Wanda Kane

2021-12-18

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 & Explanation

yotaniwc

We know that : (see the picture)
$\therefore A=\frac{1}{2}{r}^{2}\theta$
$\frac{dA}{dt}=\frac{1}{2}{r}^{2}\frac{d\theta }{dt}$ (r is constant)
The minute hand takes 60 minutes for one revolution. Then the angular velocity of one revolution per minute is
$\frac{d\theta }{dt}=\frac{2\pi }{60}=\frac{\pi }{30}\frac{rad}{min}$
Thus,
$\therefore \frac{dA}{dt}=\frac{4\pi }{15}\frac{{\in }^{2}}{min}$

raefx88y

1. Speed of the minute hand is constant = 360 degrees per hour, or $2\pi \frac{radians}{hr}$
2. Sector area $=\frac{\theta }{2\pi }\pi {r}^{2}{\in }^{2}=\frac{1}{2}{r}^{2}\theta {\in }^{2}=\frac{16}{2}\theta {\in }^{2}=8\theta {\in }^{2}$
3.$\frac{d\theta }{dt}=\frac{2\pi }{60}\frac{radians}{min}$
4. $\frac{d}{dt}$ (sector area) $=\frac{d}{dt}8\theta =8\frac{d}{dt}\theta =\frac{16\pi }{60}=\frac{4}{15}\pi \frac{{\in }^{2}}{min}$

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