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 1.9 m/s, how fast 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.

### Step-by-step solution:

1. **Define variables**:
   - Let \( x \) be the distance of the man from the wall (in meters). Since the wall is 12 m away from the spotlight, the distance from the spotlight is \( 12 - x \).
   - Let \( y \) be the length of the man’s shadow on the wall (in meters).
   - The height of the man is 2 m, and his walking speed is \( \frac{dx}{dt} = -1.9 \, \text{m/s} \) (negative because the distance to the wall is decreasing).

2. **Similar triangles**: 
   The relationship between the height of the man and the length of the shadow comes from the fact that the triangles formed by the man and his shadow and by the spotlight and the shadow are similar.
   \[
   \frac{2}{y} = \frac{12 - x}{12}
   \]
   Simplifying:
   \[
   y = \frac{12}{12 - x} \cdot 2 = \frac{24}{12 - x}
   \]

3. **Differentiate with respect to time**:
   To find how fast the length of the shadow is changing, differentiate \( y = \frac{24}{12 - x} \) with respect to \( t \). Using the chain rule:
   \[
   \frac{dy}{dt} = \frac{d}{dt} \left( \frac{24}{12 - x} \right)
   \]
   This gives:
   \[
   \frac{dy}{dt} = \frac{24 \cdot (-1)}{(12 - x)^2} \cdot \frac{dx}{dt}
   \]

4. **Substitute values**:
   When the man is 4 meters from the wall, \( x = 4 \). The rate of change of \( x \) is \( \frac{dx}{dt} = -1.9 \, \text{m/s} \).
   \[
   \frac{dy}{dt} = \frac{24 \cdot (-1)}{(12 - 4)^2} \cdot (-1.9)
   \]
   Simplifying:
   \[
   \frac{dy}{dt} = \frac{-24}{64} \cdot (-1.9) = \frac{-24 \times -1.9}{64} = \frac{45.6}{64} = 0.7125 \, \text{m/s}
   \]

5. **Answer**:
   Therefore, the length of the shadow is decreasing at a rate of approximately \( 0.7 \, \text{m/s} \).

Let me know if you need any further explanation or clarification!

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### Relative Questions:
1. How would the answer change if the man was taller or shorter?
2. What happens if the spotlight was closer to or farther from the wall?
3. How would the speed of the shadow change if the man walked faster or slower?
4. Can you derive the same result using implicit differentiation without similar triangles?
5. What if the height of the wall was limited—how would that affect the shadow?

### Tip:
In related rates problems, always identify the geometric relationships (like similar triangles) and differentiate them with respect to time to find the rates of change.

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 1.9 m/s, how fast 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.)

Solve a related rates problem involving a man walking towards a wall with a spotlight behind him. Learn how fast his shadow on the wall is decreasing when he is 4 m from the building, using concepts like similar triangles and derivatives. Suitable for high school calculus students.

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