A ladder 7.5 m long leans against a wall. The ladder slides along the floor away from the wall at the rate of 3 cm/s. How fast is the height of the ladder on the wall decreasing, when the foot of the ladder is 6 m away from the wall?
Length of the ladder $= \ell = 7.5 \; m= 750 \; cm$
Distance of the foot of the ladder from the wall $=x = 6 \; m = 600 \; cm$
Let the height of the ladder on the wall $= h \; cm$
Rate at which the ladder slips away from the wall $= \dfrac{dx}{dt} = 3 \; cm/s$
Now, $\ell = \sqrt{x^2 + h^2}$
$\implies$ $h = \sqrt{\ell^2 - x^2}$ $\;\; \cdots$ (1)
Differentiating equation (1) w.r.t time t gives
$\begin{aligned}
\dfrac{dh}{dt} & = \dfrac{1}{2 \sqrt{\ell^2 - x^2}} \times \left(-2x \dfrac{dx}{dt}\right) \\
& = - \dfrac{600}{\sqrt{750^2 - 600^2}} \times 3 \\
& = -4 \; cm/s
\end{aligned}$
$\therefore$ The height of the ladder is decreasing at the rate of 4 cm/s.