A 5.5-foot-tall woman walks at 4 ft/s toward a street light that is 27.5 ft above the ground. what is the rate of change of the length of her shadow when she is 16 ft from the street light? at what rate is the tip of her shadow moving?
Accepted Solution
A:
Answer:Rate at which the shadow is moving = -5 ft/s.Step-by-step explanation:I can do the second part for you:If the distance of the woman from the wall is Xp , the length of the shadow is Xs and the distance from the tip of the shadow to the wall is X we have the relation:X = Xp + Xs.We need to find X' (the rate that the tip of the shadow is moving). at Xp = 16 and X'p = -4 ft/s. We need a relation between X and Xp so we have to eliminate Xs. By similar triangles 5.5 / 27.5 = Xs / x1/5 = Xs/xXs = x /5 so substituting in the above relation:X = Xp + X/54X/5 = Xp X = 5Xp / 4Taking derivatives:X' = 5X'p / 4Now X'p is given as - 4 soX' = -20/4 = -5 ft/s.