[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 180
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It's a simple integration problem. Let someone else solve it :) At a glance... uh... that's a lot of minutes.
Your next problem is about multiplication.
I'll think about the muzik for now. I think it's going to crawl.
Возможны случаи:
1. Прямая L лежит вне этих окружностей = решения нет.
2. Прямая L пересекает одну из "вершиновозможных" окружностей = имеем два решения - в точках пересечения.
3. Прямая L касается одну из финальных окружностей = 1 решение в точке касания.
4. Прямая L касается обе финальные окружности = 2 решения - в точках касания.
5. Прямая L касается одну окружность и пересекает вторую = 3 решения - в точке касания + 2 в точках пересечения.
Another obvious case not prescribed:
6. The line L intersects both final circles = 4 solutions (at the intersection points).
to Mathemat: homothety rules.
I'll do my part.
Anyone who has solved this before, please don't say anything.
On the end of a 1 m long rubber hose sits a muzik. The same end is attached to a tree. The other end is pulled at a speed of 1m/s, at the same moment the mutsik starts crawling towards the opposite end of the cord at a speed of 1 cm/s. Will the mucic crawl to the end of the cord? If no, then prove, if yes, in what time?
but has it not already crawled? at what distance is it initially from the tree?
а разве он уже не дополз ? на каком расстоянии он изначально находится от дерева ?
He's sitting in a tree :)
If the mucic is somewhere in the inner point of the stretched hose (say 0<alpha<1), and the length of the hose is L, the mucic's position from the tree is L*alpha. In a second, the length of the hose will be L+1, and the distance of the mucic from the tree will be (L+1)*alpha + 0.01.
The hose has lengthened by 1 m, and the muzik has moved forward by 1*alpha + 0.01. It turns out that if alpha > 0.99, the muzik will still crawl to the end of the hose. But first it has to get to this point.
Example: The length of the hose is 1000 m, alpha = 0.995. It is separated from the end of the hose by 5 metres. After a second, the hose is 1001 metres, and his position is 1001*0.995 + 0.01 = 996.005 metres. He is now 4 metres 99.5 centimetres from the final, and his alpha has become 0.99501.
But if the alpha was 0.999 with the same length of the hose (1 meter to the end of the hose), it would be 1001*0.999 + 0.01 = 1000.009 m in a second, i.e. the distance to the end would already be 99.1 cm.
But how it will get to alpha=0.99 is still unclear. We need to integrate here as well. Or am I already crazy with these integrals...
P.S. In short, if n is the number of seconds passed, then
alpha(n+1) = alpha(n) + 0.01/(n+1)
alpha(0) = 0
Hence alpha(n) = 0.01 * (1/1 + 1/2 + 1/3 + 1/4 + ... + 1/n) ~ 0.01 * ( ln(n) + C )
C is the Euler-Mascheroni constant, equal to ~ 0.577.
Mutsik will crawl - because of the divergence of the harmonic series (when alpha gets to 1)!
And it would take him n ~ exp( 100 - C ) ~ exp(99.423) ~ 1.51*10^43 c, i.e. about 4.79*10^35 years.
Perhaps the continuous analog of the solution (with differentials) will give a more accurate answer. alsu, I haven't solved this problem before, by golly!
На конце резинового шланга длиной 1 м сидит муцик. Тот же конец прикреплен к дереву.
The muzik sits on the end of a hose. The end of the hose is attached to a tree. Consider it attached to a branch. Then the muzik sits on that branch and on the end of the hose (softer to sit), which is attached to the branch.
Of course he can also sit under a tree, but the good Dr. Aibolit likes to sit under a tree. And we have Mutsik - Kashchey the Immortal, so sits on a branch :).
Z.I. And the problem can be solved in three ways:
With integrals, without integrals, and in the coordinate system.
Z.Z.I. Above is an example of how to solve the problem in a system of coordinates :)
zxc, первые 59 стаканов нальются относительно быстро. А последний - никогда.
Mathemat, you've got the point exactly! It really has a twist, with a catch - one glass has already been poured when checking how many seconds it takes to fill it :)
For those who are going shopping at this hour:
At the shop, a customer paid for 9 kg of groats. The salesman opened the sack of 24 kg of cereal and discovered that the scale was not working. How do I let the customer go of his purchase?
Муцик доползет - из-за расходимости гармонического ряда (когда альфа доберется до 1)!
И потребуется ему для этого n ~ exp(100+C) ~ exp(100.577) ~ 4.8*10^43 c, т.е. где-то 1.52*10^36 лет.
Возможно, непрерывный аналог решения (с дифференциалами) даст более точный ответ. alsu, я эту задачу раньше не решал, ей-богу!
Score! Only quite right answer: it will crawl unless another Big Bang happens in the meantime:)))