Two solutions to a pursuit problem

Pursuit problems are a classic problem in mathematics. They involve finding the paths taken by two or more pursuers moving toward each other. Here, we outline a simple version of a pursuit problem, known sometimes as the Mice problem or Dogs in Pursuit. We’ll solve the problem in two ways, first by using the definition of a derivative and then by considering a rotating reference frame. 

The Mice Problem

Four mice are placed on the corners of a square of side length l. On command, they start moving toward each other, each pursuing its counterclockwise neighbor with constant speed v until they meet. 

Questions:

1. How long will it take for the mice to meet?
2. What distance will each mouse travel?

As each mouse starts moving, the position of its pursuee will change, so each mouse’s direction will change. Thus, each mouse will trace a spiral path before all four mice meet in the center.  

We’ll go through two illustrative approaches to solve this problem. First, we’ll brute force our way through it by applying the definition of a derivative. Then, we’ll reframe the problem using a rotating reference frame, which will let us solve it in a single step. 

Approach 1: Definition of a derivative

You may recall the definition of a derivative from an introductory calculus course: 

To be fair, you probably also pushed this definition to the deep recesses of your mind after learning more convenient rules and shortcuts for taking derivatives. Here, I hope to demonstrate how useful this definition can be. We will use it to set up a differential equation expressing how the distance between adjacent mice changes over time. 

First, let’s make note of symmetries and define variables:

• Symmetry: Since all four mice are initially separated by the same distance and travel at the same velocity, they will trace identically shaped paths. This means we can consider the path of a single mouse, knowing it’s representative of that of all four mice.
• Variables:
  • Let D represent the separation of each mouse from its neighbors.
  • Let t represent time. Initially, this would be D(t = 0) = l.
  • Let T represent the time at which the mice meet (question #1).
  • Let 𝛿 represent the distance traveled by each mouse (question #2). 

To solve the problem, we’ll note that when the mice meet, their separation will be zero. Thus, once we have an equation for D, we will be able to calculate T by finding the time at which D(t = T)= 0. 

To set up the differential equation, let’s consider how the distance between adjacent mice changes over an infinitesimally short time interval Δt. We will analyze a single pursuing mouse (mouse A), and its pursuee (mouse B), calculating their separation at the beginning of the interval (time t) and at the end of the interval (time t + Δt).

We define the x-axis using the unit vector pointing from mouse A to mouse B and we define the origin as the initial position of mouse A. At time, the mice are separated by distance D, so the position of mouse A is (0,0) and the position of mouse B is (D,0). After the interval, at time t+Δt, mouse A will have moved a distance vΔt in the positive x-direction and mouse B will have moved a distance vΔt in the positive y-direction. Thus, mouse A will be at position(vΔt,0), mouse B will be at position (D,vΔt), and the distance between them will be: 

We now have everything we need to apply the definition of a derivative! Rewriting this definition in terms of our variables, we have: 

To avoid manipulating pesky square roots, we can instead work with the distance squared: 

Computing the derivative and simplifying, we get: 

Taking the limit, our equation simplifies to: 

To solve for D instead of D², we'll apply the chain rule for derivatives: 

Substituting and simplifying: 

To find D, we can integrate and apply the initial condition (IC) we previously identified: D(t = 0) = l.

We now have an equation describing the distance between the mice as a function of time! Using this, we can solve for the time T it takes the mice to meet: 

And finally, since distance equals velocity times time, we see that the distance 𝛿 traveled by each mouse is: 

Each mouse travels a distance l before meeting in the center. In other words, the length of the spiral path traversed by each mouse is exactly equal to the side length of the initial square! Crucially, the definition of a derivative allowed us to find D(t) solely by considering how the distance between adjacent mice changes over a short time interval. 

Approach 2: Rotating reference frame 

Now, let’s take a different approach and adopt the point of view of a mouse! Here, we’ll use a purely logical argument, requiring no derivatives.  

Using the mouse’s perspective (which rotates as the mouse traces its spiral path) as our frame of reference, we can define a new set of orthogonal axes – forward/backward and left/right. From its perspective, each mouse is always moving forward (with no component in the left/right direction), and the mouse it’s pursuing is always moving left (with no component in the forward/backward direction). Since the pursuee’s movement is always orthogonal to the line connecting it to its pursuer, the motion of the pursuee does not change its distance from the pursuer. Therefore, the pursuing mouse will close in on its target as if the target weren’t moving at all!  

Thus, the mouse will simply have to travel a distance 𝛿=l(the initial separation) to reach its target, requiring time T= (l/v). 

Beyond a square 

For an extra challenge, try generalizing the solution to arbitrary n-gons! Either approach described here can get you there.