Question

g A spotlight on the ground is shining on a wall 20m away. If a woman 2m tall walks from the spotlight toward the building at a speed of 0.8m/s, how fast is the length of her shadow on the building decreasing when she is 4m from the building

Answer 1

0.3 m/s

Explanation:

`(dx)/(dt)=\text{Speed of person}=0.8\ \text{m/s}`

As the two triangles in the diagram are similar to each other we have

`(y)/(12)=(2)/(8)\\\Rightarrow y=12* (2)/(8)\\\Rightarrow y=3\ \text{m}`

Again as the triangles are similar we have

`(y)/(2)=(12)/(12-x)\\\Rightarrow x=12-(24)/(y)`

Differentiating the above equation with respect to time we get

`(dx)/(dt)=(-24*-1)/(y^2)(dy)/(dt)\\\Rightarrow (dy)/(dt)=(dx)/(dt)*(y^2)/(24)\\\Rightarrow (dy)/(dt)=0.8* (3^2)/(24)\\\Rightarrow (dy)/(dt)=0.3\ \text{m/s}`

The speed at which the shadow is changing is 0.3 m/s.