Question

Water is leaking out of an inverted conical tank at a rate of 12,000 cm3/min at the same time that water is being pumped into the tank at a constant rate. The tank has height 6 m and the diameter at the top is 4 m. If the water level is rising at a rate of 20 cm/min when the height of the water is 2 m, find the rate at which water is being pumped into the tank.

Answer 1

Answer:

$291111.1cm^3/min$

Step-by-step explanation:

We are given that

$\frac{dV}{dt}_{out}=12000 cm^3/min$

Height of tank,h=6 m

Diameter of top,d=4 m

Radius,r=$\frac{d}{2}=\frac{4}{2}=2 m$

$\frac{dh}{dt}=20 cm/min$

$\frac{r}{h}=\frac{2}{6}=\frac{1}{3}$

$r=\frac{1}{3} h$

We have to find rate at which water is being pumped into the tank.

Volume of conical ,V=$\frac{1}{3}\pi r^2 h$

$V=\frac{1}{3}\pi(\frac{1}{3} h)^2h=\frac{1}{27}\pi h^3$

$\frac{dV}{dt}=\frac{1}{9}\pi h^2(\frac{dh}{dt})$

h=2 m=200 cm

1m=100 cm

$\frac{dV}{dt}_{in}-12000=\frac{1}{9}\pi(200)^2\times 20$

$\frac{dV}{dt}_{in}=12000+\frac{1}{9}\pi (40000)\times 20$

$\frac{dV}{dt}_{in}=291111.1cm^3/min$