A 2m ladder leans against a vertical wall. It the top of the ladder begins to slide down the wall at the rate 25cm/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 1m above the ground is

Engineering

Mathematics

Rate Measure

Question

$\frac{25}{3}$

$\frac{25}{\sqrt{3}}$

$25 \sqrt{3}$

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$x^{2} + y^{2} = 4 (\frac{dy}{dt} = - 25)$

$x \frac{dx}{dt} + y \frac{dy}{dt} = 0$

$\sqrt{3} \frac{dx}{dt} - 1 (25) = 0$

$\frac{dx}{dt} = \frac{25}{\sqrt{3}} cm / \sec$