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.

A student made his Frisbee on roof of the common block in Walter Gage Residence. He finds 20-feet-long a ladder which leans against wall in order to climb on the roof. He does not set the ladder well, the bottom of the ladder slides away from the building horizontally at a rate of 3 ft/sec. So how fast is the ladder sliding down the house when the top of the ladder is 16 feet from the ground?

When the ladder leans against wall, the ladder, the wall and the ground between them form a right triangle. According the Pythagorean theorem, square of the height of the wall(h) and square of the length of the ground(l) between the ladder and the wall equals square of the length of the ladder, which is h^2+l^2=20^2. When the top of the ladder is 8 feet from the ground, 16^2+l^2=20^2, so we solved for l, the answer is 12. Then take the derivative of the equation for both sides, 2h*h’+2l*l’=0. As we known, h=16, l=12 and h’=-3 because it slides down, 2*16*(-3)+2*12*l’=0. We can get l’=4 ft/sec.