Snail Paths: How Many Times Do They Circle Their Meeting Point?

[Saitama]

Problem:
Three small snails are each at a vertex of an equilateral triangle of side $a$ units. The first sets out towards the second, the second towards the third and the third towards the first, with a uniform speed of $v$ units/sec. During their motion each of them always heads towards its respective target snail. What is the equation of their paths? If the snails are considered as point-masses, how many times does each circle their ultimate meeting point.

Attempt:
I am not sure what should be the best approach. I am thinking of trying with vectors.

Let $\vec{r_1},\vec{r_2}$ and $\vec{r_3}$ denote the position vectors of three snails at any instant. The triangle is placed such that the centroid lies on the origin and the three vertices are at $(0,a/\sqrt{3})$,$(a/2,-a/(2\sqrt{3})$ and $(-a/2,-a/(2\sqrt{3}))$. Clearly, $\vec{r_1}+\vec{r_2}+\vec{r_3}=0$ always.

By symmetry, the snails meet at origin (or the centroid).

Let $\vec{v_1}$, $\vec{v_2}$ and $\vec{v_3}$ be the velocity vectors of the three snails. So, we have the following relations:
$$\vec{v_1}=\frac{d\vec{r_1}}{dt}$$
$$\vec{v_2}=\frac{d\vec{r_2}}{dt}$$
$$\vec{v_3}=\frac{d\vec{r_3}}{dt}$$
Also, $\vec{r_2}-\vec{r_1}$ is along $\vec{v_1}$, $\vec{r_3}-\vec{r_2}$ is along $\vec{v_2}$ and $\vec{r_1}-\vec{r_3}$ is along $\vec{v_3}$. This yields the following relations:
$$\left(\vec{r_2}-\vec{r_1}\right)\times \frac{d\vec{r_1}}{dt}=0$$
$$\left(\vec{r_3}-\vec{r_2}\right)\times \frac{d\vec{r_2}}{dt}=0$$
$$\left(\vec{r_1}-\vec{r_3}\right)\times \frac{d\vec{r_3}}{dt}=0$$
I don't see where to take it from here.

Any help is appreciated. Thanks!

 

[I like Serena]

Hey Pranav! :)

How about using polar coordinates?

$$\vec{r_1} = r\binom{\cos \phi}{\sin \phi}$$
$$\vec{r_2} = r\binom{\cos(\phi + 2\pi/3)}{\sin(\phi + 2\pi/3)}$$
$$\vec{r_3} = r\binom{\cos(\phi + 4\pi/3)}{\sin(\phi + 4\pi/3)}$$

 

[Saitama]

Hi ILS! :)

View attachment 2255

 

[I like Serena]

$$\vec{r_2}=r\cos\left(\phi+\frac{\pi}{3}\right) \hat{i}+r\sin\left(\phi+\frac{\pi}{3}\right) \hat{j}$$

Perhaps use an angle $\phi+\frac{2\pi}{3}$?

 

[Saitama]

Perhaps use an angle $\phi+\frac{2\pi}{3}$?

Woops! Let me try.

But what happened to my previous post? All my equations have disappeared. (Worried)

 

[I like Serena]

Woops! Let me try.

But what happened to my previous post? All my equations have disappeared. (Worried)

Sorry! I accidentally edited it instead of responding! (Blush)

The best I could do is replace it by a screenshot from an outdated view of the reply.

 

[Saitama]

Sorry! I accidentally edited it instead of responding! (Blush)

The best I could do is replace it by a screenshot from an outdated page.

Thanks, the screenshot is good, I couldn't notice that it is a screenshot until I read your post. :)

I did the correction and got:
$$r=Ae^{\sqrt{3}\phi}$$
This again won't pass through origin.

 

[I like Serena]

Thanks, the screenshot is good, I couldn't notice that it is a screenshot until I read your post. :)

I did the correction and got:
$$r=Ae^{\sqrt{3}\phi}$$
This again won't pass through origin.

That does not look correct.
An increasing $r$? (Wondering)

For reference, I get:
$$\vec{r_2} - \vec{r_1} = r\binom{-\sin(\phi + \pi/3)}{\cos(\phi + \pi/3)}$$
which must be parallel to:
$$\frac{d}{dt}\vec{r_1} = \frac{dr}{dt}\binom{\cos \phi}{\sin\phi} + r \frac{d\phi}{dt}\binom{-\sin\phi}{\cos\phi}$$

Since we know that the speed is $v$, it follows that:
$$\frac{d}{dt}\vec{r_1} = v \binom{-\sin(\phi + \pi/3)}{\cos(\phi + \pi/3)}$$

 

[Saitama]

That does not look correct.
An increasing $r$? (Wondering)

For reference, I get:
$$\vec{r_2} - \vec{r_1} = r\binom{-\sin(\phi + \pi/3)}{\cos(\phi + \pi/3)}$$
which must be parallel to:
$$\frac{d}{dt}\vec{r_1} = \frac{dr}{dt}\binom{\cos \phi}{\sin\phi} + r \frac{d\phi}{dt}\binom{-\sin\phi}{\cos\phi}$$

I am not well versed with the notation you are using but if I interpret correctly, aren't your equations same as mine?

Since we know that the speed is $v$, it follows that:
$$\frac{d}{dt}\vec{r_1} = v \binom{-\sin(\phi + \pi/3)}{\cos(\phi + \pi/3)}$$

I can't follow this, why are you doing this? Is there something wrong with the cross product relation I used?

EDIT: I checked my working again. I used the correct angle $\phi+2\pi/3$ but wrote $\phi+\pi/3$ accidentally. The subsequent working is done using the correct angle so I guess the error is somewhere else. :(

 

[I like Serena]

I am not well versed with the notation you are using but if I interpret correctly, aren't your equations same as mine?

Yes.
I didn't know that this vector notation would be unfamiliar to you.

I can't follow this, why are you doing this? Is there something wrong with the cross product relation I used?

EDIT: I checked my working again. I used the correct angle $\phi+2\pi/3$ but wrote $\phi+\pi/3$ accidentally. The subsequent working is done using the correct angle so I guess the error is somewhere else. :(

Mostly because I felt bad for accidentally deleting your equations. (Blush)
Nevermind.

I can't really say where you are making an error when I can't see your working. (Wasntme)

 

[Saitama]

I can't really say where you are making an error when I can't see your working. (Wasntme)

I have posted my complete working in #3. Which is the step you have trouble understanding? Please let me know.

 

[Opalg]

Three small snails are each at a vertex of an equilateral triangle of side $a$ units. The first sets out towards the second, the second towards the third and the third towards the first, with a uniform speed of $v$ units/sec. During their motion each of them always heads towards its respective target snail. What is the equation of their paths? If the snails are considered as point-masses, how many times does each circle their ultimate meeting point.

On grounds of symmetry, the three snails will always form an equilateral triangle centred at the origin. If the position of one of the snails is $[r,\theta]$ (polar coordinates) then its velocity will be $[\dot{r},r\dot{\theta}]$. Its speed is given by $v^2 = \dot{r}^2 + (r\dot{\theta})^2$, and its direction of travel is at an angle $2\pi/3$ to the radial direction. Therefore $\dfrac{r\dot{\theta}}{\dot{r}} = \tan(2\pi/3) = -\sqrt3.$ It follows that $r\dot{\theta} = -\sqrt3\dot{r}$, so that $v^2 = \dot{r}^2 + (-\sqrt3\dot{r})^2 = 4\dot{r}^2$. Hence $v = \pm2\dot{r}$. But clearly $r$ must be decreasing, and so $v = -2\dot{r}$. That is a very simple differential equation, with solution $r = a-2vt$. The snails reach the origin when $t = \dfrac a{2v}.$

However, the equation for $\theta$ is $r\dot{\theta} = -\sqrt3\dot{r} = \dfrac{\sqrt3v}2$, so that $\dot{\theta} = \dfrac{\sqrt3v}{2r} = \dfrac{\sqrt3v}{2(a-2vt)}.$ That integrates (with initial condition $\theta=0$) to $\theta = -\frac{\sqrt3}4\bigl(\ln(a-2vt) - \ln a\bigr) = \frac{\sqrt3}4\ln\frac ar.$ As $r\to0$, you see that $\theta$ spins round the origin faster and faster, and (in answer to the question) each snail circles their ultimate meeting point infinitely often. If you like, the origin acts as a vortex for these spinning (infinitely small) point-masses.

This problem is reminiscent of Zeno's paradox. The snails reach their goal in a finite time. But if you break that time into intervals according to the number of times they have circled the origin, that makes it appear as though they would never actually get there.

 

[Saitama]

Hi Opalg! :)

...and its direction of travel is at an angle $2\pi/3$ to the radial direction.

Shouldn't that be $5\pi/6$?

Therefore $\dfrac{r\dot{\theta}}{\dot{r}} = \tan(2\pi/3) = -\sqrt3.$ It follows that $r\dot{\theta} = -\sqrt3\dot{r}$, so that $v^2 = \dot{r}^2 + (-\sqrt3\dot{r})^2 = 4\dot{r}^2$. Hence $v = \pm2\dot{r}$. But clearly $r$ must be decreasing, and so $v = -2\dot{r}$. That is a very simple differential equation, with solution $r = a-2vt$. The snails reach the origin when $t = \dfrac a{2v}.$

I am not very comfortable with motion in polar coordinates but looking at your post, I did something similar with my equations in #3 and I got $r$ and $\phi$ as a function of time.

Your equation $r=a-2vt$ doesn't satisfy the initial conditions, at $t=0$, $r=a/\sqrt{3}$.

Following are the equations I got:
$$r=\frac{a}{\sqrt{3}}-\frac{\sqrt{3}vt}{2}$$
$$\phi=- \frac{1}{\sqrt{3}}\ln\left(\frac{\sqrt{3}r}{a} \right)$$

However, the equation for $\theta$ is $r\dot{\theta} = -\sqrt3\dot{r} = \dfrac{\sqrt3v}2$, so that $\dot{\theta} = \dfrac{\sqrt3v}{2r} = \dfrac{\sqrt3v}{2(a-2vt)}.$ That integrates (with initial condition $\theta=0$) to $\theta = -\frac{\sqrt3}4\bigl(\ln(a-2vt) - \ln a\bigr) = \frac{\sqrt3}4\ln\frac ar.$ As $r\to0$, you see that $\theta$ spins round the origin faster and faster, and (in answer to the question) each snail circles their ultimate meeting point infinitely often. If you like, the origin acts as a vortex for these spinning (infinitely small) point-masses.

Sorry if this sounds stupid but what does "$\theta$ spins round the origin faster and faster" mean? What does spinning imply here?

 

[Opalg]

... and its direction of travel is at an angle $2\pi/3$ to the radial direction.

Shouldn't that be $5\pi/6$?

I am not very comfortable with motion in polar coordinates but looking at your post, I did something similar with my equations in #3 and I got $r$ and $\phi$ as a function of time.

Your equation $r=a-2vt$ doesn't satisfy the initial conditions, at $t=0$, $r=a/\sqrt{3}$.

Following are the equations I got:
$$r=\frac{a}{\sqrt{3}}-\frac{\sqrt{3}vt}{2}$$
$$\phi=- \frac{1}{\sqrt{3}}\ln\left(\frac{\sqrt{3}r}{a} \right)$$

Yes, you are right on both counts. It should be $5\pi/6$, and the initial value of $r$ is $a\left/\sqrt3\right.$. Your amended solution looks correct.

However, the equation for $\theta$ is $r\dot{\theta} = -\sqrt3\dot{r} = \dfrac{\sqrt3v}2$, so that $\dot{\theta} = \dfrac{\sqrt3v}{2r} = \dfrac{\sqrt3v}{2(a-2vt)}.$ That integrates (with initial condition $\theta=0$) to $\theta = -\frac{\sqrt3}4\bigl(\ln(a-2vt) - \ln a\bigr) = \frac{\sqrt3}4\ln\frac ar.$ As $r\to0$, you see that $\theta$ spins round the origin faster and faster, and (in answer to the question) each snail circles their ultimate meeting point infinitely often. If you like, the origin acts as a vortex for these spinning (infinitely small) point-masses.

Sorry if this sounds stupid but what does "$\theta$ spins round the origin faster and faster" mean? What does spinning imply here?

The question asks how many times the snails circle their ultimate meeting point. The answer is: infinitely many times. I was trying to explain how that could happen, given that the snails move at a fixed speed and arrive at the meeting point in a finite time. The answer is that as they get very close to the meeting point, the length of a circuit round that point becomes very small (in fact, exponentially small), so the time taken to complete a circuit also becomes exponentially small.

 

[Saitama]

The question asks how many times the snails circle their ultimate meeting point. The answer is: infinitely many times. I was trying to explain how that could happen, given that the snails move at a fixed speed and arrive at the meeting point in a finite time. The answer is that as they get very close to the meeting point, the length of a circuit round that point becomes very small (in fact, exponentially small), so the time taken to complete a circuit also becomes exponentially small.

Thanks Opalg! :)