Find distance between turtles

Question

I've been thinking about the following problem for some time now and was wondering if anyone could shed any light on it.

At time $t = 0$, turtle A sits at $(0,0)$ and turtle B sits at $(d,0)$. B begins moving straight upwards perpendicular to the $x$-axis with some constant velocity $v$. Simultaneously, the A turtle begins moving with the same constant velocity in such a way that its nose is always pointed at B. What is the distance between these turtles as $t \rightarrow \infty$?

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Edit

So basically I've arrived at a nasty system of differential equations. Here's my reasoning:

We know that turtle A will always point towards turtle B, so if $(x(t),y(t))$ is the curve traced out by A, then we can determine $dy/dx = \dot{y}/\dot{x}$ at each point:

$\dot{y}/\dot{x} = (vt - y)/(d - x).$

The fact that A has a velocity $v$ gives rise to a second equation:

$\dot{x}^2 + \dot{y}^2 = v^2.$

If we solve these, the problem would be reduced to taking a limit.

Anyway, this is what I have so far - not much really. Would this be a good/feasible approach to pursue so as to solve the problem? Or is there a better way?

Answer 1

The answer is $d/2$. Since this is such a nice answer, there might be a really simple way to obtain it that doesn't involve any differential equations. I don't know one, though.

Here's my way of solving it. Let's adjust the problem so that the turtle at the origin swims in the $-y$ direction, and the other turtle swims straight towards him (I did this because I thought it would make the signs easier for me). Now, let's switch the rest frame so that there is a current flowing in the $+y$ direction, the turtle at the origin sits on a little island, and the other turtle always swims towards the origin, but is pulled off-course by the current. We can now use geometry to figure out which direction the turtle moves at any point in the plane. At point $(x,y)$, the turtle swims with velocity $$\left(\frac{-x}{\sqrt{x^2+y^2}}, \frac{-y}{\sqrt{x^2+y^2}} \right).$$. Taking the current into account, the turtle's velocity is $$\left(\frac{-x}{\sqrt{x^2+y^2}}, 1-\frac{y}{\sqrt{x^2+y^2}} \right).$$ This gives us the differential equation for the turtle's path $$\frac{dy}{dx} = \frac{y - \sqrt{x^2+y^2}}{x}.$$ Maple gives us the solution to this: $$ y = \frac{1}{2} \left(C-\frac{x^2}{C} \right), $$ which can easily be checked to be a solution for $C>0$ (although I'd like to see how to solve it without a computer algebra system). Now, we know that when $y=0$, $x=d$. This gives $C=d$. When $x=0$, we have $y=d/2$.

Answer 2

I may be mistaken, but it seems that the following approach should work. Use such a (moving) polar coordinate system where turtle B is always at rest in the origin, write ordinary differential equations for $\rho$ and $\varphi$. As $\rho$ does not enter the differential equation with $\dot{\varphi}$, the latter equation can be solved (in elementary functions, as far as I can see). The solution should be used to replace $\varphi$ in the differential equation with $\dot{\rho}$ by a function of time. The resulting differential equation also seems to be solvable in elementary functions. Then the limit $t->\infty$ can be taken.