Pure maths, physics, logic (braingames.ru): non-trade-related brain games - page 156
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And by the way, should you specify the distance before the time change or after? It's kind of hard to measure the distance when asking the process "when this distance changes the fastest".
Assume that the hands are moving continuously, without jerking. This is the most logical assumption.
Somehow I can't do it without derivatives.
Seems really tricky. Intuitively it seems to be the point at which they overlap. (Another option is when they look in opposite directions.) But it's not at all obvious.
Consider that the arrows move counterclockwise from some zero point where they initially coincided in direction.
Hourly: z1 = 36*exp(i*t) = 36*cos(t) + i*36*sin(t)
Minute: z2 = 45*exp(i*12*t) = 45*cos(12*t) + i*45*sin(12*t)
The distance between the ends (or rather, its square): L^2 = (36*cos(t) - 45*cos(12*t))^2 + (36*sin(t) - 45*sin(12*t))^2 =
= 36^2 + 45^2 - 2*36*45*(cos(t)*cos(12*t) + sin(t)*sin(12*t)) =
= 36^2 + 45^2 - 2*36*45*cos(11*t) = 3321 - 3240*cos(11*t)
So L = (3321 - 3240*cos(11*t))^0.5. (***)
L' = 0.5*(3321 - 3240*cos(11*t))^(-0.5) * 11*3240*sin(11*t) -> max modulo.
That's it. I'll pass on, even Wolfram doesn't find honest extrema, it's an approximation there.
= 36^2 + 45^2 - 2*36*45*cos(11*t) = 3321 - 3240*cos(11*t)
Pfft. I was just solving the same thing myself, and it all worked out the same way. I looked at the forum and got the same idea :).
Yeah, I don't know about the derivation either. I don't remember how they're calculated. It's really not realistic to calculate the derivative from this expression. But why? There must be a solution, obviously.
It seems to be solved.
So we get this dependency function
y = (3321-3240*cos(x))^(1/2), where
y is the distance between the ends at any time
x is the angle of deflection between the arrows [0 ; 2*Pi]
From here we find the derivative and investigate for an extremum
y ' = 1/2*(3321-3240*cos(x))^(-1/2)*3240sin x = 0
sin x = 0
x1 = 0
x2 = pi
At 0 the velocity is maximum, at pi it is minimum.
So the maximum velocity is at 0gr, which means that it will be at the point where the arrows coincide, as originally assumed.
This seems to solve the problem, although if anything is wrong I will let you know.
So, maximum speed at 0g, which means that it will be at the point where the arrows coincide, as originally intended.
This seems to solve the problem, although if anything is wrong, I'll let you know.
From here, find the derivative and investigate to extremum
y ' = 1/2*(3321-3240*cos(x))^(-1/2)*3240sin x = 0
No, it isn't. I can find the derivative myself.
Here we need to find its extremum, not zero. It is the zero of the second derivative.
when this distance changes most rapidly.
That is, when the velocity is at its maximum.
So, the maximum speed at 0g, which means that it will be at the moment when the arrows coincide, as originally assumed.
This seems to solve the problem, although if anything is wrong, I'll let you know.
The numerical method gives completely different values ).
When starting at noon, the maximum speed between the arrows is at 403 seconds and repeated after 3927 seconds (the calculation is accurate to a second). Distance 27 mm
The numerical method gives completely different values ).
When starting at noon, the maximum speed between the arrows is at 403 seconds and repeated after 3927 seconds (the calculation is accurate to a second). Distance 27 mm
One more time. We remove multiplier 81 at the numbers, which solves nothing, and the frequency multiplier. We obtain function
L(t) = (41-40*cos(t))^0.5
The function is periodic. Graph:
We have to find the points where L' is maximal in modulo (on the graph we see that these points are near the minima of the function L, but they are definitely not its minima; in fact they are inflection points of the graph).
In other words, we have to choose from the zeros of the second derivative L(t). Carefully differentiate twice - and we obtain that zeros of the second derivative are the points where cos(t) = 4/5. (If you need it, you can differentiate the function L(t) twice by yourself.)
The distance (taking into account the lost multiplier sqrt(81)) is
L(t) = 9*(41-40*4/5))^0.5 = 27 mm.
I may have messed up somewhere, or not taken something into account. But the result is surprisingly "rational", indicating that the solution is probably correct.
P.S. The first time from zero (although it is not necessary to look for it) is something around pi/5, i.e. somewhere around 6 minutes after the start of the movement.
The answer turned out not at all like supposedly "intuitively obvious".
But the problem is really quite simple, but one has to be careful.
I wish I could find a solution without the upper maths...