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Dog race:

Edit 2:

I posted a possible answer below. However, I am unsure how the authors arrived at the solution.

Maybe someone can offer an explanation.

Four dogs are positioned at the corners of a square ($d= 1m$), chase each other in clockwise direction with the same constant speed . As their target is moving, they will follow a curved path, eventually colliding in the center of the square.

enter image description here

(a) Why is the total length of the path just $1 m$?

(b) Find and solve a differential equation for the radius $r(\theta)$ in polar coordinates.

This is a homework question! I just want hints, no solutions please.

The tough part is setting up an equation for the radius i.e for the motion of one of the dogs. I was thinking about an equation similar to that of an archimedean spiral.

$$r(\theta)=a+b\theta \space \space \space \text{or} \space \space \space r(\theta)=a\theta^{\frac{1}{n}}$$

However, I have no idea what values $a$ or $b$ should be. Any hints are appreciated.

Edit 1:

This is my second attempt at a solution:

If Dog 1 is positioned at $(r, \theta)$ $\implies$ Dog 2 is positioned at $(r, \theta+\frac{\pi}{2})$

Picture:

enter image description here

$$x_1=r \cos (\theta) \\ y_1= r \sin (\theta) \\ \\ \\ x_2=r \cos (\theta+\frac{\pi}{2})=-r \sin(\theta) \\y_2=r \sin (\theta+\frac{\pi}{2})=r \cos (\theta)$$

If these are the two position vectors then the vector joining the two points is my velocity vector.

$$\implies \frac{dy}{dx}=\frac{y_2-y_1}{x_2-x_1}=\frac{r \sin (\theta+\frac{\pi}{2})-r \sin (\theta)}{r \cos (\theta+\frac{\pi}{2})-r \cos (\theta)}=\frac{ \sin (\theta+\frac{\pi}{2})- \sin (\theta)}{ \cos (\theta+\frac{\pi}{2})- \cos (\theta)}=\frac{\cos(\theta)-\sin(\theta)}{- \sin(\theta)-\cos(\theta)}$$

Am I on the right path? But How do I deduce $\dfrac{dr}{d\theta}$?

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    $\begingroup$ Your attempt tries to short-circuit the entire problem by giving a finished parameterization of the curve. That's not what (b) asks you to do -- it says you must set up a differential equation for the curve first and solve that. In other words, what you should do first is come up with something that invoves $\frac{dr}{d\varphi}$ (where $\varphi$ is presumably the same as what you call $\theta$ in your attempt). $\endgroup$ Oct 22 '15 at 11:24
  • $\begingroup$ By the way, are the dogs assumed to be pointlike? Real dogs take up space that would be a significant fraction of the one-meter square arena, causing uncertainty in the precise direction each dog is moving in when it "chases" the next, and making them collide much before their centers (whatever that is) coincide. $\endgroup$ Oct 22 '15 at 11:28
  • $\begingroup$ @Henning Makholm Thanks for your answer. I will try doing that. But at the moment the only thing I can see is that $\frac{dr}{d\theta}$ must be negative since the radius is decreasing. I think we can assume that the dogs are pointlike. I will try to find some sort of differential equation and come back if i found/can't find anything. $\endgroup$
    – qmd
    Oct 22 '15 at 11:31
  • $\begingroup$ x @SuH: You need to use the information that each dog is moving straight towards the next dog. If you draw a sketch of a typical moment during the race, you ought to see that this is the same as "the dog's trajectory must make a 45° angle with the radius", which you ought to be able to work into an expression for $\frac{dr}{d\varphi}$. $\endgroup$ Oct 22 '15 at 11:34
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    $\begingroup$ Also it looks like you're still trying to do too much too soon. The part about letting the dog continue blindfolded for half the distance doesn't seem to give you any progress -- it's as if you're still trying to bypass the differential equation stage and find the shape of the entire path, instead of just aiming for the differential equation. What you need to ask is: what is the dog's velocity vector now? Not where it will be at some point in the future -- reserve that for after you have a differential equation and have solved that. $\endgroup$ Oct 22 '15 at 13:38
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For part (a), you don't have to parameterize anything, really, or set up any complicated differential equations. All you have to do is show that, thinking of yourself as one of the dogs, the dog you're chasing is always running at right angles to your line of sight and therefore, since your speed is constant and directed straight at the target, you are getting closer at a constant rate -- in short, as far as the time to overtake it is concerned, the dog you're chasing may as well not be moving at all, which is to say you'll catch it in the time it takes to travel $1$ meter.

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  • $\begingroup$ Thanks for your answer. I know you are right (because I have seen the answer) but I have a hard time getting my head around the fact that the total distance a dog has to chase is just 1 meter. It seems to me that the distance could be 1.01 meters or 1.3 meters etc.. $\endgroup$
    – qmd
    Oct 22 '15 at 19:53
  • $\begingroup$ @SuH, you are quite right that there are concepts here that can be hard to get your head around, but it's well worth making the attempt, because the same kind of reasoning can help in understanding a lot of other problems. $\endgroup$ Oct 22 '15 at 20:32
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Here is a solution for (b):

Let $$t\mapsto z(t)=r(t)e^{i\phi(t)}\qquad(t\geq0)$$ be the orbit of the dog starting at ${1\over2}(1+i)$. Then the orbit of the dog starting at ${1\over2}(-1+i)$ is simply $t\mapsto iz(t)$. It follows that at any moment the velocity vector $$\dot z=(\dot r+i r\dot\phi)e^{i\phi}$$ is parallel to $$iz-z=(-1+i)z=(-1+i)r e^{i\phi}\ .$$ This amounts to $$\dot r+ir\dot\phi=\lambda(-r+ir)\tag{1}$$ for some $\lambda>0$ changing with time. Comparing the imaginary parts in $(1)$ we see that in fact $\lambda=\dot\phi>0$, so that from looking at the real parts we obtain $\dot r=-r\dot\phi$, or $${dr\over d\phi}={\dot r\over\dot\phi}=-r\ .$$ It follows that the function $\phi\mapsto r(\phi)$ satisfies the differential equation $$r'=-r\ ,$$ with the solutions $r(\phi)=r_0 e^{-(\phi-\phi_0)}$. This implies that we see four logarithmic spirals.

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  • $\begingroup$ Thank you very much for your answer. $\endgroup$
    – qmd
    Oct 23 '15 at 14:35
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    $\begingroup$ +1: It's been quite a while since I had anything to do with differential equations (and I didn't like it much), but reading your exposition was delightful. Thanks! $\endgroup$
    – A.P.
    Oct 28 '15 at 18:12
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First off: I don't know how to do this problem the way they are asking. But I think I can do this problem my own way. Maybe this will be a hint enough for you to do it the original way.

Let's track the top dog (starts at $(0,1)$). Let its position at time $t$ be $(x(t),y(t))$. By symmetry, the dog it is chasing is at position $$(u(t),v(t)) = (\frac{x(t) + y(t)}{\sqrt{2}},\frac{-x(t) + y(t)}{\sqrt{2}})$$ which is just the original position rotated by 45 degrees.

The speed of chase is a constant $s$ but the direction is from $(x(t),y(t))$ to $(u(t),v(t))$ so the velocity vector is $$(\dot{x}(t),\dot{y}(t)) = s\frac{(u(t)-x(t),v(t)-y(t))}{|(u(t)-x(t),v(t)-y(t))|}$$

Can you change to polar coordinantes and take it from there?

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  • $\begingroup$ Thanks for your answer. I think I have some sort of idea what to do now. I will post my post my answer if I get somewhere. $\endgroup$
    – qmd
    Oct 22 '15 at 13:44
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This answer is taken from Calculus 10th Edition by Larson and Edwards

Link to solution: Solution

This is how they derive the differential equation:

If a dog is located at $(r, \theta)$ in the first quadrant, then its neighbor is at $(r, \theta+ \frac{\pi}{2})$

$$(x_1,y_1)=(r \cos\theta, r \sin\theta) \\(x_2,y_2)=(-r\sin\theta,r\cos\theta)$$

The slope joining these points is

$$\frac{r\cos\theta-r\sin\theta}{-r\sin\theta-r\cos\theta}=\frac{\sin\theta-\cos\theta}{\sin\theta+\cos\theta}=\text{slope of tangent line at} (r,\theta)$$

$$\color{}{\frac{dy}{dx}=\frac{\frac{dy}{dr}}{\frac{dx}{dr}}=\frac{\frac{dr}{d\theta}\sin\theta+r\cos\theta}{\frac{dr}{d\theta}\cos\theta-r\sin\theta}=\frac{\sin\theta-\cos\theta}{\sin\theta+\cos\theta}}$$ $$\color{blue}{\implies \frac{dr}{d\theta}=-r}$$

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  • $\begingroup$ They're using chain rule to at first. i.e. $$ \frac{ dx}{d\theta} \frac{ dy}{dx} = \frac{ dy}{d\theta} $$ and computing the derivatives using $x(\theta) = r(\theta) \cos \theta$ and $y(\theta = r(\theta) \sin \theta$. $\endgroup$
    – Jeb
    Oct 22 '15 at 19:10
  • $\begingroup$ @Jeb I am sorry but I don't get it. Wouldn't I use the product rule to differentiatie $x(\theta)$ and $y(\theta)$? $$\implies \frac{dx}{d\theta}=\frac{dr}{d\theta}\cos\theta -r\sin\theta \\ \frac{dy}{d\theta}=\frac{dr}{d\theta}\sin\theta +r\cos\theta$$ but how does that equal: $$\frac{\sin\theta-\cos\theta}{sin\theta+\cos\theta}$$ $\endgroup$
    – qmd
    Oct 22 '15 at 19:40
  • $\begingroup$ because the ratio is $\frac{dy}{dx}$ which is the slope of the tangent line at $(r,\theta)$, and you've computed that already. Then you can simplify the expression to isolate $\frac{dr}{d\theta}$ $\endgroup$
    – Jeb
    Oct 22 '15 at 19:54
  • $\begingroup$ @Jeb It's getting really late at my place so I will give it a rest for today. I will come back tomorrow and try to complete the problem. Thank you very much for your help. $\endgroup$
    – qmd
    Oct 22 '15 at 20:24

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