A man of height 'h' is walking away from a street lamp with a constant speed 'v'. The height of the street lamp is 3h. The rate at which the length of the man's shadow is increasing when he is at a distance 10 h from the base of the street lamp is

Options

A. v/2
B. v/3
C. 2v
D. v/6

Answer

To solve the problem of the man's shadow increasing as he walks away from the street lamp, we can follow these steps:

### Step 1: Understand the Geometry
Let:
- The height of the man = \( h \)
- The height of the street lamp = \( 3h \)
- The distance of the man from the base of the street lamp = \( x \)
- The length of the man's shadow = \( s \)

### Step 2: Set Up the Relationship
Using similar triangles, we can relate the height of the street lamp, the height of the man, and the lengths involved:
\[ \frac{3h}{x+s} = \frac{h}{s} \]
This equation arises because the triangles formed by the street lamp and the man's shadow are similar.

### Step 3: Cross-Multiply to Eliminate the Fractions
Cross-multiplying gives:
\[ 3h \cdot s = h \cdot (x + s) \]

### Step 4: Simplify the Equation
Expanding the right side:
\[ 3hs = hx + hs \]
Now, subtract \( hs \) from both sides:
\[ 3hs - hs = hx \]
This simplifies to:
\[ 2hs = hx \]

### Step 5: Solve for Shadow Length \( s \)
Rearranging gives:
\[ s = \frac{hx}{2h} = \frac{x}{2} \]
This means that the length of the shadow \( s \) is always half the distance \( x \) from the street lamp.

### Step 6: Differentiate with Respect to Time
Now, we need to find the rate at which the shadow length is increasing:
\[ \frac{ds}{dt} = \frac{1}{2} \frac{dx}{dt} \]
Given that the man is walking away from the lamp at speed \( v \), we have:
\[ \frac{dx}{dt} = v \]
Substituting this into the equation gives:
\[ \frac{ds}{dt} = \frac{1}{2} v \]

### Step 7: Conclusion
Thus, the rate at which the length of the man's shadow is increasing is:
\[ \frac{ds}{dt} = \frac{v}{2} \]

### Final Answer
The rate at which the length of the man's shadow is increasing when he is at a distance \( 10h \) from the base of the street lamp is \( \frac{v}{2} \).

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