A man who is 1.6 m tall walks away from a lamp which is 4 m above ground at the rate of 30 m/min. How fast is the man's shadow lengthening?

Answer

To solve the problem, we will use the concept of similar triangles and related rates. Here’s a step-by-step breakdown:

### Step 1: Understand the scenario
We have a lamp that is 4 meters above the ground and a man who is 1.6 meters tall. The man walks away from the lamp at a speed of 30 meters per minute. We need to find out how fast the length of his shadow is increasing.

### Step 2: Set up the variables
Let:
- \( x \) = distance of the man from the lamp (in meters)
- \( y \) = length of the man's shadow (in meters)
- The height of the lamp = 4 meters
- The height of the man = 1.6 meters
- The speed of the man = \( \frac{dx}{dt} = 30 \) m/min

### Step 3: Use similar triangles
From the lamp and the man, we can set up two similar triangles:
1. The triangle formed by the lamp and the tip of the shadow.
2. The triangle formed by the man and the tip of his shadow.

From the similar triangles, we can write the following proportion:
\[ \frac{4}{x + y} = \frac{1.6}{y} \]

### Step 4: Cross-multiply and simplify
Cross-multiplying gives:
\[ 4y = 1.6(x + y) \]
Expanding the right side:
\[ 4y = 1.6x + 1.6y \]
Rearranging terms:
\[ 4y - 1.6y = 1.6x \]
\[ 2.4y = 1.6x \]
Dividing both sides by 0.8:
\[ 3y = 2x \quad \text{or} \quad y = \frac{2}{3}x \]

### Step 5: Differentiate with respect to time
Now, we differentiate both sides with respect to time \( t \):
\[ \frac{dy}{dt} = \frac{2}{3} \frac{dx}{dt} \]

### Step 6: Substitute the known rate
We know \( \frac{dx}{dt} = 30 \) m/min. Substituting this into the equation:
\[ \frac{dy}{dt} = \frac{2}{3} \times 30 \]
Calculating this gives:
\[ \frac{dy}{dt} = 20 \text{ m/min} \]

### Conclusion
Thus, the man's shadow is lengthening at a rate of 20 meters per minute.

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