Question

The minute hand of a certain clock is 4 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 at any instant during the next revolution of the hand?

Answer 1

Let $A$ be the area swept out, and $θ$ the angle through which the minute hand has rotated.

$$\dfrac{d\theta}{d t}=\dfrac{2\pi\ rad}{60\ min}=\dfrac{\pi}{30} rad/min$$

$$A=\dfrac{1}{2}r^2\theta=\dfrac{1}{2}(4\ )^2\theta=8\theta$$

Then

$$\dfrac{dA}{d t}=8\dfrac{d\theta}{d t}=8(\dfrac{\pi}{30})=\dfrac{4\pi}{15} \ {in}^2/min$$