Question

The minute hand of a certain clock is 6 in long. starting from the moment when the hand is pointing straight up, how fast is the area of the sector that is swept out by the hand increasing during the next revolution of the hand. express your answer in square inches per minute.

Rate of the area change = _______ in²/min

Answer 1

The rate at which the area of the sector swept out by the minute hand of the clock is increasing during the next revolution of the hand is π square inches per minute.

Step-by-step explanation:

To find the rate at which the area of the sector is changing, we need to determine the rate at which the angle of rotation of the minute hand is changing.

Since the minute hand of the clock is 6 inches long, it sweeps out a circle with a radius of 6 inches. The area of the sector is given by the formula A = 0.5 * r^2 * θ, where r is the radius of the circle and θ is the angle of rotation.

Since the minute hand starts from the moment when it is pointing straight up, it undergoes a full revolution of 360 degrees or 2π radians in 60 minutes.

Therefore, the rate of change of the angle θ is (2π radians)/(60 minutes) = π/30 radians per minute.

Substituting this value into the formula for the rate of change of the area, we get:

Rate of the area change = 0.5 * 6^2 * (π/30) = π square inches per minute.