The rate of change of the distance between the tips of the hands of a clock The hands of a clock in some tower are approximately 5m and 2m in length. How fast is the distance between the tips of the hands changing at 9:00. (Hint: use law of cosines)
 A: Okay, so using the law of cosines:
Given that the sides of the triangle are a, b and c, respectively. Also, their corresponding angles being A, B, and C, we find that:
$$a = 2, b = 5, C = \frac{\pi}{2}$$
Also, at that instant, using the Pythagorean Theorem:
$$2^2 + 5^2 = c^2$$
$$c^2 = 29$$
$$c= \sqrt{29}$$
Law of cosines tell us that:
$$c^2 = a^2 + b^2 - 2ab*\cos C$$
We know that a and b are fixed (the lengths of the hands on the clock don't change) but C (the angle between them) constant changes.
Therefore, taking the derivative:
$$2c*c' = 0 + 0 + 2*2*5*\sin (C)*C' $$
Also, note that for a clock, the minute hand moves 6 degrees every minute, and the hour hand moves 0.5 degrees. So the angle between changes by 5.5 degrees every minute, or $\frac{11\pi}{360}$ radians.
Therefore:
$$2c*c' = 2*2*5*sin(\frac{\pi}{2})*\frac{11\pi}{360} $$
$$2*\sqrt{29}*c' = 2*2*5*1*\frac{11\pi}{360} $$
$$2*\sqrt{29}*c' = \frac{11\pi}{18} $$
$$c' = \frac{11\pi}{36\sqrt{29}} =  \frac{11\pi\sqrt{29}}{1044} \text{meters/min}$$
