Radiodrome

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 Ancient Greek: and δρόμος, drómos, '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 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.

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 be changing his heading continuously.

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 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

[math]\displaystyle{ (T_x\ ,\ T_y)\ =\ (A_x\ ,\ A_y+V_t t)~. }[/math]

(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

[math]\displaystyle{ \dot x= V_d\ \frac{T_x-x}{\sqrt{(T_x-x)^2+(T_y-y)^2}} }[/math]

(2)

[math]\displaystyle{ \dot y= V_d\ \frac{T_y-y}{\sqrt{(T_x-x)^2+(T_y-y)^2}} ~. }[/math]

(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

[math]\displaystyle{ y'(x)=\frac{T_y-y}{T_x-x} }[/math] .

(4)

Multiplying both sides with [math]\displaystyle{ T_x-x }[/math] and taking the derivative with respect to x, using that

[math]\displaystyle{ \frac{dT_y}{dx}\ =\ \frac{dT_y}{dt}\ \frac{dt}{dx}\ =\ \frac{V_t}{V_d}\ \sqrt{{y'}^2+1}~ , }[/math]

(5)

one gets

[math]\displaystyle{ y''=\frac{V_t\ \sqrt{1+{y'}^2}}{V_d(A_x-x)} }[/math]

(6)

or

[math]\displaystyle{ \frac{y''}{\sqrt{1+{y'}^2}}=\frac{V_t}{V_d(A_x-x)} ~. }[/math]

(7)

From this relation, it follows that

[math]\displaystyle{ \sinh^{-1}(y')=B-\frac{V_t}{V_d}\ \ln(A_x-x)~, }[/math]

(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.,

[math]\displaystyle{ B=\frac{V_t}{V_d}\ \ln(A_x)+\ln\left(y'(0)+\sqrt{{y'(0)}^2+1}\right) . }[/math]

(9)

From (8) and (9), it follows after some computation that

[math]\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] }[/math] .

(10)

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

[math]\displaystyle{ y'(0)= \frac{A_y}{A_x} }[/math] .

(11)

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

[math]\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], }[/math]

(12)

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

[math]\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] }[/math].

(13)

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

[math]\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\} }[/math] .

(14)

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

[math]\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] }[/math] .

(15)

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

[math]\displaystyle{ C = \frac{A_x}{4}\left(y'(0)-\sqrt{{y'(0)}^2+1}\right) . }[/math]

(16)

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

[math]\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) }[/math] .

(17)

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

[math]\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) . }[/math]

(18)

If V_t ≥ V_d, one has from (14) and (17) that [math]\displaystyle{ \lim_{x \to A_x}y(x) = \infty }[/math], which means that the hare will never be caught, whenever the chase starts.

See also

• Mice problem

References

• 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, 2, Coimbra, pp. 255, http://quod.lib.umich.edu/u/umhistmath/aat2332.0005.001/261?view=pdf