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

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To solve this problem, we need to use related rates, which involves finding the rate at which one quantity changes with respect to the rate at which another related quantity changes. In this case, we want to find how fast the length of the man's shadow on the building is changing when he is 4 m from the building.

Let's define some variables:
x = the distance between the man and the building (in meters)
y = the length of the man's shadow on the building (in meters)

Now, we are given the following information:
dx/dt = -1.5 m/s (since the man is walking towards the building, the distance x is decreasing)
x = 4 m (when the man is 4 m from the building)

We need to find dy/dt, the rate at which the length of the man's shadow on the building is changing. To do this, we will use similar triangles.

We know that the height of the man is 2 m. Therefore, we have the following proportional relationship:

y / (y + 2) = x / 12

To solve for y, we can cross-multiply and rearrange the equation:

12y = x(y + 2)

Now, we differentiate both sides of the equation with respect to time (t) using the chain rule:

12(dy/dt) = (dx/dt)(y + 2) + x(dy/dt)

Now, we can plug in the given values:

12(dy/dt) = (-1.5)(4 + 2) + (4)(dy/dt)

12(dy/dt) = -9 + 4(dy/dt)

Rearranging the equation:

12(dy/dt) - 4(dy/dt) = -9

Simplifying:

8(dy/dt) = -9

Dividing by 8:

dy/dt = -9/8

Therefore, the length of the man's shadow on the building is decreasing at a rate of 9/8 meters per second when he is 4 m from the building.