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JEE Main201912 Apr 2019Morning ShiftMathematicsApplication of DerivativesActual

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 c m / s e c , 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:

Options

  1. A25
  2. B25 3
  3. C25 3
  4. D25 3

Correct answer

D. 25 3

Step-by-step solution

d y d t = - 25 c m / s e c , d x d t = ? Now, x 2 + y 2 = 2 2 = 4 differentiating w.r.t. t both side 2 x d x d t + 2 y d y d t = 0 x 2 + y 2 = 4 when y = 1 ⇒ x 2 = 3 ⇒ x = 3 ⇒ d x d t = - y x d y d t ⇒ d x d t 3 , 1 = - 1 3 × - 25 = 25 3 c m / s e c

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