Question

A spotlight on the ground shines on a wall 12 m away. If a man 2 m tall walks from the spotlight toward the building at a speed of 1.3 m/s, how fast is the length of his shadow on the building decreasing when he is 4 m from the building

Answer 1

The length of shadow decrease at 0.48 m/s.

Step-by-step explanation:

The distance of man from building = x

Thus initially, x = 12

It is given that dx/dt= -1.3 m/s

Let the shadow height = Y

Now solve for the
`dy/dt` when x=4

from the diagram (triangle) it can be seen similar triangles with similar base/height ratios.

`(2)/(12-x) = (y)/(12) \\Y = (24)/(12 - x) \\ (dy)/(dt) = (24)/((12-x)^(2)) * (dx)/(dt) \\at x=4, dy/dt \ becomes: \\(24)/((12-4)^2) * -1.3 \\= -0.48 m/s`