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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.

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प्रश्न

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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उत्तर

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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अध्याय 12: Derivative as a Rate Measurer - Exercise 13.2 [पृष्ठ १९]

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आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 12 Derivative as a Rate Measurer
Exercise 13.2 | Q 11 | पृष्ठ १९

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