A 5’6” tall man walks at the rate of 3 ft/sec toward a streetlight that is 18 ft above the ground. At what rate is the length of his shadow changing when he is 4 feet from the base of the light?

Answer 1

Draw a diagram that shows two similar, right triangles, the smaller embedded in the larger, with vertical leg of the larger = 18 ' and the vertical leg of the smaller = 5.5 '. Label the horizontal distance between the man's feet and the base of the lamppost x, and the horizontal distance between the man's feet and the tip of his shadow y. We have dx/dt = - 3 ft/sec

By similar triangles, y / 5.5 = (x + y) / 18

18y = 5.5x + 5.5y

12.5y = 5.5x

y = .44x

dy/dt = .44dx/dt

dy/dt = -1.32 ft/sec