Now, the duck will travel a small distance epsilon out to the edge of the pond, and the wolf must choose to go in the direction of increasing θ or decreasing θ. Let us assume the wolf goes in the direction of increasing θ. The duck will now head for the point of escape (assuming there is one) at R=1, θ=φ. There is little point in the wolf changing directions now, as the duck will have gained a small distance epsilon toward the edge again before the wolf just gets back to its original location. There is also little point in the duck taking any path more complicated than a straight line, as wherever the escape point is, the duck might as well go straight there.

The wolf must get to the position φ before the duck can. The wolf must travel a distance φ+π. The duck must travel the distance D, given by the cosine law as:

D^{2}= 1+1/v^{2}-2/v cos(φ)

Now, let f(φ) be the function representing the amount of time between when the duck gets to the escape point and the wolf gets to the escape point. Specifically, f(φ) = D-φ+π)/v. Maximizing f(φ) gives one:

sin(φ) = D

So this means that the duck will travel tangental to the circle of radius 1/v. When sin(φ)=D we know cos(φ)=1/v, if this isn't obvious, you should probably be drawing a picture to follow along, I certainly needed to.

The value of f(φ) when sin(φ)=D (and thus cos(φ)=1/v) can be given as sin(φ)-(φ+π)cos(φ). If we set that to zero (to find the situation when the duck no longer beats the wolf to the edge) we find that:

tan(φ)= φ+π

Which has a solution at φ≈1.35 . Thus, v=1/cos(φ)≈4.6 is when this strategy will fail for the duck.

As far as I can tell, when the wolf speed is greater than this value, the duck can do nothing to escape, but I have found no proof of this.

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