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 2.3 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.)
  
Incorrect: Your answer is incorrect.
 m/s

This problem involves related rates, a concept from calculus, where we need to find how fast the length of the shadow on the wall is decreasing as the man walks toward the building.

### Given:
- The distance from the spotlight to the wall: \( 12 \, \text{m} \)
- The man's height: \( 2 \, \text{m} \)
- The man's walking speed: \( 2.3 \, \text{m/s} \)
- The distance of the man from the wall at the moment of interest: \( 4 \, \text{m} \)

Let’s define some variables:
- Let \( x \) be the distance of the man from the spotlight.
- Let \( s \) be the length of the shadow on the wall.

From the given information, \( x = 12 - 4 = 8 \, \text{m} \) when the man is 4 meters away from the building. We are asked to find how fast the shadow's length \( s \) is changing with respect to time when the man is 4 meters from the building.

### Step 1: Establish the relationship between the variables

Since the problem involves similar triangles (one formed by the spotlight, the top of the man's head, and his shadow, and the other formed by the spotlight, the wall, and the shadow on the wall), we can use the property of similar triangles to set up a proportion:

\[
\frac{2}{s} = \frac{1}{x}
\]

This simplifies to:

\[
s = 2x
\]

### Step 2: Differentiate with respect to time

Now, differentiate both sides of \( s = 2x \) with respect to time \( t \):

\[
\frac{ds}{dt} = 2 \frac{dx}{dt}
\]

We are given \( \frac{dx}{dt} = -2.3 \, \text{m/s} \) (negative because the man is moving towards the building, decreasing \( x \)).

Substitute \( \frac{dx}{dt} \) into the equation:

\[
\frac{ds}{dt} = 2(-2.3) = -4.6 \, \text{m/s}
\]

Thus, the length of the man's shadow on the building is decreasing at a rate of **4.6 m/s** when he is 4 meters away from the building.

### Final Answer:
The shadow's length is decreasing at a rate of \( 4.6 \, \text{m/s} \).

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Would you like any additional details or have any questions?

### Relative Questions:
1. How do similar triangles help in related rates problems?
2. What is the importance of assigning variables before starting a related rates problem?
3. How do you differentiate implicit functions in related rates problems?
4. Why do we use a negative sign for \( \frac{dx}{dt} \) in this problem?
5. How can related rates be applied to more complex real-world problems?

### Tip:
When solving related rates problems, always make sure to relate all variables to a common parameter (usually time) before differentiating.

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 2.3 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.)

Solve this calculus related rates problem where a man walking toward a building causes his shadow on the wall to decrease in length. Learn the use of similar triangles and differentiation to find how fast the shadow is shrinking when the man is 4 meters away from the building. Suitable for students in Grades 11-12.

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