Clock Hands

[BMAD]

The minute hand of a clock is twice as long as the hour hand. At what time, between 00:00 and when the hands are next aligned (just after 01:05), is the distance between the tips of the hands increasing at its greatest rate?

[TimeSpaceLightForce]

6T = 180+0.5T or 32.72727..min

[BMAD]

6T = 180+0.5T or 32.72727..min

At what time?

[TimeSpaceLightForce]

my guess above is the greatest distance but not fastest speed of separation ,

so maybe they should be perpendicular : 6T = 90+0.5T or 16.3636 min (00:16) .

or some calculus is needed for its elliptical path??

[BMAD]

my guess above is the greatest distance but not fastest speed of separation ,

so maybe they should be perpendicular : 6T = 90+0.5T or 16.3636 min (00:16) .

or some calculus is needed for its elliptical path??

elip.JPG

it can be done with calc or geometry. solution is still not the best.

[dark_magician_92]

12:10:55.45 ie 120/11 mins after 12'o clock

[BMAD]

12:10:55.45 ie 120/11 mins after 12'o clock

There is still a better answer

[dark_magician_92]

the rate of increase in distance which i got was = 5.5*l units/minute, if i am not wrong

[bonanova]

I agree with dark magician.

Let THETA be the angle between the hands.
The angle increases from 0^o to 360^o in 12/11 hours, at a rate of 5.5^o per minute.

The distance between the tips of the hands is minimal [1 unit] at 0^o and 360^o, and maximal [3 units] at 180^o.
The angle at which the tips move apart [together] at the greatest rate is 60^o [120^o].
That corresponds to 60/5.5 = 10.9090... [21.8181...] minutes after the hour.

Verified by simulation.

[BMAD]

We may regard the minute hand as fixed, and the hour hand as rotating with constant angular speed. Consider the circle with center O and radius 1. As OH sweeps around this circle, it is clear that MH increases (or decreases) at its greatest rate when MH is tangent to the circle, as it is at these points that H is moving exactly away from (or towards) M. (At other points, the component of the speed of H along MH is smaller because there is also a non-zero component of the speed that is perpendicular to MH.)

Therefore, as above, ds/dt reaches its maximum value when cos q = 1/2, which occurs at 00:10:54 6/11.

[bonanova]

We may regard the minute hand as fixed, and the hour hand as rotating with constant angular speed. Consider the circle with center O and radius 1. As OH sweeps around this circle, it is clear that MH increases (or decreases) at its greatest rate when MH is tangent to the circle, as it is at these points that H is moving exactly away from (or towards) M. (At other points, the component of the speed of H along MH is smaller because there is also a non-zero component of the speed that is perpendicular to MH.)

Therefore, as above, ds/dt reaches its maximum value when cos q = 1/2, which occurs at 00:10:54 6/11.

Nice proof. I overlooked the significance of the right angle in my solution.

[BMAD]

We may regard the hour hand as fixed, and the minute hand as rotating with a constant angular speed (equal to 11/12 of its actual angular speed.) Without loss of generality, let the hour hand have length 1 and the minute hand have length 2. Let the angle between the hands be q, and the distance between the tips of the hands be s. Clearly, for s to be increasing, the diagram must be oriented as below, with q increasing, as OM rotates clockwise.

By the law of cosines s² = 1² + 2² − 2 · 1 · 2 · cos q = 5 − 4 cos q.

Using implicit differentiation, 2s · ds/dq = 4 sin q.
Hence ds/dq = (2 sin q)/s.

By the law of sines, (sin q)/s = (sin p)/2, where p is the angle between the hour hand and the line segment joining the tips of the hands.
Hence ds/dq = (2 sin p)/2 = sin p.

Clearly, ds/dq reaches its maximum value when sin p = 1, or, since 0° < p < 180°, p = /2.
By the chain rule, ds/dt = ds/dq · dq/dt. Since dq/dt is a positive constant, ds/dt also reaches its maximum value when p = /2.
When p = /2, OHM is a right triangle, so cos q = 1/2, and, since 0° < q < 180°, we must have q = 60°.

Finally, we note that, as q tends to 0° from above, or to 180° from below, ds/dq tends to 0. Since s is clearly a smoothly varying function of q, we conclude that ds/dq = 0 if q = 0° or q = 180°, so that the maximum occurs at q = 60°.

To find the time just after 00:00 when the angle between the hour and minute hands is 60° (1/6 revolution), consider that the angular speed of the minute hand with respect to the hour hand is 11/12 revolutions per hour. Hence q = 60° at (1/6) / (11/12) = 2/11 hours after 00:00.

Therefore, the distance between the tips of the hands is increasing at its greatest rate at 00:10:54 6/11.

**whew this is why i prefer geometric solutions when possible.