A man whose eye level is 6 ft above the ground walks toward a billboard at a rate of 2 ft/s. The bottom of the billboard is 10 ft above the ground and it is 15 ft high. The man's viewing angle is the angle formed by the lines between the man's eyes and the top and bottom of the billboard. At what rate is the viewing angle changing when the man is 30 ft from the billboard?

Solution:

θ = tan-1(9/x) -  tan-1(4/x)

dθ/dt = (x2/x2+81)(-9/x2).dx/dt - (x2/x2+16)(-4/x2).dx/dt

        = (900/981)(-9/900).2 - (900/916)(-4/900).2

        = -0.0096 rad/sec