A clock's minute hand has length 4 and its hour hand length 3.
What is the distance between the tips at the moment when it is increasing most rapidly?
I wrote a long solution, but it didn't save. Grr... Rather than write it again, I'm just gonna post the important points. Hopefully, someone can follow it and make sure the math is correct.
Consider them to start at 12 o clock so both the minute and the hour are at 0 degrees. The minute hand travels at 6 degrees/min. The hour hand travels at 0.25 degrees/min. At time t, the minute hand is at 6t degrees and the hour hand is at 0.25 degrees. Converting from polar to Cartesian coordinates gives the minute hand to be at (4cos(6t), 4sin(6t)) and the hour hand to be at (3cos(0.25t), 3sin(0.25t)).
Applying the distance formula and using (cosx)^2+(sinx)^2=1 and cos(x-y)=cosx*cosy+sinx*siny simplifies the distance into d=sqrt(25-24*cos(5.5t)). d'=66*sin(5.5t)/sqrt(25-24cos(5.5t)).
Also, defining f=d^2. Then f'=2d*d'. f''=2*(d')^2+2d*d''. For the distance to increase most rapidly, d''=0. Hence f''=2*(d')^2.
f=25-24*cos(5.5t), f'=132*sin(5.5t), f''=726*cos(5.5t).
Setting f''=2*(d')^2 would simplify into
This simplifies into a quadratic in terms of cos(5.5t).
Solving yields cos(5.5t) = 3/4 or 4/3. Only 3/4 is valid. It can be verified that this gives a maximum instead of a minimum.
Plugging cos(5.5t) into the formula for d yields sqrt(7).
Posted by np_rt
on 2004-02-28 00:45:57