The tank is leaking

Relevant wiki: Related Rates of Change - Basic

By ratio and proportion,

$\frac{x}{4}=\frac{y}{16}$

$x=\frac{4y}{16}=\frac{y}{4}$

The volume of water at anytime is

$V=\frac{1}{3}\pi x^2y$

However, $x=\frac{y}{4}$

Substituting, we obtain

$V=\frac{1}{3}\pi(\frac{y}{4})^2y$

$V=\frac{1}{3}\pi(\frac{y^3}{16})$

$V=\frac{1}{48}\pi y^3$

Differentiate both sides with respect to $t$ .

$\frac{dV}{dt}=\frac{1}{48}\pi(3)(y^2)(\frac{dy}{dt})$

$\frac{dV}{dt}=\frac{1}{16}y^2\frac{dy}{dt}$

However, when $y=12~cm$ , $\frac{dy}{dt}=\frac{1}{3}$ $\frac{cm}{min}$

Substituting, we obtain

$\frac{dV}{dt}=\frac{1}{16}\pi(12^2)(\frac{1}{3})$

$\frac{dV}{dt}=3\pi=3(3.14)=9.42$ $\frac{cm^3}{min}$

Finally, the water at the bottom of the tank is leaking at the rate of $10-9.42=0.58$ $\frac{cm^3}{min}$