Answer to Question #317923 in Calculus for sach

Question

Water is being poured at the rate of 2π cubic meter/min into an inverted conical tank that is 12-meter deep with a radius of 6 meters at the top. If the water level is rising at the rate of 1/6 m/min and there is a leak at the bottom of the tank, how fast is the water leaking when the water is 6-meter deep?

Answer 1

$H=12,R=6,h=6\\The\,\,volume\,\,of\,\,the\,\,water\,\,depending\,\,of\,\,the\,\,height:\\V\left( h \right) =\left( \frac{h}{H} \right) ^3V\left( H \right) =\left( \frac{h}{H} \right) ^3\cdot \frac{1}{3}\pi R^2H=\\=\pi \cdot \frac{1}{3}\cdot \frac{6^2}{12^2}h^3=\frac{\pi}{12}h^3\\\frac{dh}{dt}=\frac{1}{6}\Rightarrow \frac{dV}{dt}=\frac{\pi}{12}\cdot 3h^2\frac{dh}{dt}=\frac{3\pi \cdot 6^2}{12}\cdot \frac{1}{6}=1.5\pi \\v_l-the\,\,speed\,\,of\,\,leaking\\1.5\pi =\frac{dV}{dt}=2\pi -v_l\Rightarrow v_l=0.5\pi$