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Question
A man 180 cm tall walks at a rate of 2 m/sec. away, from a source of light that is 9 m above the ground. How fast is the length of his shadow increasing when he is 3 m away from the base of light?
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Solution
Let AB be the lamp post. Suppose at any time t, the man CD is at a distance x km from the lamp post and y m is the length of his shadow CE.
\[\text { Since triangles ABE and CDE are similar }, \]
\[\frac{AB}{CD} = \frac{AE}{CE}\]
\[\Rightarrow \frac{9}{1 . 8} = \frac{x + y}{y}\]
\[ \Rightarrow \frac{x}{y} = \frac{9}{1 . 8} - 1\]
\[ \Rightarrow \frac{x}{y} = \frac{7 . 2}{1 . 8}\]
\[ \Rightarrow x = 4y\]
\[ \Rightarrow \frac{dy}{dt} = \frac{1}{4}\left( \frac{dx}{dt} \right)\]
\[ \Rightarrow \frac{dy}{dt} = \frac{1}{4} \times 2 \left( \because \frac{dx}{dt} = 2 \right)\]
\[ \Rightarrow \frac{dy}{dt} = 0 . 5 m/\sec\]
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