# Car washing

Discussion in 'Physics & Math' started by Pim, May 27, 2003.

1. ### PeteIt's not rocket surgeryModerator

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Man&Dog problem:

I get a set of parametric differential equations that I have no idea how to solve:

(dy/dt)<sup>2</sup> + (dx/dt)<sup>2</sup> = 1
dy/dx = y/(t-x)
Initial condition: (x,y,t)=(0,2,0)

3. ### AbsaneRocket SurgeonValued Senior Member

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Later I will try posting what I know about the problem.

As for right now, maybe knowing that the tangent line to every point on the curve passes through the point that the man is on at the time the dog is at the given point. Maybe that makes sense?

James Sibley

5. ### MacMRegistered Senior Member

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Thanks

Pim,

Got it now thanks.

7. ### PimRegistered Senior Member

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After searching a bit on the internet i found that the man-dog problem is a known problem, there was also a solution given which i never would have found. I will not post it here yet though

8. ### metacristiRegistered Senior Member

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The dog cannot continuosly adjust his position function of the movements of the man,he can do that only in a discrete manner let's say at very small intervals dt.For example initially the dog walks toward the point [0,0].At dt he observe that the man is in the point [vdt,0] so that he walks now toward it and so on.If dt is chosen as being small enough [let's say dt=0.1 seconds] then an interpolation approach [using Lagrange's method] is possible.y(x) is seeked in the form of a polynome,the number of points chosen must be high enough for a good approximation.The angle between dog's direction of movement and the 0y axis varies with time but it's tangent can be calculated as much as the step of the time dt is chosen [let's say dt=0.1 sec].As a result of this procedure the coordinates (x<sub>i</sub>,y<sub>i</sub>) of the dog can be computed as a function of v,the number of nodes 'i' being in function of the intervals of time dt chosen.Finally using Lagrange interpolation an approximation of the funtion is obtained in the form of a polynomial.
The solution is viable however it's not easy to find a general formula for y(x) when the general variable 'v' is used,the number of calculations is great enough,but for particular values of v this solution is feasible.

9. ### PimRegistered Senior Member

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Okay...this is what i found on the internet, it is somewhat different, now there is a leash between the dog and the man, but i think that doesn't change the problem?

http://mathworld.wolfram.com/Tractrix.html

Metacristi: Let's assume the dog CAN change his direction continuously

10. ### metacristiRegistered Senior Member

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Basically if dt is chosen small enough this is equivalent to a continuous changement of direction.
But in the article it is said that Leibniz found the curve otherwise...There is no clear explanation but I assume that he had not known the equations previously [otherwise he could have found the curve by simply plotting some experimental data in the derivative of the known equations].I will try further...

11. ### safXmalRegistered Member

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Three man and some hats

I have a small logical problem for you all to solve. No math involved I'm afraid.

3 man are standing in line before a wall so that the first man only sees the wall and the third one the 2 men in front of him and the wall.

Now each is put on a hat which they themselves can not see, chosen out of 3 white hats and 2 black ones.

When the first man asks the third man what color of hat he thinks he's wearing the third man replies : " I don't know"

Then the first man asks the second man to tell what color of hat he thinks he has on and he replies after thinking a while also: " I don't know"

"Oh" says the first man; " That means that I'm wearing a white hat"

He was correct but can anyone explain me the logic he used to come to this conclusion.

12. ### AbsaneRocket SurgeonValued Senior Member

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Mind at least posting the link? lol. I cannot find it myself. I just keep finding pages about dogs, even though I use keywords like "calculus" and "equation." Oh well.

James Sibley

13. ### sdeliver645Registered Member

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...three men and some hats

Good riddle. Here is the logic.

When the man at the rear responds that he doesn't know what colour hat he is wearing, this permits three possible scenarios:

I will give scenarios in the format (front | middle | rear)

i) (B | W | ?)
ii) (W | B | ?)
iii) (W | W | ?)

When he asks the gentleman in the middle, who also responds he doesn't know, this eliminates possibility number (i). This is because we know that there are three white hats, and two black hats. If the man in the middle looked forward and saw a black hat, he would know that scenarios (ii) and (iii) aren't possible. Having eliminated (ii) and (iii), scenario (i) would tell the gentleman in the middle that he is wearing a white hat.

However, because scenario (i) was discounted, both scenarios (ii) and (iii) have the gentleman in the front wearing a white hat, thus the man at the front having heard the conversation and knowing the number of hats provided knows he is wearing a white hat.

Thanks!

14. ### sdeliver645Registered Member

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Three Convergent Dogs...

In the spirit of the man/dog problem, and the man/dog problem (with a leash - see the link provided earlier - http://mathworld.wolfram.com/Tractrix.html), here is another one.

Consider three dogs, each at the corners of an equilateral triangle. Simultaneously, each dog begins to run towards the dog on his/her left (to help people out, the path of the three dogs will look like three curves spiraling in towards the centroid of the triangle).

The question is, what is the path length the dog travels before it reaches the centroid?

P.S. I forgot the answer to this one, but will see if I can recall it...

Enjoy!

15. ### AbsaneRocket SurgeonValued Senior Member

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Re: ...three men and some hats

Wow. Kick ass. I got the three senerios then I stopped. I had to get ready for work

Good work though!

James Sibley

16. ### PeteIt's not rocket surgeryModerator

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Hi Pim,
Having a fixed distance between man and dog is a different problem, as in that case the dog's speed will vary (starting at zero, and approaching the man's speed asymptotically).

17. ### PimRegistered Senior Member

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Pete: i think it is the same problem. Since there is a leash between the two, i think they must have the same speed. The man also starts at zero speed. Think about it..in one second walking the dog travels the same distance as the man because they can be seen as "one" because of the leash, so their speeds are the same. I think adding a leash does not change the problem.

18. ### sdeliver645Registered Member

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fixed distance implying the same velocity

I don't think that a fixed distance implies that the dog is travelling at the same rate as the man. This would be the case of the orientation of the leash remained unchanged, but that is not the case.

You can think of the motion (dog's velocity) as having a component due to translation, and one due to rotation (changing the angle at which the leash is oriented).

Having said that, I don't have a solution.

19. ### PimRegistered Senior Member

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Yes, maybe you're right...i have to think this over for a while..having a leash does not imply the same speed always. For example, the dog can run in a "circle""around the man, and thus have a much higher speed, but doesn't the dogs speed has to be at least the man's speed?

20. ### sdeliver645Registered Member

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Q: Does the dog's speed have to be at least the man's speed?

A: In the context of this question (tractrix), yes, the dog is always travelling AT LEAST as fast as the man. This can be seen IF the following assumption is true:

Assumption: That the dog's speed decreases monotonically.

This seems reasonable, doesn't it? One could always check this by cheating and looking at the solution (on the Wolfram site - see link previously given by someone else). Once the parametric eqn's of motion are known, one can determine the instantaneous velocity and then the speed of the dog.

If this is true, then the dog's speed decreases to that of the man's asymptotically. In the limit that the man is pulling the dog for an infinite amount of time, the dog's trajectory will become coincident with that of the man's (coincident with that of, say, the x-axis, if the man is moving along the x-axis). When this occurs, the dog and the man will be both moving at a fixed speed.

21. ### PeteIt's not rocket surgeryModerator

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In the leash situation, the dog's speed is always less than that of the man. In the first instant when the man is walking at right angles to the leash, the dog's speed is zero.

If the dog were to travel at the same speed as the man and maintain the same distance between them, the dog must walk parallel to the x-axis.

In the original problem with the dog and man moving at the same speed, the dog is always getting closer to the man.

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23. ### StrangeDaysRegistered Senior Member

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I agree with Pete, but his last sentence raises a quandry in my head. If they're travelling at the same speed, does the distance between the two asymptotically approach zero, or does it approach a nonzero number?