TS EAMCET · Maths · Application of Derivatives
A man of 5 feet height is walking away from a light fixed at a height of 15 feet at the rate of K miles/hour. If the rate of increase of his shadow is \(\frac{11}{5}\) feet/sec, then \(\mathrm{K}=\) \((\) Take 1 mile \(=5280\) feet \()\)
- A 2
- B 3
- C 4
- D 5
Answer & Solution
Correct Answer
(B) 3
Step-by-step Solution
Detailed explanation
\(\frac{15}{x+s} = \frac{5}{s}\) \(3s = x+s \implies x = 2s\) \(\frac{dx}{dt} = 2\frac{ds}{dt}\) \(\frac{dx}{dt} = 2 \times \frac{11}{5} = \frac{22}{5}\) ft/s \(K = \frac{22}{5} \times \frac{1}{5280} \times 3600\) \(K = \frac{79200}{26400} = 3\) \(K = 3\)
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