A point source of light along a straight road is at a height of 'a' metres. A boy 'b' metres in height is walking along the road. How fast is his shadow increasing if he is walking away from the light at the rate of 'c' metres per minute ?

Options

A. `(bc)/(a-b)"mt/min"`
B. `(bc)/(a+b)"mt/min"`
C. `(bc)/(2(a-b))"mt/min"`
D. `(bc)/(2(a+b))"mt/min"`

Answer

To solve the problem of how fast the boy's shadow is increasing as he walks away from the light source, we can use similar triangles and related rates. Let's break down the solution step by step.

### Step 1: Understand the Setup
We have a point source of light at a height of 'a' meters and a boy of height 'b' meters walking away from the light source along a straight road. We need to find how fast the shadow of the boy is increasing as he walks away from the light at a rate of 'c' meters per minute.

### Step 2: Define Variables
Let:
- \( x \) = distance of the boy from the point directly below the light source (on the road).
- \( y \) = length of the shadow of the boy.
- \( \frac{dx}{dt} = c \) (the speed at which the boy is walking away from the light).

### Step 3: Set Up the Similar Triangles
From the geometry of the situation, we can set up the following relationship using similar triangles:
- Triangle formed by the light source, the end of the shadow, and the point directly below the light source.
- Triangle formed by the boy, the end of his shadow, and the point directly below the light source.

From the similar triangles, we have:
\[ \frac{a}{x + y} = \frac{b}{y} \]

### Step 4: Cross Multiply
Cross-multiplying gives us:
\[ a \cdot y = b \cdot (x + y) \]
Expanding this, we get:
\[ a \cdot y = b \cdot x + b \cdot y \]
Rearranging gives:
\[ a \cdot y - b \cdot y = b \cdot x \]
Factoring out \( y \):
\[ y(a - b) = b \cdot x \]
Thus, we have:
\[ y = \frac{b \cdot x}{a - b} \]

### Step 5: Differentiate with Respect to Time
Now, we differentiate both sides with respect to time \( t \):
\[ \frac{dy}{dt} = \frac{b}{a - b} \cdot \frac{dx}{dt} \]
Substituting \( \frac{dx}{dt} = c \):
\[ \frac{dy}{dt} = \frac{b \cdot c}{a - b} \]

### Step 6: Conclusion
The rate at which the shadow is increasing is:
\[ \frac{dy}{dt} = \frac{b \cdot c}{a - b} \text{ meters per minute.} \]