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. :)
