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Question
A man 2 metres high walks at a uniform speed of 5 km/hr away from a lamp-post 6 metres high. Find the rate at which the length of his shadow increases.
Solution
Let AB be the lamp post. Suppose 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.
\[\text { Since triangles ABE and CDE are similar }, \]
\[\frac{AB}{CD} = \frac{AE}{CE}\]
\[ \Rightarrow \frac{6}{2} = \frac{x + y}{y}\]
\[ \Rightarrow 3y = x + y\]
\[ \Rightarrow x = 2y\]
\[ \Rightarrow \frac{dx}{dt} = 2\frac{dy}{dt}\]
\[ \Rightarrow \frac{dy}{dt} = \frac{1}{2}\frac{dx}{dt}\]
\[ \Rightarrow \frac{dy}{dt} = \frac{1}{2}\left( 5 \right) \left( \frac{dx}{dt} = 5 \right)\]
\[ \Rightarrow \frac{dy}{dt} = \frac{5}{2} km/hr\]
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