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A Man of Height 6 Ft Walks at a Uniform Speed of 9 Ft/Sec from a Lamp Fixed at 15 Ft Height. the Length of His Shadow is Increasing at the Rate of (A) 15 Ft/Sec (B) 9 Ft/Sec

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Question

A man of height 6 ft walks at a uniform speed of 9 ft/sec from a lamp fixed at 15 ft height. The length of his shadow is increasing at the rate of

Options

  • 15 ft/sec

  • 9 ft/sec

  •  6 ft/sec

  • none of these

MCQ
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Solution

 6 ft/sec
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 ft be the length of his shadow CE.

\[\text { Since the triangles ABE and CDE are similar }, \]

\[\frac{AB}{CD} = \frac{AE}{CE}\]

\[\Rightarrow \frac{15}{6} = \frac{x + y}{y}\]

\[ \Rightarrow \frac{x}{y} = \frac{15}{6} - 1\]

\[ \Rightarrow \frac{x}{y} = \frac{3}{2}\]

\[ \Rightarrow y = \frac{2}{3}x\]

\[ \Rightarrow \frac{dy}{dt} = \frac{2}{3}\left( \frac{dx}{dt} \right)\]

\[ \Rightarrow \frac{dy}{dt} = \frac{2}{3} \times 9\]

\[ \Rightarrow \frac{dy}{dt} = 6 \text { ft }/\sec\]

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Chapter 12: Derivative as a Rate Measurer - Exercise 13.4 [Page 26]

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R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 12 Derivative as a Rate Measurer
Exercise 13.4 | Q 23 | Page 26

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