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

Answer:

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.

$\frac{2}{12-x} = \frac{y}{12} \\Y = \frac{24}{12 - x} \\ \frac{dy}{dt} = \frac{24}{(12-x)^{2}} \times \frac{dx}{dt} \\at x=4, dy/dt \ becomes: \\\frac{24}{(12-4)^2} \times -1.3 \\= -0.48 m/s$