The minute hand on a watch is 9 mm long and the hour hand is 3 mm long. How fast is the distance between the tips of the hands changing at one o'clock?We are given that the minute hand on a watch is 9 mm long and the hour hand is 3 mm long. We need to find out how fast the distance between the tips of the hands changing at one o'clock.The position of the hands at one o'clock is shown below:Position of hands at one o'clockAt one o'clock, the minute hand is on 12 and the hour hand is on 1. Let the tip of the minute hand be A and the tip of the hour hand be B. Let the origin be the centre of the watch. Let angle AOB = θ radians. We know that the length of the minute hand (OA) is 9 mm and the length of the hour hand (OB) is 3 mm.Let the distance between A and B be L mm. From the diagram above, we can see that:L² = OA² + OB² - 2 OA OB cosθSubstituting values, we get:L² = 9² + 3² - 2 × 9 × 3 cosθL² = 81 + 9 - 54 cosθL² = 90 - 54 cosθDifferentiating with respect to time t, we get:2L dL/dt = -54 d(cosθ)/dtDifferentiating cosθ, we get:d(cosθ)/dt = -sinθ dθ/dtWe need to find dL/dt when θ = π/6 (one o'clock).At one o'clock, θ = π/6 radians.L² = 90 - 54 cos(π/6)L² = 36Therefore, L = 6√2 mmWe know that sin(π/6) = 1/2Therefore, d(cosθ)/dt = -sinθ dθ/dt = -(1/2) dθ/dtSubstituting in the formula above, we get:2L dL/dt = 54 × (1/2) dθ/dtdL/dt = (27/2) dθ/dtWe need to find dθ/dt when θ = π/6.Let's consider the hour hand. At one o'clock, it has moved through an angle of (π/6) - π/2 = -π/3 radians from 12 to 1. The number of radians moved in 1 second = (π/3)/3600 = π/10800 radians per second.Therefore, dθ/dt = π/10800 radians per secondAt one o'clock, dL/dt = (27/2) dθ/dt = (27/2) × (π/10800) = π/800 mm per second.Hence, the distance between the tips of the hands is changing at a rate of π/800 mm per second at one o'clock.