A ladder 13 m long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of 2 m/sec. How fast is its height on the wall decreasing when the foot of the ladder is 5m away from the wall?

$\frac{5}{6}\,m/sec$

The correct answer is Option (2) → $\frac{5}{6}\,m/sec$

Let the foot of the ladder be at a distance x metres from the wall and y metres be the height of the ladder on the wall at any time t, then

$x^2 + y^2 = 169$   …(i)

Diff. (i) w.r.t. t, we get

$2x\frac{dx}{dt}+2y\frac{dy}{dt}=0$

$⇒\frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}$

But $\frac{dx}{dt}= 2\, m/sec$ (given)

$⇒\frac{dy}{dt}=-\frac{x}{y}.2=-\frac{2x}{y}$   …(ii)

When $x = 5m$, from (i), $y^2 = 169-25 = 144⇒y = 12 m$.

Putting $x = 5, y = 12$ in (ii), we get $\frac{dy}{dt}=-\frac{2×5}{12}=-\frac{5}{6}$

Hence, the height of the ladder on the wall is decreasing at the rate of $\frac{5}{6}m/sec$.