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?
Answer & explanation
Correct answer: option 2
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 $\frac{5}{6}m/sec$.