Application of Derivatives — Class 12 Maths Solution

Let us assume that at any instant of time $t$, the radius of the base of the cone be $r$, its height be $h$ and the volume of the sand cone be $V$, then $h = \cfrac{1}{6}r \Rightarrow r = 6h$
and $V = \cfrac{1}{3}\pi {r^2}h = \cfrac{1}{3}\pi {(6h)^2}h = 12\pi {h^3}$ …(i)

Differentiating (i) w.r.t. $t$, we get $\cfrac{{dV}}{{dt}} = \left( {12\pi } \right)\left( {3{h^2}\cfrac{{dh}}{{dt}}} \right)$

$\Rightarrow 12{\rm{c}}{{\rm{m}}^{\rm{3}}}{\rm{/sec}} = 36\pi {\left( {4{\rm{cm}}} \right)^2}\cfrac{{dh}}{{dt}}$

$\Rightarrow \cfrac{{dh}}{{dt}} = \cfrac{{12}}{{36\pi \times 16}}{\rm{cm/sec}} = \cfrac{1}{{48\pi }}{\rm{cm/sec}}$

Therefore the rate of increase of the height of the sand cone $= \cfrac{1}{{48\pi }}$ cm/sec.