#### To determine

**To find:** The rate by which area sweeps by the minute hand of the clock.

#### Answer

The rate by which area sweeps by the minutes hand of the clock is dAdt=πr2 cm2/h.

#### Explanation

**Given:**

The length of the minute hand of the clock is r cm.

**Formula used:**

(1) Chain rule: dydx=dydu⋅dudx

(2) Area of a sector of a circle of radius r and angle θ is A=12r2θ

**Calculation:**

Let A be the area of the sector of a circle which is bring by the minute hand of a clock of length r as shown in the Figure 1 given below.

Since the radius r of the sector of a circle remains constant and only θ is varying with time t .

Therefore the area of the sector of the circle become also change with time t.

Differentiate A with respect to the time t.

ddt[A]=ddt[12r2θ]dAdt=12r2dθdt

Since the minute hand rotate 360∘=2π rad in each hour.

Therefore, dθdt=2π rad/h

Substitutes dθdt=2π in dAdt

dAdt=12r2(2π)=πr2 cm2/h

Therefore, the rate by which area sweeps by the minutes hand of the clock is dAdt=πr2 cm2/h.

This answer makes sense because the minute hand sweeps through the full area of a circle,πr2, where r is the length of the minute hand needle.