Question

A water tank has the shape of a right circular cone with its vertex down. Its altitude is $10cm$ and the radius of the base is $15cm$. Water leaks out of the bottom at a constant rate of $1cucm/sec$. Water is poured into the tank at a constant rate of $Ccu.cm/sec$. Compute $C$ so that the water level will be rising at the rate of $4cm/sec$ at the instant when the water is $2cm$ deep.

Answer 1

$r=15cm$

$h=10cm$

Let $r^{\prime }$ be radius of wooden-level at the instant when the water is $2cm$ deep

$\therefore \frac{r^{\prime }}{h^{\prime }}=\frac{r}{h}$

$\Rightarrow \frac{r^{\prime }}{h^{\prime }}=\frac{15}{10}$

$\Rightarrow r^{\prime }=\frac{3}{2}{h}^{\prime }$

Volume of the right circular cone $=\frac{1}{3}\times R^{2}H$

$=\frac{1}{3}\pi \times \frac{9}{4}{K}^2{h}^{\prime }$

$=\frac{3\pi }{4}{K}^3$

$\therefore \frac{dv}{dt}=\frac{3\pi }{4}.3K^2.\frac{dK}{dt}$

$=\frac{3\pi }{4}\times 3\times (2{)}^2\times 4Cu.cm/sec$

$=36\pi Cu.cm/sec$

$\therefore valueofc=36\pi +1Cu.cm/sec$