Question

Differential Calculus:

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

Now help me from here. I have made diagram. Area of the sector is, \(\Large\frac{\theta}{2}r^2\)

Answer 2

Kindly tell me what is problem is trying to say. actually I am not getting the question.

Answer 3

Hint: Divide the entire clock area into 60 equal sectors or units. Then area increasing at the rate of one unit/sector per second. Clear?

Answer 4

One unit or sector = ( Pi r ^2 )/60

Answer 5

\[\Large A=\frac{\theta}{2}r^2\]
\[\Large \frac{dA}{dt}=\frac{1}{2}[r^2.\frac{d\theta}{dt}+2r.\theta.\frac{dr}{dt}]\]

Is it correct?

Answer 6

@Champs
Why divide into 60 equal sectors?

Answer 7

mustafa I think what you have is correct, but note that dr/dtheta=0

Answer 8

OK OK unit/sector. That is why I should divide it? Right?

Answer 9

dr/dt=0*

Answer 10

How dr/dt= 0 ?

Answer 11

r is the length of the hand of the clock
it's not getting any longer or shorter , is it?

Answer 12

actually I am not getting this part of the question, below;

"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 13

Yes you are right. Minute hand's length is fixed i.e. 4 inch.

Answer 14

the "during the next revolution" part is poor english
I don't know who translated that part but it is confusing
a "revolution" is all the way around...

Answer 15

right, so your answer makes sense to me up to\[\frac{dA}{dt}=\frac12r^2\frac{d\theta}{dt}\]and what is \(\large \frac{d\theta}{dt}\)

Answer 16

during next revolution means, what I understood is, firstly one complete rotation and then it is taking about 2nd revolution, isn't it?

1 revolution = 2\(\Large \pi\)

Answer 17

yes, but the rate of increase in the area swept out by the hand is a constant because 1/2, r^2, and d(theta)/dt are all constants, so what is the point of phrasing the question like that?

Answer 18

We have given \(\theta\) i.e. 1 revolution=2\(\Large \pi\)
But we have not given \(\Large \frac{d\theta}{dt}\)

Answer 19

but you can figure it out...

Answer 20

it's a clock, and as far as I know \[\frac{d\theta}{dt}\]is the same for all the working clocks in the world!

Answer 21

So what it should be? Because I am not getting. If I differentiate it with respect to time\(\large2\pi\). Then it is a constant so it will go 0.

Answer 22

remember what \[\frac{d\theta}{dt}\]represents; the rate change in the angle
i.e. (angle/time)
the angle for a revolution of a clock is \(2\pi\) radians and the time it takes to make one revolution is 60seconds. Since the hand is moving at a constant rate, the average angular change is the same as the instantaneous one (the derivative).

Answer 23

Why 60sec? It is a minute hand. So it will 3600sec for one revolution.. Isn't it?

Answer 24

ah good point
still it's a constant rate so my argument persists with different numbers\[\frac{d\theta}{dt}=\frac{\Delta\theta}{\Delta t}=\frac{2\pi}{3600}=\frac\pi{1800}\]if the time is in seconds

Answer 25

OK great thinking. Let me solve what we have until do. Then I post the result. Wait then. please be online.

Answer 26

sure :)

Answer 27

\[\frac{dA}{dt}=\frac{1}{2}r^2\frac{d\theta}{dt}\]
\[\frac{dA}{dt}=\frac{1}{2}(16)(\frac{\pi}{1800})\]
\[\frac{dA}{dt}=0.014in^2/min\]

Is this right?

Answer 28

that's what I got
or about 0.84in^2/min
you never specified the units you wanted

Answer 29

your time is in seconds I think

Answer 30

theta is in radians.

Answer 31

no. time in minute.

Answer 32

see i have shown unit. it is in^2/min.

Answer 33

then your answer is wrong

Answer 34

Why? should theta be in degree? or time in sec?

Answer 35

I have made calculation above. Spot the mistake please.

Answer 36

you have time in seconds as \(you\) were actually the one to poinn out to me, lol

Answer 37

point*

Answer 38

if you want the units in radians per minute then you need to calculate \[\frac{d\theta}{dt}\]in minutes
and how many minutes per revolution of the minute hand?

Answer 39

Wait wait.

I used \(\large \frac{d\theta}{dt}\)=\(\large \frac{\pi}{1800}\).

So I think it is radian per sec. Right?

Answer 40

exactly

Answer 41

that's 'cuz we did divided 2pi by 3600 since there are 3600 seconds in a revolution
if we want it in minute we have to divide by the number of minutes in a revolution, which is...?

Answer 42

I am getting confused now.

Answer 43

I have made calculation. Did you check them?

Answer 44

if you want radians per second\[\frac{d\theta}{dt}=\frac{\Delta\theta}{\Delta t}=\frac{2\pi\text{ rad}}{3600\sec}=\frac\pi{1800}\frac{\text{rad}}{\sec}\]if you want radians per minute:\[\frac{d\theta}{dt}=\frac{\Delta\theta}{\Delta t}=\frac{2\pi\text{ rad}}{60\min}=\frac\pi{30}\frac{\text{rad}}{\min}\]you just have to be conscious of your units!

Answer 45

...because it is 3600seconds=60minutes for the minute hand to go around once

Answer 46

and yes I have checked your answers and that is your only mistake

Answer 47

OK.. That is what I don't know, that I should consider units too.

Thank you. It is done now.

Answer 48

welcome :)