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Question
A man 160 cm tall, walks away from a source of light situated at the top of a pole 6 m high, at the rate of 1.1 m/sec. How fast is the length of his shadow increasing when he is 1 m away from the pole?
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Solution
Let AB be the lamp post. Suppose at any time t, the man CD is at a distance of x km from the lamp post and y m is the length of his shadow CE.
\[\text { Since triangles ABE and CDE are similar }, \]
\[\frac{AB}{CD} = \frac{AE}{CE}\]
\[ \Rightarrow \frac{6}{1 . 6} = \frac{x + y}{y}\]
\[ \Rightarrow \frac{x}{y} = \frac{6}{1 . 6} - 1\]
\[ \Rightarrow \frac{x}{y} = \frac{4 . 4}{1 . 6}\]
\[ \Rightarrow y = \frac{16}{44}x\]
\[ \Rightarrow \frac{dy}{dt} = \frac{16}{44}\left( \frac{dx}{dt} \right)\]
\[ \Rightarrow \frac{dy}{dt} = \frac{16}{44} \times 1 . 1 \left( \because \frac{dx}{dt} = 1 . 1 \right)\]
\[ \Rightarrow \frac{dy}{dt} = 0 . 4 m/\sec\]
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