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JEE Mains · Maths · STD 12 - 6. Application of derivatives

A \(2\, m\) ladder leans against a vertical wall. If the top of the ladder begins to slide down the wall at the rate \(25\, cm/ sec\)., then the rate (in \(cm/sec\).) at which the bottom of the ladder slides away from the wall on the horizontal ground when the top of the ladder is \(1\, m\) above the ground is

  1. A \(25\)
  2. B \(\frac{{25}}{3}\)
  3. C \(25\sqrt 3 \)
  4. D \(\frac{{25}}{{\sqrt 3 }}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\frac{{25}}{{\sqrt 3 }}\)

Step-by-step Solution

Detailed explanation

\({x^2} + {y^2} = 4\left( {\frac{{dy}}{{dt}} = - 25} \right)\) \(x\frac{{dx}}{{dt}} + y\frac{{dy}}{{dt}} = 0\) \(\sqrt 3 \frac{{dx}}{{dt}} - 1\left( {25} \right) = 0\) \(\frac{{dx}}{{dt}} = \frac{{25}}{{\sqrt 3 }}\,cm/\sec \)
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