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 1.1 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.)


We are dealing with a related rates problem involving similar triangles. Here's how we can approach it:

### Step 1: Set up the problem.
Let:
- \( x \) be the distance of the man from the building (in meters).
- \( s \) be the length of the man's shadow on the building (in meters).

We know:
- The distance from the spotlight to the building is \( 12 \) meters.
- The man is \( 2 \) meters tall.
- The man is walking toward the building at a speed of \( \frac{dx}{dt} = -1.1 \) m/s (since \( x \) is decreasing).
- We are interested in finding the rate at which the length of his shadow is decreasing, \( \frac{ds}{dt} \), when the man is \( 4 \) meters from the building.

### Step 2: Establish a relationship between the variables.
We can use similar triangles to relate the height of the man and the distance of his shadow on the wall:

- The height of the man: \( 2 \) meters.
- The height of the shadow on the wall: proportional to the total distance from the spotlight to the man.

The triangles give us the ratio:
\[
\frac{s}{12 - x} = \frac{2}{x}
\]

This simplifies to:
\[
s = \frac{2(12 - x)}{x}
\]

### Step 3: Differentiate with respect to time.
We want to find \( \frac{ds}{dt} \). First, differentiate both sides of the equation with respect to \( t \):
\[
\frac{ds}{dt} = \frac{d}{dt} \left( \frac{2(12 - x)}{x} \right)
\]

Using the quotient rule:
\[
\frac{ds}{dt} = \frac{x \cdot (-2) - (12 - x) \cdot 2 \cdot (-1)}{x^2}
\]

Simplifying the expression:
\[
\frac{ds}{dt} = \frac{-2x + 2(12 - x)}{x^2}
\]
\[
\frac{ds}{dt} = \frac{-2x + 24 - 2x}{x^2}
\]
\[
\frac{ds}{dt} = \frac{24 - 4x}{x^2}
\]

### Step 4: Substitute the values.
When \( x = 4 \) meters and \( \frac{dx}{dt} = -1.1 \) m/s:
\[
\frac{ds}{dt} = \frac{24 - 4(4)}{4^2} = \frac{24 - 16}{16} = \frac{8}{16} = 0.5 \text{ m/s}.
\]

Thus, the length of the man's shadow on the building is decreasing at a rate of **0.5 m/s**.

### Final Answer:
The length of his shadow is decreasing at a rate of \( 0.5 \) meters per second when he is 4 meters from the building.

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Do you want more details on any of the steps or calculations?

Here are five related questions:
1. How does the rate of shadow length change when the man is further from the building?
2. What if the man walks at a different speed—how does that affect the shadow's rate of change?
3. How can we apply similar triangles in other related rates problems?
4. What is the relationship between rates of change in other real-life motion problems?
5. How would the problem change if the man were taller?

**Tip**: When working with related rates, always start by identifying all the variables and how they relate to each other geometrically or physically.

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 1.1 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.)

Learn how to solve a related rates problem involving a man walking toward a building and the rate at which his shadow length decreases. This example uses the concepts of similar triangles and differentiation, with a detailed solution showing the decrease in shadow length at 0.5 m/s when the man is 4 m from the building.

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