Answered: "The minute hand of a certain clock is..."

Wanda Kane

Answered question

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

Beginner2021-12-19Added 34 answers

We know that : $\frac{t h e a r e a o f t h e s e c t o r}{t h e a r e a o f t h e c i r c l e} = \frac{A}{π r^{2}} = \frac{θ}{2 π}$ (see the picture)
$∴ A = \frac{1}{2} r^{2} θ$
$\frac{d A}{d t} = \frac{1}{2} r^{2} \frac{d θ}{d t}$ (r is constant)
The minute hand takes 60 minutes for one revolution. Then the angular velocity of one revolution per minute is
$\frac{d θ}{d t} = \frac{2 π}{60} = \frac{π}{30} \frac{r a d}{min}$
Thus, $\frac{d A}{d t} = \frac{1}{2} r^{2} \frac{d θ}{d t} = \frac{1}{2} \cdot {(4)}^{2} \cdot (\frac{π}{30}) = \frac{4 π}{15} \frac{i n^{2}}{min}$
$∴ \frac{d A}{d t} = \frac{4 π}{15} \frac{\in^{2}}{min}$