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 Latin word radius (Eng. ray; spoke) and the Greek word dromos (Eng. running; racetrack), for there is a radial component in its kinematic analysis. 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 something it has spotted on the other side. Because the dog drifts with the current, it will have to change its heading; it will also have to swim further than if it had taken 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.

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 V_t. The position of the hare at time zero is (A_x, A_y) with A_x > 0 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 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 − (V_t /V_d) lnA_x), 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 computation that

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

(10)

Furthermore, since y(0)=0, it follows from (1) and (4) that

${\displaystyle y'(0)={\frac {A_{y)){A_{x))))$ .

(11)

If, now, V_t ≠ V_d, relation (10) integrates to

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

(12)

where C is the constant of integration. Since again y(0)=0, it's

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

(13)

The equations (11), (12) and (13), then, together imply

${\displaystyle y={\frac {1}{2))\left\((\frac {A_{y}+{\sqrt {A_{x}^{2}+A_{y}^{2)))){1-{\frac {V_{t)){V_{d))))}\left[1-\left(1-{\frac {x}{A_{x))}\right)^{1-{\frac {V_{t)){V_{d))))\right]+{\frac {A_{y}-{\sqrt {A_{x}^{2}+A_{y}^{2)))){1+{\frac {V_{t)){V_{d))))}\left[1-\left(1-{\frac {x}{A_{x))}\right)^{1+{\frac {V_{t)){V_{d))))\right]\right\))$ .

(14)

If V_t = V_d, relation (10) gives, instead,

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

(15)

Using y(0)=0 once again, it follows that

${\displaystyle C={\frac {A_{x)){4))\left(y'(0)-{\sqrt ((y'(0)}^{2}+1))\right).}$

(16)

The equations (11), (15) and (16), then, together imply that

${\displaystyle y={\frac {1}{4))\left(A_{y}-{\sqrt {A_{x}^{2}+A_{y}^{2))}\right)\left[1-\left(1-{\frac {x}{A_{x))}\right)^{2}\right]-{\frac {1}{2))\left(A_{y}+{\sqrt {A_{x}^{2}+A_{y}^{2))}\right)\ln \left(1-{\frac {x}{A_{x))}\right)}$ .

(17)

If V_t < V_d, it follows from (14) that

${\displaystyle \lim _{x\to A_{x))y(x)={\frac {1}{2))\left({\frac {A_{y}+{\sqrt {A_{x}^{2}+A_{y}^{2)))){1-{\frac {V_{t)){V_{d))))}+{\frac {A_{y}-{\sqrt {A_{x}^{2}+A_{y}^{2)))){1+{\frac {V_{t)){V_{d))))}\right).}$

(18)

If V_t ≥ V_d, one has from (14) and (17) that ${\displaystyle \lim _{x\to A_{x))y(x)=\infty }$, which means that the hare will never be caught, whenever the chase starts.

• Mice problem

• Nahin, Paul J. (2012), Chases and Escapes: The Mathematics of Pursuit and Evasion, Princeton: Princeton University Press, ISBN 978-0-691-12514-5.
• Gomes Teixera, Francisco (1909), Imprensa da universidade (ed.), Traité des Courbes Spéciales Remarquables, vol. 2, Coimbra, p. 255((citation)): CS1 maint: location missing publisher (link)