A man 2 metres tall walks away from a lamp post 5 metres height at the rate of 4.8 km/hr. The rate of increase of the length of his shadow, is

Options

A. 1.6 km/hr
B. 6.3 km/hr
C. 5 km/hr
D. 3.2 km/hr

Answer

To solve the problem step by step, we will analyze the situation using similar triangles and derivatives.

### Step 1: Understand the Geometry
We have a lamp post (point A) that is 5 meters tall and a man (point C) who is 2 meters tall. The man is walking away from the lamp post at a rate of 4.8 km/hr. We need to find the rate at which the length of his shadow (point D) is increasing.

### Step 2: Set Up the Variables
Let:
- \( a \) = distance of the man from the lamp post (AB)
- \( b \) = length of the shadow (CD)
- \( h_1 = 5 \) meters (height of the lamp post)
- \( h_2 = 2 \) meters (height of the man)

### Step 3: Use Similar Triangles
From the geometry, we can set up the relationship using similar triangles:
1. Triangle ABE (lamp post and shadow) and triangle CDE (man and shadow).
2. The ratios of the heights to the bases give us:
\[ \frac{h_1}{a + b} = \frac{h_2}{b} \]
Substituting the heights:
\[ \frac{5}{a + b} = \frac{2}{b} \]

### Step 4: Cross-Multiply and Simplify
Cross-multiplying gives:
\[ 5b = 2(a + b) \]
Expanding the right side:
\[ 5b = 2a + 2b \]
Rearranging terms:
\[ 5b - 2b = 2a \implies 3b = 2a \implies b = \frac{2}{3}a \]

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

### Step 6: Substitute the Given Rate
We know the man is walking away from the lamp post at a rate of \( \frac{da}{dt} = 4.8 \) km/hr. Substituting this into the equation:
\[ \frac{db}{dt} = \frac{2}{3} \times 4.8 \]

### Step 7: Calculate the Rate of Increase of the Shadow
Calculating the right side:
\[ \frac{db}{dt} = \frac{2 \times 4.8}{3} = \frac{9.6}{3} = 3.2 \text{ km/hr} \]

### Final Answer
The rate of increase of the length of his shadow is **3.2 km/hr**.

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