a Clock Problem
motivated by a discussion on the Calculus Forum
The hour, minute and second hands of a clock are 4, 6 and 8 mm respectively.
When is the distance between the hour and minute hands changing at the same rate as the distance between the minute and second hands?
H = length of Hour hand and h = the clockwise angle made with the vertical (12 o'clock) position.
M = length of Minute hand and m = the clockwise angle made with the vertical position.
S = length of Second hand and s = the clockwise angle made with the vertical position.

Then
dh/dt = 2π/12 radians per hour
dm/dt = 2π radians per hour
ds/dt = 2π*60 radians per hour

The distance between the Hour and Minute hands is given by the cosine law:
L2 = H2 + M2 - 2 H M cos(m - h)

Then 2L dL/dt = 2 H M sin(m - h) (dm/dt - dh/dt)
But dm/dt = 12 dh/dt and dh/dt = 2π/12, hence
L dL/dt = H M sin(m - h) (11) dh/dt = H M sin(m - h) (11) (2π/12) = H M sin(m - h) (22π/12)

At T hours after midnight, m = 2Tπ, h = 2Tπ/12, m - h = 22Tπ/12 and L2 = H2 + M2 - 2 H M cos(22Tπ/12), so
dL/dt = [ H M sin(22Tπ/12) (11) (2π/12) ] / [ H2 + M2 - 2 H M cos(22Tπ/12) ]1/2 = [ (22π/12) √(HM/2) sin(22Tπ/12) ] / [ (H2+M2)/ (2HM)- cos(22Tπ/12) ]1/2

Similarly, if the distance between the Minute and Second hands is K, then:
K2 = M2 + S2 - 2 M S cos(s - m)
2K dK/dt = 2 M S sin(s - m) (ds/dt - dm/dt) = 2 M S sin(s - m) (59) dm/dt = 2 M S sin(s - m) (59)(2π) = 2 M S sin(s - m) (118π)
since ds/dt =60*dm/dt and dm/dt = 2π.

At T hours after midnight:
s = 2Tπ*60, m = 2Tπ, s - m = 2Tπ*60 - 2Tπ = 118Tπ, and
dK/dt = [ M S sin(118Tπ) (59) (2π) ] / [ M2 + S2 - 2 M S cos(118Tπ) ]1/2 = [ (118π) √(MS/2) sin(118Tπ) ] / [ (M2+S2)/ (2MS) - cos(118Tπ) ]1/2

Setting dL/dt = dK/dt and solving for T means finding a solution of an equation such as:
A sin(αT) / [a - cos(αT)]1/2 = B sin(βT) / [b - cos(βT)]1/2
where a = (H2+M2)/(2HM) > 1, b = (M2+S2)/(2MS) > 1, α = 22Tπ/12 and β = 118Tπ.

For the hand lengths given, the graphs of dL/dt and dK/dt look like so:

Notes:

  1. Although T is in hours, the horizontal axis (above) is shown in seconds.
  2. The rates of change are equal at (about) 11 seconds after 12 o'clock.
  3. The lengths themselves, L and K, look like this (with the horizontal axis in seconds):

  4. Though it appears that dL/dt is monotone increasing, that's just for the first few minutes.
    Subsequently, dL/dt looks like this (where now the horizontal axis is in hours):