Final answer:
The length of the woman's shadow on the building decreases at a variable rate which can be calculated using similar triangles and derivatives. When she is 2 m from the building, the shadow's length is not decreasing since it is at the top of the building, resulting in a change rate of 0 m/s.
Step-by-step explanation:
The question deals with a related rates problem in calculus, which is a part of mathematics. The woman is walking towards the building and therefore changing the length of her shadow cast by the spotlight. To find how fast the shadow's length is decreasing, we can use similar triangles.
Let x be the distance from the woman to the building and y be the length of the shadow on the building. The height of the woman (2 m) and the total distance to the building (20 m) give us two similar triangles. We can write a proportion: (2 m) / (x) = (2 m) / (20 m - y). Simplifying, we find y = 20 - 10x. Differentiating both sides with respect to time t, we get dy/dt = -10 dx/dt. Given dx/dt = -0.8 m/s, substituting we find dy/dt = -10(-0.8) = 8 m/s. However, when the woman is 2 m from the building, her shadow is on the top of the building, and thus the rate of change is 0 m/s. The speed at which the shadow is decreasing changes from 8 m/s to 0 m/s when she goes from 2 m away from the spotlight to the wall.