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Question
A man 2 metres high walks at a uniform speed of 6 km/h away from a lamp-post 6 metres high. Find the rate at which the length of his shadow increases ?
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Solution
Let AB be the lamp post. Let at any time t, the man CD be at a distance of x km from the lamp post and y m be the length of his shadow CE.
\[\text { Since triangles ABE and CDE are similar }, \]
\[\frac{AB}{CD} = \frac{AE}{CE}\]
\[\Rightarrow \frac{6}{2} = \frac{x + y}{y}\]
\[ \Rightarrow \frac{x}{y} = \frac{6}{2} - 1 = 2\]
\[ \Rightarrow \frac{dy}{dt} = \frac{1}{2}\frac{dx}{dt}\]
\[ \Rightarrow \frac{dy}{dt} = \frac{1}{2}\left( 6 \right)\]
\[ \Rightarrow \frac{dy}{dt} = \text{3 km}/hr\]
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