ExamBro
ExamBro
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 \()\)

  1. A 2
  2. B 3
  3. C 4
  4. D 5
Verified Solution

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 \(K = \frac{22}{5} \times \frac{1}{5280} \times 3600\) \(K = \frac{79200}{26400} = 3\) \(K = 3\)
Same subject
Explore more questions on app