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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 $t \rightarrow \infty$?


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?

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  • $\begingroup$ @Qmechanic This is not homework, but if there is another reason for tagging it as such it's fine by me. $\endgroup$
    – arshajii
    Sep 30, 2012 at 20:43
  • $\begingroup$ Yes, it does not have to be actual homework for the homework tag to apply cf. meta.physics.stackexchange.com/a/715/2451 $\endgroup$
    – Qmechanic
    Sep 30, 2012 at 20:50
  • $\begingroup$ Hi A. R. S. - we prefer that you ask about the concept which is giving you trouble, rather than just posting a problem and asking for help in general. If you can narrow your question down to focus on the specific concept that is confusing you, flag it for moderator attention and I'll be happy to reopen it. See the post Qmechanic linked to for more details. (Feel free to ask in Physics Chat for help fixing up the question if you like; although the chat room is often empty, someone will eventually see your message.) $\endgroup$
    – David Z
    Oct 1, 2012 at 3:59
  • $\begingroup$ Given David Zaslavsky's comment, you really should post this problem on math.stackexchange.com; you'd probably just get a solution (my guess is 1/2) rather than no answer, and then getting the problem closed. $\endgroup$ Oct 1, 2012 at 4:03
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    $\begingroup$ @DavidZeslavsky I think what I might do is try to get a little further in it on my own and then post a more 'narrowed-down' version (or update this post). $\endgroup$
    – arshajii
    Oct 1, 2012 at 4:46

2 Answers 2

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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 $y=d/2$.

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  • $\begingroup$ Well, it is clear from the Maple solution that substitution $w=y+\sqrt{y^2+x^2}$ solves the differential equation:-) It seems that in my answer (which has a lot of similarities with yours) integration is easier. $\endgroup$
    – akhmeteli
    Oct 6, 2012 at 3:23
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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.

EDIT (Sep 22, 2023): Maybe it's silly to "exhumate" this old question, but @Peter Shor wondered if there could be a really simple way to obtain a solution, so, FWIW...

The following is close both to Peter Shor's solution and to my preliminaries. I use a polar system of coordinates $\rho,\alpha$, but $\alpha$ increases clockwise. $$\dot{\rho}=-1+\sin\alpha,$$ $$\dot{\alpha}=\frac{\cos\alpha}{\rho},$$ so $$\rho'=\frac{(-1+\sin\alpha)\rho}{\cos\alpha},$$ where $\rho'=\frac{d\rho}{d\alpha}$. Introducing $q=\ln\rho$, we get $$q'=\frac{-1+\sin\alpha}{\cos\alpha}.$$ The integration is elementary: $$q=\frac{1}{2}\ln(1-\sin\alpha)-\frac{1}{2}\ln(1+\sin\alpha)-\ln(\cos\alpha)+C.$$ Initially, $\alpha=0,\rho=1,q=0$, so we calculate that $C=0$. When the time tends to infinity, $\alpha\to\frac{\pi}{2},\sin\alpha\to 1,\cos\alpha\to 0$, and $$q=\frac{1}{2}\ln(1-\sin\alpha)-\frac{1}{2}\ln(1+\sin\alpha)-\ln(\cos\alpha)=$$ $$\frac{1}{2}\ln(1-\sin\alpha)+\frac{1}{2}\ln(1+\sin\alpha)-\ln(1+\sin\alpha)-\ln(\cos\alpha)=$$ $$\frac{1}{2}\ln(\cos^2\alpha)-\ln(1+\sin\alpha)-\ln(\cos\alpha)\to -\ln(2),$$ so $\rho\to\frac{1}{2}$.

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