A woman 5.5 ft tall walks at a rate of 6 ft/sec toward a streetlight that is 22 ft above the ground. At what rate is the length of her shadow changing when she is 15 ft from the base of the light?

Answer 1

If x is her horizontal distance from the street light, the shadow length s is given by
s = (5.5/22)*(s + x)
so
(16.5/22)s = (5.5/22) x
s = x/3

Therefore ds/dt = (1/3) dx/dt
= (2.0 ft/s)*(dx/dt)