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 1.4 m/s, how fast is the length of his shadow on the building decreasing when he is 4 m from the building?

the length of his shadow on the building is decreasing at the rate of 0.525 m/s

Step-by-step explanation:

From the diagram attached below;

the man is standing at point D with his head at point E

During that time, his shadow on the wall is y = BC

ΔABC and Δ ADE are similar in nature; thus their corresponding sides have equal ratios; i.e

8y = 24

y = 24/8

y = 3 meters

Let take an integral look  at the distance of the man from the building as x, therefore the distance from the spotlight to the man is  12 - x

∴

To find the derivatives of both sides ;we have:

During that time ;

  and y = 3

So; replacing the value into above ; we have:

Thus; the length of his shadow on the building is decreasing at the rate of 0.525 m/s

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