A man 1.5 m tall walks away from a lamp post 4.5 m high at a rate of 4 km/hr. (i) How fast is his shadow lengthening?

Answer

To solve the problem, we will use the concept of similar triangles and related rates. Let's break down the solution step by step.

### Step 1: Understand the scenario
We have a lamp post that is 4.5 m tall and a man who is 1.5 m tall. The man is walking away from the lamp post at a rate of 4 km/hr. We need to find out how fast his shadow is lengthening.

### Step 2: Set up the variables
- Let \( A \) be the distance of the man from the lamp post.
- Let \( B \) be the length of the man's shadow.
- The total distance from the lamp post to the tip of the shadow is \( A + B \).

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

Using the properties of similar triangles, we can write the following proportion:
\[ \frac{4.5}{A + B} = \frac{1.5}{B} \]

### Step 4: Cross-multiply and simplify
Cross-multiplying gives us:
\[ 4.5B = 1.5(A + B) \]
Expanding the right side:
\[ 4.5B = 1.5A + 1.5B \]
Now, rearranging the equation:
\[ 4.5B - 1.5B = 1.5A \]
\[ 3B = 1.5A \]
Dividing both sides by 3:
\[ B = \frac{1.5}{3}A = 0.5A \]

### Step 5: Differentiate with respect to time
Now we differentiate both sides with respect to time \( t \):
\[ \frac{dB}{dt} = 0.5 \frac{dA}{dt} \]

### Step 6: Substitute the known rate
We know that the man is walking away from the lamp post at a rate of \( \frac{dA}{dt} = 4 \) km/hr. Substituting this into the differentiated equation:
\[ \frac{dB}{dt} = 0.5 \times 4 = 2 \text{ km/hr} \]

### Conclusion
The rate at which the man's shadow is lengthening is \( 2 \) km/hr.

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