[bonanova]

The old Grandfather clock in the corner has a 5-inch long hour hand and a 10-inch long minute hand.

At midnight, and at twenty-two more times during the day, the hands line up.

At those times, the tips of the hands are 5 inches apart.

Midway between those times, they are 15 inches apart.

During the first hour, when does that distance increase at the greatest rate?

[Guest]

The old Grandfather clock in the corner has a 5-inch long hour hand and a 10-inch long minute hand.

At midnight, and at twenty-two more times during the day, the hands line up.

At those times, the tips of the hands are 5 inches apart.

Midway between those times, they are 15 inches apart.

During the first hour, when does that distance increase at the greatest rate?

Assuming they start together - and the rate of the hour hand is inconsiquencial then the greatest rate of distance would be at the start.

[bonanova]

Assuming they start together - and the rate of the hour hand is inconsiquencial then the greatest rate of distance would be at the start.

Hi Chronic, and welcome to the Den.

at the start - and at the points of maximum separation - the rate of change is zero.

Think of a smooth curve that goes between maximum and minimum values.

If you're familiar with a sine wave or more appropriately here a cosine wave,

you can see that the slope of that curve at the peaks and valleys is zero.

That is, the curve is horizontal at those points.

[Guest]

16 and 36/99 minutes past midnight.

[bonanova]

16 and 36/99 minutes past midnight.

It's a little earlier than that.

[Guest]

So minute hand moves 6 degrees per min, and hour hand moves at .5 degree per minute. After t minutes, you can form a triangle with those two hands with an angle between them of 5.5t degrees. The side opposite this angle, let's call it C, has property C^2 = 5^2 + 15^2 - 2*5*15*cos(5.5t). So we want to find when dC/dt is maximum, or d^2C/dt^2 = 0. After some calculus, this occurs when x = 12.82 and minimum (i.e. decreasing at the fastest rate) occurs at 52.63 min. Btw, the distance increases/decreases at 27.5 inch/min at these two time points.

Edited March 23, 2009 by sihyunie

[Yoruichi-san]

So minute hand moves 6 degrees per min, and hour hand moves at .5 degree per minute. After t minutes, you can form a triangle with those two hands with an angle between them of 5.5t degrees. The side opposite this angle, let's call it C, has property C^2 = 5^2 + 15^2 - 2*5*15*cos(5.5t). So we want to find when dC/dt is maximum, or d^2C/dt^2 = 0. After some calculus, this occurs when x = 12.82 and minimum (i.e. decreasing at the fastest rate) occurs at 52.63 min. Btw, the distance increases/decreases at 27.5 inch/min at these two time points.

I agree with your method, but the sides of the triangle are 5 and 10, not 5 and 15...so the answers I get are (for max/min in minutes):

t == 10.9091 || t == 54.5455

And the value I get for the rate is .479966 in./min, and the value of C for both times is 8.66025 in.

[bonanova]

I agree with your method, but the sides of the triangle are 5 and 10, not 5 and 15...so the answers I get are (for max/min in minutes):

t == 10.9091 || t == 54.5455

And the value I get for the rate is .479966 in./min, and the value of C for both times is 8.66025 in.

What she said ...

Fix the minute hand, and let the hour hand and clock face rotate.

A moment's thought shows the tip of the hour hand recedes fastest from the tip of the minute hand when the hour hand makes a right angle with the line between the hand tips.

The angle between the hour and minute hands is then a=60^o [cos a = 5/10 = 1/2].

From midnight, the minute hand shall thus have made 1/6 revolution with respect to the hour hand, or 1/6 [12/11] = 2/11 hours = 120/11 = 10.9091... minutes with respect to the clock face.

The distance between the hands is [100 - 25]^1/2 = 5 x 3^1/2 = 8.66... inches.

The rate of increase = r x omega = 5 [11/2 deg/min] [pi rad/180 deg] = 55 pi/360 = 0.479965544 inches/min.