Suppose that a hawk, whose initial position is (a,0)=(3000,0) on the x-axis, spo

Question

Suppose that a hawk, whose initial position is (a,0)=(3000,0) on the x-axis, spots a pigeon at (0,-4000) on the y-axis. Suppose that the pigeon flies at a constant speed of 10 ft/sec in the direction of the y-axis (oblivious to the hawk), while the hawk flies at a constant speed of 60 ft/sec, always in the direction of the pigeon.

I have already finished most of the problem, but I can not figure out how to do the last questions.

The problem is to find an equation for the flight path of the hawk (the curve of pursuit) and to find the time and place where the hawk will catch the pigeon. Assume that in this problem all distances are measured in feet and all times measured in seconds. Leave out all dimensions from your answers.

The pigeon's position Q=(0,g(t)) is given by the following function of time
g(t) = -4000+10t

The fact that the hawk is always headed in the direction of the pigeon means that the line PQ is tangent to the pursuit curve y=f(x). This tells us that dydx=h(x,y,t) where
h(x,y,t) = (y- -4000- 10*t)/x

If we solve the equation

p=h(x,y,t),
where p=dydx, for time we obtain that t=k(x,y,p) where

k(x,y,p)= = (y-p*x- -4000)/10

Again referring to the diagram above (click on it for a better view) we see that the distance that the hawk has flown in time t is given by the integral cdFdx where
c = x
d = 3000
(Hints: Note that this integral computes the length of a curve. Also recall that the hawk's initial position is at (a,0)=(3000,0). )
and
F = sqrt(1+p^2)
(Use p to denote dydx in your last answer above.)

On the other hand the hawk is flying at a constant speed of 60 for time t. Hence the total distance it has flown is 60t. If we equate this to the distance we just computed and solve for t we obtain

t=cdGdx
where G = sqrt(1+p^2)/60
(Remember to use p to represent dydx.)

Equating the two expressions for t from Problems 1 and 2 we obtain the integral equation

k(x,y,p)=cdGdx
To get rid of the integral, we differentiate both sides of the equation with respect to x. On the left hand side of the resulting equation we obtain the following expression (which might involve x, y, p=dydx and q=dpdx=d2ydx2
-x*q/10
(remember to separate different variables in a product with spaces or multiplication signs)
while on the right hand side (after applying the Fundamental Theorem of Calculus) we obtain
-sqrt(1+p^2)/60
The resulting differential equation we obtained above is a separable differential equation in the variables p and x. We can rewrite it in the form

K(p)dp=L(x)dx
(with all numerical factors moved to the right hand side of the equation so that L(1)=10/60.) where
K(p) = 1/sqrt(1+p^2)
L(x) = (10/60)*(1/x)

Integrating the left hand side of the equation K(p)dp, using the methods of sections 8.3 and 8.2, we obtain
K(p)dp = ln(p+sqrt(1+p^2))
while on the right hand side we obtain L(x)dx =(10/60)*ln(x) +C

Plugging in the initial positions of the hawk and pigeon, and recalling that p=dydx is the slope of the tangent line, we find that
C = ln((4/3)+sqrt(1+(4/3)^2))-(10/60)ln(3000)

Note that it is important to compute C to at least 5 or 6 decimal places of precision. While computing C to less precision may not affect the correctness of your answer at this point, the roundoff error tends to get magnified in your subsequent calculations in problems 5 and 6 and may lead to your answers there being unexpectedly rejected as incorrect.

Solving the equation we obtained in Problem 4 for p in terms of x, we obtain p =

(Hints for solving for p: Exponentiate to get rid of the logarithm. Then isolate the square root on one side of the equation and square both sides.)

Recalling that p=dydx and integrating. we obtain that y = +C

Plugging in the initial position of the hawk we obtain that the constant of integration is given by C =

Note: if your answer is unexpectedly rejected as incorrect, go back to problem 4 part c, recompute your constant of integration there to more decimal places and recalculate your answers here with this more accurate value. Hence the hawk catches the pigeon at the point (0,c) where c = at time t =

Explanation / Answer

Let the hawk's starting point is A(a, 0) , the pigeon's - B(0, b) and both birds flight speeds are v1 and v2 .
Let also R = v2/v1 be their ratio, y = y(x) - the equation of the pursuit curve.
At moment t>0 the pigeon will be in (0, b + v2*t) and the hawk - in point P(x, y), such that the length of the pursuit curve's arc between A and P
[x, a](1 + y'²) dx = v1*t
and the hawk's velocity vector has the direction of the tangent to y = y(x), its equation being:
- y = y'( - x),
this tangent intersecting x-axis at point T'(x - y/y', 0) and y-axis at T"(0, b + v2*t). For the slope angle we have:
tan = y' = |OT"| / |OT'| = |b + v2*t| / (x - y'/y), or
v2*t = -b + y - xy',           differentiating by x we obtain
dt/dx = (-1/v2)*xy",    differentiating the integral above:
dt/dx = (-1/v1)*(1+y'²), so
(1/v2) xy" = (1/v1)*(1 + y'²)

and we arrive to the following 2nd order differential equation for the pursuit curve:

xy" = R(1 + y'²), initial conditions
y(a) = 0 point A above,
y'(a) = -b/a /triangle OT'T"

Integrating once we obtain
ln(y' + (1 + y'²)) = R*ln x + C1 /C1 = const to be determined from the 2nd initial condition/, or
y' + (1 + y'²)) = D*x^R, here D =(-b + (a² + b²))/a^(1+R)
That leads to
y' = (D*x^R - 1/(D*x^R))/2,
integrating 2nd time, we obtain for y
y = (D*x^(1+R)/(1+R) - x^(1-R)/(D(1-R)))/2 + C2, /C2 = const to be determined from the 1st initial condition/, finally

y = (1/2)(D(x^(1+R) - a^(1+R))/(1+R) - (x^(1-R) - a^(1-R))/(D(1-R)))

The meeting point M(0, y(0)), the time of pursuit is
|BM| / v2 = (y(0) - b)/v2

We have a = 3000, b = -4000, R = 10/60 = 1/6,
D = (-b + (a² + b²))/a^(1+R)

= (4000 + 5000)/(3000^(7/6)) 0.078995,
the equation of the pursuit curve is:
y(x) =(1/2)(D(x^(1+R) - a^(1+R))/(1+R) - (x^(1-R) - a^(1-R))/(D(1-R)))

     = (1/2)*((0.078995)(x^(7/6) - 3000^(7/6)) - (x^(5/6) - 3000^(5/6))/0.078995*(5/6)),

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