Question

The hands on a clock are of length 15mm (minute hand) and 12mm (hour hand). How fast is the distance between the tips of the hands changing at 3:00?

this is a related rates problem

Answer 1

Explanation:

We know that the hour hand completes its path around the clock at:
`(1"rev")/(12h)=(2π"rad")/(12h)=(π/6)"rad"/h`

And minute hand travels in :
`(1"rev")/(1h)=(2π"rad")/(1h)=(2π"rad")/h`
Hence,
`theta` keeps changing between them at the rate of:
`color(blue)((d theta)/(dt)=(π/6-2π)=-11/6π"rad"/h)`

Now,
At `3:00`
Hour hand is at `3` and minute hand is at `12`, which is `1/4th` of revolution.
Thus, at `3` o' clock `theta=2π×1/4=π/2`

What we require is `(dD)/(dt)` when `theta=π/2`.

`color(red)"By Pythagoras law"`,
`D^2=15^2 + 12^2 =369`
`D=sqrt(369)≈19.2mm`

`color(red)"By law of cosines"`,
`D^2=(12)^2+(15)^2-2(12)(15)costheta`
`=>D^2=369-360costheta`

Differentiating wrt time,

`d/(dt)(D^2)=d/(dt)(369)-d/(dt)(360costheta)`

`=>2D(dD)/(dt)=0-360(-sinthetacolor(blue)((d theta)/(dt))) =360sinthetacolor(blue)((d theta)/(dt))`

`=>(dD)/(dt)=(180sintheta)/Dcolor(blue)((d theta)/(dt))`

Finally subbing in,

`color(red)(((dD)/(dt))_(theta=π/2))=(180sin(π/2))/(19.2)color(blue)((d theta)/(dt))`

`=(cancel((180))^(30))/(19.2)×(-11π)/cancel6`

`color(red)(≈-17.1875 " mm/h "≈-0.00477 " mm/s")`

Hope this helps. :)