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?
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.
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.
Yeah, our answers match.
To find how quickly the distance between the tips of the two hands is changing, we need to consider the rates at which the minute hand and the hour hand are moving.
Let's start by finding the rate at which the minute hand is moving. In a clock, the minute hand completes a full rotation every 60 minutes. This means that the minute hand moves 360 degrees in 60 minutes.
Next, let's find the rate at which the hour hand is moving. In a clock, the hour hand completes a full rotation every 12 hours. This means that the hour hand moves 360 degrees in 12 hours.
Now, to find how quickly the distance between the tips of the two hands is changing, we need to use the concept of rates of change. The rate at which the distance between the tips of the two hands is changing can be found by differentiating the equation that relates the positions of the two hands.
Let's call the distance between the tips of the two hands "d" and the angle between the minute hand and the hour hand "θ." The equation relating the positions of the two hands can be expressed as:
d^2 = (Length of the minute hand)^2 + (Length of the hour hand)^2 - 2 * (Length of the minute hand) * (Length of the hour hand) * cos(θ)
Taking the derivative of both sides of the equation with respect to time will give us the rate at which the distance between the tips of the two hands is changing, which we can denote as dd/dt.
Now, let's substitute the values given in the question. The length of the minute hand is 8 inches, and the length of the hour hand is 6 inches.
After differentiating the equation and substituting the values, we can find the rate at which the distance between the tips of the two hands is changing at 3 pm.