COMEDK · Maths · 27. Application of Derivatives
A man is moving away from a tower 41.6 m high at a rate of \(2 \mathrm{~m} / \mathrm{s}\). If the eyelevel of the man is 1.6 m above the ground, then the rate at which the angle of elevation of the top of the tower changes, when he is at a distance of 30 m from the foot of the tower is
- A \(\dfrac{1}{625} \mathrm{rad} / \mathrm{sec}\)
- B \(\dfrac{4}{625} \mathrm{rad} / \mathrm{sec}\)
- C \(-\dfrac{2}{125} \mathrm{rad} / \mathrm{sec}\)
- D \(-\dfrac{4}{125} \mathrm{rad} / \mathrm{sec}\)
Answer & Solution
Correct Answer
(D) \(-\dfrac{4}{125} \mathrm{rad} / \mathrm{sec}\)
Step-by-step Solution
Detailed explanation
Let \(h\) be the height of the tower, \(h = 41.6\) m. Let \(h_0\) be the height of the man's eye level, \(h_0 = 1.6\) m. The effective height of the tower above the man's eye level is \(H = h - h_0 = 41.6 - 1.6 = 40\) m. Let \(x\) be the distance of the man from the foot of the…
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