on a certain clock the minute hand is 4 in long and the hour hand is 3 in long. How fast is the distance between the tips of the hands changing at 4 PM?

Question

anonymous · Answer

Alright so we have two points which we may represent in polars....
the minute hand is at (4, pi/2) because it's on the "y" axis so to speak. 
the second hand is at (3, -pi/6) since its 1/12 of the clock (2pi) from three to four and it's below four. 
Our distance is then $$\sqrt{4^{2}+3^{2}-2*4*3\cos(\pi/2-(-\pi/6)}$$
or $$\sqrt{37}$$ (roughly 6.08inches)

anonymous · Answer

mmm i got that. i need to find the rate of change between the two hands now. I think i might have gotten it though... but thanks!

anonymous · Answer

Oh, sorry I misread ^^;... you can use that and then just do ds/dt since you know how to get distance.