There is a clock with a minute hand of 20cm and a hour hand of 10cm. At what rate is the distance between the two tips changing at 4:30?

Let the distance between two tips be c and the length of two hands be a and b. And degree of the angle between two hand is i. The two hands and the distance between tips bounded a triangle. The length of two hands are known, I need to computer the distance between two tips. c^2=a^2+b^2-2abcos(i) In order to find the rate of change, we need to take the first derivative. 2c*dc/dt=2absin(i)*di/dt. At 4:30, i=pi/4, c^2=20^2+10^2-2*20*10*sin(pi/4). Then I get c=14.74. di/dt is same as the difference between the rate of change of two hands. The rate of change of minute hand is 6 degree/minute=pi/30. The rate of change of hour hand is 1/2 degree/minute=pi/360. di/dt=pi/30-pi/360=11pi/360. Now, I get all the value I need, just plug in. 2*14.74*dc/dt=2*10*20*sin(pi/4)*11pi/360, dc/dt=0.92cm/minute.