Question

A water tank in the form of an inverted right-circular cone is being emptied at the rate of
6 m^3 /min. The altitude of the cone is 24m, and the base radius is 12m. Find how fast the water level is lowering when the water is 10m deep?

Answer 1

$v=6~\frac{m^3}{min},~h=24~m, r=12~m, l=10~m.$

$V=\frac 13 \pi r^2h=\frac 13 \pi (\frac rh)^2 h^3.$

$V-vt=\frac 13 \pi (\frac rh)^2 l^3,$

$\frac 13 \pi (\frac rh)^2( h^3-l^3)=vt, \implies$

$t=\frac {\pi}{3v} (\frac rh)^2 (h^3-l^3),$

$l=\sqrt[3]{h^3-\frac{3vt}{\pi} (\frac hr)^2}.$

$u(l)=(h-l(t))^{'}=-\frac 13\cdot (-\frac{3v}{\pi (\frac rh)^2})\cdot \frac{1}{{(h^3-\frac{3vt}{\pi} (\frac hr)^2})^{\frac 23}}=\frac{v}{\pi (\frac rh)^2}\cdot \frac{1}{{(h^3-\frac{3v}{\pi} (\frac hr)^2} \frac{\pi}{3v}(\frac rh)^2 (h^3-l^3)^{\frac 23}}=\frac{h^2 v}{\pi r^2 l^2}.$

$u(l)=\frac{24^2\cdot 6}{3.14\cdot12^2\cdot 10^2}=76.4 ~\frac{mm}{s}.$