The minute hand of a certain clock is 4 in

Wanda Kane

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

yotaniwc

Beginner2021-12-19Added 34 answers

We know that : the area of the sectorthe area of the circle=Aπr2=θ2π (see the picture)
A=12r2θ
dAdt=12r2dθdt (r is constant)
The minute hand takes 60 minutes for one revolution. Then the angular velocity of one revolution per minute is
dθdt=2π60=π30radmin
Thus,dAdt=12r2dθdt=12(4)2(π30)=4π15 in2min
dAdt=4π152min
image

raefx88y

raefx88y

Beginner2021-12-20Added 26 answers

1. Speed of the minute hand is constant = 360 degrees per hour, or 2πradianshr
2. Sector area =θ2ππr22=12r2θ2=162θ2=8θ2
3.dθdt=2π60radiansmin
4. ddt (sector area) =ddt8θ=8ddtθ=16π60=415π2min

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