Final answer:
The rate at which the area of the sector swept out by the 6-inch long minute hand of a clock is increasing at any instant is 6π in²/min.
Step-by-step explanation:
The student has asked how fast the area of the sector that is swept out by the minute hand of a clock increases at any instant during the next revolution of the hand, given that the minute hand is 6 inches long.
To find this, it's essential to understand the formula for the area of a sector of a circle:
Area of sector = (1/2)·r^2·θ, where r is the radius and θ is the central angle in radians. Since the minute hand completes a full revolution in 60 minutes, the angle θ covered by the minute hand per minute is 2π radians/60 minutes = π/30 radians per minute.
To calculate the rate of area change, we differentiate the area of the sector with respect to time t, which gives us d(Area)/dt = (1/2)·r^2·(dθ/dt).
Plugging in the values, we get d(Area)/dt = (1/2)·(6 inches)^2·(π/30 radians/minute) = 6π in²/min.