A man of height 180 cm is moving away from a lamp post at the rate of 1.2 meter per second. If the height of the lamp post is 4.5 meters, find the rate at which
(i) his shadow is lengthening. (ii) the tip of the shadow is moving.

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the given information
- Height of the man (MN) = 180 cm = 1.8 m
- Height of the lamp post (OA) = 4.5 m
- Rate at which the man is moving away from the lamp post (dy/dt) = 1.2 m/s

### Step 2: Set up the relationship using similar triangles
Let:
- MB = x (length of the shadow)
- OM = y (distance of the man from the lamp post)

From the similar triangles (triangle NMB and triangle AOB), we have:
\[ \frac{MB}{MN} = \frac{OB}{OA} \]
This can be rewritten as:
\[ \frac{x}{1.8} = \frac{x + y}{4.5} \]

### Step 3: Cross-multiply to find a relationship
Cross-multiplying gives:
\[ 4.5x = 1.8(x + y) \]
Expanding this, we get:
\[ 4.5x = 1.8x + 1.8y \]
Rearranging gives:
\[ 4.5x - 1.8x = 1.8y \]
\[ 2.7x = 1.8y \]
Thus, we can express x in terms of y:
\[ x = \frac{1.8}{2.7}y = \frac{2}{3}y \]

### Step 4: Differentiate with respect to time
Differentiating both sides with respect to time (t):
\[ \frac{dx}{dt} = \frac{2}{3} \frac{dy}{dt} \]

### Step 5: Substitute the known rate
We know that \(\frac{dy}{dt} = 1.2\) m/s. Substituting this into the differentiated equation:
\[ \frac{dx}{dt} = \frac{2}{3} \times 1.2 \]
Calculating this gives:
\[ \frac{dx}{dt} = \frac{2.4}{3} = 0.8 \text{ m/s} \]

### Step 6: Conclusion for part (i)
The rate at which the man's shadow is lengthening is:
\[ \frac{dx}{dt} = 0.8 \text{ m/s} \]

### Step 7: Find the rate at which the tip of the shadow is moving
The tip of the shadow (B) is at a distance of \(x + y\) from the lamp post. Therefore, we differentiate:
\[ \frac{d}{dt}(x + y) = \frac{dx}{dt} + \frac{dy}{dt} \]
Substituting the values we have:
\[ \frac{d}{dt}(x + y) = 0.8 + 1.2 = 2 \text{ m/s} \]

### Step 8: Conclusion for part (ii)
The rate at which the tip of the shadow is moving is:
\[ \frac{d}{dt}(x + y) = 2 \text{ m/s} \]

### Final Answers:
(i) The man's shadow is lengthening at the rate of **0.8 m/s**.
(ii) The tip of the shadow is moving at the rate of **2 m/s**.