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 (in m/s) is the length of his shadow on the building decreasing when he is 4 m from the building? (Round your answer to one decimal place.)

This problem can be solved using similar triangles and the concept of related rates in calculus.

Step 1: Set up the problem We can set up two similar triangles: one with the man and his shadow, and one with the spotlight, the man, and the top of his shadow on the wall.

Step 2: Set up the equations Let's denote:

• x as the distance between the man and the spotlight,
• y as the length of the shadow on the wall.

From the similar triangles, we have the relationship: 2/y = (12-x)/12.

Step 3: Differentiate both sides with respect to time (t) Differentiating both sides with respect to time gives us: -2/y^2 * dy/dt = -1/12 * dx/dt.

Step 4: Substitute the given values We know that dx/dt = -1.5 m/s (the man is moving towards the spotlight), and we want to find dy/dt when x = 4 m.

First, we need to find the value of y when x = 4 m. Substituting x = 4 into the equation from step 2 gives us: 2/y = (12-4)/12, 2/y = 8/12, y = 3 m.

Now, we can substitute x = 4, y = 3, and dx/dt = -1.5 into the equation from step 3 to find dy/dt: -2/3^2 * dy/dt = -1/12 * -1.5, -2/9 * dy/dt = 1.5/12, dy/dt = -0.9 m/s.

So, the length of his shadow on the building is decreasing at a rate of 0.9 m/s when he is 4 m from the building.

This problem can be solved using similar triangles and the concept of related rates in calculus.

Step 1: Set up the problem We can set up two similar triangles: one with the man and his shadow, and one with the spotlight, the man, and the top of his shadow on the wall.

Step 2: Set up the equation Let's denote:

• x as the distance between the man and the spotlight,
• y as the length of the shadow on the wall.

From the similar triangles, we have the relationship: 2/(12-x) = (2+y)/12.

Step 3: Differentiate both sides with respect to time (t) Differentiating both sides with respect to time gives us: -2/(12-x)^2 * dx/dt = 1/12 * dy/dt.

We know that dx/dt = -1.5 m/s (the negative sign indicates that the man is moving towards the spotlight), and we want to find dy/dt when x = 4 m.

Step 4: Solve for dy/dt Substitute the known values into the equation: -2/(12-4)^2 * -1.5 = 1/12 * dy/dt, dy/dt = -2/(8)^2 * -1.5 * 12, dy/dt = -2/64 * -18, dy/dt = 0.5625 m/s.

So, the length of his shadow on the building is decreasing at a rate of 0.6 m/s (rounded to one decimal place) when he is 4 m from the building.

This problem can be solved using similar triangles and the concept of related rates in calculus.

Step 1: Set up the problem We can set up two similar triangles: one with the man and his shadow, and one with the spotlight, the man, and the top of his shadow on the wall.

Step 2: Set up the equation Let's denote:

• x as the distance between the man and the spotlight,
• y as the length of the shadow on the wall.

From the similar triangles, we have the relationship: 2/(12-x) = (2+y)/12.

Step 3: Differentiate both sides with respect to time (t) Differentiating both sides with respect to time gives us: -2/(12-x)^2 * dx/dt = 1/12 * dy/dt.

We know that dx/dt = -1.5 m/s (the negative sign indicates that x is decreasing as the man walks towards the wall).

Step 4: Solve for dy/dt when x = 4 m Substitute x = 4 m and dx/dt = -1.5 m/s into the equation and solve for dy/dt: -2/(12-4)^2 * -1.5 = 1/12 * dy/dt, dy/dt = -2/(8)^2 * -1.5 * 12, dy/dt = -2/64 * -18, dy/dt = 0.5625 m/s.

So, the length of his shadow on the building is decreasing at a rate of 0.6 m/s (rounded to one decimal place) when he is 4 m from the building.