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.


*

*Does Tom catch Jerry? 

*How close does he get (in the limit t tends to infinity)?

*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.
 A: 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.
A: 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.
A: See to the essence of the problem so you can avoid needless calculations:
No.  Of course Tom will never catch Jerry.

*

*Jerry's horizontal speed is always $v_x=1$.

*Tom's horizonal speed begins $v_x<1$ and is always $v_x \le 1$.

QED.
A: Let the speed of tom be v and of jerry be u. Let tom be running at angle $\theta$ from horizontal at any instant, relative velocity of tom wrt jerry is $v-u\cos\theta$ and let them be at an initial separation $l$ so $\int_0^T(v-u\cos\theta){\rm d}t=l$ and also $\int_0^Tv\cos\theta=uT$ so $T=\frac{vl}{v^2-u^2}$. Try putting $u=0$.

From question $u=v$ so $T=vl/(v^2-v^2)\to\infty$
