Question

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. \(\frac{25}{3}\)
• B. 25
• C. \(25\sqrt{3}\)
• D. \(\frac{25}{\sqrt{3}}\)

Hint:

Let the foot be at distance x and the top at height y; then \(x^2+y^2=4\). Differentiate with respect to time.

Answer 1

The Correct Option is D

Let x = distance of the foot from the wall, y = height of the top (in m). Then \(x^2+y^2=4\).
Differentiating: \(x\frac{dx}{dt}+y\frac{dy}{dt}=0\), so \(\frac{dx}{dt}=-\frac{y}{x}\frac{dy}{dt}\).
At \(y=1\): \(x=\sqrt{4-1}=\sqrt3\) m. Given \(\frac{dy}{dt}=-25\) cm/s.
\(\frac{dx}{dt}=-\frac{1}{\sqrt3}(-25)=\frac{25}{\sqrt3}\) cm/s.
Answer: (4) \(\frac{25}{\sqrt{3}}\)