/sci/ - Science & Math

[Anonymous No.4742304]

Three small snails are each at the vertex of an equilateral triangle of side 60 cm. The first sets out toward the second,the second toward the third, and the third toward the first, each with a uniform speed of 5 cm/min. During their motion, eachsnail always heads toward its respective target. How much time elapses and how far does each snail travel before they all meet?

[Anonymous No.4742318]

Em.. They never meet?

[Anonymous No.4742323]

Doesn't matter how fast they go. If they all move from the same angle as their predecessor then they will never meet.

[Anonymous No.4742325]

probably follow one of those funny curves like a brachistochrone

can't be arsed to form the diff equation (symmetry means you only need 1, not a system)

[Anonymous No.4742326]

>>4742323
>>4742318
they meet.

initial path is lessening the distance from centre, and this will continue as shape doesn't change, only gets smaller

[Anonymous No.4742327]

>>4742318
the obviously aren't confined to the triangle


The lines connecting the snails always form an equilateral triangle at any time. That gives you the relative angles from which you can determine the rate at which the sides of the triangle shrink.

8 minutes

[Anonymous No.4742329]

>>4742318
shouldn't it go like this?

[Anonymous No.4742334]

>>4742329
They should never make it past the meeting point like the do in your picture. For example, I think 2 should never leave the red triangle in this picture. Consider that if he ever was out of it, his path would be taking him immediately back inside.

[le oldfag !!vsaJus8+pOh No.4742347]

60 / 5 = 12 minutes

[Anonymous No.4742349]

>>4742347
le xD

[Anonymous No.4742350]

Ok, if you advance each snail <span class="math">dx = 5dt[/spoiler], then the side length of the triangle becomes <span class="math">\sqrt{(x-\frac{3}{2}dx)^2 + (\frac{\sqrt{3}}{2}dx)^2}[/spoiler], so you could probably figure it out from that.

[Jimmy No.4742351]

I think the paths will look somewhat like Bézier curves: http://en.wikipedia.org/wiki/B%C3%A9zier_curve

[Anonymous No.4742355]

They meet after 8 minutes.
found it using simple components

[Anonymous No.4742361]

>>4742355
they approach each other with 3/2 v
the relative direction of those vectors stays the same over the whole time.

[Anonymous No.4742365]

>>4742327
>>4742355
>>4742361


/thread

also, everyone else should lrn2frameofreference

[Anonymous No.4742366]

>>4742350
Sorry, let that be <span class="math">\sqrt{(l-\frac{3}{2}dx)^2 + (\frac{\sqrt{3}}{2}dx)^2} = \sqrt{l^2-3dx+3dx^2}[/spoiler] where <span class="math">l[/spoiler] is the side length of the triangle. Then if we ignore higher powers of <span class="math">dx[/spoiler] we can approximate that as <span class="math">l-\frac{3}{2}dx = l - \frac{15}{2}dt[/spoiler]. So, <span class="math">dl = -\frac{15}{2}dt[/spoiler], which means that it will indeed take 8 minutes.

[Anonymous No.4742378]

>>4742366
>Then if we ignore higher powers of dx we can approximate

nigger, wtf are you doing? you can get the exact answer with a much simpler technique.

do you fucking hammer nails into your wall with a jackhammer too?

[Anonymous No.4742390]

>>4742366

[Anonymous No.4742389]

>>4742378
Sure, why not.

>Two trains, 20 miles apart, approach each other going 10 mph. A fly travels between them at 15 mph, zig-zagging just before getting smooshed by an oncoming train. Question: How far does the fly fly before meeting his ultimate demise?

Here's a famous story about the physicist and mathematician John von Neumann :

Another mathematician knew the quick solution to the Fly problem and wanted to see von Neumann struggle with it. He posed the question and von Neumann responded with the right answer in a few seconds.

"Interesting," said the first mathematician. "Most people try to sum the infinite series."

"What do you mean?" von Neumann replied. "That's how I did it."

[Anonymous No.4742399]

>>4742389
And that shows what, exactly? It would still be faster with the simpler technique.

That doesn't even apply in this case, because an approximation was made. It turns out to be right, but I doubt the approximation was made with any rigorous analysis of the problem.

If you don't see why a simpler, exact solution is better than less simple, inexact solution, then there is nothing to say to you.

Enjoy your Rube Goldberg approaches.

[Anonymous No.4742402]

>>4742389
I read a version of that problem as a kid, and did it as infinite series (which I think I guessed what it would converge to, because I didn't know the formula then). But now, I would just do it the really simple way.

Does this mean I used to be von Neumann - level clever, but now I've got less intelligent?

[Anonymous No.4742409]

>>4742399
It shows that you're getting pissy over nothing important. You probably didn't solve the problem yourself anyway - you probably just had heard it before.

And it's not really an approximation, even though I called it that. Its perfectly fine to ignore higher powers of dx. It's more rigorous than the other solutions. Note that in >>4742355
the guy separated the velocity into a component that is moving directly towards the other snail, and a component that moves away from it. He then ignores the component moving away. Are you telling me that's rigorous?

I showed that the component moving away's contribution had a higher order of dx in it, so it could safely be ignored.

[Anonymous No.4742528]

>>4742409
>lrn2frameofreference
>8 minutes is correct

I can even give you a different approach

[mtp !EKFQOBEsKo No.4742615]

wouldnt the easiest thing to do be to consider the distance from a meeting point and use the component of velocity along with distance/speed to get the time? or are we circle jerking here

[Anonymous No.4742633]

>>4742528
I bet that other approach wouldn't be rigorous either.

[Anonymous No.4742639]

>>4742615
Just circle jerking. That way is easy and gets the answer, but some guy got pissy because I posted a different way to do it that was slightly more rigorous.