\sqrt {{y'(0)}^{2}+1}}\right)}$

(9)

From (8) and (9) it follows after some computations that

${\displaystyle y'={\frac {1}{2}}\left({\frac {y'(0)+{\sqrt {{y'(0)}^{2}+1}}}{(1-{\frac {x}{A_{x}}})^{\frac {V_{t}}{V_{d}}}}}-{\frac {(1-{\frac {x}{A_{x}}})^{\frac {V_{t}}{V_{d}}}}{y'(0)+{\sqrt {{y'(0)}^{2}+1}}}}\right)}$

(10)

If, now, Vt ≠ Vd, this relation integrates to

${\displaystyle y=C-{\frac {1}{2}}\ A_{x}\left({\frac {(y'(0)+{\sqrt {{y'(0)}^{2}+1}})\ (1-{\frac {x}{A_{x}}})^{1-{\frac {V_{t}}{V_{d}}}}}{1-{\frac {V_{t}}{V_{d}}}}}-{\frac {(1-{\frac {x}{A_{x}}})^{1+{\frac {V_{t}}{V_{d}}}}}{(y'(0)+{\sqrt {{y'(0)}^{2}+1}})\ (1+{\frac {V_{t}}{V_{d}}})}}\right)}$

(11)

where C is the constant of integration.

If Vt = Vd, one gets instead

${\displaystyle y=C-{\frac {1}{2}}A_{x}\ \left(\left(y'(0)+{\sqrt {{y'(0)}^{2}+1}}\right)\ \ln(1-{\frac {x}{A_{x}}})-{\frac {(1-{\frac {x}{A_{x}}})^{2}}{(y'(0)+{\sqrt {{y'(0)}^{2}+1}})\ 2}}\right)}$

(12)

If Vt < Vd, it follows from (11) that

${\displaystyle \lim _{x\to A_{x}}y(x)=C={\frac {1}{2}}\ A_{x}\left({\frac {y'(0)+{\sqrt {{y'(0)}^{2}+1}}}{1-{\frac {V_{t}}{V_{d}}}}}-{\frac {1}{(y'(0)+{\sqrt {{y'(0)}^{2}+1}})\ (1+{\frac {V_{t}}{V_{d}}})}}\right)}$

(13)

In the case illustrated in the figure above, VtVd = 11.2 and the chase starts with the hare at position (Ax, −0.6 Ax) which means that y'(0) = −0.6. From (13) it thus follows that the hare is caught at position (Ax, 1.21688Ax), and consequently that the hare will have run the total distance (1.21688 + 0.6) Ax before being caught. The arclength of the dog's path is (13) multiplied by VdVt =1.2 .

If Vt ≥ Vd, one has from (11) and (12) that ${\displaystyle \lim _{x\to A_{x}}y(x)=\infty }$, which means that the hare will never be caught, whenever the chase starts.