Hawk Curved

How to find the flight path of a hawk?

Suppose that a hawk, whose initial position is (a,0)=(3000,0) on the -axis, spots a pigeon at (0,1000) on the -axis. Suppose that the pigeon flies at a constant speed of 60 ft/sec in the direction of the -axis (oblivious to the hawk), while the hawk flies at a constant speed of 70 ft/sec, always in the direction of the pigeon.

As I promised in my reply to the personal email with the invitation to answer, here is the detailed solution.
In problems like this the pursuit curves are complicated algebraic curves, obtaining their equations requires solving 2nd order differential equations like this one you have seen:

http://answers.yahoo.com/question/index;_ylt=ArxR1eAkdavqBJkc6YDeWIAjzKIX;_ylv=3?qid=20071201152703AAWCxon

Let H₀(a, 0) and P₀(0, b), /a, b = const > 0/ are the initial positions of the hawk and the pigeon; h and p are their velocities /h > p > 0, in your example a = 3000, b = 1000,
h = 70, p = 60/, y = y(x) is the required equation of the pursuit curve. Let in some moment of the pursuit the hawk is in the point H(x, y) in the 1st quadrant, the pigeon – in P(0, η), then the distances, covered by both birds, are proportional to their velocities:

1) Segment_Length(P₀P) : Arc_Length(H₀H) = p : h = r

(r = p/h is ratio of both velocities, 0 < r < 1 and the line HP is a tangent to y(x) - the velocity vector of the hawk points always to the pigeon). The equation of HP is
HP: η - y = y'(ξ - x) and the ordinate η of P is η = y - xy', then 1) becomes
(y - xy' - b) : ∫ [x, a] √(1 + y'²) dx = r, or, differentiating with respect to x and simplifying, we get the differential equation of the pursuit curve:

2) xy" = r√(1 + y'²), with initial conditions y(a) = 0 /point H₀/, y'(a) = -b/a /initially the velocity vector of the hawk points to P₀ - consider triangle OH₀P₀; O - origin/

Let z(x) = y'(x), then 2) implies xz' = r√(1 + z²) or dz/√(1 + z²) = r(dx/x);
integrating we get ln(z + √(1 + z²)) = ln(Cx^r) or, back to y':

3) y' + √(1 + y'²) = Cx^r

where the constant C is to be determined by the 2nd initial condition:

4) C = (√(a² + b²) - b)/a^(r+1) using y'(a) = -b/a.

Solving 3) with respect to y' and simplifying, we get the following easily integrable 1st order differential equation:
y' = (C/2) x^r - (1/(2C)) x^(-r)
y = (Cx^(1+r)/(1+r) - x^(1-r)/(C(1-r)))/2 + D, the constant D to be determined from the 1st initial condition y(a) = 0:
D = (-Ca^(1+r)/(1+r) + a^(1-r)/(C(1-r)))/2, finally the required equation is

y(x) = (1/2) * (C(x^(1+r) - a^(1+r))/(1+r) - (x^(1-r) - a^(1-r))/(C(1-r))),
with expression for C in 4) above.

The meeting point, being on the y-axis, has coordinates (0, y(0)):
y(0) = (1/2) * (a^(1-r))/(C(1-r)) - C(a^(1+r))/(1+r))
the time of the pursuit is (y(0) - b)/p.
The numeric values are: r = 60/70 ≈ 0.857, C ≈ 0.000754,
y(0) ≈ 14566 (ft), time ≈ 226 (sec.)

Cold Steel Norse Hawk (Tomahawks)

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