A Martin Gardner Arclength Problem

Question

In an $xy$ plane a mouse is placed at the origin and a cat on the $y$ axis at $y=c$. At $t=0$ the mouse begins running east along the $x$ axis at constant speed $S_m$ and the cat begins chasing the mouse at constant speed $S_c$. The cat runs so that it is always headed towards the mouse's current position. We are given $S_c > S_m$ and hence the cat will always catch the mouse.

What is the length of the cat's path between its start position and where it catches the mouse in terms of $S_c, S_m$ and $ c$ ?

Answer 1

This problem involves some work with differential equations but it is well known and studied. The curve described by the cat is a radiodrome, which is itself a particular kind of pursuit curve. You can check the first link to obtain the answers to this.

The mouse will be caught at

$$x_{\text{capture}}=\frac{c}2\left({\left(1-\frac{S_m}{S_c}\right)}^{-1}-{\left(1+\frac{S_m}{S_c}\right)}^{-1}\right)$$

which is also his displacement. The cat will then have run $\frac{S_c}{S_m}\cdot x_{\text{capture}}$.

Answer 2

Cat and Mouse Picture

A picture of the cat's trajectory has been included. For some time $t$ we've drawn a tangent line of the cat's path and where it intersects the $x$ axis is the mouse's position.

Define $D(t)$ and $x(t)$ as the distance between the cat and mouse and position of the cat at time $t$ respectively. Now

$$D(t) = c - S_ct + \int_{t=0}^t S_m\cos \theta \ \mathbb{d}t$$ In other words, $D(t)$ is the initial distance between the cat and mouse, minus the distance the cat closes between them, plus the integral of the component of the distance the mouse spends running away from the cat. Define $t_f$ as the time when the cat catches the mouse. By similar triangles $\cos \theta = \mathbb{d}x/(S_c \mathbb{d}t)$. Hence

$$D(t_f) = 0 = c - S_c t_f +\frac{S_m}{S_c}\int_{x=0}^{x(t_f)} \mathbb{d}x $$ where the limits were changed to account for the change in integration variable. Using $x(t_f) =S_m t_f$

$$D(t_f) = 0 = c - S_c t_f + \frac{S_m^2}{S_c}t_f$$ $$ \implies t_f = \frac{c}{S_c - \frac{S_m^2}{S_c}}$$ and the length of the cat's path is $$S_c t_f = \frac{c}{1 - \frac{S_m^2}{S_c^2}} $$