A clock has a minute hand that is 8 inches long, and an hour hand that is 6 inches long. How quickly is the distance between the tips of the two hands changing at 3 pm?

Answer 1

rate of hour hand = 2π/12 rads/hr = π/6 rads/hr
rate of minute hand = 2π rads/h
rate at which the angle between is changing = 2π - π/6 rads/hr = 11π/6 rad/hr

let the angle between them be θ radians
then dθ/dt = 11π/6 rad/hr

let the distance between the tips of the hands be d in
d^2 = 8^2 + 6^2 - 2(8)(6)cosθ
d^2 = 100 - 96cos θ
2d dd/dt = 96 sinθ dθ/dt
dd/dt = (48sin θ dθ/dt)/d
at 3:00 , θ = π/2 radians, (90°) and d^2 = 100 or d = 10
but at 3:00 the angle θ would be decreasing, thus dd/dt = -11π/6

dd/dt = 48sin(π/2) * (-11π/6) / 10 = appr -27.646 inches/hr = appr -.46 inches/minute

the distance between the hands is <b<decreasing at appr .46 in/min

check my arithmetic.

Answer 2

Starting at time t=0 minutes at 12:00, the minute hand's position is
8(sin(2πt/60),cos(2πt/60)) = 8(sin(πt/30),cos(πt/30))
The hour hand's speed is 1/2 as fast, so its position is
6(sin(2πt/720),cos(2πt/720)) = 6(sin(πt/360),cos(πt/360)
The distance z between the tips of the hands is thus
z^2 = (8sin(πt/30)-6sin(πt/360))^2 + (8cos(πt/30)-6cos(πt/360))^2
z dz/dt = (8sin(πt/30)-6sin(πt/360))(8cos(πt/30)(π/30)-6cos(πt/360)(π/360))
+ (8cos(πt/30)-6cos(πt/360))(-8sin(πt/30)(π/30)+6sin(πt/360)(π/360))
At 3:00, t=180 and z=10, so πt/30 = 6π and πt/360 = π/2
10 dz/dt = (8*0-6*1)(8*1*π/30-6*0*π/360) + (8*1-6*0)(-8*0*π/30+6*1*π/360)
10 dz/dt = -6(8π/30) + 8(6π/360) = -48π/30 + 48π/360 = 48π/360(-12+1)
10 dz/dt = -22π/15
dz/dt = -11π/75 in/min
Makes sense, since the minute hand is approaching the hour hand at that time.
Check my math.

Answer 3

Yeah, our answers match.