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 2.5 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 is decreasing at a rate of 0.9375 m/s

Explanation:

An image of this scenario is presented in the attached file to this solution. The image shows the exact moment when the 2m tall man is 4m away from the building.

First of, when the man is 4m away from the building, he is 8m away from the spotlight,

Using the concept of similar triangles, we can obtain the value of y at this point.

(8/2) = (12/y)

y = 3 m when the man is 8 m away from the spotlight.

From the image,

But generally, for x and y at any point,

(x/2) = (12/y)

xy = 24

Differentiating with respect to t

(d/dt)(xy) = (d/dt)12

y (dx/dt) + x (dy/dt) = 0

x (dy/dt) = - y (dx/dt)

At x = 8m, y = 3m, (dx/dt) = 2.5 m/s

8 (dy/dt) = - 3 (2.5)

(dy/dt) = - 7.5/8 = - 0.9375 m/s

The minus sign indicates that the height of the shadow decreases as the man moves towards the wall.

Hope this Helps!!!

the maxium height is 34m/s