35. If the minute hand of a clock has length $r$ (in centimeters), find the rate at which it sweeps out area as a function of $r$.
To find: The rate by which area sweeps by the minute hand of the clock.
The rate by which area sweeps by the minutes hand of the clock is dAdt=πr2 cm2/h.
The length of the minute hand of the clock is r cm.
(1) Chain rule: dydx=dydu⋅dudx
(2) Area of a sector of a circle of radius r and angle θ is A=12r2θ
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.
Since the minute hand rotate 360∘=2π rad in each hour.
Therefore, dθdt=2π rad/h
Substitutes dθdt=2π in dAdt
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.