Describing the motion of a particle which moves with constant speed, but always in the direction of another, moving particle

Imagine you are chasing someone. He is some distance away from you, and without knowing what direction he will run away in, the best option is to run directly toward his current position.

Now if we consider a situation in which the target of your running runs away in a direction perpendicular to the line that connects you to him. For example, he is up against a wall and can only run along along the wall.

If you and the other person have the same maximum running speed, and always run at this speed, my intuition tells me that you will chase him and become closer and closer to him, but never actually reach his position.

I described this as so: The escaping person/particle moves from the origin, along the y-axis in a positive direction. It's position can be described by the equations {x=0, y=t} where t is time, as it moves 1 unit per second. The chasing particle begins at the point (-1,0), and moves towards the other particle with constant speed 1 unit per second, but always in the direction of that other particle.

The situation can be broken down into discrete time units, for example 0.01 seconds. Then an assumption can be made that for the entire 0.01 seconds, the chasing particle moves in the direction of the location the escaping particle was at at the beginning of those 0.01 seconds.

The angle to the horizontal of this direction can be found using $$\frac{y_n-t}{x_n}$$ where $y_n$ is the current y co-ordinate of the chasing particle, and $x_n$ is the current x co-ordinate of the chasing particle. $t$ is the time elapsed, and represents the y co-ordinate of the escaping particle.

Then the horizontal and vertical components of the velocity of the chasing particle can be found. $$v_x=\cos(\arctan(\frac{y_n-t}{x_n}))$$ $$v_y=\sin(\arctan(\frac{y_n-t}{x_n}))$$

For acute angles, (and this is always the case) the above can also be written as $$v_x=\frac{1}{\sqrt{\frac{\left(y_n-t\right){}^2}{x_n^2}+1}}$$ $$v_y=\frac{y_n-t}{x_n \sqrt{\frac{\left(y_n-t\right){}^2}{x_n^2}+1}}$$

Then the position of the chasing particle after 0.01 seconds has elapsed can be written using these velocities, and the time elapsed. $$x_{n+1}=x_n+0.01*\cos(\arctan(\frac{y_n-t}{x_n}))$$ $$y_{n+1}=y_n+0.01*\sin(\arctan(\frac{y_n-t}{x_n}))$$

I could use Excel work out the position of the particle at time t (in multiples of 0.01 seconds). I also repeated this with a time interval of 0.0001 seconds, resulting in much smoother data.

What I wanted was to find the path traced by the chasing particle when the time interval was reduced to 0, i.e. the continuous movement of the particle. above equations are almost parametric: but they contain the other variable. If they did not contain the other variable, then I possibly I could reduce the time interval (the 0.01) to $dt$, a differential of time, and $x_{n+1}-x_n$ to $dx$, and integrate to obtain $x$ as a function of time. This I would repeat with $y$, and therefore obtain parametric equations for $x$ and $y$ in terms of $t$, which would describe the motion of the particle. Unfortunately this is not the case.

What I have appears similar to the Arithmetic-Geometric mean progression, where two series are dependent on each other to determine the next term in the series. (I'm a high school student, forgive me if there are more obvious examples of interdependent series – I don't know any.) In this case the two series do not converge to the same number however.

The problem is that I don't know where to go from here. I don't know how to separate the variables in order to solve this. If I were to substitute one equation into the other, it would still contain $dx$, $dy$, and either of $x$ or $y$. At this point it might be considered a non-linear differential equation in two variables.

Upon graphing of the data I got from Excel, the result is a curve which begins at (-1,0), and moves upwards and to the right, and approaches the y-axis but is asymptotic to it. I got the graphing program to fit a line to the data using intervals of 0.0001 seconds, and found the beautifully simple equation of $$y=\frac{1}{4} \left(x^2-2 \ln (x)-1\right)$$ to fit the data almost perfectly.

Since it was continuous compounding of a particular operation, I figured it would involve e somehow, and it does. How do you solve this system to obtain that solution? Is there a far simpler way to obtain the solution to this problem?

• You are saying that you have the escaping particle $x$ with $x(0)=(0,0),x'(t)=(0,1)$ for all $t$ (taking the agreed upon speed to be $1$), and the chasing particle $y$ with $y(0)=(-1,0),y'(t)=\frac{x(t)-y(t)}{\| x(t)-y(t) \|}$ for all $t$. Then $x(t)=(0,t)$ so $y'(t)=\frac{(-y_1(t),t-y_2(t))}{\sqrt{y_1(t)^2+(t-y_2(t))^2}}$. This is a system of two nonlinear ordinary differential equations that probably has no exact solution. – Ian May 19 '16 at 20:30
• Numerically it looks like the distance between the two of them decreases in time (as you would expect) to a nonzero limit, which appears to be close to $1/2$. – Ian May 19 '16 at 20:37

This solution is what I've come across online, but I will admit that several of the steps beyond $15$ are a bit beyond me, as it's been a while since I've done differential equations.
In my own Excel model, I assume the particle $P$ behaves like a missile (chaser) and particle $A$ behaves like the target (chased). At time $t=0$, Particle $P$ begins at the origin and moves with constant speed $p$, and particle $A$ begins at the point $(0, 1)$ and moves with velocity $(a, 0)$ (thus with speed $a$). Empirically, assuming $p>a>0$, the $x$-coordinate at impact turns out to be $\frac{pa}{p^2-a^2}$, and thus the time till impact is $\frac{p}{p^2-a^2}$.