# Calculus

If any body can solve this I would be greatly appreciative. I spent an hour in the tutoring center and no one could solve. Here goes:

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

1. When he is 4 m from the building, he is 8 m from the spotlight. The length of his shadow on the building at that time is 3 m.

The length of his shadow on the building when he is x meters from the spotlight is

Y = 2*(12/x) = 24/x

dY/dt = (dY/dx)*(dx/dt) = -(24/x^2)*1.6
= -(24/64)*1.6 = -0.6 m/s

posted by drwls

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