Putnam 1959 
Topic: Differential Equations Geometry

Problem: A sparrow, flying horizontally in a straight line, is 50 feet directly below an eagle and 100 feet directly above a hawk. Both hawk and eagle fly directly toward the sparrow, reaching it simultaneously. The hawk flies twice as fast as the sparrow. How far does the eagle fly?

Solution: Although not specifically stated, it is clearly intended that each of the three birds flies at a uniform speed. So we consider the general problem of a predator flying with speed $v$ in pursuit of a prey flying along a straight path with speed $rv$, $r < 1$.

Let the prey start at $(0, 0)$ and move along the $x$-axis so that his position at time $t$ is $(rvt, 0)$. Let the predator start at $(0, h)$, $h > 0$. If at time $t$ the predator is at $(x, y)$, then the condition that he always flies directly toward his prey is that 
\[ \frac{dx/dt}{dy/dt} = \frac{rvt - x}{-y}. \]

As long as $y$ remains positive, $dy/dt < 0$, so we may choose $y$ as the independent variable. Then (1) becomes
\[ y \frac{dx}{dy} = x - rvt. \]

The condition that the predator has constant speed $v$ is 
\[ \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} = v, \]
which we can write as
\[ \sqrt{\left(\frac{dx}{dy}\right)^2 + 1} = v \frac{dt}{dy}, \]
where we have introduced a minus sign because $dt/dy$ is negative.

We differentiate (2) to get
\[ y \frac{d^2x}{dy^2} + \frac{dx}{dy} = \frac{dx}{dy} - rv \frac{dt}{dy}, \]
and using (3) we have
\[ y \frac{d^2x}{dy^2} = r \sqrt{\left(\frac{dx}{dy}\right)^2 + 1}. \]

We can write this as the first-order differential equation
\[ y \frac{dz}{dy} = r \sqrt{1 + z^2}, \]
where $z = dx/dy$. Separating the variables we have
\[ r \frac{dy}{y} = \frac{dz}{\sqrt{1 + z^2}}, \]
whence
\[ r \log y = \log (z + \sqrt{1 + z^2}) + C. \]
Since $z = dx/dy = 0$ when $y = h$, we find $C = r \log h$. Then (4) can be rewritten
\[ \left(\frac{y}{h}\right)^r = z + \sqrt{1 + z^2}. \]

and solved for $z (= dx/dy)$ to get
\[ 2 \frac{dx}{dy} = \left(\frac{y}{h}\right)^r - \left(\frac{y}{h}\right)^{-r}. \]
This can be integrated again using the initial condition $x = 0$ when $y = h$ to give
\[ 2x = \frac{h}{1 + r} \left(\frac{y}{h}\right)^{1+r} - \frac{h}{1 - r} \left(\frac{y}{h}\right)^{1-r} + \frac{2rh}{1 - r^2}. \]

This is valid only for $y > 0$. Recalling that $0 < r < 1$, we see from (6) that
\[ x \to \frac{-rh}{1 - r^2} \]
as $y \to 0$. It follows that the predator catches his prey at the point $(rh/(1 - r^2), 0)$ when $t = h/(1 - r^2)v$. At this time the prey has flown $rh/(1 - r^2)$ and the predator, $h/(1 - r^2)$.

Now for the hawk and the sparrow, $r = \frac{1}{2}$, $h = 100$ ft., so the sparrow flies $200/3$ ft., and the hawk, $400/3$ ft. [Note that since the sparrow is above the hawk we would have to choose a coordinate system with $y$ positive in the downward direction.]

In the case of the eagle, $h = 50$ ft. and $r$ is unknown, but
\[ \frac{r50}{1 - r^2} = \frac{200}{3}; \]
hence $4r^2 + 3r - 4 = 0$ and
\[ r = \frac{-3 + \sqrt{73}}{8} \]
since $r$ must be positive.

The speed of the eagle is $1/r$ times the speed of the sparrow, or
\[ \frac{v(3 + \sqrt{73})}{16} \approx .721v \]
where $v$ is the speed of the hawk.

Since the time of capture is $400/3v$, the eagle flies a distance of
\[ \boxed{\frac{400}{3} \left(\frac{3 + \sqrt{73}}{16}\right)} \approx 96.2 \text{ ft.} \]