A ladder $5$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 $2{\rm{cm/s}}$. How fast is its height on the wall decreasing when the foot of the ladder is $4{\rm{m}}$ away from the wall ?
If the foot of the ladder is at a distance $x$ from the wall and the top is at a vertical height of $y$ at any instant of time $t$, then
${\left( 5 \right)^2} = {x^2} + {y^2}$ …(i)
Differentiating (i) w.r.t. $t$, we have
$\Rightarrow \cfrac{d}{{dt}}\left( {25} \right) = 2x\cfrac{{dx}}{{dt}} + 2y\cfrac{{dy}}{{dt}}$ …(ii)
We have $\cfrac{{dx}}{{dt}} = 0.02{\rm{m/sec}}$
$x = 4{\rm{m}}\,\,{\rm{and}}\,\,y = \sqrt {5 - {4^2}} {\rm{m}} = 3{\rm{m}}$
Hence, from (ii)
$0 = 2 \times 4{\rm{m}} \times 0.02{\rm{m/sec}} + 2 \times 3{\rm{m}}\cfrac{{dy}}{{dt}}$
$\Rightarrow \cfrac{{dy}}{{dt}} = - \cfrac{{0.16}}{6}{\rm{m/sec}}$
Therefore, rate of decrease of height of the ladder on the wall
$= \cfrac{{16}}{{600}}{\rm{m/sec}} = \cfrac{{1600}}{{600}}{\rm{cm/sec}} = \cfrac{8}{3}{\rm{cm/sec}}$
NCERT & Exemplar solution for CBSE Class 12 Mathematics, Application of Derivatives. Curated by Sachin Sharma. Free for all students.