Does Tom catch Jerry?

Tom has Jerry backed against a wall. Tom is distance 1 away (perpendicularly). At time t=0, Jerry runs along the wall. Tom runs directly towards Jerry. Tom always runs directly towards Jerry. Tom and Jerry both run at the same speed.

1. Does Tom catch Jerry?
2. How close does he get (in the limit t tends to infinity)?
3. What shaped curve does Tom run?

Edit: I made this problem up last week. Friends enjoyed it, I thought this site might too.

Hint: Take the x-axis as the wall, and assume Jerry runs to the right, without loss of generality at speed 1. Let $x(t)$ and $y(t)$ be Tom's position at time $t$. So $x(0) = 0$, and $y(0) = 1$.

Consider Tom's direction of travel at time $t$ towards Jerry at $(t, 0)$. Write $\theta$ for the (positive) angle below the horizon. Then

$$ \tan \theta = \frac{dy}{dx} = \frac {y}{t-x} $$

Tom runs at unit speed, so also

$$ \frac{dy}{dt} = - \sin \theta $$

$$ \frac{dx}{dt} = \cos \theta $$

That's as far as I got, I don't know how to solve such a complex differential equation.

Solution 1:

Let $x=1$ be the wall, let Tom start at $(0,0)$, Jerry at $(1,0)$ upwards, and assume that both have the same speed $1$. Tom's orbit is then a graph curve $$\gamma: \quad x\mapsto \bigl(x,y(x)\bigr)\qquad(0\leq x<1)\ ,$$ whereby $y(0)=y'(0)=0$. At any point $(x,y)\in\gamma$ we have $$y'={\int_0^x\sqrt{1+y'^2}\>dx -y\over 1-x}\ .\tag{1}$$ One arrives at this equation by the following argument: When Tom is at $(x,y)$ he has run the length $s:=\int_0^x\sqrt{1+y'^2}\>dx$ so far. Therefore Jerry is at $(1,s)$ now, and this enforces $(1)$.

From $(1)$ we get $$(1-x)y'+y=\int_0^x\sqrt{1+y'^2}\>dx\qquad(0\leq x<1)\ .$$ In order to get rid of the integral we take the derivative with respect to $x$ and separate variables: $${y''\over\sqrt{1+y'^2}}={1\over 1-x}\ .$$ This leads to $$\log\bigl(y'+\sqrt{1+y'^2}\bigr)=\log{1\over1-x}+C\ ,$$ and the initial condition $y'(0)=0$ immediately gives $C=0$. We solve for $y'$ and obtain $$y'={1\over2}\left({1\over 1-x}-(1-x)\right)\ .$$ One more integration then gives $$y(x)={x^2\over4}-{x\over2}+{1\over2}\log{1\over 1-x}\qquad(0\leq x<1)\ .$$ This is the explicit shape of $\gamma$. In order to compute how far Tom is staying behind in the limit we have to compute the limit for $x\to 1-$ of $$\int_0^x\left(\sqrt{1+y'^2(x)}-y'(x)\right)\>dx=\int_0^x(1-x)\>dx=x-{x^2\over2}\ .$$ It follows that Tom stays ${1\over2}$ behind in the limit.

Solution 2:

See to the essence of the problem so you can avoid needless calculations:

No. Of course Tom will never catch Jerry.

1. Jerry's horizontal speed is always $v_x=1$.
2. Tom's horizonal speed begins $v_x<1$ and is always $v_x \le 1$.

QED.

Solution 3:

The problem is simplified considerably if we consider the vector from Tom to Jerry.

So let $\vec r=\pmatrix{x\\y}$ be the vector from Tom to Jerry, with $r=|\vec r|=\sqrt{x^2+y^2}$. Then

\begin{eqnarray*} \dot x&=&-\frac xr+1\;,\\ \dot y&=&-\frac yr\;, \end{eqnarray*}

and

\begin{eqnarray*} \dot r&=&\frac{x\dot x+y\dot y}r\\ &=&\frac xr-1\\ &=&-\dot x\;. \end{eqnarray*}

Thus $r+x$ is constant. Initially $x=0$ and $r=1$, so the sum is $1$. In the limit, $x=r$, so both are $\frac12$.

This answers questions $1$ and $2$. To answer question $3$, you can introduce $\phi$ with $x=r\cos\phi$ and $y=r\sin\phi$ and calculate $\frac{\mathrm dr}{\mathrm d\phi}=\frac{\dot r}{\dot \phi}$ from $\dot x$ and $\dot y$. The result is

$$ \frac1r\frac{\mathrm dr}{\mathrm d\phi}=\tan\frac\phi2\;, $$

and integrating yields $r=\lambda\cos^{-2}\frac\phi2$. The initial condition $r=1$, $\phi=\frac\pi2$ yields $\lambda=\frac12$, so the trajectory is

$$ r=\frac1{2\cos^2\frac\phi2}\;. $$

Substituting $\phi=0$ again yields $r=\frac12$ in the limit.