Question

a six foot tall man walks at a rate of 6 feet per second away from a streetlight that is 21 feet above the ground. when he is 10 feet away from the base of the light, at what rate is the length of his shadow changing?

Answer 1

This is a related rates problem.
Given: dx/dt = 6 ft/sec. Find dy/dt when x=10.
Using similar triangles, 6/y = 21/(x+y). This leads to y=6x/15. So dy/dt = (6/15)(dx/dt) or (6/15)(6) = 36/15 = 12/5 ft/sec. Note: x=10 is irrelevant because y is a linear function of x.