A man walks along a straight path at a speed of 4ft/s. A searchlight is located on the ground 20ft from the path and is kept focused on the man. At what rate is the searchlight rotating when the man is 15 ft from the point on the path closest to the searchlight.
Now note that tan(y) = x=20, so that sec^2y *dy/dt=1/20 dx/dt At the instant described
in the problem, x = 15 and the hypotenuse of the above triangle thus has length
25, so we compute:
(25/20)^2 dy/dt=1/20*4 ==>dy/dt=1/2*400/625
