A point source of light is hung $30$ feet directly above a straight horizontal path on which a man of 6 feet in height is walking . how fast is the man 's shadow lengthening and how fast the tip of the shadow is moving when he is walking away from the light at the rate of $100 f t / m i n$ .

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Solution

Let $s$ be the point pof source of light
Rate of walking $= 100 f t / m i n$
In $△ A S C$ & $△ B D C$
$\frac{A S}{B D} = \frac{A C}{B C}$
$\frac{30}{6} = \frac{x + y}{y}$
$\Rightarrow 5 y = x + y$
$\Rightarrow 4 y = x$
$4 \frac{d y}{d x} = \frac{d x}{d t} = 100 \Rightarrow \frac{d y}{d t} = 25 f t / m i n$ (Length of shadow moves at rate $d y / d t$ )
Now Let $A C = z$
Now $B C = A C - A B = z - x$
Now Again IN $△ A S C$ & $△ B D C$
$\frac{A S}{B D} = \frac{A C}{B C}, \frac{30}{6} = \frac{z}{z - x} \Rightarrow 4 z = 5 x$
Now $\frac{d z}{d t} = 125 f t / m i n$ (Rate of change of shadow)

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