JEE PYQ: Application Of Derivatives Question 66
Question 66 - 2019 (12 Apr Shift 1)
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) $25\sqrt{3}$
(2) $\frac{25}{\sqrt{3}}$
(3) $\frac{25}{3}$
(4) $25$
Show Answer
Answer: (2)
Solution
$x^2 + y^2 = 4$. At $y = 1$: $x = \sqrt{3}$. $2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$. $\sqrt{3}\frac{dx}{dt} + (-25) = 0 \Rightarrow \frac{dx}{dt} = \frac{25}{\sqrt{3}}$.
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