The minute hand on a watch is 8mm long and the hour hand is 4mm long. How fast is the distance between the tips of the hands changing at one o'clock?
ASD
To find the rate at which the distance between the tips of the hands is changing at one o'clock, we need to use trigonometry. Let's denote the distance between the tips of the hands as 'd' and the angle between the minute hand and the hour hand as 'θ'.
First, let's find an expression for 'd' in relation to θ. Using the Law of Cosines, we have:
d² = (8² + 4²) - 2(8)(4)cos(θ)
d² = 80 - 64cos(θ)
d = √(80 - 64cos(θ))
Now, we need to find the rate of change of 'd' with respect to time (dt). To do this, we take the derivative of 'd' with respect to θ and multiply it by the rate of change of θ with respect to time (dθ/dt).
Let's assume that the minute hand has a angular speed of 1 revolution every 60 minutes, and the hour hand has an angular speed of 1 revolution every 12 hours. Since one revolution is equivalent to 2π radians, we have:
dθ/dt = (2π rad / 60 min) = π/30 rad/min (for the minute hand)
dθ/dt = (2π rad / 12 hr) = π/6 rad/hr (for the hour hand)
Now, let's differentiate 'd' with respect to θ:
dd/dθ = (d/dθ)√(80 - 64cos(θ))
dd/dθ = -32sin(θ) / √(80 - 64cos(θ))
Finally, we substitute θ = π/6 (for one o'clock) into the derivative and multiply it by dθ/dt:
dd/dt = (dd/dθ) * (dθ/dt)
dd/dt = (-32sin(π/6) / √(80 - 64cos(π/6))) * (π/6)
dd/dt = (-32 * 0.5) / √(80 - 64 * 0.866) * (π/6)
dd/dt = -16 / √(80 - 55.1) * (π/6)
dd/dt ≈ -16 / √24.9 * (π/6)
Calculating this expression will give you the rate at which the distance between the tips of the hands is changing at one o'clock.