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?
OK, this time I just started with the law of cosines getting d as a function of theta, then took d' and d" and I got the same answer as Charlie's original post. But I got it without using the solver on the spreadsheet. Call the angle x, it's too hard to figure out how to write a theta
d=sqrt(2524cos(x))
d'=12sin(x)/d
setting d" to zero gives:
0=cos(x)sqrt(2524cos(x))  12 sin^2(x) / sqrt(2524cos(x))
multiply both sides by sqrt(2524cos(x)) gives:
cos(x)(2524cos(x))  12 sin^2(x)
25 cos(x)  24 cos^2(x)  12 sin^2(x)
25 cos(x)  12 cos^2(x)  [12 sin^2(x) + 12 cos^2(x)]
25 cos(x)  12 cos^2(x)  12 or
cos^2(x)  (25/12) cos(x) + 1 = 0
solving the quadratic gives 2 values for cos(x): 3/4 and 4/3 but only 3/4 can be valid since the cosine can't be greater than 1.
so cosine is .75
the angle is arccos(.75) = .7227 radians = 41.41 degrees
distance is 2.646 = sqrt(7)
d' = 3
and there is a right angle between the hour hand and a line connecting the tips of the 2 hands
Better late than never, and that's my final story

Posted by Larry
on 20040307 10:04:01 