Offtopic: Math Puzzle

[toferc]

Originally Posted by Axel Fox

Correct, now show me how you got there and, if you have one, give us a new math puzzle.

Thought I'd allow some time for others before posting the spoiler :)

[Axel Fox]

OK, no problem.

[eepiccolo]

To solve this problem, we take advantage of the second rule of solving physics problems; chose your coordinate system wisely.

So the following picture shows the clock hands as vectors, and also the corresponding velocity vectors.

http://home.earthlink.net/~eepiccolo...0solution2.jpg

Our job is to maximize the line-of-sight velocity and find the distance between the tips of the hands when said velocity is maximized. Line-of-sight velocity is found by
vlos=vrel.nlos
where vrel is the relative velocity, anlos is the normalized line-of-sight vector from the hour hand to the minute hand, and . is the dot-product.

What follows are the equations that derive vlos.

vrel = vh - vm

vh and vm both follow uniform circular motion, and the minute hand travels at 4*2*pi/60 rad/min, and the hour hand travels at 3*2*pi/60/24 rad/min, so

vh = pi/120 î
vm = (2*pi/15)cos(theta) î - (2*pi/15)sin(theta) ĵ
(to all the physicists out there, please excuse the lack of unit definitions ;) )

vrel = (pi/120 - (2*pi/15)cos(theta)) î + (2*pi/15)sin(theta) ĵ
nlos = (rh - rm)/||rh - rm||
= (4sin(theta) î - (3 - 4cos(theta)) ĵ ) / sqrt{(4sin(theta))^2 + (3 - 4cos(theta))^2}
vlos=vrel.nlos
= {4sin(theta)*(pi/120-2*pi/15*cos(theta) + 2*pi/15*sin(theta)*(4cos(theta)-3)} / sqrt{(4sin(theta))^2 + (3 - 4cos(theta))^2}
= {pi/30*sin(theta)-8*pi/15*sin(theta)*cos(theta)-2*pi/5*sin(theta)+8*pi/15*sin(theta)*cos(theta)
/ sqrt{16sin^2(theta) + 16cos^2(theta) - 24cos(theta) + 9}

= -11*pi/30sin(theta)/sqrt(25 - 24cos(theta))

Now, to find a maximum, we take the derivative and set it equal to 0. The derivative is found using the quotient rule and chain rule, and that derivation can be supplied if anybody wants it. It simplifies to

12cos^2(theta) - 25cos(theta) + 12 = 0

And using the qudaratic formula, we find cos(theta) = 4/3, 3/4, but cos(theta) < 1, so cos(theta) = 3/4

Then, using the law of cosines, c^2 = a^2 + b^2 - 2ab*cos(theta), so

c^2 = 3^2 + 4^2 - 2*3*4*(3/4) = 9+16-18 = 7

c=sqrt(7)

[Axel Fox]

Yup, verry correct, although I find your explanation verry complicating.

The way I did it on my exam (yes, it was an exam question) was to use the cosine rule : A*A = B*B + C*C - 2*A*B*cos(theta), which gives our distance we seek as A = sqrt (B*B + C*C - 2*A*B*cos(theta)) (*)

In our case, we can already substitute B by 3 and C by 4 (or the ohter way around) and then we get A = sqrt(25 - 12*cos(theta))

We differentiate this twice and equal that to 0. We then solve that equation and we get cos(theta) = 3/4

If we substitute this into equation (*) we get sqrt(7) as the both of you stated.

And frankly, my math isn't THAT good, zo I don't understand much of your explenation, eepiccolo, sorry :(

[eepiccolo]

I see, that is certainly simpler. Having certain backgrounds can be a hinderance a times, because it can make your view of a problem more narrow. I saw it as a physics problem involving a line-of-sight velocity, and solved it that way.

I'm guessing, Axel, that you don't have much of a physics background, and the vector part was what you didn't understand. But this was an advantage in your case, because you led yourself to a simpler solution.

[Axel Fox]

Yeah maybe,

my background is more mathematical and in that not even very advanced mathematics. So that must be it.

Nevertheless, we all came to the same solution.

[toferc]

I used cosine rule and double differentiation as well, though I later thought polar coordinates might be more convenient, or elegant.

Been too busy to post my solution or find a new problem; glad someone else did.