pruhs@uwvax.UUCP (Kirk Pruhs) (02/13/85)
Here is another rather interesting clock problem that was on the Putnam last year. If you have a standard clock with hand lengths 3 and 4, how far apart are the tips of the hands when the distance between them is increasing most rapidly? A freshmen calculas student has the tools to solve this but a little care needs to be taken or you can fill a wall with the equations.
ech@spuxll.UUCP (Ned Horvath) (02/17/85)
Kirk Pruhs (pruhs@uwvax.UUCP) writes: > Here is another rather interesting clock problem that was on the Putnam >last year. If you have a standard clock with hand lengths 3 and 4, how far >apart are the tips of the hands when the distance between them is increasing >most rapidly? A freshmen calculas student has the tools to solve this but >a little care needs to be taken or you can fill a wall with the equations. One need only remember the high school formula for the third side of a triangle given two sides and the angle between them: c^2 = a^2 + b^2 - 2ab(cos C) Where C(t) is just 2pi*(11/12)t, t in hours. I mention that just to get rid of t -- C is just a linear multiple of t. But then the change in distance is given by d(c^2)/dC = 2ab(sinC) which is, of course, maximized at C=pi/2. When the hands are at right angles I can calculate c in my head... =Ned=