A classic Pursuit problem

One the most famous and classic pursuit problem goes like this: A pursuer, initially at point $(0, b)$, is trying to catch a pursuee, initially at point $(a,0)$ and running along the $x$-axis in the positive direction, by always running directly towards him. If they both run at constant speed and the ratio of pursuer's speed to that of the pursuee is $1:k$ (with $k<1$), how far do the pursuer runs in the pursuit?

This problem is discussed in The American Mathematical Monthly (P3573, Vol. 40, No. 7, (Aug. - Sep., 1933)) and Pi Mu Epsilon Journal (Vol. 6, No. 9, Fall 1978, Pg. 545) among others (for example Pursuit Curve and Pursuit).

In both AMA and PME, the solution is obtained (albeit for a special case where the pursuee starts at the origin) by solving a 'challenging' differential equation which gives the exact equation of the path traversed by the pursuer.

However, if the distance traversed by the pursuer is all we want, do we still have to go through the trouble of solving such a complicated differential equation?

A pursuit curve with $k=4/5$ (Note that Geogebra is unable to render the full curve).

One of the key points to note in this problem is that the respective distance travelled by the participants in given interval of time is directly proportional to the ratio of the speeds.

We make an assumption that the pursuer ends up catching the pursuee (say at $C$ as shown above). While this may be obvious from the full solution, for our purposes, this will remain an assumption.

Consider the infinitesimal distance $\,ds$, say between $F$ and $H$ as shown below, covered by the pursuer in time $\,dt$. Drawing tangents at these points, we know that the pursuee was at position $J$ and $K$ respectively.

As the distance covered by them are in proportion to their speeds, we know that $|JK|=k\cdot|FH|=k\,ds$.

Let $l$ denote the relative distance between the pursuer and pursuee. Then, $\,dl=|FJ|-|HK|$. To calculate this, we complete the right $\triangle FMH$ and draw a line passing through $J$ and perpendicular to $FJ$ such that it intersects line $HK$ at $L$.

We now rotate line $FJ$ about $H$ such that $F$ moves to $F'$ (not shown in image) in line $HK$. Also, because the rotation is infinitesimally small, $J$ moves to $L$ in line $HK$.

Then, $|FJ|\approx |F'L|=|FH|+|HK|-|LK|$. That is, $|FH|=|FJ|-|HK|+|LK| \implies \,ds=\,dl+|LK|$

Because both $\angle FJL$ and $\angle HLJ$ are approximately right angled, $\angle HFM = \angle LJK$. Therefore, by the similarity of $\triangle FMH$ and $\triangle JLK$ and the fact that $|JK|=k\,ds$, we have,

$|LK|=k|HM|=k\,dx$

where $\,dx$ is the infinitesimal movement of pursuer along the $x$-direction.

Combining everything we have so far, we see that $\,ds=\,dl+k\,dx$.

We have now shown that $\,d(s-l-kx)=0$ (or) $s - l - kx = \text{const.}$

We can choose $s$ to be the distance to be travelled by the pursuer to reach $C$ and $x$ to be the distance along the $x$-axis between the pursuer and $C$.

We know that at point $C$, $s=l=x=0$ which means that the constant in the equation above is $0$ giving us the relation that $s=l+kx$.

Before we celebrate our success, we notice a glaring inconvenience in the above relation. While we will know that $l=l_0$ is the relative distance between the two parties at the start of the pursuit and $s=s_0$ is the distance we are trying to find, we do not know $x_0$. However, this is simple to resolve.

Let $x_0=h_0+p_0$ where $h_0$ is the horizontal distance between the pursuer and the pursuee at the start (which should be known for the problem to be defined) and $p_0$ is the distance between the pursuee and $C$.

By our definition of point $C$, we know that the pursuer travels distance $s_0$ in the time the pursuee travels distance $p_0$. Therefore, $p_0=k\cdot s_0$. Using this and solving for $s_0$, we get (after dropping subscripts),

$\displaystyle s=\frac{l+k\cdot h}{1-k^2}$

which solves the problem of finding the distance traversed by the pursuer in a more general setting where the they are not necessarily in the same vertical.

Hope you enjoyed this post.

Until then

Yours Aye

Me