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
- A \(25\)
- B \(\frac{{25}}{3}\)
- C \(25\sqrt 3 \)
- D \(\frac{{25}}{{\sqrt 3 }}\)
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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